Algebra

Algebra

Mark SteinbergerThe University at Albany

State University of New York

August 31, 2006

Preface

The intent of this book is to introduce the student to algebra from a point of view thatstresses examples and classification. Thus, whenever possible, the main theorems aretreated as tools that may be used to construct and analyze specific types of groups,rings, fields, modules, etc. Sample constructions and classifications are given in bothtext and exercises.

It is also important to note that many beginning graduate students have not takena deep, senior-level undergraduate algebra course. For this reason, I have not assumeda great deal of sophistication on the part of the reader in the introductory portions ofthe book. Indeed, it is hoped that the reader may acquire a sophistication in algebraictechniques by reading the presentations and working the exercises. In this spirit, I haveattempted to avoid any semblance of proofs by intimidation. The intent is that theexercises should provide sufficient opportunity for the student to rise to mathematicalchallenges.

The first chapter gives a summary of the basic set theory we shall use throughoutthe text. Other prerequisites for the main body of the text are trigonometry and thedifferential calculus of one variable. We also presume that the student has seen matrixmultiplication at least once, though the requisite definitions are provided.

Chapter 2 introduces groups and homomorphisms, and gives some examples that willbe used as illustrations throughout the material on group theory.

Chapter 3 develops symmetric groups and the theory of G-sets, giving some usefulcounting arguments for classifying low order groups. We emphasize actions of one groupon another via automorphisms, with conjugation as the initial example.

Chapter 4 studies consequences of normality, beginning with the Noether Isomor-phism Theorems. We study simple groups and composition series, and then classifyall finite abelian groups via the Fundamental Theorem of Finite Abelian Groups. Theautomorphism group of a cyclic group is calculated. Then, semidirect products are intro-duced, using the calculations of automorphism groups to construct numerous examplesthat will be essential in classifying low order groups. We then study extensions of groups,developing some useful classification tools.

In Chapter 5, we develop some additional tools, such as the Sylow Theorems, andapply the theory we’ve developed to the classification of groups of many of the orders≤ 63, in both text and exercises. The methods developed are sufficient to classify therest of the orders in that range. We also study solvable and nilpotent groups.

Chapter 6 is an introduction to basic category theoretic notions, with examples drawnfrom the earlier material. The concepts developed here are useful in understanding ringsand modules, though they are used sparingly in the text on that material. Pushouts ofgroups are constructed and the technique of generators and relations is given.

Chapter 7 provides a general introduction to the theory of rings and modules. Exam-ples are introduced, including the quaternions, H, the p-adic integers, Zp, the cyclotomic

ix

Preface x

integers and rational numbers, Z[ζn] and Q[ζn], polynomials, group rings, and more. Freemodules and chain conditions are studied, and the elementary theory of vector spacesand matrices is developed. The chapter closes with the study of rings and modules offractions, which we shall apply to the study of P.I.D.s and fields.

Chapter 8 develops the theory of P.I.D.s and covers applications to field theory. Theexercises treat prime factorization in some particular Euclidean number rings. The basictheory of algebraic and transcendental field extensions is given, including many of thebasic tools to be used in Galois theory. The cyclotomic polynomials over Q are calculated,and the chapter closes with a presentation of the Fundamental Theorem of FinitelyGenerated Modules over a P.I.D.

Chapter 9 gives some of the basic tools of ring and module theory: Nakayama’sLemma, primary decomposition, tensor products, extension of rings, projectives, andthe exactness properties of tensor and hom. In Section 9.3, Hilbert’s Nullstellensatz isproven, using many of the tools so far developed, and is used to define algebraic varieties.In Section 9.5, extension of rings is used to extend some of the standard results aboutinjections and surjections between finite dimensional vector spaces to the study of mapsbetween finitely generated free modules over more general rings. Also, algebraicK-theoryis introduced with the study of K0.

Chapter 10 gives a formal development of linear algebra, culminating in the classifi-cation of matrices via canonical forms. This is followed by some more general materialabout general linear groups, followed by an introduction to K1.

Chapter 11 is devoted to Galois theory. The usual applications are given, e.g. thePrimitive Element Theorem, the Fundamental Theorem of Algebra, and the classificationof finite fields. Sample calculations of Galois groups are given in text and exercises,particularly for splitting fields of polynomials of the form Xn − a. The insolvability ofpolynomials of degree ≥ 5 is treated.

Chapter 12 gives the theory of hereditary and semisimple rings with an emphasis onDedekind domains and group algebras. Stable and unstable classifications of projectivemodules are given for Dedekind domains and semisimple rings.

The major dependencies between chapters are as follows, where A → B means thatB depends significantly on material in A.

1− 4 7 9

5 8 10

6 11 12

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Additionally there are some minor dependencies. For instance, Chapter 9 assumes anunderstanding of the basic definitions of categories and functors, as given in the first twosections in Chapter 6. Some categorical notions would also be useful in Chapter 7, butare optional there. Also, the last section in Chapter 5 assumes an understanding of someof the material on linear algebra from Chapter 10. Some material on linear algebra is alsoused in the closing section of Chapter 11. And optional references are made in Chapter10 to the material on exterior algebras from Chapter 9. Other than the references tolinear algebra in Chapter 5, the chapters may be read in order.

Preface xi

A solid box symbol ( ) is used to indicate either the end of a proof or that the proofis left to the reader.

Those exercises that are mentioned explicitly in the body of the text are labeledwith a dagger symbol (†) in the margin. This does not necessarily indicate that thereference is essential to the main flow of discussion. Similarly, a double dagger symbol(‡) indicates that the exercise is explicitly mentioned in another exercise. Of course,many more exercises will be used implicitly in other exercises.

Acknowledgments

Let me first acknowledge a debt to Dock Rim, who first introduced me to much of thismaterial. His love for the subject was obvious and catching. I also owe thanks to HaraCharalambous, SUNY at Albany , and David Webb, Dartmouth College, who have usedpreliminary versions of the book in the classroom and made some valuable suggestions.I am also indebted to Bill Hammond, SUNY at Albany , for his insights. I’ve used thismaterial in the classroom myself, and would like to thank the students for many usefulsuggestions regarding the style of presentation and for their assistance in catching typos.

Mark SteinbergerBallston Lake, NY

Contents

1 A Little Set Theory 11.1 Properties of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Factorizations of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Equivalence Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.5 Generating an Equivalence Relation . . . . . . . . . . . . . . . . . . . . . 101.6 Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.7 Formalities about Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2 Groups: Basic Definitions and Examples 152.1 Groups and Monoids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 The Subgroups of the Integers . . . . . . . . . . . . . . . . . . . . . . . . . 242.4 Finite Cyclic Groups: Modular Arithmetic . . . . . . . . . . . . . . . . . . 272.5 Homomorphisms and Isomorphisms . . . . . . . . . . . . . . . . . . . . . . 292.6 The Classification Problem . . . . . . . . . . . . . . . . . . . . . . . . . . 352.7 The Group of Rotations of the Plane . . . . . . . . . . . . . . . . . . . . . 372.8 The Dihedral Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.9 Quaternions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.10 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3 G-sets and Counting 463.1 Symmetric Groups: Cayley’s Theorem . . . . . . . . . . . . . . . . . . . . 473.2 Cosets and Index: Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . 513.3 G-sets and Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 553.4 Supports of Permutations . . . . . . . . . . . . . . . . . . . . . . . . . . . 643.5 Cycle Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 663.6 Conjugation and Other Automorphisms . . . . . . . . . . . . . . . . . . . 713.7 Conjugating Subgroups: Normality . . . . . . . . . . . . . . . . . . . . . . 77

4 Normality and Factor Groups 824.1 The Noether Isomorphism Theorems . . . . . . . . . . . . . . . . . . . . . 834.2 Simple Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 904.3 The Jordan–Holder Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 964.4 Abelian Groups: the Fundamental Theorem . . . . . . . . . . . . . . . . . 984.5 The Automorphisms of a Cyclic Group . . . . . . . . . . . . . . . . . . . . 1054.6 Semidirect Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1114.7 Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

xii

CONTENTS xiii

5 Sylow Theory, Solvability, and Classification 1345.1 Cauchy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1365.2 p-Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1375.3 Sylow Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1415.4 Commutator Subgroups and Abelianization . . . . . . . . . . . . . . . . . 1495.5 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1505.6 Hall’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1535.7 Nilpotent Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1565.8 Matrix Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158

6 Categories in Group Theory 1636.1 Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1646.2 Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1676.3 Universal Mapping Properties: Products and Coproducts . . . . . . . . . 1716.4 Pushouts and Pullbacks . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1766.5 Infinite Products and Coproducts . . . . . . . . . . . . . . . . . . . . . . . 1836.6 Free Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1866.7 Generators and Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . 1896.8 Direct and Inverse Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . 1916.9 Natural Transformations and Adjoints . . . . . . . . . . . . . . . . . . . . 1956.10 General Limits and Colimits . . . . . . . . . . . . . . . . . . . . . . . . . . 198

7 Rings and Modules 2017.1 Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2027.2 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2157.3 Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2227.4 Symmetry of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . 2347.5 Group Rings and Monoid Rings . . . . . . . . . . . . . . . . . . . . . . . . 2397.6 Ideals in Commutative Rings . . . . . . . . . . . . . . . . . . . . . . . . . 2457.7 Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2507.8 Chain Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2687.9 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2737.10 Matrices and Transformations . . . . . . . . . . . . . . . . . . . . . . . . . 2777.11 Rings of Fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

8 P.I.D.s and Field Extensions 2958.1 Euclidean Rings, P.I.D.s, and U.F.D.s . . . . . . . . . . . . . . . . . . . . 2968.2 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3088.3 Transcendence Degree . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3148.4 Algebraic Closures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3178.5 Criteria for Irreducibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3208.6 The Frobenius . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3248.7 Repeated Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3258.8 Cyclotomic Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3278.9 Modules over P.I.D.s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332

CONTENTS xiv

9 Radicals, Tensor Products, and Exactness 3419.1 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3429.2 Primary Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3469.3 The Nullstellensatz and the Prime Spectrum . . . . . . . . . . . . . . . . 3519.4 Tensor Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3659.5 Tensor Products and Exactness . . . . . . . . . . . . . . . . . . . . . . . . 3779.6 Tensor Products of Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . 3839.7 The Hom Functors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3869.8 Projective Modules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3929.9 The Grothendieck Construction: K0 . . . . . . . . . . . . . . . . . . . . . 3989.10 Tensor Algebras and Their Relatives . . . . . . . . . . . . . . . . . . . . . 406

10 Linear algebra 41610.1 Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41610.2 Multilinear alternating forms . . . . . . . . . . . . . . . . . . . . . . . . . 41810.3 Properties of determinants . . . . . . . . . . . . . . . . . . . . . . . . . . . 42410.4 The characteristic polynomial . . . . . . . . . . . . . . . . . . . . . . . . . 42910.5 Eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . 43210.6 The classification of matrices . . . . . . . . . . . . . . . . . . . . . . . . . 43410.7 Jordan canonical form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44010.8 Generators for matrix groups . . . . . . . . . . . . . . . . . . . . . . . . . 44310.9 K1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 447

11 Galois Theory 45011.1 Embeddings of Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45111.2 Normal Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45411.3 Finite Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45811.4 Separable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46011.5 Galois Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46511.6 The Fundamental Theorem of Algebra . . . . . . . . . . . . . . . . . . . . 47411.7 Cyclotomic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47611.8 n-th Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48011.9 Cyclic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48511.10Kummer Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48811.11Solvable Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49311.12The General Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49811.13Normal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50011.14Norms and Traces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502

12 Hereditary and Semisimple Rings 50612.1 Maschke’s Theorem and Projectives . . . . . . . . . . . . . . . . . . . . . 50712.2 Semisimple Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51212.3 Jacobson Semisimplicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52112.4 Homological Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52612.5 Hereditary Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52912.6 Dedekind Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53112.7 Integral Dependence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 539

Bibliography 550

Chapter 1

A Little Set Theory

Here, we discuss the basic properties of functions, relations and cartesian products.The first section discusses injective and surjective functions and their behavior under

composition. This is followed in Section 1.2 by a discussion of commutative diagramsand factorization. Here, duality is introduced, and injective and surjective maps arecharacterized in terms of their properties with respect to factorization. In the lattercase, the role of the Axiom of Choice is discussed. Some introductory remarks about theaxiom are given, to be continued in Chapter 6, where infinite products are introduced.

In the next three sections, we discuss relations, including equivalence relations andpartial orderings. Particular attention is given to equivalence relations. For an equiva-lence relation ∼, the set of equivalence classes, X/∼, is studied, along with the canonicalmap π : X → X/∼. In Section 1.5, we analyze the process by which an arbitrary re-lation may be used to generate an equivalence relation. We shall make use of this inconstructing free products of groups.

We then describe the basic properties of cartesian products, showing that they sat-isfy the universal mapping property for the categorical product. We close with someapplications to the theory of functions and to the basic structure of the category of sets.

With the exception of the discussion of the Axiom of Choice, we shall not delve intoaxiomatics here. We shall assume a familiarity with basic naive set theory, and shall digdeeper only occasionally, to elucidate particular points.

1.1 Properties of Functions

We use the notation f : X → Y to denote that f is a function from X to Y , and say thatX is its domain, and Y its codomain. We shall assume an intuitive understanding offunctions at this point in the discussion, deferring a more formal discussion to Section 1.7.

We shall use the word “map” as a synonym for function.

Definitions 1.1.1. Let X be a set. The identity map 1X : X → X is the functiondefined by 1X(x) = x for all x ∈ X.

If Y is a subset of X, then the inclusion map i : Y → X is obtained by settingi(y) = y for all y ∈ Y .

We give names for some important properties of functions:

Definitions 1.1.2. A function f : X → Y is one-to-one, or injective, if f(x) �= f(x′) forx �= x′. An injective function is called an injection.

1

CHAPTER 1. A LITTLE SET THEORY 2

A function f : X → Y is onto, or surjective, if for each y ∈ Y , there is at least onex ∈ X with f(x) = y. A surjective function is called a surjection.1

A function is bijective if it is both one-to-one and onto. Such a function may also becalled a one-to-one correspondence, or a bijection.

Composition of functions is useful in studying the relationships between sets. Thereader should verify the following properties of composition.

Lemma 1.1.3. Suppose given functions f : X → Y , g : Y → Z, and h : Z →W . Then

1. (h ◦ g) ◦ f = h ◦ (g ◦ f).

2. f ◦ 1X = f .

3. 1Y ◦ f = f .

Here are some more interesting facts to verify regarding composition.

Lemma 1.1.4. Suppose given functions f : X → Y and g : Y → Z. Then the followingrelationships hold.

1. If f and g are injective, so is g ◦ f .2. If f and g are surjective, so is g ◦ f .3. If g ◦ f is injective, so is f .

4. If g ◦ f is surjective, so is g.

Two sets which are in one-to-one correspondence may be thought of as being identicalas far as set theory is concerned. One way of seeing this is in terms of inverse functions.

Definition 1.1.5. Let f : X → Y . We say that a function g : Y → X is an inversefunction for f if the composite f ◦ g is the identity map of Y and the composite g ◦ f isthe identity map of X.

Proposition 1.1.6. A function f : X → Y has an inverse if and only if it is bijective.If f does have an inverse, the inverse is unique, and is defined by setting f−1(y) to bethe unique element x ∈ X such that f(x) = y.

Proof If f is bijective, then for each y ∈ Y , there is a unique x ∈ X such that f(x) = y.Thus, there is a function f−1 : Y → X defined by setting f−1(y) to be equal to thisunique x. The reader may now easily verify that f−1 is an inverse function for f .

Conversely, suppose that f has an inverse function g. Note that identity maps arealways bijections. Thus, f◦g = 1Y is surjective, and hence f is surjective by Lemma 1.1.4.Also, g ◦ f = 1X is injective, so that f is injective as well.

Finally, since f(g(y)) = y for all y, we see that g(y) is indeed the unique x ∈ X suchthat f(x) = y.

Definition 1.1.7. Let f : X → Y be a function. Then the image of f , written im f , isthe following subset of Y : im f = {y ∈ Y | y = f(x) for some x ∈ X}.

1“One-to-one” and “onto” are the terms native to the development of mathematical language inEnglish, while “injection” and “surjection” come from French mathematics.


Thus, the image of f is the set that classically would be called the range of f .Since the values f(x) of the function f : X → Y all lie in im f , f defines a function,

which we shall call by the same name, f : X → im f . Notice that by the definition ofthe image, f : X → im f is surjective. Notice also that f factors as the composite

Xf �� im f ⊂ Y,

where the second map is the inclusion of im f in Y . Since inclusion maps are alwaysinjections, the next lemma is immediate.

Lemma 1.1.8. Every function may be written as a composite g ◦ f , where g is injectiveand f is surjective.

Exercises 1.1.9.1. Give an example of a pair of functions f : X → Y and g : Y → X, where X and Y

are finite sets, such that g ◦ f = 1X , but f is not surjective, and g is not injective.

2. Show that if X and Y are finite sets with the same number of elements, then anyinjection f : X → Y is a bijection. Show also that any surjection f : X → Y is abijection.

3. Give an example of a function f : X → X which is injective but not bijective.

4. Give an example of a function f : X → X which is surjective but not bijective.

1.2 Factorizations of Functions

Commutative diagrams are a useful way to summarize the relationships between differentsets and functions.

Definition 1.2.1. Suppose given a diagram of sets and functions.

Xh ��

f ��

��

Z.

Y

g

��

We say that the diagram commutes if h = g ◦ f . Similarly, given a diagram

X Y

Z W

��f

��

g

��g′

��f ′

we say the diagram commutes if g′ ◦ f = f ′ ◦ g.More generally, given any diagram of maps, we say the diagram commutes if, when

we follow two different paths of arrows which start and end at the same points (followingthe directions of the arrows, of course) and compose the maps along these paths, thenthe compositions from the two paths agree.

Commutative diagrams are frequently used to model the factorization of functions.


Definitions 1.2.2. Let f : X → Y be a function. We say that a function g : X → Zfactors through f if there is a function g′ : Y → Z such that the following diagramcommutes.

Xg ��

f ��

��

Z

Y

g′

��

We say that g factors uniquely through f if there is exactly one function g′ : Y → Z thatmakes the diagram commute.

The language of factorization can be confusing, as there is a totally different meaningto the words “factors through,” which is also used frequently:

Definitions 1.2.3. Let f : X → Y be a function. We say that a function g : Z → Yfactors through f if there is a function g′ : Z → X such that the following diagramcommutes.

X

Z Y��

�� f��

g′

��g

We say that g factors uniquely through f if there is exactly one function g′ : Z → Xthat makes the diagram commute.

Notice that despite the fact that there are two possible meanings to the statementthat a function factors through f : X → Y , the meaning in any given context is almostalways unique. Thus, if we say that a map g : X → Z factors through f : X → Y , thenthe only possible interpretation is that g = h◦f for some h : Y → Z. Similarly, if we saythat a map g : Z → Y factors through f : X → Y , then the only possible interpretationis that g = f ◦ h for some h : Z → X. Indeed, the only way that ambiguity can enter isif we ask whether a function g : X → X factors through another function f : X → X.In this last case, one must be precise as to which type of factorization is being discussed.

Factorization properties may be used to characterize injective and surjective maps.

Proposition 1.2.4. Let f : X → Y . Then the following two statements are equivalent.

1. The map f is injective.

2. Any function g : Z → Y with the property that im g ⊂ im f factors uniquely throughf .

Proof Suppose that f is injective. Let g : Z → Y , and suppose that im g ⊂ im f . Thenfor each z ∈ Z there is an x ∈ X such that g(z) = f(x). Since f is injective, this x isunique, and hence there is a function g′ : Z → X defined by setting g′(z) equal to theunique x ∈ X such that g(z) = f(x). But then f(g′(z)) = g(z), and hence f ◦ g′ = g, sothat g factors through f .

The factorization is unique, since if f ◦ h = g, we have f(h(z)) = g(z), and hencef(h(z)) = f(g′(z)) for all z ∈ Z. Since f is injective, this says that h(z) = g′(z) for allz ∈ Z, and hence h = g′.


Conversely, suppose that any function g : Z → Y with im g ⊂ im f factors uniquelythrough f . Let Z be the set with one element: Z = {z}. Then if f(x) = f(x′), defineg : Z → Y by g(z) = f(x). Then if we define h, h′ : Z → X by h(z) = x and h′(z) = x′,respectively, we have f ◦ h = f ◦ h′ = g. By uniqueness of factorization, h = h′, andhence x = x′. Thus, f is injective.

The uniqueness of the factorization is the key in the above result. There is an en-tirely different characterization of injective functions in terms of factorizations, wherethe factorization is not unique.

Proposition 1.2.5. Let f : X → Y with X �= ∅. Then the following two statements areequivalent.

1. The map f is injective.

2. Any function g : X → Z factors through f .

Proof Suppose that f is injective and that g : X → Z. Choose any z ∈ Z,2 and defineg′ : Y → Z by

g′(y) ={g(x) if y = f(x)z if y �∈ im f .

Since f is injective, g′ is a well defined function. By construction, g′ ◦ f = g, and henceg′ provides the desired factorization of g. (Note that if Z has more than one element,then g′ is not unique. For instance, we could choose a different element z ∈ Z.)

Conversely, suppose that every map g : X → Z factors through f . We apply this bysetting g equal to the identity map of X. Thus, there is a function g′ : Y → X such thatg′ ◦ f = 1X . But then f is injective by Lemma 1.1.4.

There is a notion in mathematics called dualization. The dual of a statement is thestatement obtained by turning around all of the arrows in it. It is often interesting tosee what happens when you dualize a statement, but one should bear in mind that thedual of a true statement is not always true.

The dual of Proposition 1.2.4 is true, but its proof is best deferred to our discussionof equivalence relations below. The dual of Proposition 1.2.5 is also true. In fact, it turnsout to be equivalent to the famous Axiom of Choice.

Axiom of Choice (first form) Let f : X → Y be a surjective function. Then there isa function s : Y → X such that f ◦ s = 1Y .

In this form, the Axiom of Choice seems to be obviously true. In order to define s,all we have to do is choose, for each y ∈ Y , an element x ∈ X with f(x) = y. Since suchan x is known to exist for each y ∈ Y (because f is surjective), intuition says we shouldbe able to make these choices.

Sometimes the desired function s : Y → X may be constructed by techniques thatdepend only on the other axioms of set theory. One example of this is when f is abijection. Here, s is the inverse function of f , and is constructed in the preceding section.Alternatively, if Y is finite, we may construct s by induction on the number of elements.Other examples exist where we may construct s by a particular formula.

2Since X �= ∅ and g : X → Z, we must have Z �= ∅ as well. (See Proposition 1.7.2.)


Nevertheless, intuition aside, for a generic surjection f : X → Y , we cannot constructthe desired function s via the other axioms, and, if we wish to be able to construct sin all cases, we must simply accept the Axiom of Choice as part of the foundations ofmathematics. Most mathematicians are happy to do this without question, but somepeople prefer models of set theory in which the Axiom of Choice does not hold. Thus,while we shall use it freely, we shall give some discussion of the ways in which it entersour arguments.3

We now give the dual of Proposition 1.2.5.


1. The map f is surjective.

2. Any function g : Z → Y factors through f .

Proof Suppose that f : X → Y is surjective. Then the Axiom of Choice provides afunction s : Y → X with f ◦ s = 1Y . But then if g : Z → Y is any map, s ◦ g : Z → Xgives the desired factorization of g through f .

Conversely, suppose that any function g : Z → Y factors through f . Applying thisfor g = 1Y , we obtain a function s : Y → X with f ◦ s = 1Y . But then f is surjective byLemma 1.1.4.

1.3 Relations

We shall make extensive use of both equivalence relations and order relations in our work.We first consider the general notion of a relation on a set X.

Definitions 1.3.1. A relation R on a set X is a subset R ⊂ X × X of the cartesianproduct of X with itself. We shall use xR y as a shorthand for the statement that thepair (x, y) lies in the relation R.

We shall often discuss a relation entirely in terms of the expressions xR y, and dis-pense with discussion of the associated subset of the product. This is an example ofwhat’s known as “abuse of notation”: we can think and talk about a relation as a prop-erty xR y which holds for certain pairs (x, y) of elements of X. We can then, if we choose,define an associated subset R ⊂ X ×X by letting R equal the set of all pairs such thatxR y.

For instance, there is a relation on the real numbers R, called ≤, which is defined inthe usual way: we say that x ≤ y if y − x is a non-negative number. We can discuss allthe properties of the relation ≤ without once referring to the subset of R ×R which itdefines.

Not all relations on a set X have a whole lot of value to us. We shall enumeratesome properties which could hold in a relation that might make it interesting for certainpurposes.

3The desired functions s : Y → X often exist for reasons that do not require the assumption of thechoice axiom. In the study of the particular examples of groups, rings, fields, topological spaces, etc.,that mathematicians tend to be most interested in, the theorems that we know and love tend to be truewithout assuming the axiom. However, the proofs of these theorems are often less appealing withoutthe choice axiom. Moreover, if we assume the axiom, we can prove very general theorems, which thenmay be shown to hold without the axiom, for different reasons in different special cases, in the casesof greatest interest. Thus, the Axiom of Choice permits us to look at mathematics from a broader andmore conceptual viewpoint than we could without it. And indeed, in the opinion of the author, a worldwithout the Axiom of Choice is a poorer one, with less functions and narrower horizons.


Definitions 1.3.2. Let R be a relation on a set X. If xRx for each x ∈ X, then we saythat R is reflexive.

Let R be a relation on a set X. Suppose that y Rx if and only if xR y. Then we saythat R is symmetric.

Alternatively, we say that R is antisymmetric if xR y and y Rx implies that x = y.Let R be a relation on a set X. Then we say that R is transitive if xR y and y R z

together imply that xR z.

These definitions can be used to define some useful kinds of relations.

Definitions 1.3.3. A relation R on X is an equivalence relation if it is reflexive, sym-metric, and transitive.

A relation R on X is a partial ordering if it is reflexive, antisymmetric, and transitive.A set X together with a partial ordering R on X is called a partially ordered set.

A relation R on X is a total ordering if it is a partial ordering with the additionalproperty that for all x, y ∈ X, either xR y or y Rx. A set X together with a totalordering R on X is called a totally ordered set.

Examples 1.3.4.1. The usual ordering ≤ on R is easily seen to be a total ordering.

2. There is a partial ordering, ≤, on R×R obtained by setting (x, y) ≤ (z, w) if bothx ≤ z and y ≤ w. Note that this example differs from the last one in that it is nota total ordering.

3. There is a different partial ordering, called the lexicographic ordering, on R ×Robtained by setting (x, y) ≤ (z, w) if either x < z, or both x ≤ z and y ≤ w.(Thus, both (1, 5) and (2, 3) are ≤ (2, 4).) Note that in this case, we do get a totalordering.

4. There is an equivalence relation, ≡, on the integers, Z, obtained by setting m ≡ nif m− n is even.

5. For any set X, there is an equivalence relation, ∼, on X defined by setting x ∼ yfor all x, y ∈ X (i.e., the subset of X ×X in question is all of X ×X).

6. For any set X, there is exactly one relation which is both a partial ordering andan equivalence relation: here xR y if and only if x = y.

It is customary to use symbols such as ≤ and � for a partial ordering, and to usesymbols such as ∼, , and ≡ for an equivalence relation.

1.4 Equivalence Relations

We now look at equivalence relations, as defined in the preceding section, in greaterdepth.

Definition 1.4.1. Let∼ be an equivalence relation onX and let x ∈ X. The equivalenceclass of x in X (under the relation ∼), denoted E(x), is defined by

E(x) = {y ∈ X |x ∼ y}.

In words, E(x) is the set of all y ∈ X which are equivalent to x under the relation ∼.


Lemma 1.4.2. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then y is inthe equivalence class E(x) of x if and only if E(x) = E(y).

Proof Suppose y ∈ E(x). This just says x ∼ y. But if z ∈ E(y), then y ∼ z, and hencex ∼ z by the transitivity of ∼. Therefore, E(y) ⊂ E(x).

But equivalence relations are also symmetric. Thus, x ∼ y if and only if y ∼ x, andhence y ∈ E(x) if and only if x ∈ E(y). In particular, the preceding argument thenshows that x ∼ y also implies that E(x) ⊂ E(y), and hence E(x) = E(y).

Conversely, if E(x) = E(y), then reflexivity shows that y ∈ E(y) = E(x).

In consequence, we see that equivalence classes partition the set X into distinctsubsets:

Corollary 1.4.3. Let ∼ be an equivalence relation on X and let x, y ∈ X. Then theequivalence classes of x and y have a common element if and only if they are equal. Insymbols,

E(x) ∩ E(y) �= ∅ if and only if E(x) = E(y).

Proof If z ∈ E(x)∩E(y), then both E(x) and E(y) are equal to E(z) by Lemma 1.4.2.

The global picture is now as follows.

Corollary 1.4.4. Let ∼ be an equivalence relation on X. Then every element of Xbelongs to exactly one equivalence class under ∼.

Proof By Corollary 1.4.3, every element belongs to at most one equivalence class. Butreflexivity shows that x ∈ E(x) for each x ∈ X, and the result follows.

Given a set with an equivalence relation, the preceding observation allows us to definea new set from the equivalence classes.

Definitions 1.4.5. Let ∼ be an equivalence relation on a set X. By the set of equiva-lence classes under ∼, written X/∼, we mean the set whose elements are the equivalenceclasses under ∼ in X. In particular, each element of X/∼ is a subset of X.

We define π : X → X/∼ by π(x) = E(x). Thus, π takes each element of X to itsequivalence class. We call π the canonical map from X to X/∼.

Example 1.4.6. Let X = {1, 2, 3, 4} and let ∼ be the equivalence relation on X inwhich 1 ∼ 3, 2 ∼ 4, and the other related pairs are given by reflexivity and symmetry.Then there are two elements in X/∼: {1, 3} and {2, 4}. We have π(1) = π(3) = {1, 3}and π(2) = π(4) = {2, 4}.

We shall show that the canonical map π characterizes the equivalence relation com-pletely.

Lemma 1.4.7. Let ∼ be an equivalence relation on X and let π : X → X/∼ be thecanonical map. Then π is surjective, and satisfies the property that π(x) = π(y) if andonly if x ∼ y.Proof Surjectivity of π follows from the fact that every equivalence class has the formE(x) for some x ∈ X, and E(x) = π(x). Now π(x) = π(y) if and only if E(x) = E(y).By Lemma 1.4.2, the latter condition is equivalent to the statement that y ∈ E(x), whichis defined to mean that x ∼ y.


Intuitively, this says that X/∼ is the set obtained from X by identifying two elementsif they lie in the same equivalence class.

The canonical map has an important universal property.4

Proposition 1.4.8. Let ∼ be an equivalence relation on X. Then a function g : X → Yfactors through the canonical map π : X → X/∼ if and only if x ∼ x′ implies thatg(x) = g(x′) for all x, x′ ∈ X. Moreover, if g factors through π, then the factorization isunique.

Proof Suppose that g = g′ ◦ π for some g′ : X/∼ → Y . If x ∼ x′ ∈ X, thenπ(x) = π(x′), and hence g′(π(x)) = g′(π(x′)). Since g = g′ ◦ π, this gives g(x) = g(x′),as desired. Also, since π is surjective, every element of X/∼ has the form π(x) for somex ∈ X. Since g′(π(x)) = g(x), the effect of g′ on the elements of X/∼ is determined byg. In other words, the factorization g′, if it exists, is unique.

Conversely, suppose given a function g : X → Y with the property that x ∼ x′ impliesthat g(x) = g(x′). Now define g′ : X/∼ → Y by setting g′(π(x)) = g(x) for all x ∈ X.Since π(x) = π(x′) if and only if x ∼ x′, this is well defined by the property assumedfor the function g. By the construction of g′, g′ ◦ π = g, and hence g′ gives the desiredfactorization of g through π.

We have seen that an equivalence relation determines a surjective function: the canon-ical map. We shall give a converse to this, and show that specifying an equivalencerelation on X is essentially equivalent to specifying a surjective function out of X.

Definition 1.4.9. Let f : X → Y be any function. By the equivalence relation deter-mined by f , we mean the relation ∼ on X defined by setting x ∼ x′ if f(x) = f(x′).

The reader may easily verify that the equivalence relation determined by f is indeedan equivalence relation.

We now show that if ∼ is the equivalence relation determined by f : X → Y , thenthere is an important relationship between the canonical map π : X → X/∼ and f .

Proposition 1.4.10. Let f : X → Y be any function and let ∼ be the equivalencerelation determined by f . Then f factors uniquely through the canonical map π : X →X/∼, via a commutative diagram

Xf ��

π ��

��

� Y.

X/∼f ′

��

Moreover, the map f ′ : X/∼ → Y gives a bijection from X/∼ to the image of f . Inparticular, if f is surjective, then f ′ gives a bijection from X/∼ to Y .

Proof Recall that the definition of ∼ was that x ∼ x′ if and only if f(x) = f(x′). Thus,the universal property of the canonical map (i.e., Proposition 1.4.8) provides the uniquefactorization of f through π: just set f ′(π(x)) = f(x) for all x ∈ X.

Since every element ofX/∼ has the form π(x), the image of f ′ is the same as the imageof f . Thus, it suffices to show that f ′ is injective. Suppose that f ′(π(x)) = f ′(π(x′)).Then f(x) = f(x′), so by the definition of ∼, x ∼ x′. But then π(x) = π(x′), and hencef ′ is injective as claimed.

4The expression “universal property” has a technical meaning which we shall explore in greater depthin Chapter 6 on category theory.


In particular, if f : X → Y is surjective, and if ∼ is the equivalence relation deter-mined by f , then we may identify Y with X/∼ and identify f with π. In particular, thedual of Proposition 1.2.4 is now almost immediate from the universal property of thecanonical map π.


1. The map f is surjective.

2. Let g : X → Z be such that f(x) = f(x′) implies that g(x) = g(x′) for all x, x′ ∈ X.Then g factors uniquely through f .

Proof Suppose that f is surjective, and let ∼ be the equivalence relation determinedby f . Then we may identify f with the canonical map π : X → X/∼. But under thisidentification, the second statement is precisely the universal property of π.

Conversely, suppose that the second condition holds. Then the hypothesis says thatthere is a unique map g′ : Y → Y such that the following diagram commutes.

Xf ��

f ��

��

Y

Y

g′

��

Of course, g′ = 1Y will do, but the key is the uniqueness: if f were not surjective, wecould alter g′ on those points of Y which are not in im f without changing the fact thatthe diagram commutes, contradicting the uniqueness of the factorization.

Exercises 1.4.12.† 1. Let n be a positive integer. Define a relation ≡ on the integers, Z, by setting i ≡ j

if j − i is a multiple of n. Show that ≡ is an equivalence relation. (We call itcongruence modulo n.)

2. Let ∼ be any equivalence relation. Show that the equivalence relation determinedby the canonical map π : X → X/∼ is precisely ∼.

1.5 Generating an Equivalence Relation

There are many situations in mathematics where we are given a relation R which isnot an equivalence relation, and asked to form the “smallest” equivalence relation, ∼,with the property that xR y implies that x ∼ y. Here, “smallest” means that if ≡ isanother equivalence relation with the property that xR y implies that x ≡ y, then x ∼ yalso implies that x ≡ y. In other words, the equivalence classes of ∼ are subsets of theequivalence classes of ≡.

This smallest equivalence relation is called the equivalence relation generated by R,and its canonical map π : X → X/∼ will be seen to have a useful universal property withrespect to R. Of course, it is not immediately obvious that such an equivalence relationexists, so we shall have to construct it.

Definition 1.5.1. Let R be any relation on X. The equivalence relation, ∼, generatedby R is the relation obtained by setting x ∼ x′ if there is a sequence x1, . . . , xn ∈ X suchthat x1 = x, xn = x′, and for each i with 1 ≤ i < n, either xi = xi+1, or xiRxi+1, orxi+1Rxi.


The intuition is that we take the relation R to be a collection of “basic equivalences,”and declare that x ∼ x′ if they may be connected by a chain of basic equivalences, somein one direction and some in the other.

Example 1.5.2. Let n be a positive integer, and define a relation R on the integers, Z,by setting i R j if j = i+ n. Note then that if ∼ is the equivalence relation generated byR and if i ∼ j, then j = i + kn for some k ∈ Z. But the converse is also clear: in fact,i ∼ j if and only if j = i + kn for k ∈ Z. Thus, ∼ is the equivalence relation known as“congruence modulo n,” which was given in Problem 1 of Exercises 1.4.12.

We now verify that the equivalence relation generated by R has the desired properties.

Proposition 1.5.3. Let R be any relation and let ∼ be the equivalence relation generatedby R. Then ∼ is, in fact, an equivalence relation, and has the property that xRx′ impliesthat x ∼ x′. Moreover, if ≡ is another equivalence relation with the property that xRx′

implies that x ≡ x′, then x ∼ x′ also implies that x ≡ x′.Proof If xRx′, then the sequence x1 = x, x2 = x′ shows that x ∼ x′, so that xRx′

implies that x ∼ x′, as desired. Similarly, we see that ∼ is reflexive. Symmetry of ∼ maybe obtained by reversing the sequence x1, . . . , xn that’s used to show that x ∼ x′, whiletransitivity is obtained by concatenating two such sequences.

Now let ≡ be another equivalence relation with the property that xRx′ implies thatx ≡ x′, and suppose that x ∼ x′. Thus, we are given a sequence x1, . . . , xn ∈ X suchthat x1 = x, x2 = x′, and for each i with 1 ≤ i < n, either xi = xi+1, or xiRxi+1,or xi+1Rxi. But in any of the three situations, this says that xi ≡ xi+1, so thatx = x1 ≡ x2 ≡ · · · ≡ xn = x′. Since ≡ is transitive, this gives x ≡ x′, as desired.

The fact that ∼ is the smallest equivalence relation containing R translates almostimmediately into a statement about the canonical map π : X → X/∼.

Corollary 1.5.4. Let R be a relation on X and let ∼ be the equivalence relation gener-ated by R. Then a map f : X → Y factors through the canonical map π : X → X/∼if and only if xRx′ implies that f(x) = f(x′) for all x, x′ ∈ X. The factorization, if itexists, is unique.

Proof Suppose that f : X → Y has the property that xRx′ implies that f(x) = f(x′)for all x, x′ ∈ X. Let be the equivalence relation induced by f (i.e., x x′ if and onlyif f(x) = f(x′)). Then xRx′ implies that x x′.

Thus, by Proposition 1.5.3, x ∼ x′ implies that x x′. By the definition of , thisjust says that x ∼ x′ implies that f(x) = f(x′). But now the universal property of thecanonical map π : X → X/∼ (Proposition 1.4.8) shows that f factors uniquely throughπ.

For the converse, if f factors through π, then Proposition 1.4.8 shows that x ∼ x′

implies that f(x) = f(x′). But xRx′ implies that x ∼ x′, so the result follows.

1.6 Cartesian Products

The next definition should be familiar.

Definition 1.6.1. Let X and Y be sets. The cartesian product X × Y is the set of allordered pairs (x, y) with x ∈ X and y ∈ Y .


When describing a function f : X × Y → Z, it is cumbersome to use multipleparentheses, as in f((x, y)). Instead, we shall write f(x, y) for the image of the orderedpair (x, y) under f .

Definition 1.6.2. There are projection maps π1 : X × Y → X and π2 : X × Y → Ydefined by π1(x, y) = x, and π2(x, y) = y, respectively.

An understanding of products with the null set is useful for the theory of functions.

Lemma 1.6.3. Let X be any set and let ∅ be the null set. Then X × ∅ = ∅ ×X = ∅.

Proof If one of the factors of a product is empty, then there are no ordered pairs: onecannot choose an element from the empty factor.

We shall make extensive use of products in studying groups and modules. The nextproposition is useful for studying finite groups. Here, if X is finite, we write |X| for thenumber of elements in X.

Proposition 1.6.4. Let X and Y be finite sets. Then X × Y is finite, and |X × Y | =|X| · |Y |, i.e., the number of elements in X × Y is the product of the number of elementsin X and the number of elements in Y .

Proof If either X or Y is empty, the result follows from Lemma 1.6.3. Thus, we assumethat both X and Y are nonempty.

Consider the projection map π1 : X×Y → X. For each x0 ∈ X, write π−11 (x0) for the

set of ordered pairs that are mapped onto x0 by π1. Thus, π−11 (x0) = {(x0, y) | y ∈ Y }.5

For each x ∈ X, there is an obvious bijection between π−11 (x) and Y , so that

each π−11 (x) has |Y | elements. Now π−1

1 (x) ∩ π−11 (x′) = ∅ for x �= x′, and X × Y =⋃

x∈X π−11 (x). Thus, X × Y is the union of |X| disjoint subsets, each of which has |Y |

elements, and hence |X × Y | = |X| · |Y | as claimed.

Products have an important universal property. Let f : Z → X × Y be a map, andwrite f(z) = (f1(z), f2(z)) for z ∈ Z. In the usual parlance, f1 : Z → X and f2 : Z → Yare the component functions of f . Note that if π1 : X × Y → X and π2 : X × Y → Yare the projection maps, then fi = πi ◦ f , for i = 1, 2.

Since a pair (x, y) ∈ X × Y is determined by its coordinates, we see that a functionf : Z → X×Y is determined by its component functions. In other words, if g : Z → X×Yis a map, and if g1 : Z → X and g2 : Z → Y are its component functions, then f = g ifand only if f1 = g1 and f2 = g2.

Finally, if we are given a pair of functions f1 : Z → X and f2 : Z → Y , then we candefine a function f : Z → X × Y by setting f(z) = (f1(z), f2(z)) for z ∈ Z. In otherwords, we have shown the following.

Proposition 1.6.5. For each pair of functions f1 : Z → X and f2 : Z → Y , there is aunique function f : Z → X ×Y such that fi = πi ◦ f for i = 1, 2. Here, π1 : X ×Y → Xand π2 : X × Y → Y are the projection maps.

5Note that the subsets π−11 (x), as x ranges over the elements of X, are precisely the equivalence

classes for the equivalence relation determined by π1.


1.7 Formalities about Functions

Intuitively, a function f : X → Y is a correspondence which assigns to each elementx ∈ X an element f(x) ∈ Y . The formal definition of function is generally given in termsof its graph.

Definitions 1.7.1. A subset f ⊂ X × Y is the graph of a function from X to Y if foreach x ∈ X there is exactly one ordered pair in f whose first coordinate is x.

If f ⊂ X × Y is the graph of a function and if (x, y) ∈ f , we write y = f(x). Thus,f = {(x, f(x)) |x ∈ X}. By the function determined by f we mean the correspondencef : X → Y that assigns to each x ∈ X the element f(x) ∈ Y .

Two functions from X to Y are equal if and only if their graphs are equal.

Thus, every function is determined uniquely by its graph.We obtain some immediate consequences.

Proposition 1.7.2. Let Y be any set. Then there is one and only one function fromthe null set, ∅, to Y .

On the other hand, if X is a set with X �= ∅, then there are no functions from X to∅.

Proof For a set Y , we have ∅ × Y = ∅. Thus, ∅ × Y has exactly one subset, and hencethere is only one candidate for a function. The subset in question is the graph of afunction, because, since the null set has no elements, the condition that must be satisfiedis automatically true.

On the other hand, if X �= ∅, then the unique subset of X × ∅ is not a function: forany given x ∈ X, there is no ordered pair in this subset whose first coordinate is x.

The next proposition is important in the foundations of category theory.

Proposition 1.7.3. Let X and Y be sets. Then the collection of all functions from Xto Y forms a set.

Proof The collection of all functions from X to Y is in one-to-one correspondence withthe collection of their graphs. But the latter is a subcollection of the collection of allsubsets of X × Y . The fact that the collection of all subsets of a set, W , is a set (calledthe power set of W ) is one of the standard axioms of set theory.

An important special case of this is when X is finite. For simplicity of discussion, letus establish a notation.

Notation 1.7.4. Let X and Y be sets. We write F (X,Y ) for the set of functions fromX to Y .

Proposition 1.7.5. Let X1 and X2 be subsets of X such that X = X1 ∪X2 and X1 ∩X2 = ∅. Write ι1 : X1 ⊂ X and ι2 : X2 ⊂ X for the inclusions of these two subsets inX. Let Y be any set, and define

α : F (X,Y )→ F (X1, Y )× F (X2, Y )

by α(f) = (f ◦ ι1, f ◦ ι2). Then α is a bijection.6

6In the language of category theory, this says that the disjoint union of sets is the coproduct in thecategory of sets.


Proof Suppose that α(f) = α(g). Then f ◦ ι1 = g ◦ ι1, and hence f and g have thesame effect on the elements of X1. Similarly, f ◦ ι2 = g ◦ ι2, and hence f and g have thesame effect on the elements of X2. Since a function is determined by its effect on theelements of its domain, f = g, and hence α is injective.

Now suppose given an element (f1, f2) ∈ F (X1, Y ) × F (X2, Y ). Thus, f1 : X1 → Yand f2 : X2 → Y are functions. Define f : X → Y by

f(x) ={f1(x) if x ∈ X1

f2(x) if x ∈ X2.

Then f ◦ ιi = fi for i = 1, 2, and hence α(f) = (f1, f2). Thus, α is surjective.

As a result, we can count the number of distinct functions between two finite sets.

Corollary 1.7.6. Let X and Y be finite. Then the set of functions from X to Y isfinite. Specifically, if X has n elements and Y has |Y | elements, then the set F (X,Y ) offunctions from X to Y has |Y |n elements. (Here, we take 00 to be equal to 1.)

Proof If Y = ∅, the result follows from Proposition 1.7.2. Thus, we assume that Y �= ∅,and argue by induction on n. If n = 0, then the result is given by Proposition 1.7.2. Forn = 1, say X = {x}, there is a one-to-one correspondence ν : F (X,Y ) → Y given byν(f) = f(x). Thus, we assume n > 1.

Write X = X ′ ∪ {x}, with x �∈ X ′. Then we have a bijection α : F (X,Y ) →F (X ′, Y )×F ({x}, Y ). By induction, F (X ′, Y ) has |Y |n−1 elements, while F ({x}, Y ) has|Y | elements by the case n = 1, above. Thus, the result follows from Proposition 1.6.4.

Chapter 2

Groups: Basic Definitions andExamples

In the first four sections, we define groups, monoids, and subgroups, and develop themost basic properties of the integers and the cyclic groups Zn.

We then discuss homomorphisms and classification. The homomorphisms out ofcyclic groups are determined, and the abstract cyclic groups classified. Classification isemphasized strongly in this volume, and we formalize the notion as soon as possible. Wethen discuss isomorphism classes and show that there are only finitely many isomorphismclasses of groups of any given finite order.

We then give some examples of groups, in particular the dihedral and quaternionicgroups, which we shall use throughout the book as illustrations of various concepts. Weclose with a section on cartesian products, giving a way to create new groups out of oldones.

2.1 Groups and Monoids

A binary operation on a set X is a function

μ : X ×X → X.

That is, μ assigns to each ordered pair (x, y) of elements of X an element μ(x, y) ∈ X.We think of these as multiplications, and generally write x · y, or just xy (or sometimesx + y), instead of μ(x, y). Note that the order of x and y is important. Generally x · yand y · x are different elements of X.

Binary operations without additional properties, while they do arise in various situ-ations, are too general to study in any depth. The condition we shall find most valuableis the associative property: our binary operation is associative if

(x · y) · z = x · (y · z)for every x, y, and z in X.

Another important property is the existence of an identity element. An identityelement for the operation is an element e ∈ X such that

e · x = x = x · efor every x in X. The next result is basic.

15

CHAPTER 2. GROUPS: BASIC DEFINITIONS AND EXAMPLES 16

Lemma 2.1.1. A binary operation can have at most one identity element.

Proof Suppose that e and e′ are both identity elements for the binary operation ·.Since e is an identity element, we have e · e′ = e′. But the fact that e′ is also an identityelement shows that e · e′ = e. Thus, e = e′.

Definition 2.1.2. A monoid is a set with an associative binary operation that has anidentity element.

Examples 2.1.3.1. Let Z+ be the set of positive integers: Z+ = {1, 2, . . . }. Then ordinary multiplica-

tion gives Z+ the structure of a monoid, with identity element 1.

2. Let N ⊂ Z be the set of non-negative integers: N = {0, 1, 2, . . . }. Then ordinaryaddition gives N the structure of a monoid, with identity element 0.

Monoids occur frequently in mathematics. They are sufficiently general that theirstructure theory is extremely difficult.

Definition 2.1.4. Let X be a monoid. We say that an element x ∈ X is invertible ifthere is an element y ∈ X with

x · y = e = y · x,

where e is the identity. We call y the inverse for x and write y = x−1.

The next result shows that the above definition uniquely defines y.

Lemma 2.1.5. Let X be a monoid. Then any element x ∈ X which is invertible has aunique inverse.

Proof Suppose that y and z are both inverses for x. Then x · y = e = y · x, andx · z = e = z · x. Thus,

y = y · e = y · (x · z) = (y · x) · z = e · z = z.

Note that the statement that x · y = e = y · x, used in defining inverses, is symmetricin x and y, establishing the following lemma.

Lemma 2.1.6. Let X be a monoid, and let x ∈ X be invertible, with inverse y. Then yis also invertible, and y−1 = x.

Finally, we come to the object of our study.

Definition 2.1.7. A group is a monoid in which every element is invertible.

Notice that the monoid, Z+, of positive integers under multiplication is not a group,as, in fact, no element other than 1 is invertible: if n ∈ Z+ is not equal to 1, then n > 1.Thus, for any m ∈ Z+, n ·m > 1 ·m, and hence n ·m > 1.

The monoid, N, of the non-negative integers under addition also fails to be a group,as the sum of two positive integers is always positive.

We shall generally writeG for a group, rather thanX. We do have numerous examplesof groups which come from numbers.


Examples 2.1.8. The simplest example of a group is the group with one element, e.There is only one binary operation possible: e · e = e. We call this group the trivialgroup and denote the group itself by e.

Let Z be the integers: Z = {0,±1,±2, . . . }. Then addition of numbers gives Z thestructure of a group: the identity element is 0, and the inverse of n ∈ Z is −n. Similarly,the rational numbers, Q, and the real numbers, R, form groups under addition.

We also know examples of groups obtained from multiplication of numbers. WriteQ× and R× for the nonzero numbers in Q and R, respectively. Since the product of twononzero real numbers is nonzero, multiplication provides Q× and R× with an associativeoperation with identity element 1. And if x ∈ R×, the reciprocal, 1/x, provides an inverseto x in R×. Moreover, if x = m/n ∈ Q×, with m,n ∈ Z, then 1/x = n/m ∈ Q×, andhence x has an inverse in Q×.

Now write Q×+ and R×

+ for the positive rational numbers and the positive real num-bers, respectively. Since the product of two positive numbers is positive and the reciprocalof a positive number is positive, the above arguments show that Q×

+ and R×+ are groups

under multiplication.

The examples above are more than just groups. They also satisfy the commutativelaw : x · y = y · x for all x and y in G. In fact, the ubiquity of the commutative law inour background can exert a pernicious influence on our understanding of groups, as mostgroups don’t satisfy it. The ones that do merit a special name:

Definitions 2.1.9. A group or monoid is abelian if it satisfies the commutative law. Agroup or monoid in which the commutative law is known to fail is said to be nonabelian.

We will often use + for the operation in an abelian group. In fact, we shall restrictthe use of + to these groups.

Notice that the groups in the examples above all have infinitely many elements. Ourmain focus here will be on finite groups: those with finitely many elements.

Definition 2.1.10. If G is a finite group, then the order of G, written |G|, is the numberof elements in G. If G is infinite, we say |G| =∞.

Let X be a monoid. Write Inv(X) ⊂ X for the collection of invertible elements in X.

Lemma 2.1.11. Inv(X) is a group under the multiplication inherited from that of X.

Proof Let x and y be elements of Inv(X). Thus, x has an inverse, x−1, and y has aninverse y−1. Now the element y−1x−1 may easily be seen to be an inverse for xy, andhence xy ∈ Inv(X), also.

Since the operation on Inv(X) is inherited from that on X, it must be associative.Since e−1 = e, e is an element of Inv(X), and acts as an identity element there. And forx ∈ Inv(X), x−1 is invertible, with inverse x, and hence x−1 ∈ Inv(X), and is the inverseof x in Inv(X). In particular, every element of Inv(X) is invertible, so that Inv(X) is agroup.

A primary source for nonabelian groups is from matrices. We assume familiaritywith the basic properties of matrix multiplication shown in a beginning course in linearalgebra. We shall state the properties we are to use, and leave them as exercises to thereader.


Definitions 2.1.12. We writeMn(R) for the set of n×nmatrices over the real numbers,R. We shall make use of the binary operation of matrix multiplication, which is definedas follows. If

A =

⎛⎜⎝ a11 . . . a1n

...an1 . . . ann

⎞⎟⎠ and B =

⎛⎜⎝ b11 . . . b1n...

bn1 . . . bnn

⎞⎟⎠ ,

then the product AB is the matrix whose ij-th coordinate is

cij =n∑k=1

aikbkj = ai1b1j + · · ·+ ainbnj .

We write I = In for the n×n identity matrix: the matrix whose diagonal entries areall equal to 1 and whose off-diagonal entries are all equal to 0.

Thus, I2 =(

1 00 1

), I3 =

(1 0 00 1 00 0 1

), etc.

The properties we need here regarding matrix multiplication are few. We shall discussmatrices in greater detail in Section 7.10 and Chapter 10.

Lemma 2.1.13. Multiplication of n × n matrices gives an associative binary operationon Mn(R). Moreover, In is an identity element for this operation. Thus, Mn(R) is amonoid under matrix multiplication.

The invertible elements in this monoid structure onMn(R) are precisely the invertiblematrices in the usual sense. As such, they merit a name.

Definition 2.1.14. We write Gln(R) for Inv(Mn(R)), the group of invertible elementsof Mn(R). We call it the n-th general linear group of R.

Later in this chapter, we shall construct two infinite families of finite nonabeliangroups, the dihedral groups and the quaternionic groups, as explicit subgroups of Gln(R)for n = 2 and 4, respectively. We shall show in Chapter 10 that every finite group is asubgroup of Gln(R) for some value of n.

It is sometimes useful to study partial inverses in a monoid.

Definitions 2.1.15. Let X be a monoid with identity element e. If x, y ∈ X withxy = e, then we say that x is a left inverse for y and that y is a right inverse for x.

Exercises 2.1.16.1. Let M be the set of all nonzero integers. Then M is a monoid under multiplication.

What is Inv(M)?

† 2. Let G be a group and let x, y ∈ G. Show that x and y commute if and only ifx2y2 = (xy)2.

† 3. Let G be a group such that x2 = e for all x ∈ G. Show that G is abelian.

4. Let G be a group and let x1, . . . , xk ∈ G. Show that (x1 . . . xk)−1 = x−1k . . . x−1

1 .

5. Let G be a group and let x, y ∈ G. Show that x and y commute if and only if(xy)−1 = x−1y−1.

6. Let G be a group and let x, y ∈ G. Suppose there are three consecutive integers nsuch that xnyn = (xy)n. Show that x and y commute.


7. Let G be an abelian group and let x, y ∈ G. Show that xnyn = (xy)n for all n ∈ Z.

8. Verify that multiplication of n× n matrices is associative.

9. Show that Gl1(R) is isomorphic to R×, the group of non-zero real numbers undermultiplication.

10. Show that Gln(R) is nonabelian for n ≥ 2.

11. Let

G ={(

a b0 c

) ∣∣∣∣ b ∈ R, a, c ∈ Q and ac �= 0}.

Show that G is a group under matrix multiplication. Is G abelian?

12. Let X be a monoid. Suppose that x ∈ X has a left inverse, y, and a right inverse,z. Show that y = z, and that x is invertible with inverse y.

† 13. Let X be a monoid with the property that every element of X has a left inverse.Show that X is a group.

14. Let X be a finite monoid with the left cancellation property: if xy = xz, theny = z. Show that X is a group.

15. Let X be a finite set with an associative binary operation. Suppose this operationhas both the left and the right cancellation properties. Show that X is a group.

2.2 Subgroups

One can tell quite a bit about about a group by knowing its subgroups.

Definitions 2.2.1. A subset S of a group G is said to be closed under multiplication iffor x and y in S, the product xy is also in S.

A subset H of G is said to be a subgroup if it is nonempty and closed under multi-plication, and if for each x ∈ H, the inverse element x−1 is also in H.

Examples 2.2.2.1. The groups Z and Q are subgroups of R.

2. The inclusions below are all inclusions of subgroups.

Q×+ ⊂ Q×

∩ ∩R×

+ ⊂ R×

3. Z+ ⊂ R×+ is closed under multiplication, but is not a subgroup, because the inverses

in R×+ of the non-identity elements of Z+ do not lie in Z+.

4. Any group G is a subgroup of itself.

5. For any group G, consider the subset {e} ⊂ G, consisting of the identity elementalone. Because e · e = e and e−1 = e, {e} is a subgroup of G, called the trivialsubgroup, or identity subgroup. By abuse of notation (i.e., for convenience), weshall generally write e in place of {e} for this subgroup. (When the identity elementis called 1 or 0, we shall write 1 or 0 for the trivial subgroup as well.)


The next lemma is immediate from the definitions.

Lemma 2.2.3. Let H ⊂ G be a subgroup. Then H is a group under the operationinherited from that of G.

We also have what is sometimes called Gertrude Stein’s Theorem.1 We leave theproof as an exercise to the reader.

Lemma 2.2.4. A subgroup of a subgroup is a subgroup. (I.e., if H is a subgroup of Gand K is a subgroup of H, then K is a subgroup of G.)

Identifying the subgroups of a given group is a deep and interesting activity, and theresult says a lot about the group. As we shall see presently, the easiest subgroups to findare the smallest ones.

Definition 2.2.5. Let G be a group and let g ∈ G. For any integer n, we define then-th power of g, gn, as follows. For n = 0, 1, gn is given by e and g, respectively. Forn > 1, we define gn by induction on n: gn = gn−1 · g. For negative exponents, we requirethat g−1 be the inverse of g, and if n is positive, then g−n = (g−1)n.

If G is abelian and written in additive notation, we write n · g, or ng for the n-thpower of G in the above sense.

The power laws for a group are fundamental:

Lemma 2.2.6. Let G be a group and let g ∈ G. Then for any integers m and n,

gm · gn = gm+n, and(gm)n = gmn.

If the group operation in G is written additively, this translates to the following:

mg + ng = (m+ n)g, andn(mg) = (nm)g.

Proof We use the multiplicative notation. Consider the first statement first. For n = 0,it reduces to gm · e = gm which is certainly true.

For n = 1, the statement depends on the value of m: if m > 0 it just gives thedefinition of gm+1, and hence is true. For m = 0, it says e · g = g. For m < 0, we havem = −k, say, and gm = (g−1)k = (g−1)k−1 · g−1. Thus,

gm · g = (g−1)k−1(g−1 · g) = (g−1)k−1 = g−(k−1) = gm+1,

because −k = m.For n > 1, we argue by induction on n. We have gn = gn−1 · g, so that gm · gn =

(gm · gn−1) · g. By induction, this is gm+n−1 · g, which is gm+n by the case n = 1 above.The case of n < 0 follows from that of n > 0 by exchanging the roles of g and g−1.We also use induction for the second statement. As with the first, we may assume

n ≥ 0. For n = 0, 1 it says nothing more than e = e, or gm = gm, respectively. Forn > 1, we have (gm)n = (gm)n−1 · gm. By induction, this is gmn−m · gm, which is gmn

by an application of the first statement.1This usage was coined by Arunas Liulevicius.


Notice that there’s a potential conflict between the additive notation for powers andthe notation for ordinary multiplication if G = Z. This is resolved by the followinglemma.

Lemma 2.2.7. Let m and n be integers. Then mn (as defined by ordinary multiplicationof integers) is the m-th power of n in the group Z.

Proof First note the definition of multiplication in Z: if m and n are both positive,then mn is the sum of m copies of n and (−1)km · (−1)ln = (−1)k+lmn.

The sum of m copies of n is by definition the m-th power of n. The remaining casesfollow from Lemma 2.2.6.

Lemma 2.2.6 gives us a convenient source of subgroups.

Definition 2.2.8. Let G be a group and let g ∈ G. Define 〈g〉 ⊂ G to be the set of allinteger powers of g:

〈g〉 = {gn |n ∈ Z}.

Our next result will show that 〈g〉 is a subgroup of G. We call it the cyclic subgroupgenerated by g.

Note that if the group operation in G is written additively, then 〈g〉 is defined by

〈g〉 = {ng |n ∈ Z}.

In particular, in the case G = Z, for n ∈ Z, 〈n〉 = {nk | k ∈ Z} is the set of all integermultiples of n. In this case, there are a couple of alternate notations that come out ofring theory: for n ∈ Z we may write either (n) or nZ for 〈n〉.Lemma 2.2.9. Let G be a group and let g ∈ G. Then 〈g〉 is a subgroup of G. Moreover,〈g〉 has the property that it is the smallest subgroup of G which contains g. In otherwords, if H ⊂ G is a subgroup, and if g ∈ H, then 〈g〉 ⊂ H.

Proof Since gm · gn = gm+n, 〈g〉 is closed under multiplication in G. Also, the inverseof gm is g−m ∈ 〈g〉, so that 〈g〉 is a subgroup of G.

If g ∈ H ⊂ G, with H a subgroup of G, then gn must lie in H for all n ∈ Z by theclosure properties of a subgroup. Thus, 〈g〉 ⊂ H. Indeed, viewing g as an element of H,〈g〉 is the subgroup of H generated by g.

We shall see in the next section that the cyclic subgroups 〈n〉 ⊂ Z give all of thesubgroups of Z.

Cyclic subgroups have a very special kind of group structure. They are cyclic groups:

Definition 2.2.10. A group G is cyclic if there is an element g ∈ G such that G = 〈g〉.Cyclic groups are simplest kind of group there is. For this reason, we can answer

some very complicated questions about them, questions we could not answer for morecomplicated groups. Nevertheless, there are many deep and interesting questions aboutcyclic groups (some of them connected with algebraic number theory) which have yet tobe answered.

We can also talk about the subgroup generated by a family of elements.


Definitions 2.2.11. Let G be a group. For any subset S ⊂ G, the subgroup, 〈S〉,generated by S is the set of all products (of arbitrary length) of powers of elements of S:

〈S〉 = {gn11 . . . gnk

k | k ≥ 1, gi ∈ S and ni ∈ Z for 1 ≤ i ≤ k}.

Warning: In the products gn11 . . . gnk

k , we do not, and cannot, assume that the g1, . . . , gkare distinct. Indeed, if g, h ∈ S do not commute, then, for instance, g2h2 and ghgh aredistinct elements of 〈S〉, by Problem 2 of Exercises 2.1.16.

If S is finite, say S = {x1, . . . , xn}, then by abuse of notation, we shall write〈x1, . . . , xn〉 for 〈S〉.

If 〈S〉 = G, we say that G is generated by S. If G is generated by a finite set, we saythat G is finitely generated.

The following lemma is straightforward.

Lemma 2.2.12. 〈S〉 is the smallest subgroup of G which contains S.

We now give a method for detecting finite subgroups.

Proposition 2.2.13. Let G be a group and let H be a nonempty finite subset of G whichis closed under multiplication. Then H is a subgroup of G.

Proof We must show H is closed under inverses. Let x ∈ H and let S = {xk | 1 ≤ k ∈Z}. Then S is contained in H, and hence S must be finite. Thus, the elements xk withk ≥ 1 cannot all be distinct. We must have xk = xl for some k �= l. Say l > k. Butthen multiplying both sides by x−k, we get e = xl−k = xl−k−1 · x. Thus, xl−k−1 = x−1.But l − k − 1 ≥ 0 since l > k. If l − k − 1 = 0, we have x = e = x−1 ∈ H. Otherwise,xl−k−1 ∈ S ⊂ H, and hence H is closed under inverses.

We can give one useful operation on subgroups at this point. The proof is left to thereader.

Lemma 2.2.14. Let H and K be subgroups of G. Then H ∩K is also a subgroup of G.

Warning: The union of two subgroups is almost never a subgroup.

The collection of all subgroups of a group G is easily seen to form a partially orderedset as defined in Definition 1.3.3. Here the order relation is given by inclusion as subsetsof G; explicitly, H is less than or equal to K if H ⊂ K.

Definition 2.2.15. Let G be a group. The partially ordered set of subgroups of G isknown as the lattice of subgroups of G.

It is often useful to diagram the lattice of subgroups of a group G. Right now wehave neither enough examples of groups nor enough theory about subgroups to be ableto give a good example and verify its accuracy. However, we shall give one anyway, fora group we shall encounter in this chapter. The reader should reconsider the followingexample as the theory develops.


Example 2.2.16. We write D6 for the dihedral group of order 6. As will become clearlater, the following is the lattice of subgroups of D6.

D6

〈b〉〈a〉〈ab〉〈ab2〉

e

��

��

��

��

��

��

Here, 〈b〉 has order 3, while 〈a〉, 〈ab〉, and 〈ab2〉 have order 2. The upward-slanted linesrepresent the inclusions of subgroups.

Because the elements of monoids don’t necessarily have inverses, the definition of asubmonoid will have to be different from that of a subgroup.

Definition 2.2.17. A submonoid of a monoid M is a subset which is closed undermultiplication and contains the identity element.

Exercises 2.2.18.1. Show that in the real numbers R, the cyclic subgroup generated by 1 is the integers.

In particular, Z is cyclic.

2. In Z, show that 〈n〉 = Z if and only if n = ±1.

3. Consider the group, Q×+, of positive rational numbers under multiplication. What

are the elements of 〈2〉 ⊂ Q×+?

4. Consider the group Q× of nonzero rational numbers under multiplication. Whatare the elements of 〈−1〉 ⊂ Q×? What are the elements of 〈−2〉?

† 5. Let M be a monoid. We can define the positive powers of the elements in M inexactly the same way that positive powers in a group are defined. We have m1 = mfor all m ∈ M , and the higher powers are defined by induction: mk = mk−1m.Show that for m ∈M and for i, j ≥ 1, we have

(a) mi ·mj = mi+j , and

(b) (mi)j = mij .

6. In Z, show that 〈2, 3〉 = Z.

7. In Z, show that 〈3n, 5n〉 = 〈n〉 for any n ∈ Z.

8. In the group, Q×+, of positive rational numbers under multiplication, show that

〈2, 3〉 is not a cyclic subgroup. In other words, there is no rational number q suchthat 〈2, 3〉 = 〈q〉.

9. Show that Q×+ is generated by the set of all prime numbers.

10. Show that the group Q× of nonzero rational numbers is generated by the setconsisting of −1 and all of the prime numbers.


11. Let G be a group and let a, b ∈ G such that aba−1 ∈ 〈b〉. Show that H ={aibj | i, j ∈ Z} is a subgroup of G. Deduce that H = 〈a, b〉.

12. In Z, show that 〈2〉 ∩ 〈3〉 = 〈6〉.13. In Z, show that 〈m〉∩ 〈n〉 = 〈k〉, where k is the least common multiple of m and n.

14. In the group Q×+ of positive rational numbers under multiplication, show that

〈2〉 ∩ 〈3〉 = 1. Here, 1 is the trivial subgroup of Q×+.

15. Let M be a monoid. Show that the group of invertible elements Inv(M) is asubmonoid of M .

16. Show that not every submonoid of a group is a group.

17. Show that every submonoid of a finite group is a group.

2.3 The Subgroups of the Integers

One of the simplest yet most powerful results in mathematics is the Euclidean Algorithm.We shall use it here to identify all subgroups of Z and to derive the properties of primedecomposition in Z.

Theorem 2.3.1. (The Euclidean Algorithm2) Let m and n be integers, with n > 0.Then there are integers q and r, with 0 ≤ r < n, such that m = qn+ r.

Proof First, we assume that m ≥ 0, and argue by induction on m. If m < n, we maytake q = 0 and r = m. If m = n, we take q = 1 and r = 0. Thus, we may assume thatm > n and that the result holds for all non-negative integers less than m.

In particular, the induction hypothesis gives

m− 1 = q′n+ r′,

for integers q′ and r′ with 0 ≤ r′ < n. If r′ < n− 1, we may take q = q′ and r = r′ + 1for the desired result. Otherwise, m = (q′ + 1)n, and the proof for m ≥ 0 is complete.

If m < 0, then −m is positive, and hence −m = q′n + r′ with 0 ≤ r′ < n. If r′ = 0,this gives m = −q′n. Otherwise, we have m = −q′n− r′. Subtracting and adding a copyof n on the right of the equation gives m = (−q′ − 1)n+ (n− r′), and since 0 < r′ < n,we have 0 < n− r′ < n as well.

We shall use this to characterize the subgroups of Z. The subgroups we already knoware the cyclic ones:

〈n〉 = {qn | q ∈ Z}.

We shall next show that these are all the subgroups of Z. First, note that ifH is a nonzerosubgroup of Z, then there must be a nonzero element in it, and hence, by closure underinverses, a positive element. Since the set of positive integers less than or equal to agiven one is finite, there is a unique smallest positive element in it.

2The terminology that we’ve chosen here is not universal. There are some mathematicians who referto Theorem 2.3.1 as the Division Algorithm, and use the term Euclidean Algorithm for the procedurefor calculating greatest common divisors, outlined in Problem 3 of Exercises 2.3.18.


Proposition 2.3.2. Let H ⊂ Z be a subgroup, and suppose that H �= 0. Let n be thesmallest positive element of H. Then H = 〈n〉.

Proof Since n ∈ H, 〈n〉 ⊂ H. Thus, it suffices to show that H ⊂ 〈n〉.For m ∈ H, write m = qn + r with q and r integers and 0 ≤ r < n. Now m is in

H, as is qn. Thus, m− qn = r must be in H. But r is non-negative and is less than n.Since n is the smallest positive integer in H, we must have r = 0. Thus, m = qn ∈ 〈n〉.

The language of divisibility is useful for discussing subgroups of Z:

Definition 2.3.3. Let m and n be integers. We say n divides m (or that m is divisibleby n) if there is an integer q such that m = nq.

The next lemma is now immediate from the fact that 〈n〉 is the set of all multiplesof n.

Lemma 2.3.4. Let m and n be integers. Then m ∈ 〈n〉 if and only if n divides m.

By Lemma 2.2.9, Lemma 2.3.4 may be restated this way.

Lemma 2.3.5. Let m and n be integers. Then n divides m if and only if 〈m〉 ⊂ 〈n〉.We shall also need a uniqueness statement about generators of cyclic subgroups.

Proposition 2.3.6. Let m and n be integers such that 〈m〉 = 〈n〉. Then m = ±n.

Proof Clearly, if either of m and n is 0, so is the other. By passage to negatives, ifnecessary, we may assume that m and n are both positive. By Lemma 2.3.5, they mustdivide each other. Say m = qn and n = rm. But then q and r are positive. If q > 1,then m > n, and hence n = rm > rn ≥ n, i.e., n > n, which is impossible. Thus, q = 1,and hence m = n.

Corollary 2.3.7. Let H be a subgroup of Z. Then there is a unique non-negative integerwhich generates H.

Definition 2.3.8. Let m and n be integers. We write

〈m〉+ 〈n〉 = {rm+ sn | r, s ∈ Z}.

Lemma 2.3.9. 〈m〉+ 〈n〉 is a subgroup of Z.

Proof The inverse of rn+ sm is (−r)m+ (−s)n. The result follows since (rm+ sn) +(r′m+ s′n) = (r + r′)m+ (s+ s′)n ∈ 〈m〉+ 〈n〉.

In fact, it is easy to see that 〈m〉 + 〈n〉 is the subgroup of Z generated by the set{m,n}.Definition 2.3.10. Let m,n ∈ Z. We write (m,n) for the unique non-negative integerwhich generates the subgroup 〈m〉+ 〈n〉 of Z:

〈m〉+ 〈n〉 = 〈(m,n)〉.

We shall refer to (m,n) as the greatest common divisor, or g.c.d., of m and n.


We shall now derive the familiar properties of the g.c.d. and of prime decompositionfrom the above results.

Proposition 2.3.11. Let m and n be integers. Then an integer k divides both m and nif and only if it divides (m,n). In particular, if m and n are not both 0, then (m,n) isthe largest integer which divides both m and n.

In addition, there exist integers r and s with rm+ sn = (m,n).

Proof The last statement follows because

(m,n) ∈ 〈m〉+ 〈n〉 = {rm+ sn | r, s ∈ Z}.

Now let k divide both m and n. Then it must divide any number of the form rm+sn,including (m,n). Conversely, to show that the divisors of (m,n) must divide both m andn, it suffices to show that (m,n) divides both m and n. But m = 1 · m + 0 · n andn = 0 ·m+ 1 · n both lie in 〈(m,n)〉, so the result follows from Lemma 2.3.4.

Definition 2.3.12. We say that the integers m and n are relatively prime if (m,n) = 1.

Note that if 1 ∈ 〈m〉+ 〈n〉, then 〈m〉+ 〈n〉 = Z. The following lemma is immediate.

Lemma 2.3.13. The integers m and n are relatively prime if and only if there existintegers r and s with rm+ sn = 1.

Definition 2.3.14. We say that an integer p > 1 is prime if its only divisors are ±1 and±p. By Lemma 2.3.5, this says that 〈p〉 ⊂ 〈k〉 if and only if either 〈k〉 = 〈p〉 or 〈k〉 = Z.

Lemma 2.3.15. Let p and a be integers, with p prime. Then p and a are relativelyprime if and only if p does not divide a.

Proof The greatest common divisor of p and a must be either p or 1. In the formercase, p divides a. In the latter, it does not.

Proposition 2.3.16. Let p be a prime number. Then p divides a product ab if and onlyif it divides at least one of a and b.

Proof Clearly, if p divides a or b, it must divide their product. Conversely, suppose pdivides ab, but does not divide a. We shall show that it must divide b.

Since p does not divide a, we have (p, a) = 1, so that qp+ ra = 1 for some integers qand r. But then qpb+ rab = b. Since p divides both qpb and rab, it must divide b.

We can now give the fundamental theorem regarding prime factorization.

Theorem 2.3.17. Every integer m > 1 has a unique prime decomposition: there is afactorization

m = pr11 . . . prk

k ,

where pi is prime and ri > 0 for 1 ≤ i ≤ k, and pi < pj for i < j. Uniqueness meansthat if m = qs11 . . . qsl

l with qi prime and si > 0 for 1 ≤ i ≤ l and qi < qj for i < j, thenwe must have k = l and have pi = qi and ri = si for 1 ≤ i ≤ k.


Proof The existence of a prime decomposition amounts to showing that any integerm > 1 is a product of primes. Clearly, this is true for m = 2. We argue by induction onm, assuming that any integer k with 1 < k < m has a prime decomposition.

If m is prime, we are done. Otherwise, m = kl, where both k and l are greater than1 and less than m. But then k and l are products of primes by induction, and hence sois m.

For uniqueness, suppose given decompositions pr11 . . . prk

k and qs11 . . . qsl

l of m as above.We argue by induction on t = r1 + · · ·+ rk.

If t = 1, thenm = p1, a prime. But then p1 is divisible by q1, and since both are prime,they must be equal. We then have p1(1−qs1−1

1 . . . qsl

l ) = 0, so that (1−qs1−11 . . . qsl

l ) = 0,and hence qs1−1

1 . . . qsl

l = 1. Since the qi are prime and since no positive number divides1 other than 1 itself, we must have l = 1 and s1 = 1.

Suppose t > 1. Since p1 divides m, Proposition 2.3.16, together with an inductionon s1 + · · ·+ sl, shows that p1 must divide qi for some i. But since p1 and qi are prime,we must have p1 = qi. A similar argument shows that q1 = pj for some j, and ourorder assumption then shows that i = j = 1, so that p1 = q1. Now p1(pr1−1

1 . . . prk

k −qs1−11 . . . qsl

l ) = 0, so that pr1−11 . . . prk

k = qs1−11 . . . qsl

l . But our inductive hypothesisapplies here, and the resulting equality is exactly what we wanted.

Exercises 2.3.18.1. Let p1, . . . , pk be distinct primes and let ri ≥ 0 for 1 ≤ i ≤ k. Show that an integern divides pr11 . . . prk

k if and only if n = ±ps11 . . . psk

k with 0 ≤ si ≤ ri for 1 ≤ i ≤ k.2. Show that there are infinitely many prime numbers.

† 3. Let n = qm+ r with 0 ≤ r < m. Show that (m,n) = (r,m). Use this to obtain aniterative procedure to calculate (m,n).

4. Implement the iterative procedure above by a computer program to calculate great-est common divisors.

† 5. A subgroup G ⊂ R is called discrete if for each g ∈ G there is an open interval(a, b) ⊂ R such that (a, b) ∩ G = {g}. Show that every discrete subgroup of R iscyclic.

2.4 Finite Cyclic Groups: Modular Arithmetic

Here, we develop Zn, the group of integers modulo n.

Definition 2.4.1. Let n > 0 be a positive integer. We say that integers k and l arecongruent modulo n, written k ≡ l mod n, or sometimes k ≡ l (n), if k− l is divisibleby n (i.e., k − l ∈ 〈n〉).Lemma 2.4.2. Congruence modulo n is an equivalence relation.

Proof Clearly, k − k ∈ 〈n〉 for any n, so congruence modulo n is reflexive. It is alsosymmetric, since l − k = −(k − l) is contained in any subgroup containing k − l. Ifk ≡ l mod n and l ≡ m mod n, then both k − l and l −m are in 〈n〉, and hence so isk −m = (k − l) + (l −m). Thus, the relation is transitive as well.


Definition 2.4.3. We write Zn for the set of equivalence classes of integers modulo n,and write m ∈ Zn for the equivalence class containing m ∈ Z. 3 We shall also use 0 todenote 0.

Zn is called the group of integers modulo n.

Lemma 2.4.4. There are exactly n elements in Zn, represented by the elements 0 =0, 1, . . . , n− 1.

Proof The elements listed are precisely the classes r for 0 ≤ r < n. For m ∈ Z, theEuclidean algorithm provides an equation m = qn+ r, with 0 ≤ r < n. But then m = r,and hence m is in the stated list of elements.

To see that these classes are all distinct, suppose that 0 ≤ k < l < n. Then 0 <l − k < n, so that l − k is not divisible by n, and hence k �= l in Zn.

There are two important binary operations on Zn.

Definitions 2.4.5. The operations of addition and multiplication in Zn are defined bysetting k + l = k + l and k · l = kl, respectively, for all k, l ∈ Z.

We must show that these operations are well defined.

Lemma 2.4.6. The operations of addition and multiplication are well defined binaryoperations on Zn.

Proof Suppose that k = q and l = r in Zn. In other words, k ≡ q mod n and l ≡ rmod n. We must show that k + l ≡ q + r mod n and kl ≡ qr mod n.

Addition is easy: k− q ∈ 〈n〉 and l−r ∈ 〈n〉. Now, (k+ l)− (q+r) = (k− q)+(l−r).This lies in 〈n〉, as 〈n〉 closed under addition.

For multiplication, we have kl − qr = kl − kr + kr − qr = k(l − r) + r(k − q). Thislast is divisible by n because l − r and k − q are.

The chief properties of addition and multiplication are inherited from those in theintegers:

Proposition 2.4.7. Addition gives Zn the structure of an abelian group with identity0. Multiplication makes Zn into an abelian monoid with identity 1. The two operationssatisfy the distributive law: a(b+ c) = ab+ a c.

Proof

(a+ b) + c = a+ b+ c = a+ b+ c.

Since a + (b + c) has the same expansion, addition is associative. The other assertionsfollow by similar arguments.

In the language of abstract algebra, Proposition 2.4.7 says, simply, that the operationsof addition and multiplication give Zn the structure of a commutative ring.

When we refer to Zn as a group, we, of course, mean with respect to addition. ByLemma 2.4.4, Zn has order n.

Corollary 2.4.8. Let n be a positive integer. Then there are groups of order n.3In the notation of Section 1.4, Zn = Z/ ≡, where ≡ denotes equivalence modulo n.


Definition 2.4.9. The canonical map π : Z→ Zn is defined by setting π(k) = k for allk ∈ Z.

Given that Zn is the set of equivalence classes of integers with respect to the equiva-lence relation of congruence modulo n, and that π takes each integer to its equivalenceclass, we see that this coincides with the canonical map defined in Section 1.4 for anarbitrary equivalence relation.

We shall see in the next section that the canonical map π : Z→ Zn is what’s knownas a homomorphism of groups.

Exercises 2.4.10.1. Show that Zn is generated as a group by 1. Thus, Zn is a cyclic group.

2. Show that 2 does not generate Z4. Deduce that there are finite cyclic groups thatpossess nontrivial proper subgroups.

3. Let p be a prime. Show that the nonzero elements of Zp form a group undermultiplication modulo p.

4. If n is not prime, show that the nonzero elements of Zn do not form a group undermultiplication modulo n.

2.5 Homomorphisms and Isomorphisms

There are many ways to construct a finite set, but the thing that describes the essenceof the structure of such a set is the number of elements in it. Two sets with the samenumber of elements may be placed in one-to-one correspondence, at which point theirproperties as sets are indistinguishable.

In the case of groups, the structure is too complicated to be captured by the numberof elements alone. We shall see examples of finite groups with the same order whosegroup properties are quite different. For instance, one may be abelian and the other not.

But the idea of a one-to-one correspondence makes a good start at deciding when twogroups may be identified, provided that the correspondence respects the multiplicationsof the groups.

Definition 2.5.1. An isomorphism between two groups is a function f : G→ G′ whichis one-to-one and onto, such that f(x · y) = f(x) · f(y) for all x, y ∈ G. Here, · is usedfor the multiplications of both G and G′.

We say that the groups G and G′ are isomorphic if there exists an isomorphism fromG to G′, and write G ∼= G′.

This turns out to be enough to capture the full structure of the groups in question.A glance at the definition shows that there is a weaker notion which is part of it:

Definition 2.5.2. Let G and G′ be groups. A homomorphism from G to G′ is a functionf : G→ G′ such that f(x · y) = f(x) · f(y) for all x, y ∈ G.

Thus, an isomorphism is a homomorphism which is one-to-one and onto.Recall that one-to-one functions are sometimes called injective functions, or injections,

and that onto functions are known as surjective functions, or surjections. Also, a functionwhich is both one-to-one and onto is called bijective.

Clearly, an understanding of homomorphisms will contribute to our understandingof isomorphisms, but in fact, homomorphisms will turn out to say a lot more about the


structure of a group than may be apparent at this point. Before we dive into theory, let’sconsider some examples.

Examples 2.5.3.1. For any two groups, G and G′, there is a homomorphism f : G → G′ obtained by

setting f(g) = e for all g ∈ G. We call this the trivial homomorphism from G toG′.

2. Let H be a subgroup of the group G. Then the inclusion i : H ⊂ G is a homomor-phism, because the multiplication in H is inherited from that of G.

3. Recall that the canonical map π : Z → Zn is defined by π(k) = k for all k ∈ Z.Thus, π(k + l) = k + l = k + l = π(k) + π(l), and hence π is a homomorphism.

For some pairs G, G′ of groups, the trivial homomorphism is the only homomorphismfrom G to G′.

The relationship between a homomorphism and the power laws in a group is animportant one:

Lemma 2.5.4. Let f : G → G′ be a homomorphism of groups and let g ∈ G. Thenf(gn) = f(g)n for all n ∈ Z. In particular, f(e) = e (recall that e = g0) and f(g−1) =f(g)−1.

Proof For n > 0, this is a quick induction from the definition of powers. For n = 0,we have f(e) = f(e · e) = f(e) · f(e). Multiplying both sides of the resulting equationf(e) = f(e) · f(e) by f(e)−1, we get e = f(e). (Any group element which is its ownsquare must be the identity element of the group.) For n < 0, we have e = f(e) =f(gng−n) = f(gn)f(g−n) = f(gn)f(g)−n by the two cases above, so that f(gn) is theinverse of f(g)−n.

Since the canonical map π : Z → Zn is a homomorphism, Lemma 2.2.7 gives us thenext corollary.

Corollary 2.5.5. Let m ∈ Zn and let k be any integer. Then the k-th power of m withrespect to the group operation in Zn is km.

Another consequence of Lemma 2.5.4 is a determination of all of the homomorphismsfrom Z to an arbitrary group G.

Proposition 2.5.6. Let G be a group and let g ∈ G. Then there is a unique homomor-phism fg : Z→ G with fg(1) = g.

Proof Define fg by fg(n) = gn for all n ∈ Z. This is a homomorphism by the powerlaws of Lemma 2.2.6.

For uniqueness, suppose that f : Z→ G is an arbitrary homomorphism with f(1) = g.We’d like to show that f(n) = gn. Since n is the n-th power of 1 with respect to thegroup operation in Z, this follows from Lemma 2.5.4.

The most basic properties of homomorphisms are the following. We leave the proofto the reader.

Lemma 2.5.7. Let f : G → G′ and g : G′ → G′′ be homomorphisms. Then so is thecomposite g ◦ f : G→ G′′. If f and g are both isomorphisms, then so is g ◦ f . If f is anisomorphism, then so is the inverse map f−1 : G′ → G.


There are two subgroups associated with a homomorphism that can be useful indetecting its deviation from being an isomorphism. One of them is the kernel:

Definition 2.5.8. Let f : G → G′ be a homomorphism. Then the kernel of f , writtenker f , is given by

ker f = {x ∈ G | f(x) = e}.

Recall that the image of a function f : X → Y is the subset im f ⊂ Y given by

im f = {y ∈ Y | y = f(x) for some x ∈ X}.

An easy check, using Lemma 2.5.4, shows that the kernel and image of a homomor-phism f : G→ G′ are subgroups of G and G′, respectively.

Lemma 2.5.9. Let f : G→ G′ be a homomorphism. Then ker f detects the deviation off from being injective: we have f(x) = f(x′) if and only if x−1x′ ∈ ker f . In particular,f is injective if and only if the kernel of f is the trivial subgroup, e, of G.

The image detects whether f is onto: it is onto if and only if im f = G′.

Proof Suppose that f(x) = f(x′). Then f(x−1x′) = f(x)−1f(x′) = e, and hencex−1x′ ∈ ker f . Conversely, if x−1x′ ∈ ker f , then f(x′) = f(xx−1x′) = f(x)f(x−1x′) =f(x) · e, and hence f(x′) = f(x).

Suppose that f is injective. Since f(e) = e, no element of G other than e may becarried to e by f , and hence ker f = e. Conversely, if ker f = e, then x−1x′ ∈ ker fimplies that x−1x′ = e, so that x = x′. But then x = x′ whenever f(x) = f(x′), andhence f is injective.

The statement regarding im f is immediate.

We shall refer to an injective homomorphism as an embedding. The following lemmais immediate:

Lemma 2.5.10. Let f : G → G′ be an embedding. Then f induces an isomorphism ofG onto im f .

Definitions 2.5.11. Let G be a group and let n ∈ Z. We say that g ∈ G has exponentn if gn = e.

If there is a positive integer n which is an exponent for g, we say that g has finiteorder, and define the order, o(g), of g to be the smallest positive integer which is anexponent for g.

If g does not have a positive exponent, we say that o(g) =∞.

If one knows the order of an element, one can determine all possible exponents for it.

Lemma 2.5.12. Let G be a group and let g ∈ G be an element of finite order. Let n beany integer. Then g has exponent n if and only if o(g) divides n.

Proof Of course, gmo(g) = (go(g))m = em = e, so if o(g) divides n, then g has exponentn.

Conversely, if g has order k and exponent n, write n = qk + r with 0 ≤ r < k.Then e = gn = gqkgr = gr, since both n and qk are exponents for g. But this says ghas exponent r. If r were positive, this would contradict the fact that k is the smallestpositive exponent for g. Thus, r = 0, and n = qk, and hence k divides n, as claimed.


Information about exponents sometimes leads to a determination that a particularelement must be the identity.

Corollary 2.5.13. Let m and n be integers that are relatively prime. Suppose that mand n are both exponents for an element g ∈ G. Then g = e.

Proof The order of g has to divide both m and n, and hence it must also divide theirg.c.d. Since the order of an element is positive, g has order 1, and hence g1 = e.

Let fg : Z → G be the unique homomorphism with fg(1) = g. Then fg(n) = gn forall n ∈ Z. Thus, an integer n is in the kernel of fg if and only if g has exponent n. Also,im fg is precisely 〈g〉. The next result now follows immediately from Proposition 2.3.2.It includes, in part, a slicker argument for Lemma 2.5.12.

Proposition 2.5.14. Let G be a group and let g ∈ G. Let fg : Z → G be the uniquehomomorphism with fg(1) = g. Then the kernel of fg is 〈o(g)〉, if o(g) is finite, and is 0otherwise.4 The image of fg is 〈g〉, the cyclic subgroup of G generated by g.

Lemma 2.5.10 now gives us the following corollary.

Corollary 2.5.15. If o(g) =∞, then fg : Z→ 〈g〉 is an isomorphism.

Corollary 2.5.16. Every nonzero subgroup of Z is isomorphic to Z.

Proof Since mn = 0 if and only if either m or n is 0, no nonzero element of Z has finiteorder. But every subgroup of Z is cyclic.

Next, we study the homomorphisms out of Zn. We shall make use of the notion ofcommutative diagrams, as defined in Definition 1.2.1.

Proposition 2.5.17. Let G be a group and let g ∈ G. Then there is a homomorphismhg : Zn → G with hg(1) = g if and only if g has exponent n.

If g does have exponent n, then there is only one such homomorphism, and it has theproperty that the following diagram commutes

Zfg ��

π��

��

�� G,

Zn

hg

��

meaning that hg ◦ π = fg. Here π : Z → Zn is the canonical map, while fg : Z → Gis the unique homomorphism that takes 1 to g. The homomorphism hg must satisfy theformula

hg(m) = gm

for all m ∈ Zn.If the homomorphism hg exists, then its image is 〈g〉. Moreover, the induced map

hg : Zn → 〈g〉 is an isomorphism if and only if o(g) = n.In particular, if g ∈ G has finite order, then 〈g〉 ∼= Zo(g).

4Recall that, since we are using additive notation for Z, we write 0 for the trivial subgroup consistingof the identity element alone.


Proof Suppose there exists a homomorphism hg : Zn → G with hg(1) = g. Thenhg ◦ π is a homomorphism from Z to G which takes 1 to hg(π(1)) = hg(1) = g. Thus,hg ◦ π = fg, and hence hg(m) = hg(π(m)) = fg(m) = gm for all m ∈ Zn. Thus, hg isunique if it exists. Moreover, the existence of hg gives e = hg(0) = hg(n) = gn, sincen = 0 in Zn, and hence g has exponent n. Clearly, imhg = 〈g〉.

Conversely, if g has exponent n, we wish to show that hg(m) = gm gives a well definedfunction from Zn to G. Ifm = l, thenm−l = sn for some integer s, and hencem = l+sn.But then gm = gl(gn)s = gles = gl by the rules of exponents. Thus, hg is well defined.But hg(q + r) = gq+r = gqgr = hg(q)hg(r), and hence hg is a homomorphism.

It suffices to show that hg is injective if and only if o(g) = n. Suppose first thato(g) �= n. Since g has exponent n, o(g) divides n. Thus, 0 < o(g) < n, and hence0 �= o(g) ∈ Zn. Since o(g) ∈ kerhg, hg is not injective.

On the other hand, if o(g) = n, let k ∈ kerhg. But this says that gk = e, and henceg has exponent k. But that means that k is divisible by o(g) = n, and hence k = 0 inZn. In particular, kerhg = 0, as desired.

Since Zn has order n, we obtain the following corollary.

Corollary 2.5.18. Let G be a group and let g ∈ G. Then the order of g is equal to theorder of the group 〈g〉.Corollary 2.5.19. Let f : G → H be a homomorphism, and suppose that g ∈ G hasfinite order. Then the order of f(g) divides the order of g.

Proof Let hg : Zo(g) → G be the homomorphism which carries 1 to g. Then f ◦ hg :Zo(g) → H is a homomorphism that takes 1 to f(g). By Proposition 2.5.17, f(g) hasexponent o(g), so the order of f(g) divides o(g).

Because the elements of monoids do not all have inverses, the definition of homomor-phism must be modified for it to be useful for monoids.

Definitions 2.5.20. A homomorphism of monoids is a function f : M →M ′ such that

1. f carries the identity element of M to that of M ′.

2. For m,m′ ∈M , f(mm′) = f(m)f(m′).

An isomorphism of monoids is a homomorphism f : M →M ′ which is bijective.

Exercises 2.5.21.1. Show that any group with one element is isomorphic to the trivial group.

2. Show that any group with two elements is isomorphic to Z2.

3. List the homomorphisms from Z9 to Z6.

4. List the homomorphisms from Z5 to Z6.

5. Show that the order of m in Zn is n/(m,n). Deduce that the order of each elementdivides the order of Zn. Deduce that every non-identity element of Zp has order p,for any prime p.

† 6. Let G be a group and let x ∈ G. Show that o(x−1) = o(x). (Hint : We may assumeG = 〈x〉.)


7. Show that a group G is cyclic if and only if there is a surjective homomorphismf : Z→ G.

† 8. Let f : G→ G′ be a homomorphism.

(a) Let H ′ ⊂ G′ be a subgroup. Define f−1(H ′) ⊂ G by

f−1(H ′) = {x ∈ G | f(x) ∈ H ′}.Show that f−1(H ′) is a subgroup of G which contains ker f .

(b) Let H ⊂ G be a subgroup. Define f(H) ⊂ G′ by

f(H) = {f(x) |x ∈ H}.Show that f(H) is a subgroup of G′. If ker f ⊂ H, show that f−1(f(H)) = H.

(c) Suppose that H ′ ⊂ im f . Show that f(f−1(H ′)) = H ′. Deduce that there is aone-to-one correspondence between the subgroups of im f and those subgroupsof G that contain ker f . What does this tell us when f is onto?

9. Let f : G→ G′ be a homomorphism and let x ∈ G. Show that f(〈x〉) = 〈f(x)〉.10. Show that every subgroup of Zn is cyclic.

11. Show that Zn has exactly one subgroup of order d for each d dividing n, and hasno other subgroups. (In particular, among many other important implications, thisshows that the order of every subgroup of Zn divides the order of Zn.) Deduce thefollowing consequences.

(a) Show that a group containing a subgroup isomorphic to Zk × Zk for k > 1cannot be cyclic.

(b) Show that if a prime p divides n, then Zn has exactly p− 1 elements of orderp.

(c) Let H and K be subgroups of Zn. Show that H ⊂ K if and only if |H| divides|K|.

(d) Show that if d divides n, then Zn has exactly d elements of exponent d.

12. Let H and K be subgroups of Zpr , where p is prime. Show that either H ⊂ K orK ⊂ H.

13. Let p and q be distinct prime numbers and let n be divisible by both p and q.Find a pair of subgroups H,K ⊂ Zn such that neither H nor K is contained in theother.

14. Show that a group G is abelian if and only if the function f : G → G given byf(g) = g2 is a homomorphism.

15. Show that a group G is abelian if and only if the function f : G → G given byf(g) = g−1 is a homomorphism.

16. Let G be an abelian group and let n be an integer. Show that the function f : G→G given by f(g) = gn is a homomorphism.

17. Show that the exponential function exp(x) = ex gives an isomorphism from theadditive group of real numbers to R×

+.


18. Show that a group is finite if and only if it has finitely many subgroups.

19. Let i : Gln(R)→Mn+1(R) be defined by setting i(A) to be the matrix whose firstn rows consist of the matrix A with a column of 0’s added on the right, and whoselast row consists of n 0’s followed by a 1, for all A ∈ Gln(R). Pictorially,

i(A) =

⎛⎜⎜⎜⎝0

A...0

0 . . . 0 1

⎞⎟⎟⎟⎠ .

Show that i(A) lies in Gln+1(R), and that i defines an embedding from Gln(R)into Gln+1(R).

20. Show that there is a monoid with two elements which is not a group. Show thatany other monoid with two elements which is not a group is isomorphic to this one.

21. Let f : M →M ′ be a homomorphism of monoids. Show that f restricts to a grouphomomorphism f : Inv(M)→ Inv(M ′) between their groups of invertible elements.In particular, if m ∈M is invertible, then so is f(m).

22. Give an example to show that condition 1 in the definition of monoid homomor-phism does not follow from condition 2.

2.6 The Classification Problem

Definition 2.6.1. By a classification of the groups with a certain property, we meanthe identification of a set of groups {Gi | i ∈ I}5 such that the following conditions hold.

1. Each group Gi has the stated property.

2. If Gi is isomorphic to Gj , then i = j.

3. Each group with the stated property is isomorphic to one of the Gi.

A classification of groups with a given property is often referred to as a classification“up to isomorphism” of the groups in question.

The property by which we collect the groups to be classified is often their order.This is a natural criterion, as isomorphic groups must have the same order. However,other criteria are often used. For instance, the Fundamental Theorem of Finite AbelianGroups, which we give in Section 4.4, gives a classification of all finite abelian groups.

Examples 2.6.2.1. Problem 2 of Exercises 2.5.21 shows that every group of order 2 is isomorphic to

Z2. This classifies the groups of order 2.

2. Corollary 2.5.15 and Proposition 2.5.17 show that every cyclic group is isomorphicto exactly one of the groups in {Z, e,Zn |n ≥ 2}. Thus, we obtain a classificationof the cyclic groups.

5Here the set I is called the indexing set for the set {Gi | i ∈ I} of groups.


Classification can be phrased in more general terms using the language of categorytheory. (See Chapter 6.) In general, questions of classification are the most fundamentaland important questions that can be asked in mathematics.

It may come as a surprise to some readers that the finite groups have not yet beenclassified. It is often astounding that questions which may be so easily phrased can beso difficult to answer.

Modern mathematics abounds with such questions. A beginning student of math-ematics has no conception of the number of important, yet really basic, mathematicalquestions that have not yet been solved, despite the number of astute mathematiciansthat have been working on them over the years. The surprise that comes later is howmany interesting open questions are amenable to solutions, given the right mixture ofideas, luck, and persistence.

The main goal of the group theory portion of this book is to be able to obtain asmuch information as possible about the classification of groups of relatively small order.In particular, for a given integer n, say with n ≤ 63, we’d like to classify all groups oforder n.

There is some language we can use to make our discussion clearer.

Definition 2.6.3. We say that two groups are in the same isomorphism class if thegroups are isomorphic to one another.

Isomorphism classes are analogous to equivalence classes, where the equivalence rela-tion used is that of isomorphism. The point is that the collection of all groups of a givenorder is not a set, as it has too many elements. (There are too many groups of order 1,for instance, to form a set, even though there is only one isomorphism class of groupsof order 1.) Thus, we are looking at an analogue for classes (as opposed to sets) of anequivalence relation.

The interesting thing is that classification, as defined above, can only occur in situ-ations where the isomorphism classes themselves form a set. Indeed, the next lemma isan immediate consequence of the definition of a classification.

Lemma 2.6.4. The set {Gi | i ∈ I} classifies the collection of groups with a givenproperty if and only if each isomorphism class of the groups with the stated propertyincludes exactly one of the groups Gi. In particular, the collection of isomorphismclasses of the groups in question forms a set which is in one-to-one correspondence with{Gi | i ∈ I}.

Notice that if X is a finite set, then the set of possible group multiplications we couldput on X is a subset of the set of functions from X×X to X. But the set of all functionsfrom one finite set to another is finite, by Corollary 1.7.6. Thus, there are finitely manypossible group multiplications on a finite set. Different multiplications will in many casesgive rise to isomorphic groups, but the number of isomorphism classes of groups with|X| elements must be finite.

Proposition 2.6.5. Let n > 0. Then there are finitely many isomorphism classes ofgroups of order n.

A countable union of finite sets is countable.

Corollary 2.6.6. The set of isomorphism classes of finite groups is countable.

Exercises 2.6.7.1. Let X = {x, y} be a set with two elements. Which of the functions from X ×X toX are group multiplications? Which are monoid multiplications?

2. Classify the groups of order 3.


2.7 The Group of Rotations of the Plane

Here, we describe an important subgroup of Gl2(R), the group, SO(2), of rotations ofthe plane.

Definition 2.7.1. For θ ∈ R, we define the rotation matrix Rθ by

Rθ =(

cos θ − sin θsin θ cos θ

).

Here, the sine and cosine are taken with respect to radian measure.

Elements of M2(R) act as transformations of the plane in the usual way. We identifythe plane with the space of column vectors �v = ( xy ). A matrix A =

(a bc d

)acts on the

plane by the usual multiplication of matrices on columns:

A · �v =(ax+ bycx+ dy

).

To display the geometric effect of the transformation induced by a rotation matrix, wemake use of polar coordinates and the sum formulæ for sines and cosines, as developedin calculus.

Proposition 2.7.2. The transformation of the plane induced by the rotation matrix Rθis the counterclockwise rotation of the plane through the angle θ.

Proof We write the vector �v = ( xy ) in polar coordinates, setting x = r cosφ andy = r sinφ, where r =

√x2 + y2 is the length of �v and φ is the angle that starts at the

positive x-axis and ends at �v. The product formula now gives

Rθ · �v =(r(cos θ cosφ− sin θ sinφ)r(cos θ sinφ+ sin θ cosφ)

)=

(r cos(θ + φ)r sin(θ + φ)

),

by the sum formulæ for the sine and cosine. Thus, the length of Rθ ·�v is the same as thatof �v, but the angle that it makes with the positive x-axis has been increased by θ.

As geometric intuition would suggest, the product (composite) of two rotations isagain a rotation:

Lemma 2.7.3. For any real numbers θ and φ, we have

Rθ ·Rφ = Rθ+φ.

Proof The definition of matrix multiplication gives

Rθ ·Rφ =(

cos θ cosφ− sin θ sinφ − cos θ sinφ− sin θ cosφcos θ sinφ+ sin θ cosφ cos θ cosφ− sin θ sinφ

).

Once again, the result follows from the sum formulæ for the sine and cosine.

Definition 2.7.4. Let SO(2) ⊂ Gl2(R) be the set of all rotation matrices Rθ, as θranges over all real numbers. We call it the second special orthogonal group.

By Lemma 2.7.3, SO(2) is closed under multiplication in Gl2(R). Moreover, sincecos 0 = 1 and sin 0 = 0, we have R0 = I2, the identity element of Gl2(R). SinceRθ ·R−θ = R0, R−θ is the inverse of Rθ.


Lemma 2.7.5. SO(2) is a subgroup of Gl2(R).

Notice that since addition of real numbers is commutative, Lemma 2.7.3 shows thatRθ ·Rφ = Rφ ·Rθ. Thus, unlike Gl2(R), SO(2) is an abelian group. Lemma 2.7.3 actuallyshows more:

Proposition 2.7.6. There is a homomorphism, exp, from R onto SO(2), defined byexp(θ) = Rθ. The kernel of exp is 〈2π〉 = {2πk | k ∈ Z}. Thus, Rθ = Rφ if and only ifθ − φ = 2πk for some k ∈ Z.

Proof That exp is a homomorphism is the content of Lemma 2.7.3. Suppose thatθ ∈ ker exp. Since the identity element of SO(2) is the identity matrix, the definition ofRθ forces cos θ to be 1 and sin θ to be 0. From calculus, we know this happens if andonly if θ is a multiple of 2π.6 By Lemma 2.5.9, we have that exp θ = expφ if and only if−φ+ θ ∈ ker(exp), so the result follows.

We conclude with a calculation of the order of the elements of SO(2).

Proposition 2.7.7. The rotation Rθ has finite order if and only if θ = 2πk/n for someintegers k and n. The order of R2πk/n is n/(k, n).

Proof Suppose that Rθ has finite order. Then (Rθ)n = Rnθ = I2 for some integer n.Then Proposition 2.7.6 shows that nθ = 2πk for some integer k, so that θ = 2πk/n.

If θ = 2πk/n, the order of Rθ will be the smallest positive integer m for which mθis an integer multiple of 2π. Thus, m is the smallest positive integer such that mk isdivisible by n. But a glance at the prime decompositions of n and k shows that thissmallest positive integer is precisely n/(n, k).

In particular, R2π/n has order n for all n ≥ 1.

Corollary 2.7.8. There are embeddings of Zn in SO(2) for all n ≥ 1.

Since SO(2) gives transformations of the plane, this says that Zn may be used toproduce geometric transformations of a familiar space. A certain school of thoughtwould summarize this by saying that Zn “exists in nature.”

2.8 The Dihedral Groups

Let n > 0, we construct a group, D2n, called the dihedral group of order 2n. Thesegroups are nonabelian for n > 2, and provide our first examples of nonisomorphic groupsof the same order.

We give a definition here via matrices, which relies on calculus. A direct definitionas an abstract group is also possible, but is a little awkward notationally without somefurther discussion. We shall give such a derivation in Section 4.6.

Let b, or bn, if more than one dihedral group is being discussed, be the rotation ofthe plane through the angle 2π/n: b = R2π/n ∈ SO(2) in the notation of the last section.Then by Proposition 2.7.7, b has order n in Gl2(R). Let a ∈ Gl2(R) be the matrix

a =(

1 00 −1

).

6Actually, one of the nicer ways to prove this fact from calculus is to use derivatives to show that thekernel of exp is a discrete subgroup of R and apply Problem 5 of Exercises 2.3.18. One may then define2π as the smallest positive element in ker(exp) and use numerical methods to compute its value. Here,we may take the sine and cosine to be defined by their Taylor series, and derive their basic properties,such as their derivatives and their sum formulæ, by manipulating the series.


The matrix a is said to reflect the plane through the x-axis, and we say that a matrixis a reflection matrix if it is a product a · c with c ∈ SO(2). (It is possible to give adirect geometric definition of reflection using dot products, and then to prove that thesegeometric operations are induced by precisely the matrices just stated. We shall treatthis approach in exercises below, and content ourselves here with the algebraic definitionjust given.)

The most important piece of information about rotations and reflections is the waythey interrelate. They don’t commute with each other, but they come close to it:

Lemma 2.8.1. For θ ∈ R, we have

Rθ · a = a ·R−θ.

Thus, for b = R2π/n as above, bk · a = a · b−k for any integer k.

Proof Multiplying matrices, we get

Rθ · a =(

cos θ sin θsin θ − cos θ

).

This is equal to the matrix product a ·R−θ, since cos(−θ) = cos θ and sin(−θ) = − sin θ.The statement about powers of b follows since b±k = R±k·2π/n.

The following lemma may be verified by direct computation.

Lemma 2.8.2. The matrix a has order 2.

Definition 2.8.3. Let n ≥ 1. Then the dihedral group of order 2n, D2n ⊂ Gl2(R), isgiven as follows:

D2n = {bk, abk | 0 ≤ k < n},

with a and b = bn as above.

Of course, this definition needs a bit of justification:

Proposition 2.8.4. The dihedral group D2n is a subgroup of Gl2(R). Moreover, theelements {bk, abk | 0 ≤ k < n} are all distinct, so that D2n has order 2n.

Proof We first show that the 2n elements bk and abk with 0 ≤ k < n are all distinct. Ifabk = bl, then multiplication by b−k on the right gives a = bl−k, a rotation matrix. Butrotation matrices all commute with one another, and Lemma 2.8.1 shows that a fails tocommute with most rotations. Thus, abk �= bl.

If abk = abl, then multiplying both sides on the left by a−1 gives bk = bl, whichimplies that k ≡ l mod n by Proposition 2.5.17. With 0 ≤ k, l < n, this gives k = l byLemma 2.4.4.

We now show that D2n is closed under multiplication in Gl2(R). By Lemma 2.8.1,we have abk · abl = a2bl−k = bl−k, as a has order 2. Since b has order n, any power of bis equal to one in the list.

Similarly, bk · abl = abl−k is in D2n. The remaining cases are covered by the fact that〈b〉 is closed under multiplication.

The fact that D2n is closed under multiplication implies that it is a subgroup ofGl2(R), since D2n is finite (Proposition 2.2.13).


The dihedral groups have geometric meaning as groups of transformations of regularpolygons in the plane. In particular, for n ≥ 3, there is a regular n-gon whose verticesare the points

�vk =(

cos(2πk/n)sin(2πk/n)

)for 1 ≤ k ≤ n. This means that the n-gon is the set of convex combinations of the vertices,where a convex combination of �v1, . . . , �vn is a vector sum

∑ni=1 ai�vi with 0 ≤ ai ∈ R for

1 ≤ i ≤ n and∑ni=1 ai = 1.

If c is a matrix in D2n, then it is easy to see that for each i with 1 ≤ i ≤ n, the matrixproduct c · �vi is equal to �vj for some j with 1 ≤ j ≤ n. In other words, multiplicationby c carries each vertex of our n-gon to another vertex of the n-gon. But the linearity ofmatrix multiplication now shows that multiplication by c carries convex combinations ofthe vertices to convex combinations of the vertices, and hence c gives a transformationof the n-gon. In fact, it can be shown that the elements of D2n give all possible lineartransformations from the n-gon onto itself.

Exercises 2.8.5.1. What is the order of abk ∈ D2n?

2. Show that D4 and Z4 are non-isomorphic abelian groups of order 4.

3. Show that D2n is nonabelian if and only if n ≥ 3. Deduce that D2n and Z2n arenon-isomorphic groups of order 2n for all n ≥ 3.

4. Show that there is a subgroup of Gl2(R) consisting of all rotation matrices Rθ,together with the matrices a · Rθ. (This subgroup is generally called O(2), thesecond orthogonal group.) What is the order of a ·Rθ?

5. Show that if k divides n, then there is a subgroup of D2n which is isomorphic toD2k.

† 6. Let G be a group. Show that there is a one-to-one correspondence between thehomomorphisms from D2n to G and pairs of elements x, y ∈ G such that o(x)divides 2, o(y) divides n, and x−1yx = y−1.

7. Show that if k divides n, then there is a homomorphism of D2n onto D2k.

8. Let be a line through the origin in the plane. Standard vector geometry showsthere is a vector �u of length 1 such that is the set of all vectors perpendicular to�u. (Note this does not define �u uniquely: −�u would do as well.) Then the reflectionthrough is the function f� defined by f�(�w) = �w − 2(�w · �u)�u, where · is the dotproduct. Show that the reflection through is the transformation induced by thematrix aRθ for some θ. Show also that the transformation induced by aRθ is thereflection through some line for any θ ∈ R.

2.9 Quaternions

We construct yet another family of groups via matrices. They are the quaternionic groupsQ4n ⊂ Gl4(R) of order 4n. First, we give a generalization of the rotation matrices. For


θ ∈ R, let

Sθ =

⎛⎜⎜⎝cos θ − sin θ 0 0sin θ cos θ 0 0

0 0 cos θ − sin θ0 0 sin θ cos θ

⎞⎟⎟⎠ .

The geometric effect of Sθ on an element of the space R4 of column vectors is to simul-taneously rotate the first two coordinates and rotate the last two coordinates, treatingthese pairs of coordinates as independent planes.

The following lemma may be verified directly by multiplying the matrices.

Lemma 2.9.1. Let θ and φ be real numbers. Then Sθ · Sφ = Sθ+φ. Thus, there is ahomomorphism f : R → Gl4(R) defined by f(θ) = Sθ. Moreover, as in the case ofrotations of the plane, Sθ = Sφ if and only if θ − φ = 2πk for some k ∈ Z.

Once again, we shall give a matrix a that will interact with these “rotations” in aninteresting way. It is given by

a =

⎛⎜⎜⎝0 0 1 00 0 0 −1−1 0 0 00 1 0 0

⎞⎟⎟⎠ .

Lemma 2.9.2. The matrix a has order 4. The lower powers of a are given by

a2 = −I4 =

⎛⎜⎜⎝−1 0 0 0

0 −1 0 00 0 −1 00 0 0 −1

⎞⎟⎟⎠ and a3 =

⎛⎜⎜⎝0 0 −1 00 0 0 11 0 0 00 −1 0 0

⎞⎟⎟⎠ .

Proof The square of a is as stated by direct calculation. We then have that a3 =a2a = −a, which is the displayed matrix. Clearly, this is not the identity matrix, but(−I4)2 = a4 is the identity matrix. Thus, the 4-th power of a is the smallest that willproduce the identity, so that a has order 4 and a3 = a−1.

Once again, there is a deviation from commutativity between the Sθ and a. Thefollowing lemma may be verified by direct computation.

Lemma 2.9.3. We have Sθ · a = a · S−θ.

Because Sθ and Sφ commute for any real numbers θ and φ, we obtain the followingcorollary.

Corollary 2.9.4. The matrix a is not equal to Sθ for any θ.

Write b (or b2n if more than one quaternionic group is under discussion) for S2π/2n.Then b has order 2n. The quaternionic group of order 4n is the subgroup of Gl4(R)generated by a and b:

Definition 2.9.5. The quaternionic group of order 4n, Q4n ⊂ Gl4(R), is given as fol-lows:

Q4n = {bk, abk | 0 ≤ k < 2n},with a and b as above.


Once again we need some justification.

Proposition 2.9.6. For each n ≥ 1, Q4n is a subgroup of Gl4(R). Moreover, the ma-trices {bk, abk | 0 ≤ k < 2n} are all distinct, so that Q4n does indeed have order 4n.

Proof The key here is that a2 = −I4 = Sπ = bn, which is why b was chosen to haveeven order. Thus, abk · abl = a2bl−k = bnbl−k = bn+l−k. Since bk · abl = abl−k and sinceb has order 2n, Q4n is closed under multiplication. Since it is finite, it is a subgroup byProposition 2.2.13.

It remains to show that the stated elements are all distinct. As in the dihedral case,it suffices to show that abk �= bl, and hence that a �= bk−l for any k and l. But once againthis follows from the fact that a cannot equal Sθ for any θ.

Exercises 2.9.7.1. What is the order of abk ∈ Q4n?

2. Show that Q4 is isomorphic to Z4.

3. For n ≥ 2, show that Q4n is isomorphic to neither D4n nor Z4n.

4. Show that if k divides n, then there is a subgroup of Q4n which is isomorphic toQ4k.

† 5. Let G be a group. Show that there is a one-to-one correspondence between thehomomorphisms from Q4n to G and pairs of elements x, y ∈ G such that o(x)divides 4, o(y) divides 2n, x2 = yn, and x−1yx = y−1.

6. Show that there is a homomorphism of Q4n onto D2n.

7. Show that if n is an odd multiple of k, then there is a homomorphism of Q4n ontoQ4k.

8. Show that every subgroup of Q8 other than e contains 〈a2〉. (Hint : The cyclicsubgroups are always the smallest ones in a group.)

9. Let n be any power of 2. Show that every subgroup of Q4n other than e contains〈a2〉.

2.10 Direct Products

We give a way of constructing new groups from old. It satisfies an important universalproperty with respect to homomorphisms.

Definition 2.10.1. Let G and H be groups. Then the product (or direct product) ofG and H is the group structure on the cartesian product G ×H which is given by thebinary operation

(g1, h1) · (g2, h2) = (g1g2, h1h2).

We denote this group by G×H.

Once again, we must justify our definition:

Lemma 2.10.2. The direct product G×H is a group.


Proof Associativity follows directly from the associativity of G and H. The identityelement is the ordered pair (e, e). The inverse of (g, h) is (g−1, h−1).

The reader should note that while this is not the only interesting group structure wecan put on the product of G and H (we shall discuss semidirect products in Section 4.6),it is the only one we shall refer to as G×H.

The next result is immediate from Proposition 1.6.4.

Corollary 2.10.3. Let G and H be finite groups. Then G×H is finite, and

|G×H| = |G| · |H|.

The reader should supply the verification for the next lemma.

Lemma 2.10.4. There are subgroups G× e = {(g, e) | g ∈ G} and e×H = {(e, h) |h ∈H} of G ×H. Any element of G × e commutes with any element of e ×H. There arehomomorphisms ι1 : G → G × H and ι2 : H → G × H defined by ι1(g) = (g, e) andι2(h) = (e, h). These induce isomorphisms G ∼= G× e and H ∼= e×H.

There are also homomorphisms π1 : G × H → G and π2 : G × H → H defined byπ1(g, h) = g and π2(g, h) = h.

The homomorphisms ιi are called the canonical inclusions of G and H in G×H. Thehomomorphisms π1 and π2 are called the projections of G×H onto the first and secondfactors, respectively.

Recall that a function into a product is defined uniquely by its component functions:f : X → G×H is defined by f(x) = (f1(x), f2(x)), where f1 : X → G and f2 : X → H.In fact, the component functions are nothing other than the composites fi = πi ◦ f ,where the πi are the projections in the above lemma. The group product in G × H isthe unique product that makes the following statement true:

Lemma 2.10.5. Let G, H, and K be groups. A function f : K → G × H is a homo-morphism if and only if its component functions are homomorphisms.

We can also study homomorphisms out of a product. If f : G × H → K is ahomomorphism, so are f ◦ ι1 : G→ K and f ◦ ι2 : H → K.

Proposition 2.10.6. Let f : G×H → K be a homomorphism. Then every element ofthe image of f ◦ ι1 commutes with every element of the image of f ◦ ι2. Moreover, f isuniquely determined by its restrictions to G× e and e×H.

Conversely, given homomorphisms g1 : G → K and g2 : H → K such that eachelement in the image of g1 commutes with each element in the image of g2, then there isa unique homomorphism f : G×H → K with f ◦ ιi = gi for i = 1, 2.

Proof Since (g, h) = (g, e) · (e, h), f(g, h) = f ◦ ι1(g) · f ◦ ι2(h). Thus, f is determinedby the f ◦ ιi, which are determined by the restriction of f to the stated subgroups. Thecommutativity assertion follows since if two elements commute, so must their imagesunder any homomorphism.

Given the gi as above, define f : G × H → K by f(g, h) = g1(g) · g2(h). Thenf(g, h) · f(g′, h′) = g1(g)g2(h)g1(g′)g2(h′) = g1(g)g1(g′)g2(h)g2(h′) = g1(gg′)g2(hh′),with the key step being given by the fact that g1(g′) commutes with g2(h). But the lastterm is clearly f((g, h) · (g′, h′)).


It is valuable to know when a group breaks up as the product of two of its subgroups.

Definition 2.10.7. Let H and K be subgroups of the group G. We say that G is theinternal direct product of H and K if there is an isomorphism μ : H × K → G suchthat μ ◦ ι1 and μ ◦ ι2 are the inclusions of H and K, respectively, as subgroups of G.Explicitly, this says μ(h, k) = hk.

We now give a characterization of internal direct products.

Proposition 2.10.8. Let H and K be subgroups of the group G. Then G is the internaldirect product of H and K if and only if the following conditions hold.

1. H ∩K = e.

2. If h ∈ H and k ∈ K, then h and k commute in G.7

3. G is generated by the elements of H and K.

Proof Suppose that G is the internal direct product of H and K. Then the functionμ : H × K → G defined by μ(h, k) = hk is an isomorphism of groups. Since μ is ahomomorphism and since μ ◦ ι1(h) = h and μ ◦ ι2(k) = k, each h ∈ H must commutewith each k ∈ K by Proposition 2.10.6. Since μ is onto, each element of G must havethe form hk, with h ∈ H and k ∈ K, so the elements of H and K must generate G.

Finally, let x ∈ H ∩K. Then (x, x−1) ∈ H ×K, and μ(x, x−1) = xx−1 = e. Sinceμ is injective, this says that (x, x−1) is the identity element of H × K, which is (e, e).Thus, x = e, and hence H ∩K = e.

Conversely, suppose that the three stated conditions hold. Since the elements of Hcommute with those of K, Proposition 2.10.6 shows that the function μ : H ×K → Ggiven by μ((h, k)) = hk is a homomorphism. Suppose that (h, k) is in the kernel of μ.Then hk = e, and hence h = k−1 is in H ∩K. But H ∩K = e, and hence h = e, andhence k = h−1 = e also. Thus, (h, k) = (e, e), and hence μ is injective.

Since H and K generate G, any element in G may be written as a product g1 . . . gnwhere gi is an element of either H or K for i = 1, . . . , n. Since elements of H commutewith those of K, we can rewrite this as a product h1 . . . hrk1 . . . ks with hi ∈ H fori = 1, . . . , r and ki ∈ K for i = 1, . . . , s, by moving the elements in H past those in K.But this is just μ(h1 . . . hr, k1 . . . ks), so μ is onto.

We shall see in Proposition 3.7.12 that the second condition of Proposition 2.10.8may be replaced with the statement that both H and K are what’s known as normalsubgroups (to be defined in Section 3.7) of G.

Now let us generalize the material in this section to the study of products of morethan two groups.

Definition 2.10.9. Suppose given groups G1, . . . , Gk, for k ≥ 3. The product (or directproduct), G1 × · · · × Gk, of the groups G1, . . . , Gk is the set of all k-tuples (g1, . . . , gk)with gi ∈ Gi for 1 ≤ i ≤ k, with the following multiplication:

(g1, . . . , gk) · (g′1, . . . , g′k) = (g1g′1, . . . , gkg′k).

The reader should be able to supply the proof of the following proposition.7Of course, neither H nor K need be abelian. Thus, if h, h′ ∈ H, then h and h′ must commute with

every element of K, but they need not commute with each other.


Proposition 2.10.10. The direct product G1 × · · · × Gk is a group under the statedmultiplication. There are homomorphisms πi : G1 × · · · ×Gk → Gi and ιi : Gi → G1 ×· · · × Gk for 1 ≤ i ≤ k, defined by πi(g1, . . . , gk) = gi, and ιi(g) = (e, . . . , e, g, e, . . . , e),where g appears in the i-th coordinate.

Given homomorphisms fi : H → Gi for 1 ≤ i ≤ k, there is a unique homomorphismf : H → G1 × · · · ×Gk such that πi ◦ f = fi for 1 ≤ i ≤ k. Similarly, for each collectionhi : Gi → H of homomorphisms such that each element of the image of hi commuteswith each element of the image of hj whenever i �= j, there is a unique homomorphismh : G1 × · · · ×Gk → H with h ◦ ιi = hi for 1 ≤ i ≤ k.

Similarly, we can discuss internal direct products of several subgroups.

Definition 2.10.11. Let H1, . . . , Hk be subgroups of G. We say that G is the inter-nal direct product of H1, . . . , Hk if the function μ : H1 × · · · × Hk → G defined byμ(h1, . . . , hk) = h1 . . . hk is an isomorphism of groups.

We shall not give a characterization here of the internal direct product of multiplesubgroups. (I.e., we shall not give a generalization of Proposition 2.10.8 to the case ofmore than two subgroups.) The reader may wish to formulate one after reading the proofof Proposition 4.4.9.

Exercises 2.10.12.1. Show that the order of (g, h) ∈ G ×H is the least common multiple of the orders

of g and h.

† 2. More generally, show that the order of (g1, . . . , gk) ∈ G1 × · · · × Gk is the leastcommon multiple of the orders of g1, . . . , gk.

3. Show that Z6 is isomorphic to Z3 × Z2.

† 4. Let m and n be relatively prime. Show that Zmn is isomorphic to Zm × Zn.

5. Let n = pr11 . . . prk

k , where p1, . . . , pk are distinct primes. Show that

Zn ∼= Zpr11× · · · × Zprk

k.

6. Show that if m and n are not relatively prime, then Zm × Zn is not cyclic.

7. Show that D4 is isomorphic to Z2 × Z2.

8. Show that D12 is isomorphic to D6 × Z2.

9. Show that if k is odd, then D4k is isomorphic to D2k × Z2.

10. Show that G×H ∼= H ×G.

11. Show that (G×H)×K ∼= G× (H ×K) ∼= G×H ×K.

12. Show that G× e ∼= G, where e is the trivial group.

13. Show that the group multiplication μ : G×G→ G is a homomorphism if and onlyif G is abelian.

14. Let fi : Gi → Hi be a homomorphism for 1 ≤ i ≤ k. Define f1 × · · · × fk :G1 × · · · ×Gk → H1 × · · · ×Hk by f1 × · · · × fk(g1, . . . , gk) = (f1(g1), . . . , fk(gk)).Show that f1 × · · · × fk is a homomorphism. In addition, show that f1 × · · · × fkis an isomorphism if and only if each fi is an isomorphism.

15. Show that not every homomorphism f : G1×G2 → H1×H2 has the form f1× f2.

Chapter 3

G-sets and Counting

Many of the examples in Chapter 2 were obtained from matrices, and hence give sym-metries of n-dimensional Euclidean space, Rn. Here, we study the symmetries of sets,obtaining a more combinatorial view of symmetry.

In the first section, we define the symmetric group on n letters, Sn, and prove Cayley’sTheorem, which shows that every finite group is a subgroup of Sn for some n. Later,in Chapter 10, we will show that Sn embeds in Gln(R). In this manner, we obtain theassertion of the last chapter that every finite group is a subgroup of Gln(R).

The symmetric groups are interesting in their own right as examples of groups. Viathe technique of cycle decomposition, which we give in Section 3.5, one can say quite abit about its structure.

Also, there are natural subgroups An ⊂ Sn, called the alternating groups, which arethe easiest examples of nonabelian simple groups. We define An in Section 3.5 in termsof the sign homomorphism ε : Sn → {±1} to be defined there. We shall define simplegroups in Section 4.2, and show that An is simple for n ≥ 5. We may then deduce inChapter 5 that Sn is not what’s known as a solvable group for n ≥ 5.

The non-solvability of Sn will allow us, in Chapter 11, to prove Galois’s famoustheorem that there is no formula for solving polynomials of degree ≥ 5. This should beenough to arouse one’s interest in the symmetric groups.

We also introduce a very important concept in group theory: that of the set G/H ofleft cosets of a subgroup H in G. One of the first consequences of the study of cosets isLagrange’s Theorem: if G is finite, then the order of any subgroup of G must divide theorder of G. We give Lagrange’s Theorem in Section 3.2, and give many more applicationsof the study of cosets throughout the material on group theory.

The sets of cosetsG/H form an important example of the notion of a group acting on aset. We formulate the notion of G-sets in Section 3.3 and give a generalization of Cayley’sTheorem in which a one-to-one correspondence is established between the actions of Gon a set X and the homomorphisms from G to the group of symmetries of X. Thisallows us to associate to a finite G-set (such as G/H, if G is finite) a homomorphismfrom G into Sn for appropriate n. These homomorphisms are quite useful in determiningthe structure of a group G, and will figure prominently in the classification results weget in Chapter 5. Section 3.3 also contains a couple of important counting arguments,including what’s known as the G-set Counting Formula.

In the final two sections of the chapter, we study actions of one group on another.The point is that not all symmetries of a group H preserve the group structure. The onesthat do are the automorphisms of H, which are defined to be the group isomorphisms

46

CHAPTER 3. G-SETS AND COUNTING 47

H → H. These form a subgroup, Aut(H), of the group of symmetries of H. We studythe actions of G on H such that the action of each g ∈ G induces an automorphism ofH.

Actions through automorphisms play an important role in constructing and classifyingthe extensions of one group by another (Section 4.7), and therefore, as we shall see, inthe classification of groups.

Closer at hand, there is a very important action of a group G on itself throughautomorphisms, called conjugation. Many of the important results in group theory followfrom the study of conjugation, including the Noether Isomorphism Theorems of Chapter 4and the Sylow Theorems of Chapter 5. We shall begin that study in this chapter.

3.1 Symmetric Groups: Cayley’s Theorem

Here, we show that every finite group embeds in the group of symmetries of a finite set.Recall that a function is bijective if it is one-to-one and onto.

Definitions 3.1.1. Let X be a set. By a permutation of X we mean a bijective functionσ : X → X.

We write S(X) for the set of all permutations of X. Note that S(X) has a binaryoperation given by composition of functions: for σ, τ ∈ S(X), σ · τ is the compositeσ ◦ τ . As we shall see shortly, S(X) is a group under this operation, called the group ofsymmetries of X.

Recall that the identity function 1X of a set X is the function defined by 1X(x) = xfor all x ∈ X.

Recall from Proposition 1.1.6 that every bijective function σ : X → Y has an inversefunction σ−1 : Y → X, with the property that σ−1 ◦ σ = 1X and σ ◦ σ−1 = 1Y . Here,the inverse function σ−1 is defined by setting σ−1(y) to be the unique x ∈ X such thatσ(x) = y.

Lemma 3.1.2. Let X be a set. Then S(X) forms a group under the operation of com-position of functions. The identity element of S(X) is the identity function 1X , and theinverse of an element σ ∈ S(X) is its inverse function, σ−1.

We shall see below that if X is infinite, then not only is S(X) an infinite group, italso contains elements of infinite order. On the other hand, if X is finite, then there areonly finitely many elements in S(X), as, in fact, there are only finitely many functionsfrom X to X by Corollary 1.7.6.

We shall calculate the order of S(X), for X finite, in Corollary 3.3.11.

Definition 3.1.3. Let f : X → Y be a bijection of sets. Define the induced mapf∗ : S(X)→ S(Y ) by

f∗(σ) = f ◦ σ ◦ f−1

for σ ∈ S(X).

Since two sets are considered equivalent if they may be placed in bijective correspon-dence, the following lemma should come as no surprise.

Lemma 3.1.4. Let f : X → Y be a bijection of sets. Then the induced map f∗ : S(X)→S(Y ) is an isomorphism of groups.


Proof It is easy to see that f∗ gives a group homomorphism from S(X) to S(Y ). More-over, (f−1)∗ : S(Y )→ S(X) gives an inverse function for f∗, so f∗ is an isomorphism.

Note that there is no preferred isomorphism between S(X) and S(Y ), but rather anisomorphism for each choice of bijection from X to Y .

Of special interest here is the case when X is finite. Our preferred model for a setwith n elements is {1, . . . , n}.Definition 3.1.5. We write Sn for S({1, . . . , n}), and call it the symmetric group on nletters.

Fixed points are an important feature of a permutation:

Definition 3.1.6. Let σ be a permutation of X. Then σ is said to fix an element x ∈ Xif σ(x) = x. We call such an x a fixed point of σ.

We should now construct some examples of elements of Sn. Recall that if Y is asubset of X, then the complement of Y in X is the set of all elements of X that are notin Y .

Definition 3.1.7. Let i1, . . . , ik be k distinct elements of {1, . . . , n}, k ≥ 2. We write(i1 i2 · · · ik) for the permutation of {1, . . . , n} which carries ij to ij+1 for 1 ≤ j < k,carries ik to i1, and is the identity map on the complement of {i1, . . . , ik} in {1, . . . , n}.We call (i1 i2 · · · ik) a k-cycle, or a cycle of length k.

Note that a 2-cycle (i j) just interchanges i and j and fixes all other elements of{1, . . . , n}. Another name for a 2-cycle is a transposition.

Some people consider a 1-cycle (i) to be another way of writing the identity map.While this is consistent with the notation above, our cycles will all have length ≥ 2.

Note that the permutations (i1 i2 · · · ik) and (ik i1 · · · ik−1) are equal, as they havethe same effect on all elements of {1, . . . , n}. Thus, there is more than one way to denotea given k-cycle.

We can calculate the order and inverse of a k-cycle.

Lemma 3.1.8. A k-cycle (i1 i2 · · · ik) has order k. Its inverse is (ik · · · i2 i1).

Proof Let σ = (i1 i2 · · · ik). An easy induction shows that for 1 ≤ j < k, σj(i1) =ij+1 �= i1, while σk does fix i1. Thus, σj �= e for 1 ≤ j < k. But since σ may also bewritten as (i2 · · · ik i1), σk fixes i2 by the preceding argument. Another induction showsthat σk fixes each of the ij . Since it also fixes the complement of {i1, . . . , ik}, we haveσk = e.

The permutation (ik · · · i1) is easily seen to reverse the effect of σ.

One of the reasons for studying the symmetric groups is that every finite group isisomorphic to a subgroup of Sn for some n. Recall that an embedding of groups is aninjective homomorphism.

Theorem 3.1.9. (Cayley’s Theorem) Let G be a group. For g ∈ G write g : G → Gfor the function g(g′) = gg′. Then g is a permutation of G. Moreover, if we defineΛ : G→ S(G) by Λ(g) = g, then Λ is an embedding. In consequence, if G is finite, thenit is isomorphic to a subgroup of S|G|.


Proof To show that g is injective, suppose g(g′) = g(g′′), and hence gg′ = gg′′.Multiplying both sides on the left by g−1 and cancelling, we get g′ = g′′. And g issurjective because g′ = g(g−1g′) for all g′ ∈ G. Thus, g is a permutation of G.

Now Λ(g) · Λ(g′) is the permutation g ◦ g′ . We wish to show this coincides withΛ(gg′), which is the permutation gg′ . We show they have the same value on each g′′ ∈ G.

(g ◦ g′)(g′′) = g(g′g′′) = (gg′)g′′ = gg′(g′′).

Thus, Λ is a homomorphism. To show it is an injection, we consider its kernel.Suppose that g = e. Thus, g(g′) = g′ for all g′ ∈ G, i.e., gg′ = g′ for all g′ ∈ G.But multiplying both sides of the equation on the right by (g′)−1, we get g = e. Thus,ker Λ = e, the trivial subgroup, so Λ is an embedding by Lemma 2.5.9.

Recall that a set is countably infinite if it may be put in one-to-one correspondencewith the positive integers, Z+. It is easy to show that Z is countably infinite. Sinceevery nonzero element of Z has infinite order, the next corollary follows from Cayley’sTheorem.

Corollary 3.1.10. Let X be a countably infinite set. Then S(X) contains elements ofinfinite order.

Cayley’s Theorem shows that any finite group G is isomorphic to a subgroup of S|G|.But most groups are actually isomorphic to a group of permutations of a smaller numberof letters.

As we shall see in a moment, there is a natural choice of embedding of Sn in Sn+1.Thus, if G embeds in Sn, then it also embeds in Sn+1.

In fact, if Y is a subset of X, there is a natural embedding of S(Y ) in S(X):

Definition 3.1.11. Let Y be a subset of X, and let i : Y ⊂ X be the inclusion map.Define i∗ : S(Y )→ S(X) by

(i∗(σ))(x) ={σ(x) if x ∈ Yx otherwise.

We shall refer to i∗ as the canonical inclusion of S(Y ) in S(X).

Lemma 3.1.12. Let i : Y ⊂ X be the inclusion of a subset, Y , of a set X. Thenthe canonical inclusion i∗ : S(Y ) → S(X) is an embedding onto the subgroup of S(X)consisting of those elements of S(X) which fix all elements of the complement of Y inX.

Proof Write X − Y for the complement of Y in X. It is easy to see that i∗ is anembedding and that elements in the image of i∗ fix all elements of Y − X. Thus, itsuffices to show that any element of S(X) that fixes the elements of Y − X lies in theimage of i∗.

Thus, suppose that τ ∈ S(X) fixes the elements of of X − Y . Note that it sufficesto show that τ restricts to a bijection from Y to itself, as then this restriction definesan element of S(Y ), which we shall denote by τ ′. Since τ is the identity on X − Y andagrees with τ ′ on Y , we obtain that i∗(τ ′) = τ , as desired.

We first show that τ(Y ) ⊂ Y . Suppose not. Then we can find y ∈ Y such thatτ(y) = x ∈ Y −X. But since τ fixes Y −X, we have τ(x) = x, so x and y have the sameimage under τ . This is impossible, as τ is injective.


Thus, τ restricts to a function from Y to Y , which is injective, because restrictionsof injective functions are always injective. But note that if x ∈ X − Y , then τ(x) cannotlie in Y . Thus, since τ is onto, we must have τ(Y ) = Y . Thus, τ restricts to a bijectionfrom Y to itself, and the result follows.

Taking i to be the inclusion map {1, . . . , n − 1} ⊂ {1, . . . , n}, we get the canonicalinclusion Sn−1 ⊂ Sn. The next corollary is now immediate from Lemma 3.1.12.

Corollary 3.1.13. The canonical inclusion of Sn−1 in Sn identifies Sn−1 with the sub-group of Sn consisting of those permutations that fix n.

Note that the inclusion of {1, . . . , n} in the positive integers, Z+, induces a canonicalinclusion Sn ⊂ S(Z+) which is compatible with the inclusion of Sn in Sn+1.

Definition 3.1.14. Write i∗ : Sn ⊂ S(Z+) for the canonical inclusion, and define S∞ ⊂S(Z+) by

S∞ =⋃n≥1

i∗(Sn).

Lemma 3.1.15. S∞ is a subgroup of S(Z+).1

Proof Since each (i∗(x))−1 = i∗(x−1), S∞ is closed under inverses, and it suffices toshow that S∞ is closed under multiplication.

For x, y ∈ S∞, there are m,n ≥ 1 such that x ∈ i∗(Sm) and y ∈ i∗(Sn). Suppose,then, that m ≤ n. Since the canonical inclusions of the Sk in S(Z+) are compatiblewith their canonical inclusions in one another, we have i∗(Sm) ⊂ i∗(Sn), and hencex, y ∈ i∗(Sn). But i∗(Sn) is a subgroup of S(Z+), and hence xy ∈ i∗(Sn) ⊂ S∞.

In practice, we identify each Sn with its image under i∗, and treat Sn as a subgroupof S∞ (and hence also of S(Z+)).

Exercises 3.1.16.1. Show that S2 is isomorphic to Z2.

2. Show that S(X) is nonabelian whenever X has more than two elements.

3. Show that the square of a 4-cycle is not a 4-cycle.

4. What is the order of (1 2) · (3 4) in S4?

5. What is the order of (1 2 3) · (4 5) in S5?

6. Show that there is an embedding f : D6 → S3 induced by setting f(a) = (1 3) andf(b) = (1 2 3). We shall show in Section 3.3 that S3 has order 6. We may thendeduce that f is an isomorphism.

7. Define x, y ∈ S4 by x = (1 4) · (2 3) and y = (1 2 3 4). Show that there is anembedding f : D8 → S4 induced by setting f(a) = x and f(b) = y.

8. Show that D12 embeds in S5.

9. Show that D2n embeds in Sn for all n ≥ 3.1Those who are acquainted with category theory should note that we can identify S∞ with the direct

limit lim−→Sn.


10. Show that Z2 × Z2 embeds in S4.

11. Show that Zm × Zn embeds in Sm+n.

12. Show that the subgroup S∞ ⊂ S(Z+) is an infinite group whose elements all havefinite order.

13. Show that an element of S(Z+) lies in S∞ if and only if it fixes all but finitelymany of the elements of Z+.

14. Let G be a group and let g ∈ G. Define rg : G → G by rg(x) = xg. Show thatrg ∈ S(G). Show that the function R : G → S(G) defined by R(g) = rg−1 is anembedding.

‡ 15. Let f : Y → Z be any injective function, and define f∗ : S(Y )→ S(Z) by

(f∗(σ))(z) ={f(σ(y)) if z = f(y)z if z is not in the image of f .

(a) Show that f∗ is an embedding onto the subgroup of S(Z) consisting of thosepermutations that fix all elements of Z which are not in f(Y ).

(b) If f is a bijection, show that this f∗ coincides with the isomorphism f∗ givenin Definition 3.1.3.

(c) If f is the inclusion of a subset Y ⊂ Z, show that this f∗ coincides with theembedding f∗ given in Definition 3.1.11.

(d) Let g : Z →W be another injection. Show that g∗ ◦ f∗ = (g ◦ f)∗.

(e) If X is finite, and if f : X → Z+ is injective, show that the image of f∗ lies inS∞. Thus, there is an induced embedding f∗ : S(X)→ S∞.

16. Let X be a set. Show that the set of all functions f : X → X forms a monoidunder the operation of composition of functions. Show that S(X) is its group ofinvertible elements.

17. Show that the set of functions from {1, 2} to itself which take 1 to itself gives, viacomposition of functions, the unique monoid with two elements that is not a group.

3.2 Cosets and Index: Lagrange’s Theorem

We here begin the study of a very important concept in algebra: the left cosets of asubgroup H ⊂ G. One of the consequences of this consideration is the famous theoremof Lagrange: If H is a subgroup of the finite group G, then the order of H divides theorder of G.

Lagrange’s Theorem is the very first step in trying to classify groups up to isomor-phism.

Definition 3.2.1. Let H be a subgroup of G and let x ∈ G. Then the left coset xH ofH in G is the following subset of G:

xH = {xh |h ∈ H}.


Thus, as x ranges over the elements of G, we get a collection of different subsetsxH ⊂ G, which are the left cosets of H in G.

Note that eH is just H ⊂ G. The coset H is special in that it is the only left cosetof H in G which is a subgroup. (The next lemma shows that e ∈ xH if and only ifxH = H.) We study the cosets to understand how H relates to the other elements of G.

Example 3.2.2. Let G be a cyclic group of order 4, written multiplicatively, with gen-erator b: G = {e, b, b2, b3}. Let H be the cyclic subgroup generated by b2: H = 〈b2〉 ={e, b2}. Then an easy inspection shows that eH = b2H = {e, b2}, while bH = b3H ={b, b3}. Since we have inspected xH for each element x ∈ G, we see that these are all ofthe left cosets of H in G.

Recall that cosets are defined to be subsets of G. Thus, since bH and b3H have thesame elements, they are equal as cosets. In particular, for this choice of H and G, thereare exactly two left cosets of H in G.

We now show that two left cosets that have a single element in common must beequal.

Lemma 3.2.3. Let H be a subgroup of G and let x, y ∈ G. Then the following statementsare equivalent.

1. xH = yH.

2. xH ∩ yH �= ∅.3. x−1y ∈ H.

4. y−1x ∈ H.

Proof Clearly, the first statement implies the second. Suppose, then, that the secondholds. Let z ∈ xH ∩ yH. Then there are h, h′ ∈ H with z = xh = yh′. But thenx−1y = h(h′)−1 ∈ H. Thus, the second statement implies the third.

The third and fourth statements are equivalent, as y−1x is the inverse of x−1y. Sup-pose, then, that the fourth statement holds. Thus, y−1x = k ∈ H, and hence x = yk.But then xh = ykh ∈ yH for all h ∈ H, and hence xH ⊂ yH. But also, y = xk−1, sothat yh = xk−1h for all h ∈ H, and hence yH ⊂ xH. Thus, xH = yH, and we see thatthe fourth statement implies the first.

Since x = x · e we have x ∈ xH for each x ∈ G. But any two left cosets of H thathave an element in common must be equal by Lemma 3.2.3.

Corollary 3.2.4. Let H be a subgroup of G. Then each element x ∈ G is contained inexactly one left coset of H, and that coset is xH. In particular, the elements x and y liein the same left coset of H if and only if xH = yH.

If the operation of G is written additively, we have an additive notation for cosets.

Notation 3.2.5. Suppose that G is abelian and that we use additive notation for thegroup operation in G. Then we also use additive notation for the left cosets of a subgroupH ⊂ G. For x ∈ G, we write x+H for the coset which contains x:

x+H = {x+ h |h ∈ H}.

It is very useful in group theory to consider the set of all left cosets of H in G.


Definitions 3.2.6. Let H be a subgroup of G. Write G/H for the set whose elementsare the cosets xH of H in G. Thus, if xH = yH as subsets of G, then xH and yH areequal as elements of G/H. If the number of elements of G/H (which is equal to thenumber of distinct left cosets of H in G) is finite, we say that H has finite index in G,and write [G : H], the index of H in G, for the number of elements in G/H. If the setG/H is infinite, we write [G : H] =∞.

We define the canonical map, π : G→ G/H, by π(x) = xH for x ∈ G.

Remarks 3.2.7. Let H be a subgroup of G. We put a relation ∼ on G by setting x ∼ yif x−1y ∈ H. Then ∼ is an equivalence relation: it is reflexive because (x−1y)−1 = y−1x,and is transitive because x−1y · y−1z = x−1z.

But x−1y ∈ H if and only if y ∈ xH, so the equivalence class of x under ∼ is preciselyxH. Thus, G/H is the set of equivalence classes of G under ∼, and the canonical mapπ : G → G/H, as defined above, takes each x ∈ H to its equivalence class under ∼.Thus, π agrees with the canonical map of Definition 1.4.5.

The following corollary is immediate from Corollary 3.2.4.

Corollary 3.2.8. Let H be a subgroup of G and let π : G→ G/H be the canonical map.Then π(x) = π(y) if and only if x and y are in the same left coset of H.

Proposition 3.2.9. (Lagrange’s Theorem) Let H be a subgroup of G and suppose thatH is finite. Then every left coset of H has |H| elements.

Given that H is finite, if either of G or [G : H] is finite, then so is the other, and wehave

|G| = [G : H] · |H|.

In particular, if G is a finite group and H is a subgroup of G, then the order of Hdivides the order of G.

Proof For x ∈ G, the permutation x defined in the last section (by x(g) = xg)restricts to a bijection from H to xH (with inverse induced by x−1). Thus, if H is finite,each coset xH has the same number of elements as H.

Since each element of G lies in exactly one left coset of H, the order of G, if finite, isequal to the sum of the “orders” of the distinct left cosets of H. Since each left coset has|H| elements, this says that |G| must be equal to the number of left cosets of H timesthe order of H.

Of course, if both H and [G : H] are finite, then G is the union of a finite number ofthe finite sets xH, and hence G is finite as well.

Recall that the order of a given element of a group is equal to the order of the cyclicsubgroup generated by that element (Corollary 2.5.18).

Corollary 3.2.10. Let G be a finite group. Then the order of any element of G mustdivide the order of G.

We should point out that the converse of Lagrange’s Theorem is false: if k dividesthe order of G, it is not necessarily the case that G will have a subgroup of order k. Forinstance, in Problem 2 of Exercises 4.2.18, we shall see that the alternating group A4,which has order 12, has no subgroup of order 6.

We shall spend some time investigating partial converses to Lagrange’s Theorem, asthey are quite useful in obtaining classification results. In Cauchy’s Theorem, we will


show that if a prime p divides the order of a group, then G has a subgroup of order p.Then, in the Sylow Theorems, we will show that G has a subgroup whose order is thehighest power of p that divides the order of G.

We can get even better partial converses to Lagrange’s Theorem if we put extraassumptions on the group G. For instance, Hall’s Theorem (Theorem 5.6.2) gives aresult that holds for what’s known as solvable groups: if G is solvable and |G| = nkwith (n, k) = 1, then G has a subgroup of order n. Even here, we do not get a fullconverse, as the alternating group A4 is solvable. But the full converse does hold infinite abelian groups, or, more generally, in the finite nilpotent groups, which we shallstudy in Section 5.7.

Lagrange’s Theorem allows us to classify the groups of order 4.

Corollary 3.2.11. Let G be a group of order 4 which is not cyclic. Then G is isomorphicto Z2 × Z2.

Proof Let a �= b be elements of G, neither of which is the identity element. Neithera nor b can have order 4, as then G would be cyclic. Thus, both have order 2. Nowab �= a, as that would imply that b = e, and ab �= b, because then a would be the identityelement. And ab �= e, as then a and b would be inverse to one another. But an elementof order 2 is its own inverse, so this would mean a = b. Thus, the four elements of G aree, a, b, and ab.

A similar argument shows that ba cannot equal e, a, or b. But this forces ab = ba, sothat a and b commute. But now Proposition 2.10.8 shows that G is the internal directproduct of 〈a〉 and 〈b〉.

There are a couple of very important consequences of Lagrange’s Theorem.

Corollary 3.2.12. Let H and K be subgroups of G such that |H| and |K| are relativelyprime. Then H ∩K = e.

Proof Let x ∈ H ∩ K. Then Corollary 3.2.10 shows that o(x) must divide both |H|and |K|. Since |H| and |K| are relatively prime, this forces o(x) = 1.

Corollary 3.2.13. Let G and G′ be groups such that |G| and |G′| are relatively prime.Then the only homomorphism from G to G′ is the trivial homomorphism, which sendseach x ∈ G to e ∈ G′.

Proof Let x ∈ G. Then o(f(x)) divides o(x) (Corollary 2.5.19), which in turn divides|G|. But o(f(x)) also divides |G′|, as f(x) ∈ G′. Since |G| and |G′| are relatively prime,we must have o(f(x)) = 1.

So far, we have studied the left cosets of a subgroup H of G. There is an analogousnotion of right cosets, which we shall study in the exercises below:

Definition 3.2.14. Let H be a subgroup of the group G. The right cosets of H in Gare the subsets

Hx = {hx |h ∈ H}for x ∈ G.

Clearly, if G is abelian, then xH = Hx, and the right and left cosets of H coincide.For some nonabelian groups, the left and right cosets of a given subgroup can be totallydifferent subsets of G. However, we shall see that the left and right cosets of H agree ifH is what’s known as a normal subgroup of G.


Exercises 3.2.15.1. Let p be a prime number. Show that any group of order p is isomorphic to Zp.

2. For a prime number p, show that the only subgroups of Zp are e and Zp itself.

3. Let H be a subgroup of the finite group G and let K be a subgroup of H. Showthat [G : K] = [G : H] · [H : K].

† 4. Generalize the preceding problem to the case where G is infinite, but [G : K] isfinite.

5. State and prove the analogue of Lemma 3.2.3 for right cosets.

6. Show that every subgroup of Q8 other than Q8 itself is cyclic. List all of thesubgroups.

7. Find all of the left cosets of 〈a〉 ⊂ D6. Now find all of the right cosets and showthat they are different subsets of D6.

8. Repeat the preceding exercise with D8 in place of D6.

9. Repeat the preceding exercise for D2n with n arbitrary.

10. What are the right cosets of 〈b〉 ⊂ D2n?

11. Find all of the left cosets of 〈a〉 ⊂ Q12.

12. Let H be a subgroup of G. Show that there is a one-to-one correspondence betweenthe left cosets of H and the right cosets of H. Deduce that H has finite index in Gif and only if there are finitely many distinct right cosets of H in G, and that thenumber of right cosets, if finite, is [G : H].

13. Show that the order of a submonoid of a finite monoid need not divide the orderof the larger monoid.

3.3 G-sets and Orbits

Definitions 3.3.1. Let G be a group. An action of G on a set X is a rule that assignsto each ordered pair (g, x) with g ∈ G and x ∈ X an element g · x ∈ X,2 such that

1. e · x = x for all x ∈ X2. The following analogue of the associative law holds:

g1 · (g2 · x) = (g1g2) · x

for all g1, g2 ∈ G and x ∈ X.

A set X together with an action of G on X is called a G-set. If X and Y are G-sets,then a G-map, or G-equivariant map, f : X → Y is a function f from X to Y such thatf(g · x) = g · f(x) for all g ∈ G and x ∈ X.

A G-isomorphism is a G-map which is a bijection.2In other words, the passage from the ordered pair (g, x) to the product g · x defines a function from

G × X to X.


We will often abbreviate g · x by gx. We now note that we may identify two G-setswhich are G-isomorphic.

Lemma 3.3.2. Let f : X → Y be a G-isomorphism. Then so is the inverse functionf−1 : Y → X.

Examples 3.3.3.

1. G itself is a G-set, under the action obtained by setting g · g′ equal to the productgg′ of g and g′ in G. We call this the standard action of G on G.

2. Let X be a set with one element: X = {x}. Then there is a unique action of G onX: g · x = x for all g ∈ G.

3. Let H be a subgroup of G. We let G act on G/H by g · xH = (gx)H for g, x ∈ G.As the reader may verify, this gives a well defined action. Moreover, the canonicalmap π : G → G/H is easily seen to be a G-map, where G is given the action ofthe first example. The G-sets G/H are sometimes called “homogeneous spaces” ofG. Here, “spaces” comes from the fact that a similar construction is studied fortopological groups G, where H is a subgroup of G which is also a closed subspacewith respect to the topology of G.

4. Let G be any group and let X be any set. Then there is at least one action of Gon X: just set g · x = x for all g ∈ G and x ∈ X. We call this the trivial action ofG on X. Clearly, it does not tell us very much about either G or X.

5. Let X be a set. Then S(X) acts on X by σ · x = σ(x). Indeed, the facts that1X is the identity of S(X) and that the group product in S(X) is composition offunctions are precisely what is needed to show that this specifies an action of S(X)on X. We give a generalization of this example in Proposition 3.3.16.

6. Let X be any set. Let G be a cyclic group of order 2, written multiplicatively:G = {e, a}, where a has order 2. Then there is an action of G on X ×X, obtainedby setting a · (x, y) = (y, x) for each ordered pair (x, y) ∈ X ×X.

As Example 3 suggests, the study of G-sets encodes information about the subgroupsof G. We will use G-sets as one of our main tools in understanding the structure ofgroups.

Definitions 3.3.4. Let X be a G-set and let x ∈ X. Then the isotropy subgroup of x,denoted Gx, is the set of elements of G which fix x:

Gx = {g ∈ G | g · x = x}.The orbit of x, denoted G ·x, is the set of all elements obtained from x via multiplicationby elements of G:

G · x = {g · x | g ∈ G} ⊂ X.We say that G acts transitively on X if X = G · x for some x ∈ X.

The next lemma is needed to justify the preceding definitions.

Lemma 3.3.5. Let X be a G-set and let x ∈ X. Then the isotropy subgroup Gx is asubgroup of G.

The orbit, G · x, of x is the smallest sub-G-set of X which contains x.


Proof Clearly, Gx is a submonoid of G, and we must show that Gx is closed underinverses. But if g ∈ Gx, then g−1 ·x = g−1 · (g ·x) = (g−1g) ·x = x, and hence g−1 ∈ Gx.

We leave the rest to the reader.

As a first example of these concepts, let us consider their application to the “homo-geneous spaces” G/H.

Lemma 3.3.6. In the G-set G/H, the isotropy subgroup of eH is H. Also, the orbit ofeH is all of G/H, and hence G/H is a transitive G-set.

Proof Here g · eH = gH, so that every element in G/H is in the orbit of eH. Also, gis in the isotropy subgroup of eH if and only if gH = eH. By Lemma 3.2.3, this is thecase if and only if e−1g ∈ H. Thus, the isotropy subgroup of eH is H, as claimed.

G-maps are of course important in the understanding of G-sets. The reader shouldverify the following lemma.

Lemma 3.3.7. Let f : X → Y be a G-map and let x ∈ X. Then we have an inclusionof isotropy subgroups Gx ⊂ Gf(x). If, in addition, f is injective, then Gx = Gf(x).

In the study of G-sets, orbits play a role similar to (but simpler than) the role playedby cyclic subgroups in the study of groups. Just as the cyclic groups Zn form modelsfor the cyclic subgroups, every orbit turns out to be G-isomorphic to one of the G-setsG/H:

Lemma 3.3.8. Let X be a G-set and let x ∈ X. Let H be a subgroup of G. Then thereis a G-map fx : G/H → X with fx(eH) = x if and only if H ⊂ Gx. If fx exists, it isunique, and has image equal to the orbit G · x. Moreover, fx induces a G-isomorphismfrom G/H to G · x if and only if H = Gx.

In particular, for X any G-set and x ∈ X, G · x is G-isomorphic to G/Gx. Thus, aG-set is transitive if and only if it is G-isomorphic to G/H for some subgroup H ⊂ G.

Proof Suppose given a G-map fx : G/H → X with fx(eH) = x. Then since H isthe isotropy subgroup of eH in G/H, it must be contained in the isotropy subgroupof fx(eH) = x. In other words, we must have H ⊂ Gx. Also, since gH = g · eH,fx(gH) = g · fx(eH) = gx, so fx is uniquely determined by its effect on eH. Clearly, itsimage is G · x.

Conversely, if H ⊂ Gx, we use the above formula to define fx : G/H → X: fx(gH) =gx. We must show this is well defined: i.e. if gH = g′H, we must show that gx = g′x. ByLemma 3.2.3, gH = g′H if and only if g = g′h for some h ∈ H. But then gx = g′hx = g′x,since hx = x. Thus, fx is a well defined function. To see that fx is a G-map, note thatfx(g′ · gH) = fx((g′g)H) = g′gx = g′fx(gH).

If fx is a G-isomorphism, then the isotropy subgroups of eH and fx(eH) must beequal. But the isotropy subgroup of eH in G/H is H, while the isotropy subgroup of xis Gx, so if fx is a G-isomorphism, then H = Gx.

Conversely, suppose that H = Gx. To show that fx induces an isomorphism of G/Gxonto G · x, it suffices to show that fx is injective. Suppose that fx(gH) = fx(g′H),and hence that gx = g′x. Multiplying both sides by g−1 and cancelling, we see thatx = g−1g′x, so that g−1g′ is in the isotropy subgroup of x. Thus g−1g′ ∈ Gx = H, sothat gH = g′H by Lemma 3.2.3.

The next corollary now follows from Lagrange’s Theorem.


Corollary 3.3.9. Let X be a finite G-set and let x ∈ X. Then the number of elementsin the orbit G · x is equal to [G : Gx]. In particular, since [G : Gx] is finite, if either Gor Gx is finite, so is the other, and G · x has |G|/|Gx| elements.

As noted in Example 5 of Examples 3.3.3, S(X) acts on X for any set X, via σ · x =σ(x), for σ ∈ S(X) and x ∈ X. Thus, we may use the notions of orbit and isotropy toobtain information about S(X). We are primarily concerned with the case where X isfinite.

Recall from Corollary 3.1.13 that the canonical inclusion of Sn−1 in Sn identifies Sn−1

with the subgroup of Sn consisting of those elements that fix n. We restate this in termsof isotropy groups:

Lemma 3.3.10. Consider the standard action of Sn on {1, . . . , n}. Under this action,the isotropy subgroup of n is the image of the canonical inclusion Sn−1 ⊂ Sn.

Corollary 3.3.9 now allows us to give an inductive calculation of the order of Sn.

Corollary 3.3.11. The order of Sn is n! (n-factorial).

Proof Since the 2-cycle (i n) takes n to i for each 1 ≤ i < n, each element of {1, . . . , n}is in the orbit of n under the standard action of Sn. In other words, Sn acts transitivelyon {1, . . . , n}. Since this orbit has n elements, and since the isotropy subgroup of n isSn−1, we have n = |Sn|/|Sn−1| by Corollary 3.3.9. The result follows by induction on n,starting with the obvious fact that S1 is the trivial group.

It is often desirable to count the elements in a G-set X by adding up the numbers ofelements in its orbits. To do that, we shall need the following lemma.

Lemma 3.3.12. Let X be a G-set and let x, y ∈ X. Then y ∈ G · x if and only ifG ·y = G ·x. In consequence, if x, z ∈ X, then G ·x = G ·z if and only if G ·x∩G ·z �= ∅.Proof Since y ∈ G ·x, y = gx for some g ∈ G. Thus, g′y = (g′g)x ∈ G ·x for all g′ ∈ G,so G · y ⊂ G · x. But x = g−1y, so G · x ⊂ G · y by the same reasoning.

Notice that the analogy between the orbits in a G-set and the cyclic subgroups of agroup has just been left behind, as we can find lots of examples (e.g., in cyclic groups)of inclusions K ⊂ H of cyclic subgroups which are not equal.

Definition 3.3.13. Let X be a G-set. Then a set of orbit representatives in X is asubset Y ⊂ X such that

1. If x, y ∈ Y with x �= y, then G · x ∩G · y = ∅.2. X =

⋃x∈Y G · x.

For a general G-set, existence of a set of orbit representatives turns out to be animportant special case of the Axiom of Choice. But the case for finite G-sets is mucheasier.

Lemma 3.3.14. Any finite G-set has a set of orbit representatives.

Proof Let X be our G-set. We argue by induction on the number of elements in X. Letx ∈ X. If G · x = X, then {x} is the desired set. Otherwise, let X ′ be the complementof G · x in X. By Lemma 3.3.12, if g ∈ G and x′ ∈ X ′, gx′ must lie in X ′. Thus, X ′

is a sub-G-set of X. By induction, X ′ has a set, Y ′, of orbit representatives. But thenY = Y ′ ∪ {x} is a set of orbit representatives for X.


For infinite sets, one can prove the existence of a set of orbit representatives by usingZorn’s Lemma (a form of the Axiom of Choice) to perform what amounts to an infiniteinduction argument. We shall use similar techniques in the material on ring theory inChapters 7 through 12.

The following corollary is now immediate from Corollary 3.3.9.

Corollary 3.3.15. (G-set Counting Formula) Let Y be a set of orbit representatives forthe finite G-set X. Then the number, |X|, of elements in X is given by

|X| =∑x∈Y

[G : Gx].

An understanding of G-sets is vital to an understanding of permutation groups be-cause of the following generalization of Cayley’s Theorem.

Proposition 3.3.16. Let X be a G-set. For g ∈ G, define g : X → X by g(x) = gx.Then g is a permutation of X. Moreover, the function Λ : G → S(X) defined byΛ(g) = g is a group homomorphism, whose kernel is the intersection of the isotropysubgroups of the elements of X:

ker Λ =⋂x∈X

Gx.

Conversely, if Λ : G → S(X) is any group homomorphism, then X is a G-set underthe action defined by g · x = Λ(g)(x).

In fact, the correspondence which takes a G-set X to the homomorphism Λ : G →S(X) gives a one-to-one correspondence between the actions of G on X and the homo-morphisms from G to S(X).

Proof Let X be a G-set and let g ∈ G. We wish to show that g is a permutation ofX. It is onto, because x = g(g−1x) for all x ∈ X. It is one-to-one, since if g(x) = g(y),then gx = gy, and, multiplying both sides by g−1, we see that x = y.

Now Λ is a homomorphism if and only if g ◦ g′ = gg′ for all g, g′ ∈ G. But clearly,(g ◦ g′)(x) = gg′x = gg′(x) for all x ∈ X, so Λ is indeed a homomorphism. The kernelof Λ is the set of elements in G which act on X by the identity map, and hence fix eachelement in X. These then are the group elements which lie in the isotropy group of everyelement of X.

Conversely, given the homomorphism Λ, and defining g ·x to be equal to Λ(g)(x), wehave that ex = x because Λ(e) is the identity permutation, while g1(g2x) = Λ(g1)(Λ(g2)(x)) =(Λ(g1) ◦ Λ(g2))(x) = Λ(g1g2)(x) = (g1g2)x.

Finally, the passage from the G-set X to the homomorphism Λ is easily seen to beinverse to the passage from a homomorphism Λ : G→ S(X) to an action on X.

Note that if the G-set X is G (with the usual action), then the map Λ : G → S(G)given above is precisely the one given in Cayley’s Theorem.

Definitions 3.3.17. Let X be a G-set.

1. We say that the action of G on X is effective, or that X is an effective G-set, if⋂x∈X Gx = e. In words, G acts effectively on X if the only element of G which

fixes every element of X is e.


2. We say that the action of G on X is free, or that X is a free G-set, if the isotropysubgroup of every element of X is e. In other words, X is free if for any g ∈ Gand x ∈ X, gx = x implies that g = e.

Clearly, any free action is effective. However, there are many effective actions thatare not free.

Examples 3.3.18.1. The standard action of G on G is free. Indeed, if gx = x, then right multiplication

by x−1 gives g = e.

2. The standard action of S(X) on X is effective, since the only function X → Xwhich fixes every element of X is the identity function. Note, however, that if Xhas more than two elements, then this action is not free. For instance, a 2-cycle(i j) fixes every element other than i and j.

Via the map Λ : G→ S(X) of Proposition 3.3.16, we may interpret effective actionsin terms of embeddings into symmetric groups.

Corollary 3.3.19. An action of a group G on a set X is effective if and only if theinduced map Λ : G→ S(X) is an embedding.

Thus, there is an embedding of G into Sn if and only if there is an effective action ofG on a set with exactly n elements.

Proof For any G-set X, Proposition 3.3.16 tells us that the kernel of the induced mapΛ : G → S(X) is

⋂x∈X Gx. Since effective actions are those for which

⋂x∈X Gx = e,

they are precisely those actions for which Λ is an embedding.

Thus, the key to discovering whether a given group embeds in Sn lies in understandingthe isotropy groups that appear in the various orbits of G-sets. We shall take an ad hocapproach to these questions at this time, but the reader may want to come back to thismaterial after reading Section 3.7.

Definition 3.3.20. By a minimal subgroup of G we mean a subgroup H �= e with theproperty that the only subgroups of H are e and H.

This last is, of course, an abuse of language. Such a subgroup should more properlybe called a minimal nontrivial subgroup. But the trivial subgroup is rather uninteresting,so we make the above definition for the purposes of economy of language. (Similarly,in our discussion of commutative rings, we shall use the expression “maximal ideal” foran ideal which is maximal in the collection of all ideals which are not equal to the ringitself.)

Note that not all groups have minimal subgroups. For instance, it is easy to seethat every nontrivial subgroup of Z contains a subgroup which is strictly smaller, butnontrivial. But this cannot happen in a finite group.

Lemma 3.3.21. Every nontrivial finite group G has a minimal subgroup.

Proof We argue by induction on the order of G. If G has no subgroups other than eand itself (e.g. if |G| = 2), then G is a minimal subgroup of itself. Otherwise, there is anontrivial subgroup H �= G. But then |H| < |G|, so that H has a minimal subgroup byinduction. However, a minimal subgroup of H is also minimal in G.


Most finite groups have several different minimal subgroups, which are not necessarilyisomorphic to each other. However, in the case that there is a unique minimal subgroup,we obtain a strong consequence for the embedding problem.

Proposition 3.3.22. Let G be a finite group with exactly one minimal subgroup. Thenany effective G-set has at least |G| elements. In particular, n = |G| is the smallest integersuch that G embeds in Sn.

Proof Let H be the minimal subgroup of G. We claim that every nontrivial subgroupof G contains H.

To see this, let K be a nontrivial subgroup of G. Then there is a minimal subgroupK ′ ⊂ K. But any minimal subgroup of K is also a minimal subgroup of G. Since G hasonly one minimal subgroup, we must have K ′ = H, and hence H ⊂ K.

Now suppose given an effective G-set X. We claim there must be at least one x ∈ Xwith Gx = e.

Suppose not. Then each Gx is a nontrivial subgroup, and hence H ⊂ Gx for each x.But then H ⊂ ⋂

x∈X Gx, and hence X is not effective.Thus, there is an x ∈ X with Gx = e. But then G · x is G-isomorphic to G/e, and

hence X contains an orbit which has |G| elements in it. In particular, X has at least |G|elements.

Exercises 3.3.23.1. Show that the standard action on G is G-isomorphic to G/e and that the one-pointG-set is G-isomorphic to G/G.

‡ 2. Let G = D6. Show that G acts effectively (but not freely) on G/〈a〉, and hencethat G embeds in S3. Deduce from Corollary 3.3.11 that this embedding is anisomorphism.

3. Generalize the preceding problem to obtain an embedding from D2n into Sn. Sincethe order of the two groups is different for n > 3, this will not be an isomorphismfor these values of n.

4. Let p be a prime and let r ≥ 1. Show that Zpr cannot embed in Sn for n < pr.

5. Show that Q8 cannot embed in Sn for n < 8.

6. Show that Q2r cannot embed in Sn for n < 2r.

7. Find the minimal value of n such that D2r embeds in Sn. Do the same for D2pr ,where p is an odd prime.

8. Let X and Y be G-sets. By X × Y , we mean the G-set given by the action of Gon the cartesian product of X and Y in which g · (x, y) = (gx, gy), for all g ∈ G,x ∈ X, and y ∈ Y . What is the isotropy subgroup of (x, y) ∈ X × Y ?

9. Let G = Z6. Let H,K ⊂ G be the subgroups of order 2 and 3, respectively. Showthat G/H ×G/K is a free G-set.

10. Show that the action ofG onX×Y need not be transitive, even ifG acts transitivelyon both X and Y .

11. Show that if G acts freely on X, then it also acts freely on X × Y for any Y .


12. There is a construction in set theory called the disjoint union. The disjoint unionof X and Y , written X

∐Y , is a set containing both X and Y in the following

way: in X∐Y we have X ∩ Y = ∅, and X ∪ Y = X

∐Y . We shall give a formal

construction of the disjoint union in Chapter 6.

Show that if X and Y are G-sets, then G acts on X∐Y in such a way that X and

Y are G-subsets.

13. Let G = Z6. Let H,K ⊂ G be the subgroups of order 2 and 3, respectively. Showthat G acts effectively (but not freely) on G/H

∐G/K. Deduce that Z6 embeds

in S5.

14. Let m and n be relatively prime. Show that Zmn embeds in Sm+n.

15. Show that Q12 embeds in S7.

16. What is the isotropy subgroup of gH ∈ G/H for an arbitrary g ∈ G?

17. Let X be a G-set and let H be a subgroup of G. The set of H-fixed points of X,written XH , is defined to be the set of x ∈ X such that hx = x for all h ∈ H.Show that x ∈ XH if and only if H ⊂ Gx.

‡ 18. Let X be a G-set and let H be a subgroup of G. Show that there is a one-to-onecorrespondence between the G-maps from G/H to X and the elements of XH .

‡ 19. Let X be a G-set. Show that there is an equivalence relation on X defined bysetting x ∼ y if y = gx for some g ∈ G. Show that the equivalence classes of thisrelation are precisely the orbits G · x for x ∈ X.

The set of equivalence classes thus obtained is called the orbit space of X, anddenoted X/G. (Thus, the elements of X/G are the orbits G · x in X.) We writeπ : X → X/G for the canonical map, which takes x to its equivalence class.

20. A G-map f : X → Y is said to be isovariant if Gx = Gf(x) for each x ∈ X. Showthat a G-map f : X → Y is isovariant if and only if its restriction to each orbit isinjective.

21. Recall that a G-set Y has the trivial action if gy = y for all y ∈ Y and g ∈ G.Show that if X is any G-set, and if we give the orbit space X/G the trivial action,then the canonical map π : X → X/G is a G-map. Show also that if Y is any G-setwith the trivial action, and if f : X → Y is a G-map, then there is a unique G-mapf : X/G→ Y such that f ◦ π = f .

22. Show that if G acts trivially on X and if Y is any G-set, then the G-maps fromG × X to Y are in one-to-one correspondence with the functions from X to Y .(Here, G in the above product is the standard free G-set G/e.) In this context, wecall G×X the free G-set on the set X.

‡ 23. Let X be any set and write Xn for the product X×· · ·×X of n copies of X. Showthat Xn is an Sn-set via σ · (x1, . . . , xn) = (xσ−1(1), . . . , xσ−1(n)), for σ ∈ Sn and(x1, . . . , xn) ∈ Xn.

† 24. Let G be a finite group whose order is divisible by a prime number p. In thisexercise, we shall prove Cauchy’s Theorem (also given by a different proof as The-orem 5.1.1), which says that such a group G must have an element of order p.


(a) Define σ : Gp → Gp by σ(g1, . . . , gp) = (gp, g1, . . . , gp−1). Show that σ givesa permutation of Gp of order p (i.e., σ ∈ S(Gp) with σp = e). Deduce thatthere is an action of Zp on Gp, given by k · (g1, . . . , gp) = σk(g1, . . . , gp) forall k ∈ Zp. (In fact, if we identify Zp with the subgroup of Sn generated bythe p-cycle (1 2 · · · p), this action is just the restriction to Zp of the Sn actiongiven in Problem 23.)

(b) Define X ⊂ Gp by

X = {(g1, . . . , gp) | g1 · · · gp = e}.

In other words, the product (in order) of the gi is the identity element of G.Show that X is a sub-Zp-set (but not a subgroup, unless G is abelian) of Gp.

(c) Show that the isotropy group of (g1, . . . , gp) ∈ X is the trivial subgroup of Zpunless g1 = g2 = · · · = gp = g for some g ∈ G with gp = e. Deduce that thereis a one-to-one correspondence between the fixed-point set XZp and the set{g ∈ G | gp = e}. Deduce that G has an element of order p if and only if XZp

has more than one element.

(d) Note that if x ∈ X is not in XZp , then the orbit Zp · x has p elements. Sincetwo orbits are either disjoint or equal, X is the disjoint union of XZp with theorbits which lie outside of XZp . Deduce that

|X| = |XZp |+ some multiple of p.

Show that |X| = |G|p−1, so that |X| is divisible by p. Deduce that |XZp | isdivisible by p, and hence G has an element of order p.

25. We say that G acts on the right on a set X (or that X is a right G-set) if for allx ∈ X and g ∈ G there is an element xg ∈ X such that

(xg)g′ = x(gg′) and xe = x

for all x ∈ X and all g, g′ ∈ G. Show that if X is a right G-set, then there isan induced left action of G on X given by g · x = xg−1. Deduce that there is aone-to-one correspondence between the left G-sets and the right G-sets.

26. LetH be a subgroup ofG. ThenH acts on the right ofG by ordinary multiplication.Show that we may identify the set of left cosets G/H with the orbit space (seeProblem 19) of this action.

27. Let G act on the right on X and act on the left on Y . Then G acts on the left onX×Y by g · (x, y) = (xg−1, gy). We call the orbit space of this action the balancedproduct of X and Y , which we denote X ×G Y .

Let H be a subgroup of G and let X be a (left) H-set. Then using the right actionof H on G by ordinary multiplication, we can form the balanced product G×H X.Show that the standard left action of G on itself induces a left action of G onG×H X.

Write [g, x] for the element ofG×HX represented by the ordered pair (g, x) ∈ G×X.(In other words, [g, x] is the orbit of (g, x) under the above action of H on G×X.)Show that the isotropy subgroup G[e,x] of [e, x] under the above G-action coincideswith the isotropy subgroup Hx of x under the original action of H on X.


28. Let H be a subgroup of G and let X be a (left) H-set. Let Y be a (left) G-set.Then H acts on Y by restricting the action of G. Show that there is a one-to-onecorrespondence between the H-maps from X to Y and the G-maps from G×H Xto Y .

29. Let M be a monoid. Define an M -set by substituting M for G in the definitionof G-set. Find a monoid M and an M -set X such that the function m : X → Xdefined by m(x) = m · x need not be a permutation for all values of m.

30. Show that if you delete the requirement that e · x = x from the definition of G-set,then the functions g need not be permutations.

3.4 Supports of Permutations

We isolate here some of the more elementary considerations for the study of cycle struc-ture in Section 3.5. We give some consequences in the exercises.

Definition 3.4.1. Let σ ∈ S(X). The support of σ, Supp(σ), is the following subset ofX:

Supp(σ) = {x ∈ X |σ(x) �= x}In particular, σ is the identity outside of its support.

It is easy to calculate the support of a k-cycle.

Lemma 3.4.2. The support of the k-cycle (i1 · · · ik) is {i1, . . . , ik}.Here is another way to think of supports.

Lemma 3.4.3. Let Y be a subset of X and let σ ∈ S(X). Then Supp(σ) ⊂ Y if andonly if σ is the identity on the complement of Y . In particular, if i : Y → X is theinclusion of Y in X, then Supp(σ) ⊂ Y if and only if σ is in the image of the canonicalinclusion i∗ : S(Y ) ⊂ S(X).

Proof The support is the complement of the subset of points fixed by σ, hence the firststatement. For the second, recall from Lemma 3.1.12 that the image of i∗ is the set ofpermutations that fix the complement of Y .

Applying Lemma 3.4.3 to Y = Supp(σ), we obtain the following corollary.

Corollary 3.4.4. Let σ be a permutation of the set X. Then σ restricts to a bijectionσ : Supp(σ)→ Supp(σ).

Definition 3.4.5. We say the permutations σ, τ ∈ S(X) are disjoint if Supp(σ) ∩Supp(τ) = ∅.

Disjoint permutations interact well with one another:

Proposition 3.4.6. Let σ and τ be disjoint permutations of X. Then σ and τ commutewith each other.

Moreover, the product στ is given as follows:

(στ)(x) =

⎧⎨⎩ σ(x) if x ∈ Supp(σ)τ(x) if x ∈ Supp(τ)x otherwise.


In particular, the support of στ is the union Supp(σ) ∪ Supp(τ).Finally, στ has finite order if and only if both σ and τ have finite order, in which

case the order of στ is the least common multiple of the orders of σ and τ .

Proof We wish to show that both στ and τσ have the effect specified by the displayedformula.

Away from Supp(σ)∪Supp(τ), both σ and τ are the identity, and hence both στ andτσ are also.

On Supp(σ), τ is the identity, as Supp(σ) ∩ Supp(τ) = ∅. Thus, for x ∈ Supp(σ),(στ)(x) = σ(x). Since x ∈ Supp(σ), σ(x) is also in Supp(σ) by Corollary 3.4.4, andhence (τσ)(x) = σ(x) as well.

A similar argument shows that both στ and τσ act as τ on Supp(τ). Thus, στ = τσ,and the displayed formula holds.

An easy induction based on the displayed formula now shows:

For k > 0, (στ)k(x) =

⎧⎨⎩ σk(x) if x ∈ Supp(σ).τk(x) if x ∈ Supp(τ).x otherwise.

Thus, (στ)k = e if and only if both σk = e and τk = e, and the result follows.

Exercises 3.4.7.1. Show that an element σ ∈ S(Z+) lies in S∞ if and only if Supp(σ) is finite.

‡ 2. Suppose given subsets Y,Z ⊂ X, with Y ∩Z = ∅. Show that there is an embeddingμ : S(Y )×S(Z)→ S(X) whose restrictions to S(Y ) = S(Y )×e and S(Z) = e×S(Z)are the canonical inclusions of S(Y ) and S(Z), respectively, in S(X). Show thatthe image of μ is the set of permutations σ ∈ S(X) such that the following twoproperties hold:

(a) Supp(σ) ⊂ Y ∪ Z(b) σ(Y ) = Y .

‡ 3. Combinatorics Consider the action of Sn on {0, 1}n obtained by permuting thefactors (see Problem 23 of Exercises 3.3.23). We can use it to derive some of thestandard properties of the binomial coefficients. First, for 0 ≤ k ≤ n, let us write

xk = (1, . . . , 1︸︷︷︸k times

, 0, . . . , 0︸︷︷︸n−k times

) ∈ {0, 1}n.

(a) Show that the number of elements in the orbit Sn · xk is equal to the numberof ways to choose k things out of n things (which we may identify with thenumber of k-element subsets of an n-element set), which we shall denote bythe binomial coefficient

(nk

).

(b) Show that the isotropy subgroup of xk is isomorphic to the direct productSk × Sn−k. Deduce that (

n

k

)=

n!k! · (n− k)! ,

and that the right-hand side must therefore be an integer.


(c) Show that the elements x0, . . . , xn form a set of orbit representatives for theSn action on {0, 1}n. Deduce that

2n =n∑k=0

(n

k

).

(d) Deduce that there are exactly 2n subsets of an n-element set.

(e) Show that the number of ordered k-element subsets of an n-element set is k!times the number of unordered k-element subsets of that set. Deduce that thenumber of ordered k-element subsets is n!/(n− k)!.

4. It can be shown that if X is any infinite set, then the disjoint union X∐X may

be placed in bijective correspondence with X. Deduce from Problem 2 that for anyinfinite set X, we may embed S(X)× S(X) in S(X).

5. Reconsider the preceding problem forX = Z+. Show that for any choice of injectionZ+

∐Z+ → Z+, the induced embedding S(Z+) × S(Z+) ⊂ S(Z+) restricts to an

embedding of S∞ × S∞ in S∞. (These embeddings turn out to be importantin studying the cohomology of the group S∞, an endeavor which falls under theheading of homological algebra, but has applications to topology.)

6. In this problem we assume a familiarity with infinite disjoint unions and withinfinite products, as discussed in Chapter 6.

(a) Suppose given a family of sets {Xi | i ∈ I}. Show that we may embed theinfinite product

∏i∈I S(Xi) in S(

∐i∈I Xi). Deduce that there is a permu-

tation of Z+ which may be thought of as the disjoint product of infinitelymany cycles, with one k-cycle being given for each k ≥ 2. Deduce that thereare permutations in S(Z+) whose orbits are all finite, but which have infiniteorder.

(b) Construct a permutation of Z+ which is a product of infinitely many cyclesof the same order. Deduce that there are elements of S(Z+) which have finiteorder but do not lie in S∞.

3.5 Cycle Structure

Here, using the ideas of G-sets and orbits, we develop the notion of cycle decompositionsfor permutations. This gives a powerful tool for studying permutation groups that hassome analogies to prime decompositions in Z.

Using cycle decomposition, we shall construct the sign homomorphism ε : Sn → {±1}and define the alternating groups, An. We shall encounter further applications of cycledecomposition in all of our subsequent study of permutation groups, particularly inSection 4.2.

Recall that the permutations σ and τ are disjoint if Supp(σ) ∩ Supp(τ) = ∅, whereSupp(σ) is the complement of the set of points fixed by σ.

Definition 3.5.1. By a product of disjoint permutations, we mean a product σ1 · · ·σk,such that σi and σj are disjoint for i �= j.

We are particularly interested in products of disjoint cycles: i.e., products σ1 · · ·σkof disjoint permutations in Sn, such that each σi is a cycle.


We now generalize Proposition 3.4.6.

Corollary 3.5.2. Let σ1 · · ·σk ∈ S(X) be a product of disjoint permutations. Then theeffect of σ1 · · ·σk is given as follows: on Supp(σi), the effect of σ1 · · ·σk is the sameas that of σi for 1 ≤ i ≤ k. Away from

⋃ki=1 Supp(σi), σ1 · · ·σk is the identity. In

particular, Supp(σ1 · · ·σk) =⋃ki=1 Supp(σi).

In addition, if each σi has finite order, then the order of σ1 · · ·σk is the least commonmultiple of the orders of the σi.

Proof We argue by induction on k. The case k = 2 is given by Proposition 3.4.6.For k > 2, the induction hypothesis gives Supp(σ1 · · ·σk−1) =

⋃k−1i=1 Supp(σi), and hence

σ1 · · ·σk−1 is disjoint from σk. The result now follows by applying Proposition 3.4.6 (andthe induction hypothesis) to the product (σ1 · · ·σk−1) · σk.

The following proposition gives the cycle decomposition for a permutation, σ.

Proposition 3.5.3. Let σ ∈ Sn for n ≥ 2, with σ �= e. Then σ may be written uniquelyas a product of disjoint cycles: σ = σ1 · · ·σk. Here, uniqueness means that if σ may alsobe written as the product τ1 · · · τl of disjoint cycles, then k = l, and there is a permutationφ ∈ Sk such that σi = τφ(i) for 1 ≤ i ≤ k.

Proof Let G = 〈σ〉, the cyclic subgroup of Sn generated by σ. We wish to understandX = {1, . . . , n} as a G-set. Note that if the orbit of x ∈ X has only one element, thenthe isotropy subgroup of x has index 1 in G by Corollary 3.3.9. This means that theisotropy subgroup of x must be all of G. If this were true for all x ∈ X, then G wouldact as the identity on X. Since σ �= e, this cannot be the case.

Suppose, then, that y ∈ X has an orbit with more than one element. Let r be thesmallest positive integer with the property that σr(y) = y. (Such an integer exists,because σo(σ)(y) = y.) Note that r > 1, as otherwise y is fixed under σ and all of itspowers, and hence the orbit of y under G would have only one element.

We claim first that the elements y, σ(y), . . . , σr−1(y) are all distinct. To see this, notethat otherwise there are integers s and t with 0 ≤ s < t < r such that σs(y) = σt(y) =σs(σt−s(y)). Thus, y and σt−s(y) have the same image under σs. Since σs is injective,this gives σt−s(y) = y. Since 0 < t− s < r, this contradicts the minimality of r.

Thus, there is an r-cycle σ1 = (y σ(y) · · · σr−1(y)). Let Y1 = Supp(σ1) = {y, σ(y), . . . , σr−1(y)}.Then, by inspection, we see that σ has the same effect as σ1 on each element of Y1. Thus,σ restricts to a bijection Y1 → Y1. By passage to powers, we see that Y1 is a sub-G-setof X. Since G acts transitively on Y1, we see that Y1 is the orbit of y under the actionof G.

Now let Y1, . . . , Yk be the set of all distinct orbits in X with more than one element.By Lemma 3.3.12, Yi ∩ Yj = ∅ for i �= j. Repeating the argument above, we can findcycles σi for 2 ≤ i ≤ k such that σi has the same effect on Yi is σ does, and Supp(σi) = Yi.In particular, σ1 · · ·σk is a product of disjoint cycles.

We claim that σ = σ1 · · ·σk. By Corollary 3.5.2, the effect of σ1 · · ·σk on Yi =Supp(σi) is the same as that of σi. But σi was chosen to have the same effect on Yi asσ. Thus, σ agrees with σ1 · · ·σk on

⋃ki=1 Yi = Supp(σ1 · · ·σk). Since σ is the identity on

all one-element orbits, it is the identity off⋃ki=1 Yi, and σ = σ1 · · ·σk, as claimed.

We show uniqueness by reversing the above argument. In a product τ = τ1 . . . τl ofdisjoint cycles, the effect of τ agrees with that of τi on Supp(τi). As the cycle notationshows, 〈τi〉 acts transitively on Supp(τi), so 〈τ〉 acts transitively on Supp(τi) as well.


Thus, there are l distinct orbits under the action of 〈τ〉 which have more than oneelement.

Thus, if τ is equal to the permutation σ, above, then k = l and, equating the orbitsof 〈σ〉 and 〈τ〉, we may relabel the τi so that Supp(τi) = Supp(σi) for i = 1, . . . , k. Butsince σ = σi and τ = τi on this common support, τi and σi are equal as permutations.

Remarks 3.5.4. Note the concluding statement of the last proof. A cycle is determinedby its effect as a permutation, not by its notation. However, it’s easy to see that thereare precisely r different ways to write an r-cycle in cycle notation. It depends solely onwhich element of its support is written first.

Proposition 3.5.3 shows that every permutation is a product of cycle permutations.But we can actually write a permutation as a product of even simpler terms as long aswe no longer insist that these terms be disjoint. Recall that a transposition is anotherword for a 2-cycle. The following lemma may be verified by checking the effect of bothsides of the proposed equality on the elements {i1, . . . , ik}. Here, we should emphasizethe fact that if τ, τ ′ ∈ Sn and j ∈ {1, . . . , n}, then (ττ ′)(j) = τ(τ ′(j)).

Lemma 3.5.5. Let σ be the k-cycle σ = (i1 · · · ik). Then σ is the product of k − 1transpositions:

σ = (i1 i2)(i2 i3) · · · (ik−1 ik).

Since the cycle permutations generate Sn, we see that every element of Sn is a productof transpositions.

Corollary 3.5.6. Sn is generated by the transpositions.

We wish to define the sign of a permutation to be±1 in such a way that the sign will be+1 if the permutation may be written as a product of an even number of transpositions,and is −1 otherwise. Our initial definition will be more rigid than this.

Definition 3.5.7. The sign of a k-cycle is (−1)k−1. Thus, the sign is −1 if k is evenand is +1 if k is odd.

The sign of a product of disjoint cycles is the product of the signs of the cycles. Thesign of the identity map is +1.

Thus, the sign of every permutation is defined, via Proposition 3.5.3. Notice therelationship between the sign of a cycle and the number of transpositions in the productgiven in Lemma 3.5.5. Multiplying out the signs over a product of disjoint cycles, we seethat if the sign of a permutation, σ, is +1, then there is at least one way to write σ asthe product of an even number of transpositions. If the sign of σ is −1, then there is atleast one way to write σ as the product of an odd number of transpositions.

The next proposition is the key to the utility of signs. Notice that {±1} forms agroup under multiplication. Since it has two elements, it is isomorphic to Z2.

Proposition 3.5.8. For any σ, τ ∈ Sn, n ≥ 2, the sign of στ is the product of the signsof σ and τ .

Thus, there is a homomorphism ε : Sn → {±1} obtained by setting ε(σ) equal to thesign of σ. In particular, ε has the property that ε(τ) = −1 for any transposition τ .


Proof Since we may write τ as a product of m transpositions in such a way that thesign of τ is (−1)m, we may argue by induction, assuming that τ is a transposition, sayτ = (i j).

Write σ as a product of disjoint cycles: σ = σ1 · · ·σk. There are four cases to consider,where a non-trivial orbit of σ is one which has more than one element:

1. Neither i nor j lies in a non-trivial orbit of σ.

2. j lies in a non-trivial orbit of σ and i does not.

3. i and j lie in different non-trivial orbits of σ.

4. Both i and j lie in the same non-trivial orbit of σ.

We analyze the decomposition of στ as a product of disjoint cycles in each of thesecases. The first case is simplest, as then σ and τ are disjoint. Thus, τ is one of the cyclesin the decomposition of στ , and the others are those from the decomposition of σ. Theresult follows in this case from the definition of sign.

In case 2, we may assume that σ is an r-cycle, σ = (i1 · · · ir), with ir = j. Thenστ = (i i1 · · · ir), an (r + 1)-cycle.

In case 3, we may assume that σ is the product of two disjoint cycles, σ1 = (i1 · · · ir),with ir = i, and σ2 = (j1 · · · js), with js = j. Here, we see that στ is the (r + s)-cycleστ = (i1 · · · ir j1 · · · js). The signs work out correctly once again.

Case 4 is the most complicated. We may assume that σ is a single cycle. Theanalysis falls naturally into sub-cases. In the first, σ = τ = (i j), and here στ = e, astranspositions have order 2. The next case is where σ is an r-cycle with r ≥ 3 and iand j appear next to each other in the cycle representation of σ, say σ = (i1 · · · ir) withir−1 = i and ir = j. Here στ is the (r − 1)-cycle στ = (i1 · · · ir−1).

In the final case, i and j cannot be written next to each other in the cycle represen-tation of σ. Thus, we may write σ = (i1 · · · ir j1 · · · js), with ir = i, js = j, and with rand s both greater than or equal to 2. Here, στ is given by στ = (i1 · · · ir)(j1 · · · js).

The sign homomorphism gives rise to an important subgroup of Sn.

Definitions 3.5.9. The n-th alternating group, An, is the kernel of the sign homomor-phism ε : Sn → {±1}.

We say that a permutation is even if it lies in An (i.e., if its sign is 1). A permutationis said to be odd if its sign is −1.

The fact that ε is a homomorphism allows us to determine the sign of a permutationfrom any representation of it as a product of transpositions.

Corollary 3.5.10. Let σ ∈ Sn. Then σ ∈ An if and only if σ may be written as aproduct of an even number of transpositions. If σ ∈ An, then whenever σ is written asa product of transpositions, the number of transpositions must be even.

Proof The homomorphism ε carries any product of an odd number of transpositionsto −1 and carries any product of an even number of transpositions to 1. Thus, givenany two presentations of a permutation σ as a product of transpositions, the number oftranspositions in the two presentations must be congruent mod 2.

We have made use of the fact that Sn is generated by the transpositions. UsingCorollary 3.5.10, we may obtain a similar result for An.


Corollary 3.5.11. For n ≥ 3, the alternating group An is generated by 3-cycles. Thus,since each 3-cycle lies in An, a permutation σ ∈ Sn lies in An if and only if σ is aproduct of 3-cycles.

Proof Corollary 3.5.10 shows that a permutation is in An if and only if it is the productof an even number of transpositions. Thus, it suffices to show that the product of anytwo transpositions is a product of 3-cycles.

If the two transpositions are not disjoint, then either they are equal, in which casetheir product is the identity, or the product has the form

(a b) · (b c) = (a b c),

a 3-cycle.The remaining case is the product of two disjoint cycles, (a b) and (c d). But then we

have

(a b) · (c d) = (a b c) · (b c d),

the product of two 3-cycles.

Exercises 3.5.12.1. Write the product (1 2 3 4) ·(3 4 5) as a product of disjoint cycles. What is the order

of this product?

2. Consider the function from Z7 to itself induced by multiplication by 2. (Recall thatmultiplication mod n gives a well-defined binary operation on Zn.) Identifying theelements of Z7 with {1, . . . , 7} (by identifying k with k for 1 ≤ k ≤ 7), showthat multiplication by 2 induces a permutation of {1, . . . , 7}. Write down its cyclestructure.

3. Repeat the preceding problem using multiplication by 3 in place of multiplicationby 2.

4. What are the elements of A3?

5. What are the elements of A4?

6. Recall from Problem 2 of Exercises 3.3.23 that S3 is isomorphic to the dihedralgroup D6. Which subgroup of D6 corresponds to A3 under this isomorphism?

7. Let p be prime and let r ≥ 1. Show that Sk has an element of order pr if and onlyif k ≥ pr. (Hint : Write the element in question as a product of disjoint cycles.What can you say about the length of the cycles?)

8. For a given n, what is the smallest k such that Zn embeds in Sk? (Hint : Theanswer depends on the prime decomposition of n.)

9. What is the largest value of n such that Zn embeds in S8?

‡ 10. Expanding on Remarks 3.5.4, show that the number of k-cycles in Sn is 1/k timesthe number of ordered k-element subsets of {1, . . . , n}. Deduce from Problem 3 ofExercises 3.4.7 that there are n!

k·(n−k)! k-cycles in Sn.


3.6 Conjugation and Other Automorphisms

In addition to the study of actions of the group G on sets, it is useful to study actions ofother groups on G. Of course, S(G) acts on G, but we have seen that the structure of asymmetric group S(X) depends only on the size of the set X. In particular, the actionof S(G) on G loses all the information about the group structure of G, and captures onlyits size.

To capture the group structure on G, a group should act on it through permutationsσ : G→ G which are group homomorphisms. These are precisely the group isomorphismsfrom G to itself.

Definitions 3.6.1. An automorphism of a group G is a group isomorphism from G toitself. We write Aut(G) ⊂ S(G) for the set of all automorphisms of G.

The reader may check the details of the following examples, using Problem 6 of Exer-cises 2.8.5 for the examples involving dihedral groups, and Problem 5 of Exercises 2.9.7for the examples involving quaternionic groups.

Examples 3.6.2.1. Let G be an abelian group. Then there is an automorphism f : G → G given byf(g) = g−1 for all g ∈ G.

2. Let 0 ≤ k < n. Then there is an automorphism f ∈ Aut(D2n) determined byf(b) = b, f(a) = abk.

3. There is an automorphism f ∈ Aut(D2n) determined by f(b) = b−1, f(a) = a.

4. Let 0 ≤ k < 2n. Then there is an automorphism f ∈ Aut(Q4n) determined byf(b) = b, f(a) = abk.

5. There is an automorphism f ∈ Aut(Q4n) determined by f(b) = b−1, f(a) = a.

We now state an immediate consequence of Lemma 2.5.7.

Lemma 3.6.3. Let G be a group. Then Aut(G) is a subgroup of S(G).

Definition 3.6.4. We say that the group K acts on G via automorphisms if G is aK-set in such a way that the action of each k ∈ K, k : G → G (defined as usual byk(g) = k · g), is an automorphism of G.

The proof of the following proposition is identical to that of Proposition 3.3.16.

Proposition 3.6.5. There is a one-to-one correspondence between the actions of K onG via automorphisms and the homomorphisms Λ : K → Aut(G). Here, a given actioncorresponds to the homomorphism defined by Λ(k) = k.

This formalism is useful, as we are about to define a very important action of G onitself through automorphisms, called conjugation. But we cannot denote the resultingG-set by “G,” as we reserve this symbol for the standard action induced by left multi-plication. So it is useful to be able to refer to conjugation in terms of a homomorphismfrom G to Aut(G).

Definition 3.6.6. Let G be a group and let x ∈ G. By conjugation by x, we mean thefunction cx : G→ G specified by cx(g) = xgx−1.


Lemma 3.6.7. Conjugation by x ∈ G is an automorphism of G. There is an action ofG on itself by automorphisms given by the homomorphism Γ : G → Aut(G) defined byΓ(x) = cx.

Proof To see that cx is a homomorphism, we have

cx(g)cx(g′) = xgx−1xg′x−1 = xgg′x−1 = cx(gg′).

It is a bijection for the same reason that left or right multiplication by any element ofG is a bijection: There is an inverse function given by an appropriate multiplication. Inthis case the inverse function is conjugation by x−1.

Thus, there is a well-defined function Γ : G→ Aut(G) with Γ(x) = cx. To see that Γis a group homomorphism, we have

(cx ◦ cy)(g) = cx(ygy−1) = xygy−1x−1 = (xy)g(xy)−1 = cxy(g),

because (xy)−1 = y−1x−1.

The elements in the image of Γ : G→ Aut(G) have a special name.

Definitions 3.6.8. An automorphism f ∈ Aut(G) is said to be an inner automorphismof G if f = cx, the conjugation by x, for some x ∈ G. We write Inn(G) ⊂ Aut(G) forthe set of inner automorphisms. Note that since Inn(G) = im Γ, Inn(G) is a subgroup ofAut(G).

The isotropy subgroup of y ∈ G under the conjugation action is by definition the set{x ∈ G |xyx−1 = y}. It is an important enough subgroup that it merits a name:

Definition 3.6.9. Let y be an element of the group G. Then the centralizer of y in Gis the subgroup

CG(y) = {x ∈ G |xyx−1 = y}.

Of course, if x ∈ H ⊂ G, then we write CH(x) for the centralizer of x in H. Whenthere’s no ambiguity, we shall simply refer to CG(x) as the centralizer of x.

The next lemma is useful in understanding the effect of conjugation. The proof is leftto the reader.

Lemma 3.6.10. Let x, y ∈ G. Then the following are equivalent.

1. x and y commute.

2. x ∈ CG(y), i.e., xyx−1 = y.

3. xyx−1y−1 = e.

We now give two useful consequences of Lemma 3.6.10.

Corollary 3.6.11. Suppose that the elements x, y ∈ G commute with each other. Thenevery element of 〈x〉 commutes with every element of 〈y〉.

Proof Since x ∈ CG(y) and since CG(y) is a subgroup of G, 〈x〉 ⊂ CG(y). Thus, ycommutes with xk for all k ∈ Z, and hence y ∈ CG(xk) as well. So 〈y〉 ⊂ CG(xk), andhence yl commutes with xk for all k, l ∈ Z.


Corollary 3.6.12. The element x ∈ G is in the kernel of the conjugation-induced ho-momorphism Γ : G→ Aut(G) if and only if x commutes with every element of G.

Proof We have x ∈ ker Γ if and only if x is in the isotropy subgroup of every y ∈ G.But this says that x ∈ CG(y) for every y ∈ G, and hence x commutes with every elementof G.

We shall refer to the kernel of Γ as the center, Z(G), of G:3

Definition 3.6.13. The center, Z(G), of a group G is given as follows:

Z(G) = {x ∈ G |xy = yx for all y ∈ G}.Definition 3.6.14. The conjugacy class of g ∈ G is {xgx−1 |x ∈ G}, which is the orbitof g under the action of G on itself by conjugation. Its elements are called the conjugatesof g.

Since the isotropy subgroup of g is CG(g), the next result follows from Corollary 3.3.9.

Corollary 3.6.15. The number of conjugates of g ∈ G is equal to [G : CG(g)], the indexin G of the centralizer, CG(g), of g in G.

The fact that CG(g) is the set of all elements that commute with g gives a character-ization of Z(G) in terms of conjugacy classes.

Corollary 3.6.16. The conjugacy class of g ∈ G consists of a single element if and onlyif g ∈ Z(G).

Proof [G : CG(g)] = 1 if and only if CG(g) = G.

The G-set Counting Formula (Corollary 3.3.15) for the conjugation action on G isextremely useful. We shall give a variant on it, called the Class Formula. We shall needsome setup for it first.

Definitions 3.6.17. We say that a conjugacy class of G is nontrivial if it has more thanone element.

By a set of class representatives for the nontrivial conjugacy classes we mean a subsetX ⊂ G such that

1. The conjugacy class of each x ∈ X is nontrivial.

2. If x, y ∈ X with x �= y, then the conjugacy classes of x and y intersect in the nullset.

3. Every element of G that is not in Z(G) lies in the conjugacy class of some x ∈ X.

The desired Class Formula is now immediate from the G-set Counting Formula.

Corollary 3.6.18. (Class Formula) Let G be a finite group and let X be a set of classrepresentatives for the nontrivial conjugacy classes in G. Then

|G| = |Z(G)|+∑x∈X

[G : CG(x)].

3The German word for center is Zentrum.


We can compute the conjugates and the centralizers of the elements in the dihedraland quaternionic groups by hand, using the explicit information we have about thesegroups. We shall leave these calculations for the next set of exercises. The next resultcomputes the conjugates of a k-cycle in Sn.

Lemma 3.6.19. Let σ be any permutation and let (i1 · · · ik) be any k-cycle in Sn. Thenσ · (i1 · · · ik) · σ−1 is a k-cycle:

σ · (i1 · · · ik) · σ−1 = (σ(i1) · · · σ(ik)).

Proof If j is not in {σ(i1), . . . , σ(ik)}, then σ−1(j) is not in {i1, . . . , ik}, and hence(i1 · · · ik) acts as the identity on σ−1(j). But then σ · (i1 · · · ik) · σ−1(j) = j, so thatσ · (i1 · · · ik) · σ−1 acts as the identity outside {σ(i1), . . . , σ(ik)}.

For 1 ≤ j < k, σ · (i1 · · · ik) · σ−1(σ(ij)) = σ · (i1 · · · ik)(ij) = σ(ij+1). Similarly,σ · (i1 · · · ik) · σ−1(σ(ik)) = σ(i1), so that σ · (i1 · · · ik) · σ−1 = (σ(i1) · · · σ(ik)), asclaimed.

Lemma 3.6.19 gives important information about the behavior of the symmetricgroups. Here is a sample.

Corollary 3.6.20. Suppose that σ ∈ Sn lies in the centralizer of the k-cycle (i1 · · · ik).Then σ({i1, . . . , ik}) = {i1, . . . , ik}.

Proof Let σ ∈ Sn Then

σ · (i1 · · · ik) · σ−1 = (σ(i1) · · · σ(ik)).

Thus,

Supp(σ · (i1 · · · ik) · σ−1) = {σ(i1), . . . , σ(ik)}= σ({i1, . . . , ik}).

Thus, if σ · (i1 · · · ik) · σ−1 = (i1 · · · ik), then

σ({i1, . . . , ik}) = Supp((i1 · · · ik)) = {i1, . . . , ik}.

There is an important action related to conjugation, in which G acts on the set ofhomomorphisms from another group into G. We shall use a notation consistent withthat used in our discussion of category theory in Chapter 6.

Notation 3.6.21. Let K and G be groups. We write MorGp(K,G) for the set of grouphomomorphisms from K to G.

Definition 3.6.22. Let K and G be groups. For g ∈ G and f ∈ MorGp(K,G), theconjugate of f by g, written g · f , is the homomorphism (g · f) ∈ MorGp(K,G) definedby (g · f)(k) = gf(k)g−1 for k ∈ K.

Two homomorphisms f, f ′ ∈ MorGp(K,G) are said to be conjugate if f ′ = g · f forsome g ∈ G.

Note that g · f is equal to the composite cg ◦ f , where cg : G→ G is conjugation byg. Thus, g · f is a homomorphism, as claimed. The following lemma is now immediate.


Lemma 3.6.23. Let K and G be groups. Then there is an action of G on MorGp(K,G)obtained by setting the action of g ∈ G on f ∈ MorGp(K,G) to be equal to the conjugatehomomorphism g · f defined above.

Conjugate homomorphisms are important in a number of questions, including theclassification of semidirect products and the cohomology of groups.

Exercises 3.6.24.1. What are all the conjugacy classes of elements of D6? Write down all the elements

in each class.

2. What are all the conjugacy classes of elements of D8? Write down all the elementsin each class.

3. What are all the conjugacy classes of elements of Q8? Write down all the elementsin each class.

4. What are all the conjugacy classes of elements of D2n for n ≥ 3? (Hint : The answerdepends in a crucial way on the value of n mod 2.)

5. What are all the conjugacy classes of elements of Q4n?

6. Find the centralizers of each of the elements of Q8.

7. Find the centralizers of the elements a, b ∈ D2n.

8. Find the centralizers of the elements a, b ∈ Q4n.

9. Let H be a subgroup of G and let x ∈ H. Show that

CH(x) = H ∩ CG(x).

10. Suppose that Sn has an element of order k. Show that D2k embeds in Sn.

11. Consider the 2-cycle (n− 1n) ∈ Sn. Show that the centralizer of (n− 1n) in Sn isisomorphic to Sn−2 × Z2.

‡ 12. Let a, b, c, d be distinct elements of {1, . . . , n}. Show that any element of Sn whichcentralizes both (a b c) and (b c d) must fix both a and d. Deduce that if n ≥ 4,then the only element which centralizes every 3-cycle is the identity.

13. Show that the centralizer of any k-cycle in Sn is isomorphic to Sn−k×Zk. We shallsee in Section 4.2 (using a straightforward application of Lemma 3.6.19 above) thatany two k-cycles in Sn are conjugate. Assuming this, deduce that there are n!

k·(n−k)!k-cycles in Sn. (See Problem 10 of Exercises 3.5.12 for a different line of argument.)

14. What is the centralizer of (1 2 3) in A4? How many conjugacy classes of 3-cyclesare there in A4?

15. Let σ, τ ∈ S(X). Show that Supp(τστ−1) = τ(Supp(σ)).

16. What is the center of D8?

17. What is the center of D2n for n ≥ 3? (Hint : The answer depends in a crucial wayon the value of n mod 2.)


18. What is the center of Q8?

19. What is the center of Q4n?

20. What is the center of Sn? What is the center of An?

‡ 21. Let H be any group and write Hn for the product H × · · · × H of n copies ofH. Show that Sn acts on Hn through automorphisms via σ · (h1, . . . , hn) =(hσ−1(1), . . . , hσ−1(n)).

22. Verify that the specifications given in Examples 3.6.2 do in fact give automorphismsof the groups in question.

23. Show that the only automorphism of Z4 other than the identity is the one whichtakes each element to its inverse.

24. Let G be a finite abelian group and let f : G → G be given by f(g) = gn for allg ∈ G, where n is an integer that is relatively prime to |G|. Show that f is anautomorphism of G.

If G is cyclic of order 7 and n = 2, what is the order of f in Aut(G)?

25. Show that not every automorphism of D8 is an inner automorphism.

26. Show that every automorphism of D6 is an inner automorphism. (Hint : Thecomposite of two inner automorphisms is an inner automorphism.) Deduce thatAut(D6) ∼= D6.

27. Show that any two nontrivial homomorphisms from Z2 to D6 are conjugate.

28. Show that any two nontrivial homomorphisms from Z3 to D6 are conjugate.

29. What are the conjugacy classes of homomorphisms from Z4 to D8?

30. Let G be a group and let f : G→ Sn be a homomorphism. Write {1, . . . , n}f for theG-set obtained from the action on {1, . . . , n} induced by f under the correspondenceof Proposition 3.3.16: for g ∈ G and i ∈ {1, . . . , n}f , g · i = f(g)(i).

Let σ ∈ Sn and let f ′ = σ · f , the conjugate homomorphism obtained by conju-gating f by σ. Show that σ induces a G-isomorphism of G-sets, σ : {1, . . . , n}f →{1, . . . , n}f ′ .

31. Let G be a group. Show that there is a one-to-one correspondence between theconjugacy classes of homomorphisms from G to Sn and the G-isomorphism classesof the G-sets with n elements. Here, two G-sets are in the same G-isomorphismclass if and only if they are G-isomorphic.

32. Let X and Y be sets, with X finite. Show that if i : X → Y and j : X → Y are anytwo injective functions, then the induced homomorphisms i∗ : S(X) → S(Y ) andj∗ : S(X) → S(Y ) (as defined in Problem 15 of Exercises 3.1.16) are conjugate.(Hint : In the case that Y is infinite, set theory shows that Y may be put inone-to-one correspondence with the complement of any of its finite subsets.)

33. Let i and j be injective functions from {1, . . . , n} to Z+. Show that the inducedhomomorphisms i∗, j∗ : Sn → S∞ are conjugate.


3.7 Conjugating Subgroups: Normality

We haven’t yet made full use of the fact that conjugation is an action by automorphisms.The key point is that if f is an automorphism of G and if H ⊂ G is a subgroup, thenf(H) = {f(h) |h ∈ H} is also a subgroup. We get an induced action of Aut(G) on theset Sub(G) of all subgroups of G, via f ·H = f(H). The following lemma is immediate.

Lemma 3.7.1. If K acts on G via automorphisms, then there is an induced action ofK on Sub(G), the set of all subgroups of G.

Let’s consider the action of G on Sub(G) induced by conjugation. Here, for x ∈ G andH ∈ Sub(G), the subgroup obtained by letting x act on H is xHx−1 = {xhx−1 |h ∈ H}:Definition 3.7.2. Let H be a subgroup of G, and let x ∈ G. The conjugate of H by x isxHx−1 = {xhx−1 |h ∈ H}, the subgroup obtained by letting x act on H by conjugation.

Of course, the set of all conjugates of H is the orbit of H under the action of G onSub(G) by conjugation.

The isotropy subgroup of H ∈ Sub(G) under the conjugation action is called NG(H),the normalizer of H in G:

Definition 3.7.3. Let H be a subgroup of G. The normalizer in G of H, NG(H), isgiven as

NG(H) = {x ∈ G |xHx−1 = H}.

It is the isotropy subgroup of H under the action of G on Sub(G) by conjugation.

Our study of G-sets now allows us to calculate the size of the orbit of H underconjugation.

Corollary 3.7.4. The number of conjugates of H ⊂ G is [G : NG(H)], the index of thenormalizer of H in G.

Of course, if x ∈ H, then conjugation by x gives an automorphism of H, and hencexHx−1 = H.

Corollary 3.7.5. Let H be a subgroup of G. Then H ⊂ NG(H). In particular, thenumber of conjugates of H in G divides [G : H].

Proof We just saw that xHx−1 = H for x ∈ H, so H ⊂ NG(H), as claimed. Thelast statement follows from Problem 4 of Exercises 3.2.15: If [G : H] is finite, then[G : H] = [G : NG(H)] · [NG(H) : H]. (The reader should be able to verify this easily inthe case that G is finite, which is the most important case for our purposes.)

Thus, we have H ⊂ NG(H) ⊂ G. In practice, we can find examples where NG(H) =H, and others where NG(H) strictly contains H and is strictly contained in G. Finally,there is the case where NG(H) = G, which is important enough to merit a name.

Definitions 3.7.6. A subgroupH ⊂ G is called normal ifNG(H) = G (i.e., xHx−1 = Hfor all x ∈ G). We write H � G to indicate that H is a normal subgroup of G.

A subgroup H ⊂ G is called characteristic if f(H) = H for all automorphisms f ofG.


Clearly, a characteristic subgroup of G is normal in G. However, not all normalsubgroups are characteristic.

For instance, conjugation is trivial in an abelian group, and hence every subgroup ofan abelian group is normal. However, as we shall see in the exercises below, the onlycharacteristic subgroups of Z2 ×Z2 are e and the group itself. Thus, abelian groups canhave subgroups which are not characteristic.

We shall consider many more examples of both normal and characteristic subgroupsin the exercises.

If H � G, then we get an action of G on H through automorphisms:

Lemma 3.7.7. Let H be a normal subgroup of G. Then for x ∈ G, conjugation by xgives an automorphism, cx, of H: cx(h) = xhx−1 for all h ∈ H. Moreover, there is ahomomorphism Γ : G→ Aut(H) defined by Γ(x) = cx for all x ∈ G.

Proof Since cx gives an automorphism of G, it restricts to a group isomorphism fromH to cx(H) = xHx−1. Since xHx−1 = H, this restriction is precisely an automorphismof H. Now clearly, cx ◦ cy = cxy, and the result follows.

Essentially the same proof gives the following lemma.

Lemma 3.7.8. Let H be a characteristic subgroup of G. Then every automorphism f :G→ G restricts to an automorphism f : H → H. We obtain a restriction homomorphismρ : Aut(G) → Aut(H), by setting ρ(f) to be the restriction f : H → H for each f ∈Aut(G).

As we shall begin to see in the next chapter, the study of normal subgroups is oneof the most important subjects in group theory. It is also a very complicated subject.For instance, as we shall see in the exercises below, Gertrude Stein’s Theorem fails fornormal subgroups: we can find subgroups K ⊂ H ⊂ G such that K � H and H � G,but K is not normal in G. This failure of Gertrude Stein’s Theorem helps motivate thestudy of characteristic subgroups:

Corollary 3.7.9. Suppose given subgroups K ⊂ H ⊂ G such that H � G and K is acharacteristic subgroup of H. Then K � G.

Proof Let x ∈ G. Then conjugation by x restricts to an automorphism of H, andhence preserves K.

Recognition of normal subgroups is an important endeavor. Of course, if H is finite,then any injection from H to H is a bijection, so that xHx−1 ⊂ H if and only ifxHx−1 = H. However, there are examples of infinite groups H ⊂ G and x ∈ G wherexHx−1 is contained in but not equal to H. However, if xHx−1 ⊂ H for all x ∈ G, thingswork out nicely:

Proposition 3.7.10. Let H be a subgroup of G. Then H � G if and only if xHx−1 ⊂ Hfor all x ∈ G. Also, H is characteristic if and only if f(H) ⊂ H for all automorphismsf of G.

Proof Suppose that xHx−1 ⊂ H for each x ∈ G. To show that H � G, we must showthat H ⊂ xHx−1 for each x ∈ G. But x−1Hx ⊂ H, and hence H = x(x−1Hx)x−1 ⊂xHx−1, as desired.

The proof for the case of characteristic subgroups is nearly identical: if f(H) ⊂ Hfor all f ∈ Aut(G), then f−1(H) ⊂ H, so H = f(f−1(H)) ⊂ f(H).


Notice that the proof of Proposition 3.7.10 actually shows more than was stated. Forinstance, in the case of normality, it shows that if xHx−1 ⊂ H and x−1Hx ⊂ H, thenxHx−1 = H, and hence x ∈ NG(H). Thus, we have the following proposition.

Proposition 3.7.11. Let H and K be subgroups of G. Suppose that xHx−1 ⊂ H forall x ∈ K. Then K ⊂ NG(H).

One application of the concept of normality is the following variant of Proposi-tion 2.10.8.

Proposition 3.7.12. Let H and K be subgroups of the group G. Then G is the internaldirect product of H and K if and only if the following conditions hold.

1. H ∩K = e.

2. Both H and K are normal in G.

3. G is generated by the elements of H and K.

Proof Suppose that G is the internal direct product ofH andK. By Proposition 2.10.8,it suffices to show that H and K are normal in G. But by definition of the internal directproduct, this will follow if we know that the subgroups H × e and e×K are normal inH ×K. The verification of this is left to the reader.

Conversely, suppose the three conditions hold. By Proposition 2.10.8, it suffices toshow that for each h ∈ H and k ∈ K, the elements h and k commute.

By Lemma 3.6.10, it suffices to show that for h ∈ H and k ∈ K, hkh−1k−1 = e. Bycondition 1, this says that hkh−1k−1 ∈ H ∩K.

Now K � G, so hkh−1 ∈ K, and hence hkh−1k−1 ∈ K. Similarly, the normality ofH in G gives kh−1k−1 ∈ H, and hence hkh−1k−1 ∈ H, so the result follows.

Normality is intimately connected with the relationship between left and right cosets.The reader may verify the following lemma.

Lemma 3.7.13. Let H be a subgroup of G. Then the following conditions are equivalent.

1. H is normal in G.

2. For each x ∈ G, we have xH = Hx.

3. Every left coset of H in G is equal to a right coset of H in G; i.e., for each x ∈ Gthere is an element y ∈ G such that xH = Hy.

Exercises 3.7.14.1. Let G be a group and let H ⊂ G×G be given by

H = {(g, g)|g ∈ G}.

(H is called the diagonal subgroup of G ×G.) Show that H � G ×G if and onlyif G is abelian.

2. Let G be a group and let y ∈ G. Show that 〈y〉 � G if and only if xyx−1 ∈ 〈y〉 forall x ∈ G. Show that 〈y〉 is a characteristic subgroup of G if and only if f(y) ∈ 〈y〉for each f ∈ Aut(G).

3. Show that the subgroup 〈b〉 ⊂ D2n is normal.


‡ 4. Show that the subgroup 〈b〉 ⊂ D2n is characteristic for n ≥ 3.

5. Find a non-normal subgroup of D8.

6. Show that every subgroup of Q8 is normal.

7. Find the unique characteristic subgroup of Q8 which is equal to neither e nor Q8.

‡ 8. Show that if n ≥ 3, then 〈b〉 ⊂ Q4n is characteristic.

‡ 9. Let H be a subgroup of Zn and let f : Zn → Zn be any homomorphism. Showthat f(H) ⊂ H. Deduce that every subgroup of Zn is characteristic. (Hint : Showthat H = 〈d〉 = {dk | k ∈ Zn} for some d that divides n.)

† 10. Using the fact that An is the subgroup of Sn generated by the 3-cycles, show thatAn � Sn. Thus Lemma 3.7.7 provides a homomorphism Γ : Sn → Aut(An), whereΓ(σ) is induced by conjugation by σ for all σ ∈ Sn. Deduce from Problem 12 ofExercises 3.6.24 that Γ is an injection for n ≥ 4.

11. Show that the only characteristic subgroups of G = Z2 × Z2 are the identity andG itself.

† 12. Let H be an index 2 subgroup of the finite group G. Show that H � G.

13. Show that any index 2 subgroup of an infinite group is normal.

14. For x ∈ G, show that CG(x) ⊂ NG(〈x〉).15. If x ∈ G has order 2, show that CG(x) = NG(〈x〉). Deduce that 〈x〉 � G if and

only if x ∈ Z(G).

16. Give an example of a group G and an x ∈ G with CG(x) �= NG(〈x〉).17. List the distinct subgroups of D2n that are conjugate to 〈a〉.18. What is the normalizer of 〈a〉 ⊂ Q16?

19. List the distinct subgroups of Q16 that are conjugate to 〈a〉.20. What is the normalizer of 〈a〉 ⊂ Q4n? List the distinct subgroups of Q4n that are

conjugate to 〈a〉. (Hint : The answer depends in a crucial way on the value of nmod 2.)

21. Show that characteristic subgroups satisfy Gertrude Stein’s Theorem: If H is acharacteristic subgroup of G, and K is a characteristic subgroup of H, show thatK is a characteristic subgroup of G.

22. Show that S∞ � S(Z+).

23. Show that S({2, 3, . . . }) and S({3, 4, . . . }) are not conjugate as subgroups of S(Z+).

24. Let H be a subgroup of G. Show that H is a normal subgroup of NG(H). Showthat NG(H) is the largest subgroup of G that contains H as a normal subgroup;i.e., if H ⊂ K ⊂ G and H � K, then K ⊂ NG(H).

† 25. Let H and K be subgroups of G. Show that

NG(H) ∩NG(K) ⊂ NG(H ∩K).


26. Let H be a subgroup of G and let x ∈ G. Show that

NG(xHx−1) = xNG(H)x−1.

27. Show that Gertrude Stein’s Theorem fails for normal subgroups: i.e., find a groupG with a normal subgroup K � G and a subgroup H of K such that H � K, butH is not normal in G. (Hint : Find a group G with a subgroup H such that NG(H)is normal in G but not equal to G.)

28. Let H be a subgroup of the finite group G. Show that the number of elements in⋃x∈G xHx

−1 is less than or equal to [G : NG(H)] · (|H| − 1) + 1. Deduce that noproper subgroup of G contains an element in every conjugacy class of G.

29. Let X be a G-set and let x ∈ X. Show that the isotropy subgroup of gx is gGxg−1.

30. (See Problem 18 of Exercises 3.3.23.) What are the elements of the H-fixed-pointset (G/H)H? For which elements yH ∈ (G/H)H is the G-map from G/H to itselfwhich takes eH to yH a G-isomorphism (i.e., which elements of (G/H)H haveisotropy subgroup exactly equal to H)?

‡ 31. Let H be a subgroup of G and consider the standard action of G on G/H. LetΛ : G→ S(G/H) be the homomorphism induced by this action. Show that

ker Λ =⋂x∈G

xHx−1.

32. Let H be a subgroup of G and let X be the set of all subgroups conjugate to H:X = {xHx−1 |x ∈ G}. Thus, X is the orbit of H under the action of G on Sub(G)(the set of subgroups of G) by conjugation. Since X is a G-set, we get an inducedhomomorphism Λ : G→ S(X). Show that

ker Λ =⋂x∈G

xNG(H)x−1.

Chapter 4

Normality and Factor Groups

The closing section of Chapter 3 gives the definition of a normal subgroup:

A subgroup H ⊂ G is normal if xHx−1 = H for all x ∈ G. We write H � G to denotethat H is a normal subgroup of G.

The normal subgroups of a given group are absolutely vital in understanding itsstructure.

Some groups G have no normal subgroups other than the trivial subgroup e andthe group G itself. Such groups are called simple. We shall see, via the techniqueof composition series below, that all finite groups are built up out of the finite simplegroups.

The finite simple groups are all known. In one of the triumphs of mathematicalhistory, the finite simple groups have been classified, according to a plan of attack devel-oped by Daniel Gorenstein that played out over a sequence of papers by several authors.One of the final steps was the construction, by Robert Griess, of a group that was thenknown as the “monster group,” as its order is nearly 1054. In current parlance, themonster group is the sporadic simple group F1.1

We shall not need the classification by Gorenstein et. al., as we are concentratinghere on the groups of relatively low order. As will become apparent in the exercises fromthe next two chapters, the only simple groups whose order is ≤ 60 are the cyclic groupsZp, where p is a prime < 60, together with the alternating group A5.

In any case, the study of the ways in which finite groups can be put together out ofthe simple groups is the most important remaining problem for the classification of finitegroups. It is known as the Extension Problem.

Thus, the Extension Problem is, in the opinion of this author at least, one of theimportant unsolved problems in mathematics. The study of normal subgroups and factorgroups is at its core.

We do not yet have the language with which to give a precise statement of theExtension Problem, but we shall do so within this chapter. In fact, we shall developtechniques to study certain special cases of the Extension Problem. These special caseswill turn out to be enough, together with the Sylow theorems and the material on G-setsfrom Chapter 3, to classify the groups in a fair number of the lower orders.

1See Finite Simple Groups, An Introduction to Their Classification, by Daniel Gorenstein, PlenumPress, 1982.

82

CHAPTER 4. NORMALITY AND FACTOR GROUPS 83

The most important aspect of normal subgroups is that if H is normal in G, thenthere is a group structure on the set of left cosets G/H such that the canonical mapπ : G → G/H is a homomorphism. If G is finite and H is nontrivial, then G/H has alower order than G, and hence is, in general, easier to analyze. If G has enough normalsubgroups (e.g. if G is abelian or solvable), then induction arguments on the order ofG may be used to deduce facts about G from facts about the lower order factor groupsG/H.

In Section 4.1, we give the Noether Isomorphism Theorems, which study the proper-ties of the factor groups G/H when H is normal in G. These are truly fundamental inthe understanding of groups, and will open the door to some new classification results.As an application, we shall show that if G is finite and p is the smallest prime dividingthe order of G, then any subgroup of G of index p is normal in G.

In Section 4.2, we give the technique of composition series, which shows that finitegroups are built up out of simple groups. We also show that the alternating groups An

are simple for n ≥ 5. Then, in Section 4.3, we give the Jordan–Holder Theorem, whichshows that the simple groups out of which a given group is built are unique, though theway in which they are put together may not be.

Then, by an induction on order, using the fact that every subgroup of an abelian groupis normal, we give a complete classification of the finite abelian groups, the FundamentalTheorem of Finite Abelian Groups.

In Section 4.5, we calculate the automorphism group of a cyclic group. This cal-culation will be of essential value in constructing some new groups by the technique ofsemidirect products and in classifying the extensions of a given cyclic group.

Section 4.6 develops the technique of semidirect products, by which new groups maybe constructed from old. Unlike the direct product, many of the semidirect products oftwo abelian groups are nonabelian. We shall construct some interesting new groups inthis manner.

Section 4.7 defines and studies the extensions of one group by another. Extensionsdescribe some of the ways that groups are built up out of smaller groups. We shall statethe Extension Problem here and shall develop some classification techniques, particularlyfor groups that are semidirect products. We shall make some significant progress inobtaining actual classification results.

4.1 The Noether Isomorphism Theorems

Here, we study normal subgroups H � G and the group structure that’s induced on theset of left cosets G/H.

One convenient source of normal subgroups (which will also turn out to give all ofthem) is from the kernels of homomorphisms:

Lemma 4.1.1. Let f : G → G′ be a homomorphism. Then ker f is a normal subgroupof G.

Proof Let K = ker f . By Proposition 3.7.10, it suffices to show that xKx−1 ⊂ K foreach x ∈ G, i.e., that xkx−1 ∈ K for all x ∈ G and all k ∈ K. But if k ∈ K, thenf(xkx−1) = f(x) · f(k) · f(x)−1 = f(x) · e · f(x)−1 = e, and hence xkx−1 ∈ K.

We shall see that im f need not be normal in G′. We next explore the relationshipbetween normality of a subgroup H ⊂ G and group structures on the set of cosets G/H:


Theorem 4.1.2. Let H be a subgroup of G. Then H is normal in G if and only ifthere is a group structure on the set G/H of left cosets of H with the property that thecanonical map π : G→ G/H is a homomorphism.

If H � G, then the group structure on G/H which makes π a homomorphism isunique: we must have gH · g′H = gg′H for all g, g′ ∈ G. Moreover, the kernel ofπ : G→ G/H is H. Thus, a subgroup of G is normal if and only if it is the kernel of ahomomorphism out of G.

Proof Suppose that there is a group structure on G/H such that π is a homomorphism.Then the identity element of G/H is π(e) = eH. So g ∈ kerπ if and only if gH = eH.By Lemma 3.2.3, this is true if and only if g ∈ H. Thus, H is the kernel of π, and henceis normal by Lemma 4.1.1.

Clearly, there is at most one group structure onG/H which makes π a homomorphism,as, since π(g) = gH, we must have gH · g′H = gg′H for all g, g′ ∈ G.

Suppose then that H is normal in G. We wish to show that gH · g′H = gg′H definesa group structure on G/H. We first show that it gives a well defined binary operation.Thus, if gH = xH and g′H = x′H, we wish to show that gg′H = xx′H. But Lemma 3.2.3shows that x = gh and x′ = g′h′ for some h, h′ ∈ H. Thus, xx′H = ghg′h′H = ghg′H.By Lemma 3.2.3, this is equal to gg′H provided that (gg′)−1ghg′ ∈ H. But (gg′)−1ghg′ =(g′)−1g−1ghg′ = (g′)−1hg′. And (g′)−1hg′ ∈ H because H � G. Thus, the operation iswell-defined.

That this operation gives a group structure now follows immediately from the grouplaws in G: associativity in G gives associativity in G/H, and π(e) = eH is an identityelement for G/H. Finally, the inverse of gH is g−1H, so G/H is indeed a group.

Definition 4.1.3. When H � G, we shall simply write G/H for the group structure onG/H which makes π a homomorphism (i.e., with the multiplication gH · g′H = gg′H),and call it the factor group, or quotient group, of G modulo H.

Example 4.1.4. We’ve already seen examples of factor groups, as the groups Zn areprecisely the factor groups Z/〈n〉. The point is that for k ∈ Z, the congruence class ofk modulo n is precisely the coset k + 〈n〉. (To see this, note that l ≡ k mod n if andonly if l − k ∈ 〈n〉, i.e., l ∈ k + 〈n〉.) The canonical map π : Z → Zn that we’ve beenusing coincides with the canonical map above, as it takes each element in Z to its cosetmodulo 〈n〉. And, indeed, the group operation in Zn is defined precisely as in the moregeneral case above.

Another way of seeing that Z/〈n〉 ∼= Zn will be given using the Noether IsomorphismTheorems, below.

Since |G/H| = [G : H], Lagrange’s Theorem gives the following formula.

Lemma 4.1.5. Let H � G with G finite. Then

|G| = |H| · |G/H|.

Notation 4.1.6. Let H � G. For brevity, we shall generally write g in place of gH ∈G/H. Thus, every element of G/H has the form g = π(g) for some g ∈ G, and haveg · g′ = gg′. The identity element is e = e = h for any h ∈ H.


Note that |G/H| = [G : H] is less than |G| whenever G is finite and H �= e. Inthis case, the factor group G/H is less complicated than G, and we may obtain infor-mation about G from an understanding of the groups G/H as H ranges over the normalsubgroups of G.

We can characterize the homomorphisms out of G/H in terms of the homomorphismsout of G.

Proposition 4.1.7. Let H � G. Then for any group G′ and any homomorphism f :G→ G′ whose kernel contains H, there is a unique homomorphism f : G/H → G′ thatmakes the following diagram commute.

Gf ��

π ��

��

� G′

G/Hf

��

Conversely, if f : G → G′ factors by a commutative diagram as above, then H mustbe contained in ker f . Thus, the passage from a homomorphism f : G/H → G′ to thecomposite (f◦π) : G→ G′ gives a one-to-one correspondence between the homomorphismsfrom G/H to G′ and those homomorphisms from G to G′ whose kernels contain H.

Proof Suppose given f : G → G′. If a homomorphism f : G/H → G′ exists withf ◦π = f , then for each h ∈ H, f(h) = f(h) = f(e) = e, and hence H ⊂ ker f . Moreover,f(g) = f ◦π(g) = f(g), so f is unique if it exists. It suffices to show that such an f existswhenever H ⊂ ker f .

Thus, suppose H ⊂ ker f . As we’ve just seen, f , if it exists, must be given byf(g) = f(g). The key is to show this is a well-defined function from G/H to G′. Thus,suppose that g = g′ in G/H. By Lemma 3.2.3, this means g′ = gh for some h ∈ H.Thus, f(g′) = f(g)f(h) = f(g), since h ∈ ker f . So f is well-defined.

But f(g)f(g′) = f(g)f(g′) = f(gg′) = f(gg′) = f(gg′), and hence f is a homomor-phism.

We call f the homomorphism induced by f . In the counting argument below, weagain make use of the fact that |G/H| = [G : H].

Theorem 4.1.8. (First Noether Isomorphism Theorem) Let f : G → G′ be a homo-morphism. Then f induces an isomorphism from G/ ker f to im f . Thus, if G is finite,|im f | = [G : ker f ], so Lagrange’s Theorem gives

|G| = |ker f | · |im f |.

Proof Let f : G/ ker f → im f be the homomorphism induced by f . Since f(g) = f(g),f is onto. Thus, it suffices (Lemma 2.5.9) to show that ker f = e. Let g ∈ ker f . Thenf(g) = f(g) = e, so g ∈ ker f . But then g = e in G/ ker f .

Example 4.1.9. Recall the rotation matrices Rθ ∈ Gl2(R). We saw in Section 2.7that the collection of all rotation matrices, {Rθ | θ ∈ R}, forms a subgroup, SO(2), ofGl2(R). Moreover, as shown in Proposition 2.7.6, there is a surjective homomorphismexp : R → SO(2) given by exp(θ) = Rθ. The kernel of exp is shown there to be thecyclic subgroup 〈2π〉 ⊂ R. Thus, by the first Noether theorem, SO(2) is isomorphic toR/〈2π〉.


The Noether Theorem gives us another way to recognize Zn as Z/〈n〉.Corollary 4.1.10. Let n be a positive integer. Then Zn is isomorphic to the factorgroup Z/〈n〉.

Proof The canonical map π : Z→ Zn is a surjection, with kernel 〈n〉.

Recall that there is a homomorphism ε : Sn → {±1} defined by setting ε(σ) equal tothe sign of σ for σ ∈ Sn. Since ε is onto, its image has order 2. But An is the kernel ofε, so we may calculate its order:

Corollary 4.1.11. An has index 2 in Sn, and hence |An| = n!/2.

The next result foreshadows the techniques we shall use in Chapter 5. Recall fromProblem 12 of Exercises 3.7.14 that any index 2 subgroup of a finite group is normal.Using the First Noether Isomorphism Theorem, we shall give a generalization of thisresult. First, note from the example of 〈a〉 ⊂ D6 that a subgroup of prime index in afinite group need not be normal. Of course, 2 is the smallest prime, so the next resultgives the desired generalization.

Proposition 4.1.12. Let G be a finite group and let p be the smallest prime dividingthe order of G. Then any subgroup of G of index p is normal.

Proof Let H be a subgroup of index p in G, and consider the usual action of G on theset of left cosets G/H. As in Proposition 3.3.16, we get an induced homomorphism

Λ : G→ S(G/H) ∼= Sp

by setting Λ(g)(xH) = gxH for all g ∈ G and xH ∈ G/H.As shown in Proposition 3.3.16, the kernel of Λ is the intersection of the isotropy

subgroups of the elements of G/H. Since the isotropy subgroup of eH is H, we see thatker Λ ⊂ H.

Since |Sp| = p! and since p is the smallest prime dividing |G|, the greatest commondivisor of |G| and |Sp| is precisely p. The First Noether Isomorphism Theorem showsthat |im Λ| must divide |G|. Since |im Λ| also divides |Sp|, the order of im Λ must beeither 1 or p.

Since ker Λ ⊂ H, Λ cannot be the trivial homomorphism, so we must have |im Λ| = p.By the First Noether Isomorphism Theorem, we see that |ker Λ| = |G|/p = |H|. Sinceker Λ ⊂ H, this forces ker Λ = H. Since the kernel of a homomorphism is always normal,this gives H � G, as desired.

The remaining isomorphism theorems are concerned with a deeper analysis of thecanonical maps. The reader may easily verify the following lemma.

Lemma 4.1.13. Let f : G → G′ be a homomorphism and let H ⊂ G. Let f ′ : H → G′

be the restriction of f to H. Then ker f ′ = H ∩ ker f .

Recall from Problem 8 of Exercises 2.5.21 that if f : G → G′ is a surjective homo-morphism, then there is a one-to-one correspondence between the subgroups of G′ andthose subgroups of G that contain ker f . Under this correspondence, the subgroup K ′ ofG′ corresponds to f−1(K ′) ⊂ G, while a subgroup K ⊂ G containing ker f correspondsto f(K) ⊂ G′. Let us reconsider this correspondence in the case where f is the canonicalmap π : G→ G/H.


Lemma 4.1.14. Let H � G. Then the subgroups of G/H all have the form K/H, whereK is a subgroup of G that contains H.

Proof The one-to-one correspondence between the subgroups of G/H and those sub-groups of G that contain H is based on the following facts (which we assume were verifiedby the reader).

1. If K ′ is a subgroup of G/H, then π(π−1(K ′)) = K ′.

2. If K is a subgroup of G which contains H, then π−1(π(K)) = K.

Given K ′ ⊂ G/H, let K = π−1(K ′). Then K contains H, and K ′ = π(K). Now πrestricts to give a homomorphism, which we shall call π′, from K onto π(K).

Now, the kernel of π′ : K → π(K) is K ∩ kerπ = H (Lemma 4.1.13). The firstNoether theorem now allows us to identify π(K) with K/H.

Example 4.1.15. Let n and k be positive numbers such that k divides n. Then theunique subgroup of Zn of order n/k is 〈k〉.

Let π : Z → Zn be the canonical map. Clearly, 〈k〉 = π(〈k〉). But since k divides n,we have kerπ = 〈n〉 ⊂ 〈k〉. Thus, 〈k〉 = π−1〈k〉, and the above gives an isomorphismfrom 〈k〉/〈n〉 to 〈k〉 ∼= Zn/k.

We wish to generalize Lemma 4.1.14 to an analysis of the subgroups π(K) whereK ⊂ G does not contain H. First consider this:

Definition 4.1.16. Let H and K be subgroups of G. Then there is a subset KH of Gdefined by

KH = {kh | k ∈ K,h ∈ H}.

(Of course, if G is abelian, we write K +H = {k + h | k ∈ K,h ∈ H} for this subset.)

Examples 4.1.17. Since any element of D2n can be written as either bk or as abk foran appropriate value of k, we see that D2n = 〈a〉〈b〉.

Similarly, Q4n = 〈a〉〈b〉. The fact that some elements may be written in more thanone way as a product arbs with 0 ≤ r ≤ 3 and 0 ≤ s ≤ 2n doesn’t change this: the set{arbs | r, s ∈ Z} is 〈a〉〈b〉, and it includes every element in Q4n.

Lemma 4.1.18. Let H be a subgroup of G and let K be a subgroup of NG(H). ThenKH is a subgroup of G, and H ⊂ KH ⊂ NG(H). In addition, in this case, KH is equalto the subset HK of G, and is the smallest subgroup of G that contains both H and K.

Proof Since K ⊂ NG(H), for any k ∈ K and h ∈ H, we have k−1hk = h′ ∈ H, andhence hk = kh′. This is what we need to show that KH is closed under multiplication:for k1, k2 ∈ K and h1, h2 ∈ H, we have k1h1k2h2 = k1k2h

′1h2 for some h′1 ∈ H.

Similarly, (kh)−1 = h−1k−1 = k−1h′ for some h′ ∈ H, and hence KH a subgroup ofG.

Since h = eh and k = ke, both H and K are contained in KH. And any subgroupcontaining both H and K must contain each product kh, so KH is indeed the smallestsubgroup containing H and K. Since K ⊂ NG(H), this implies HK ⊂ NG(H).

Finally, if h ∈ H and k ∈ K, we have khk−1 = h′ ∈ H, so that kh = h′k ∈ HK, andhence KH ⊂ HK. Similarly, HK ⊂ KH.


Example 4.1.19. Let K be any subgroup of Sn that is not contained in An (e.g., K =〈(1 2)〉). Since An � Sn, KAn is a subgroup of Sn. Moreover, since K is not containedin An, [KAn : An] > 1. But [KAn : An] divides [Sn : An]. Since [Sn : An] = 2, we seethat KAn = Sn.

Lemma 4.1.18 sets up the second Noether Theorem.

Theorem 4.1.20. (Second Noether Isomorphism Theorem) Let H and K be subgroupsof G with H � G. Write π′ : K → G/H for the restriction to K of the canonical mapπ : G→ G/H. Then the image of π′ is KH/H, while the kernel of π′ is K ∩H. Thus,the First Noether Isomorphism Theorem provides an isomorphism

π′ : K/(K ∩H) KH/H.��∼=

Proof By Lemma 4.1.13, the kernel of π′ is K ∩ kerπ = K ∩ H. Thus, it suffices toshow that the image of π′ is KH/H.

Of course, the image of π′ is just π(K). Now for k ∈ K and h ∈ H, kh = k ∈ G/H,and hence π(kh) = π(k). But this says that π(KH) = π(K). Since H ⊂ KH, π(KH) =KH/H by Lemma 4.1.14, and the result follows.

Corollary 4.1.21. Let H and K be subgroups of G with H � G. Then [K : H ∩K]divides [G : H].

Proof [K : H ∩K] is the order of K/(H ∩K), which is thus equal to the order of thesubgroup KH/H of G/H. By Lagrange’s Theorem, this divides [G : H], the order ofG/H.

Here is a sample application.

Corollary 4.1.22. Let K be a subgroup of Sn and suppose that K is not contained inAn. Then [K : K ∩An] = 2.

Proof Corollary 4.1.21 shows that [K : K ∩An] divides [Sn : An], which is 2. But[K : K ∩An] �= 1, as K is not contained in An.

The Second Noether Theorem may be used for yet other calculations of indices.

Corollary 4.1.23. Let f : G→ G′ be a surjective homomorphism between finite groupsand let K be a subgroup of G. Then [G′ : f(K)] divides [G : K].

Proof Let H be the kernel of f . By the First Noether Isomorphism Theorem, thereis an isomorphism f : G/H → G′ such that f ◦ π = f , where π : G → G/H is thecanonical map. But then f induces an isomorphism from π(K) to f(K). Thus, we seethat [G′ : f(K)] = [G/H : π(K)].

By the Second Noether Isomorphism Theorem, π(K) ∼= K/(H ∩K). Thus, we obtain

[G′ : f(K)] =|G|/|H|

|K|/|H ∩K|=

|G|/|K||H|/|H ∩K|

=[G : K]

[H : H ∩K].

Since these indices are all integers, the result follows.


We now give a slight refinement of the Second Noether Theorem.

Corollary 4.1.24. Let H and K be subgroups of G, with K ⊂ NG(H). Then H∩K � Kand K/(H ∩K) ∼= KH/H.

Proof Just replace G by NG(H) and apply the second Noether Theorem.

Now, we give the final Noether Isomorphism Theorem.

Theorem 4.1.25. (Third Noether Isomorphism Theorem) Suppose given two subgroups,H and K, of G, which are both normal in G, with H ⊂ K. Then K/H � G/H, and thefactor group (G/H)/(K/H) is isomorphic to G/K.

Proof Write πH : G → G/H and πK : G → G/K for the two canonical maps. ThenH ⊂ K = kerπK . By Proposition 4.1.7, there is a homomorphism f : G/H → G/K withf ◦ πH = πK . Clearly, f is onto. It suffices to show that ker f = K/H.

Let g = gH ∈ G/H. Since f(g) = πK(g) ∈ G/K, we see that g ∈ ker f if and only ifg ∈ K = kerπK . Thus, ker f is precisely K/H.

Exercises 4.1.26.1. Give an example of a homomorphism f : G → G′ such that im f is not normal inG′.

† 2. Show that every factor group of a cyclic group is cyclic. (Suggestion: Give at leasttwo different proofs.)

3. Recall that 〈b〉 � D8. For each subgroup H ⊂ 〈b〉, list all the subgroups K ⊂ D8

with K ∩ 〈b〉 = H. (Hint : What is the order of K/H = K/(K ∩ 〈b〉)?)4. List all the normal subgroups of D8. What is the resulting factor group in each

case?

5. List all the groups (up to isomorphism) that appear as subgroups of D8.

6. Recall that 〈b〉 � D2n. Let H be a subgroup of 〈b〉. How many subgroups of D2n arethere whose intersection with 〈b〉 is H? (Hint : The answer depends on [〈b〉 : H].)What are these subgroups?

7. List all the normal subgroups of D2n. What is the resulting factor group in eachcase?

8. List all the groups (up to isomorphism) that appear as subgroups of D2n.

9. Recall that 〈b〉 � Q4n. Let H be a subgroup of 〈b〉. How many subgroups ofQ4n are there whose intersection with 〈b〉 is H? (Hint : The answer depends on[〈b〉 : H].) What are these subgroups?

10. List all the normal subgroups of Q4n. What is the resulting factor group in eachcase?

11. List all the groups (up to isomorphism) that appear as subgroups of Q4n.

† 12. Show that every element of Q/Z has finite order. Show also that Q/Z has elementsof order n for every n ≥ 1.


† 13. Show that there is an isomorphism between R/Z and R/〈x〉 for any 0 �= x ∈ R.Show also that the elements of finite order in R/Z are precisely those lying in thesubgroup Q/Z.

† 14. Let Hi � Gi for 1 ≤ i ≤ k. Show that H1 × · · · × Hk � G1 × · · · × Gk and that(G1 × · · · ×Gk)/(H1 × · · · ×Hk) ∼= (G1/H1)× · · · × (Gk/Hk).

15. What are all the subgroups of Z2×Z4? What is the resulting factor group in eachcase?

16. Suppose given an index three subgroup H ⊂ G such that H is not normal in G.Show that there is a surjective homomorphism from G to S3.

† 17. Let H and K be subgroups of G with K ⊂ NG(H). Show that NG(K)∩NG(H) ⊂NG(KH).

18. Generalize Corollary 4.1.23 as follows: Show that if f : G → G′ is a surjectivehomomorphism and if [G : K] is finite, then [G′ : f(K)] is finite and divides[G : K]. (Hint : Show that [G′ : f(K)] = [G : KH], where H = ker f .)

19. Let H be a subgroup of G. Show that the factor group NG(H)/H acts on the rightof the set of left cosets G/H via gH · x = gxH for all g ∈ G and x ∈ NG(H).Show also that the action of each x ∈ NG(H)/H gives a G-isomorphism from G/Hto itself. Finally, show that every G-isomorphism from G/H to itself arises inthis manner, and deduce that the group of G-isomorphisms from G/H to itself isisomorphic to the factor group NG(H)/H.

20. Let X be any G-set and let H be a subgroup of G. Show that there is an inducedaction of the factor group NG(H)/H on the H-fixed point set XH .

4.2 Simple Groups

The finite simple groups, which we define in this section, are the building blocks out ofwhich finite groups are made. We show this here via the technique of composition series.We then show that the alternating groups An are simple for n ≥ 5.

Definition 4.2.1. A group G is simple if it has no normal subgroups other than e andG.

We can characterize the abelian simple groups immediately. We shall give morecomplicated examples below.

Lemma 4.2.2. An abelian group is simple if and only if it is a cyclic group of primeorder.

Proof Since every subgroup of an abelian group is normal, we see that an abelian groupis simple if and only if it has no subgroups other than e and itself. Thus, an abeliangroup G is simple if and only if every non-identity element in it generates G.

In particular, G must be cyclic, and, since Z has nontrivial proper subgroups, itmust be finite. But our calculation of the orders of the elements in Zn (Problem 5 ofExercises 2.5.21) shows that Zn is simple if and only if n is prime.


The choice of the word “simple” for the simple groups is perhaps unfortunate, as thenonabelian simple groups can be much more complicated than, say, solvable ones (seebelow). Nevertheless, all finite groups are built up out of simple groups, hence the name.

Lemma 4.2.3. Let G �= e be a finite group. Then G has a factor group which is simpleand is not equal to the trivial group.

Proof By the first Noether theorem, the statement is equivalent to saying that eachgroup G �= e admits a surjective homomorphism onto a nontrivial simple group. Weargue by induction on |G|. If |G| = 2, then G itself is simple. Thus, suppose |G| > 2.If G is simple, we’re done, as G = G/e is a factor group of itself. Otherwise, G has anormal subgroup H, with H not equal to either e or G. But then 1 < |G/H| < |G|, so byinduction, there is a surjective homomorphism f : G/H → G′, where G′ is a nontrivialsimple group. But then the composite f ◦ π : G→ G′ gives the desired surjection.

The way in which finite groups are built up from simple groups is as follows.

Definition 4.2.4. Let G be a finite group. A composition series for G is a sequencee = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 � Hi, and Hi/Hi−1 is a nontrivial simplegroup for 1 ≤ i ≤ k.

Recall that this does not imply that the groups Hi are normal in G.

Proposition 4.2.5. Every nontrivial finite group has a composition series.

Proof Again, we argue by induction on the order of G. Once again, if G is simple, thenthere is nothing to prove. Otherwise, Lemma 4.2.3 provides a subgroup H � G such thatG/H is a nontrivial simple group. But then |H| < |G|, and hence H has a compositionseries e = H0 ⊂ · · · ⊂ Hk−1 = H. But then tacking on Hk = G gives a compositionseries for G.

The groups Hi/Hi−1 appearing in a composition series are subquotient groups of G:

Definition 4.2.6. A subquotient of a group G is a group of the form H/K, whereK � H ⊂ G.

Composition series are far from unique in many cases. We shall discuss their unique-ness properties in Section 4.3, and shall utilize them here simply to motivate the discus-sion of simple groups.

Our remaining goal for this section is to prove the following theorem, which producesour first examples of nonabelian simple groups.

Theorem 4.2.7. The alternating groups An are simple for all n ≥ 5.

In the process of proving this, we shall derive some useful information about sym-metric groups. We shall start by identifying the conjugacy classes in Sn. First considerthis:

Definition 4.2.8. We say that the permutations σ, τ ∈ Sn have the same cycle structureif when you write each of them as a product of disjoint cycles, both have the same numberof k-cycles for each k ≥ 2.

Recall that Lemma 3.6.19 computes the conjugates of a k-cycle.

Proposition 4.2.9. Two permutations in Sn are conjugate if and only if they have thesame cycle structure.


Proof Write σ ∈ Sn as a product of disjoint cycles: σ = σ1 . . . σk. Then for τ ∈ Sn,τστ−1 = τσ1τ

−1 . . . τσkτ−1. By Lemma 3.6.19, each τσiτ−1 is a cycle of the same length

as σi. Thus, σ and its conjugate τστ−1 have the same cycle structure, provided we showthat for two disjoint cycles σ1 and σ2, and any permutation τ , the cycles τσ1τ

−1 andτσ2τ

−1 are disjoint. But that follows from the explicit formula of Lemma 3.6.19.Thus, it suffices to show that any two permutations with the same cycle structure are

conjugate. Let σ and τ be permutations with the same cycle structure. Thus, we canwrite σ and τ as products of disjoint cycles, σ = σ1 . . . σk, and τ = τ1 . . . τk, in such away that σj and τj have the same length, length rj , for 1 ≤ j ≤ k. Say σj = (ij1 . . . ijrj

),and τj = (i′j1 . . . i

′jrj

), for 1 ≤ j ≤ k.Suppose we can find a permutation μ with μ(ijs) = i′js for 1 ≤ j ≤ k and 1 ≤ s ≤ rj .

Then μσjμ−1 = τj by Lemma 3.6.19 for all j. Thus, μσμ−1 = τ , so it will suffice to findsuch a μ.

Since the permutations σj are disjoint for 1 ≤ j ≤ k, an element of {1, . . . , n} mayoccur in at most one of the sets {iji, . . . , ijrj}. Thus, setting μ(ijs) = i′js for all of theindices js gives a well-defined function μ :

⋃1≤j≤k{ij1, . . . , ijrj

} → ⋃1≤j≤k{i′j1, . . . , i′jrj

}.Moreover, the disjointness of the τj shows that this function is bijective.

Now the complement of⋃

1≤j≤k{ij1, . . . , ijrj} in {1, . . . , n} may be placed in one-to-one correspondence with the complement of

⋃1≤j≤k{i′j1, . . . , i′jrj

} in {1, . . . , n}. But anysuch one-to-one correspondence extends μ to a permutation of {1, . . . , n}, which thengives the desired conjugation.

It would make our lives easier in the arguments below if we knew that any two elementsof An with the same cycle structure were conjugates in An (i.e., if the conjugatingpermutation μ could be chosen to lie in An). Unfortunately, this is not always true, sosome care will be required.

Lemma 4.2.10. Let σ, σ′ ∈ An be conjugate in Sn. Suppose that σ commutes with anelement of Sn which is not in An. Then σ and σ′ are conjugate in An.

Proof Since σ and σ′ are conjugate in Sn, there is a permutation τ with τστ−1 = σ′.If τ ∈ An, there’s nothing to prove. Otherwise, the hypothesis provides a τ ′ ∈ Sn, withτ ′ not in An (hence ε(τ ′) = −1), which commutes with σ. Now ε(ττ ′) = ε(τ)ε(τ ′) = +1,so ττ ′ ∈ An. And (ττ ′)σ(ττ ′)−1 = τ(τ ′στ ′−1)τ−1 = τστ−1 = σ′.

And here are some situations where the hypothesis is satisfied.

Lemma 4.2.11. Let σ ∈ An. Suppose that either of the following conditions hold.

1. There are at least two elements of {1, . . . , n} which are left fixed by σ.

2. At least one of the cycles in the cycle structure of σ has even length.

Then σ commutes with an element of Sn which is not in An.

Proof Suppose the first conditions holds, and, say, σ fixes i and j. Then σ is disjointfrom the transposition (i j), and hence commutes with it.

Suppose now that the second condition holds. Then σ commutes with the evenlength cycle in question. (Said cycle is disjoint from all of the other cycles in the cycledecomposition of σ, and hence commutes with them, but it also commutes with itself.)But even length cycles have sign −1.


Our strategy for proving Theorem 4.2.7 may now be laid out. Recall from Corol-lary 3.5.11 that An is generated by the 3-cycles. Since n ≥ 5, any 3-cycle commutes witha transposition in Sn. Thus, by Lemma 4.2.10, any two 3-cycles are conjugate in An. Inparticular, if H � An and if H contains a 3-cycle, it must contain all of its conjugatesin An, and hence must contain every 3-cycle. Since the 3-cycles generate An, we musthave H = An. Thus, we have verified the following lemma.

Lemma 4.2.12. Let H � An for n ≥ 5. If H contains a 3-cycle, then H = An.2

Thus, it suffices to show that if n ≥ 5, then any normal subgroup of An which is notthe identity must contain a 3-cycle. Given a subgroup e �= H � G, the only informationwe have for free is that there is a non-identity element in H. We shall consider variouspossibilities for the cycle structure of such an element, and verify, in the end, that thepresence of any non-identity element in H implies the existence of a 3-cycle in H.

We may now generalize Lemma 4.2.12.

Lemma 4.2.13. Let H � An for n ≥ 5. Suppose that H contains an element of theform στ such that

1. σ is a cycle of length ≥ 3.

2. σ and τ are disjoint. (Note that τ may be the identity here.)

3. στ commutes with an element of Sn which is not in An.

Then H = An.

Proof Let σ = (i1 . . . ik). Since the inverse of an r-cycle is an r-cycle, τ−1 has the samecycle structure as τ . Thus, στ is conjugate to σ′τ−1, where σ′ = (ik−1 ik ik−2 . . . i1). Butthen σ′τ−1στ is in H. Since σ and τ are disjoint, they commute, so this is just σ′σ, whichis easily seen to be equal to the 3-cycle (ik−2 ik ik−1). So H = An by Lemma 4.2.12.

We would like to be able to eliminate the condition in the preceding lemma whichrequires that στ commutes with an element of Sn which is not in An. We shall firstconsider another special case, which is actually already covered by Lemma 4.2.13 exceptin the cases of n = 5 or 6.

Lemma 4.2.14. Let H � An, and suppose that H contains a 5-cycle. Then H = An.

Proof Let σ = (a b c d e) be a 5-cycle in H. Then μ = (a b)(c d) is in An. Since H isnormal in An, μσμ−1 ∈ H, and hence σ · μσμ−1 is also in H. By Lemma 3.6.19,

σ · μσμ−1 = (a b c d e) · (b a d c e)= (a e c),

a 3-cycle. So H = An by Lemma 4.2.12.

We can now improve on Lemma 4.2.13, provided the cycle σ in the statement therehas length > 3.

Lemma 4.2.15. Let H � An. Suppose that H contains an element of the form στ ,where σ is a cycle of length > 3 and σ and τ are disjoint. Then H = An.

2This result is still true for n < 5, but with different argumentation. We shall treat these cases inthe exercises below.


Proof First note that the hypothesis forces n ≥ 5, as if σ has length 4, then τ cannotbe the identity element, as 4-cycles are not in An.

Note also that if σ is a 4-cycle, then στ commutes with an element (σ) which is inSn but not An, and hence the hypothesis of Lemma 4.2.13 is satisfied.

Thus, we may assume that σ has length ≥ 5. Write σ = (i1 . . . ik), and let μ =(i1 i2)(i3 i4). Then μ ∈ An. Moreover, since the support of μ is contained in the supportof σ and since σ and τ are disjoint, μ commutes with τ .

Since μ ∈ An and H � An, στ · μ(στ)−1μ−1 is in H. We shall show that στ ·μ(στ)−1μ−1 is a 5-cycle. The result will then follow from Lemma 4.2.14.

Expanding (στ)−1 and associating, we get

στ · μ(στ)−1μ−1 = σ(τμτ−1)σ−1μ−1.

But τ commutes with μ, so that τμτ−1 = μ. Simplifying, we see that

στ · μ(στ)−1μ−1 = σ · μσ−1μ−1

= (i1 . . . ik)(ik . . . i5 i3 i4 i1 i2)= (i5 i4 i2 i1 i3).

Here, the second equality follows from Lemma 3.6.19, and the last by direct calculation.Thus, there is a 5-cycle in H as claimed, and the result follows.

We can now go one step farther and give a version of Lemma 4.2.13 that eliminatesthe third condition.

Proposition 4.2.16. Let H � An for n ≥ 5. Suppose there is an element σ ∈ H suchthat the decomposition of σ as a product of disjoint cycles contains a cycle of length ≥ 3.Then H = An.

Proof Suppose that the decomposition of σ as a product of disjoint cycles contains acycle of length > 3. Then the hypothesis of Lemma 4.2.15 is satisfied, and H = An asdesired. Thus, we may assume that σ is the product of disjoint cycles whose lengths areall ≤ 3. But if any of them is a 2-cycle, then σ commutes with an element of Sn whichis not in An, and hence the hypothesis of Lemma 4.2.13 is satisfied.

Thus, we may assume that σ is a product of disjoint 3-cycles. If σ consists of a single3-cycle, then H = An by Lemma 4.2.12. Thus, we may assume there are at least two3-cycles in the decomposition of σ as a product of disjoint cycles.

Write σ = (a b c)(d e f)τ , where τ is disjoint from the stated 3-cycles. But an im-mediate application of Lemma 3.6.19 shows that σ commutes with the permutation(a d)(b e)(c f), which lies in Sn but not in An. Thus, σ satisfies the hypothesis ofLemma 4.2.13, and hence H = An.

Thus, the only remaining possibility for an H � An with H �= An is for H to consistonly of products of disjoint transpositions (2-cycles). Indeed, A4 has such a normalsubgroup, as we shall see in the exercises below. So n ≥ 5 will be essential in what is tofollow. We begin with an important special case.

Lemma 4.2.17. Let H � An for n ≥ 5. Suppose that H contains a product of twodisjoint transpositions. Then H = An.


Proof Let σ = (a b)(c d) ∈ H. Since n ≥ 5, there is a fifth letter, e, to play with.Since σ commutes with a 2-cycle, it is conjugate in An to any element of the same cyclestructure, by Lemma 4.2.10. In particular, it is conjugate to (b c)(d e).

Thus,

(a b)(c d)(b c)(d e) = (a b d e c)

is in H. The result now follows from Lemma 4.2.14.

We may now complete the proof of the theorem.

Proof of Theorem 4.2.7 Let e �= H � An for n ≥ 5. Let σ be any non-identityelement of H. If the decomposition of σ as a product of disjoint cycles contains a cycleof length ≥ 3, then H = An by Proposition 4.2.16. Thus, we may assume that σ is aproduct of disjoint transpositions.

Since transpositions have sign −1, there must be an even number of disjoint transpo-sitions in the decomposition of σ. But if σ is the product of two disjoint transpositions,then H = An by Lemma 4.2.17. Thus, we may assume that σ is the product of four ormore disjoint transpositions.

Thus, we can write σ = (a b)(c d)(e f)τ , where τ is disjoint from the displayed trans-positions. Once again, since the decomposition of σ contains a cycle of even length, it isconjugate in An to any permutation with the same cycle structure. In particular, σ isconjugate to (b c)(d e)(f a)τ−1. Thus,

(a b)(c d)(e f)(b c)(d e)(f a)ττ−1 = (a e c)(b d f)

lies in H. The result now follows from Proposition 4.2.16.

Exercises 4.2.18.† 1. Show that A4 has exactly four elements whose order is a power of 2. Show that

these elements form a normal subgroup of A4.

† 2. Show that A4 has no subgroup of order 6. Deduce that A4 is the only subgroupof S4 which has order 12. (Hint : One way to proceed is as follows. Use theNoether Theorems to show that if H ⊂ A4 has order 6, then H must have anormal subgroup, K � H of order 2. Show that every element of H outside of Kmust have order 3 or 6, and that K ⊂ Z(H). Deduce that H must have an elementof order 6 and derive a contradiction.)

‡ 3. Show that An is a characteristic subgroup of Sn for n ≥ 3. (Hint : Use the simplicityof An for n ≥ 5. For n = 3 or 4, give a separate argument.)

4. Show that any normal subgroup of A4 containing a 3-cycle must be all of A4.

5. Show that any normal subgroup of A3 containing a 3-cycle must be all of A3.

6. Let σ ∈ An. Show that the conjugates of σ in Sn are all conjugate to σ in An ifand only if the centralizer of σ in Sn is not contained in An. (Hint : Recall thatthe centralizer, CAn(σ), of σ in An satisfies

CAn(σ) = An ∩ CSn(σ).

What are the possible values for [CSn(σ) : CAn(σ)]? What does this say about[Sn : CSn(σ)]/[An : CAn(σ)]?)


4.3 The Jordan–Holder Theorem

Recall that a composition series for a group G is a sequence

e = H0 ⊂ · · · ⊂ Hk = G

such that Hi−1 � Hi for i = 1, . . . , k and the factor groups Hi/Hi−1 are all nontrivialsimple groups.

Proposition 4.2.5 shows that every nontrivial finite group has a composition series.But composition series are not unique. However, the Jordan–Holder Theorem shows thatthe simple subquotient groups Hi/Hi−1 that appear in a composition series for G areindependent of the choice of composition series, though they may occur in a differentorder in different composition series:

Theorem 4.3.1. (Jordan–Holder Theorem) Suppose given two different composition se-ries,

e = H0 ⊂ · · · ⊂ Hk = G and

e = G0 ⊂ · · · ⊂ Gl = G

of the finite group G. Then k = l and there is a permutation σ ∈ Sk such that Hi/Hi−1

is isomorphic to Gσ(i)/Gσ(i)−1 for 1 ≤ i ≤ k.Example 4.3.2. We have two different composition series for Z6:

e ⊂ 〈2〉 ⊂ Z6 and

e ⊂ 〈3〉 ⊂ Z6.

The successive subquotient groups are Z3 followed by Z2 in the first case, and Z2 followedby Z3 in the second.

Rather than present the proof from the bottom up, which would lose the intuitionin this case, we shall argue conceptually, leaving the proof (and statement) of the keylemma for the end.

We first generalize the notion of composition series:

Definition 4.3.3. A subnormal series for a group G is a sequence e = H0 ⊂ · · · ⊂ Hk =G such that Hi−1 � Hi for 1 ≤ i ≤ k.

Some treatments would call this a normal series, but we prefer subnormal since it isnot required that Hi � G.

Note that a composition series is a subnormal series for which each of the successivefactor groups Hi/Hi−1 is a nontrivial simple group. Note also that it is possible in asubnormal series (but not a composition series) to have Hi−1 = Hi.

Definition 4.3.4. A refinement of a subnormal series is a subnormal series obtained byinserting more subgroups into the sequence.

Thus, every subnormal series is a refinement of the trivial series e ⊂ G.


Definition 4.3.5. Two subnormal series

e = H0 ⊂ · · · ⊂ Hk = G and

e = G0 ⊂ · · · ⊂ Gl = G

are equivalent if k = l and there is a permutation σ ∈ Sk such that Hi/Hi−1 is isomorphicto Gσ(i)/Gσ(i)−1 for 1 ≤ i ≤ k.

Thus, the Jordan–Holder Theorem simply states that any two composition series areequivalent. It will follow rather quickly from the next result.

Theorem 4.3.6. (Schreier’s Theorem) Suppose given two subnormal series for the groupG. Then we may refine the two series in such a way that the refined series are equivalent.

This immediately implies the Jordan–Holder Theorem because refining a compositionseries does nothing other than adding trivial groups to the list of factor groups. So twocomposition series equivalent after refinement must have been equivalent in the firstplace.

Let us now consider the proof of Schreier’s Theorem. We are given subnormal series

e = H0 ⊂ · · · ⊂ Hk = G and

e = G0 ⊂ · · · ⊂ Gl = G

of G. We shall insert between each pair Hi−1 ⊂ Hi a collection of inclusions derivedfrom the G0 ⊂ · · · ⊂ Gl in such a way that the resulting sequence of k · l inclusions is asubnormal series. Similarly, we shall insert groups between Gi−1 and Gi.

Since Hi−1 � Hi, the group Hi ∩Gj normalizes Hi−1 for each j. Thus, the productHi−1(Hi ∩Gj) is a subgroup of G, and we have inclusions

Hi−1 = Hi−1(Hi ∩G0) ⊂ · · · ⊂ Hi−1(Hi ∩Gl) = Hi.

Similarly, we have inclusions of subgroups

Gj−1 = Gj−1(H0 ∩Gj) ⊂ · · · ⊂ Gj−1(Hk ∩Gj) = Gj .

Schreier’s Theorem now follows by setting Gj−1 = B, Hi−1 = A, Gj = K, andHi = H in the next lemma.

Lemma 4.3.7. (Zassenhaus’ Butterfly Lemma) Let H and K be subgroups of G, andsuppose given normal subgroups A � H and B � K of H and K. Then A(H ∩ B) �A(H ∩K), B(A ∩K) � B(H ∩K) and there is an isomorphism

A(H ∩K)/A(H ∩B) ∼= B(H ∩K)/B(A ∩K).

Proof We shall show that both of the quotient groups are isomorphic to a third quotient:H∩K/(A∩K)(H∩B). (We’ll justify presently that this is a group.) The diagram below,


which gives this result its name, illustrates the relations between the various subgroupswe shall consider. All arrows are the natural inclusions.

A(H ∩K) B(H ∩K)

H ∩K

A(H ∩B) B(A ∩K)

A (A ∩K)(H ∩B) B

A ∩K H ∩B

��

��

��

��

��

��

��

��

��

��

We shall give the proof that H ∩K/(A ∩K)(H ∩ B) ∼= A(H ∩K)/A(H ∩ B). Theargument that H ∩K/(A ∩K)(H ∩B) ∼= B(H ∩K)/B(A ∩K) is analogous.

First note that since H ∩ B ⊂ H ∩ K ⊂ H ⊂ NG(A), the subsets A(H ∩ K) andA(H ∩B) are subgroups of G. We argue by a straightforward application of the SecondNoether Isomorphism Theorem, which we restate as follows (using the form given inCorollary 4.1.24) to avoid conflicting notation:

If X and Y are subgroups of G with X ⊂ NG(Y ), then X∩Y � X, and X/(X∩Y ) ∼=XY/Y .

We apply this with X = H ∩K and Y = A(H ∩ B). We must first verify that forthis choice of X and Y that X ⊂ NG(Y ), and then must show that the quotients statedin the Noether theorem are the ones that concern us.

First recall from Problem 25 of Exercises 3.7.14 that for subgroups H1, H2 ⊂ G, wehave NG(H1) ∩ NG(H2) ⊂ NG(H1 ∩H2). Also, Problem 17 of Exercises 4.1.26 showsthat if H1 ⊂ NG(H2), then NG(H1) ∩NG(H2) ⊂ NG(H1H2).

Thus, since Y = A(H ∩ B), the inclusion of X in NG(Y ) will follow if we show thatX ⊂ NG(A) ∩ NG(H) ∩ NG(B). But X ⊂ H, which is contained in both NG(A) andNG(H), and X ⊂ K, which is contained in NG(B), so X normalizes Y as desired.

It suffices to show that X ∩ Y = (A ∩K)(H ∩B) and that XY = A(H ∩K).For the former equality, we have X ∩Y = [H ∩K]∩ [A(H ∩B)]. Clearly, this contains

(A∩K)(H ∩B), so it suffices to show the opposite inclusion, i.e., that [H ∩K]∩ [A(H ∩B)] ⊂ (A ∩K)(H ∩ B). So let a ∈ A and let b ∈ H ∩ B, and suppose that ab ∈ H ∩K.It suffices to show that a ∈ K. But since ab ∈ H ∩K ⊂ K, and since b ∈ B ⊂ K, wemust have a ∈ K.

Finally, we have XY = [H ∩K][A(H ∩ B)]. This group clearly contains A(H ∩K),so it suffices to show the opposite inclusion. Let a ∈ A, b ∈ H ∩B, and c ∈ H ∩K. SinceH ∩K normalizes A, cab = a′cb with a′ ∈ A. But cb ∈ H ∩K and the result follows.

4.4 Abelian Groups: the Fundamental Theorem

Here, we classify all finite abelian groups.

Recall that an element x of a group G has exponent n if xn = e. Recall also that ifx has exponent n, then the order of x divides n.


Definitions 4.4.1. We say that a group G has exponent n if every element of G hasexponent n.

Given an arbitrary element x ∈ G, we say that x is a torsion element if x has finiteorder. If every element of G has finite order, we call G a torsion group. At the otherextreme, if no element of G other than the identity has finite order, we call G torsion-free.

Finally, let p be a prime number. We say that x ∈ G is a p-torsion element if theorder of x is a power of p. If every element of G is a p-torsion element, we call G ap-group.

Recall that an element has finite order if and only if it has exponent n for some n > 0.Since the order then divides all exponents for x, we see that x is a p-torsion element ifand only if it has exponent pr for some r ≥ 0.

Examples 4.4.2.1. Let G be a finite group. By the corollary to Lagrange’s Theorem (Corollary 3.2.10),

the order of any element of G divides |G|. Thus, the group G has exponent |G|.2. S∞ (the ascending union of the symmetric groups Sn for all n ≥ 1) has been shown

to be an infinite torsion group. But it does not have an exponent, as it containselements of every finite order.

3. As shown in Problem 12 of Exercises 4.1.26, Q/Z is an example of an infiniteabelian torsion group which does not have an exponent.

4. The reader acquainted with infinite products should have no trouble verifying that∏∞i=1 Z2, the product of infinitely many copies of Z2, has exponent 2. Thus, in

addition to having an exponent, it is an infinite abelian 2-group.

Let us now specialize to the study of abelian groups. To emphasize this specialization,we shall use additive notation for the group operation in the remainder of this section.

Lemma 4.4.3. Let G be abelian and let x, y ∈ G be torsion elements. Then the leastcommon multiple of the orders of x and y is an exponent for x+ y.

Proof Let n be the least common multiple of the orders of x and y. Then n is anexponent for both x and y: nx = ny = 0. But since G is abelian, n(x+y) = nx+ny = 0.

It is immediate from Lemma 4.4.3 and the fact that the inverse of an element of finiteorder also has finite order (Problem 6 of Exercises 2.5.21) that the torsion elements ofan abelian group form a subgroup of it:

Definition 4.4.4. Let G be an abelian group. Write Tors(G) ⊂ G for the set of alltorsion elements of G. We call it the torsion subgroup of G.

Example 4.4.5. As shown in Problem 13 of Exercises 4.1.26, the torsion subgroup ofR/Z is Q/Z.

Surprisingly, the set of torsion elements in a nonabelian group is not necessarily asubgroup. For instance, Problem 2 of Exercises 10.8.13 shows that Sl2(Z), the group of2× 2 matrices with integer coefficients that have determinant 1, is generated by torsionelements, but the matrix (

1 10 1

)is easily seen to have infinite order.


Definition 4.4.6. Let G be an abelian group and let p be prime. The p-torsion sub-group, Gp, of G is the set of p-torsion elements of G.

The justification is, in this case, immediate from Lemma 4.4.3:

Corollary 4.4.7. Let G be an abelian group and let p be prime. Then the p-torsionsubgroup Gp is, in fact, a subgroup of G.

The analogue for nonabelian groups of the preceding corollary is false even for finitenonabelian groups:

Example 4.4.8. Recall from Corollary 3.5.11 that for n ≥ 3, An is generated by 3-cycles. But while 3-cycles are 3-torsion elements, if n ≥ 4 there are elements of An (e.g.,(1 2)(3 4)) that are not 3-torsion elements. Thus, the 3-torsion elements of An do notform a subgroup if n ≥ 4.

Let us now specialize to the study of finite abelian groups.

Proposition 4.4.9. Let G be a finite abelian group. Let p1, . . . , pk be the prime numbersthat divide |G|. Then G is the internal direct product of Gp1 , . . . , Gpk

, i.e., the function

μ : Gp1 × · · · ×Gpk→ G

defined by μ(g1, . . . , gk) = g1 + · · ·+ gk is an isomorphism of groups.

Proof Of course, μ restricts on each Gpi to its inclusion into G. Since G is abelian, theimages of these inclusions commute with each other, so that μ is a homomorphism byProposition 2.10.10. Suppose that g = (g1, . . . , gk) lies in the kernel of μ. If each gi = 0,then g is the identity element of Gp1 × · · · ×Gpk

. Otherwise, let i be the smallest indexwith gi �= 0. Then gi + (gi+1 + · · · + gk) = 0, and we must have i < k, as otherwisegi = μ(g) = 0. By inductive application of Lemma 4.4.3, the element gi+1 + · · · + gkhas an exponent which is a product of powers of pi+1, . . . , pk. But gi+1 + · · ·+ gk is theinverse of gi, and hence has order equal to a power of pi. This forces the order of −gi tobe 1 = p0

i , and hence gi = 0. So g had to be 0 in the first place, and hence μ is injective.It suffices to show μ is surjective. Let g ∈ G, and let n = o(g). We argue by induction

on the number of primes that divide n. If there is only one prime dividing n, then n isa power of pi for some i, and hence g ∈ Gpi ⊂ imμ. Otherwise, n = pri

i m for some i,where ri ≥ 1, and (pi,m) = 1. Thus, we can find integers s and t with spri

i + tm = 1.But then

g = (sprii + tm)g = s(pri

i g) + t(mg).

By Problem 5 of Exercises 2.5.21, mg has order prii , and hence lies in Gpi ⊂ imμ, and

prii g has order m. Since m has one less prime divisor than n, pri

i g ∈ imμ by induction.Thus, g is the sum of two elements of imμ, and hence lies in imμ as claimed.

Note that if p doesn’t divide the order of G, then Gp = 0 by Lagrange’s Theorem.Since the cartesian product of any group H with the trivial group is canonically iso-morphic to H, we obtain the following immediate extension of Proposition 4.4.9, whichis useful when considering homomorphisms between abelian groups whose orders havediffering prime divisors.


Corollary 4.4.10. Let G be a finite abelian group. Let p1, . . . , pk be a collection ofprimes including all the prime divisors of |G|. Let

μG : Gp1 × · · · ×Gpk→ G

be defined by μG(g1, . . . , gk) = g1 + · · ·+ gk. Then μG is an isomorphism.

The proof of Proposition 4.4.9 is an elaboration of the proof of the next result, whichcould have been used inductively to obtain it. We include both proofs as it may beinstructive to the beginner.

Proposition 4.4.11. Let G be a finite group of order n. Suppose that n = mk, wherem and k are relatively prime. Let H ⊂ G be the set of all elements of exponent m andlet K ⊂ G be the set of all elements of exponent k. Then H and K are subgroups of G,and G is the internal direct product of H and K.

Proof That H and K are subgroups is a consequence of Lemma 4.4.3. For instance,if x, y ∈ H, then the prime divisors of o(x+ y) must all divide m. Since o(x+ y) mustdivide |G|, this forces o(x+ y) to divide m.

By Proposition 2.10.8, since G is abelian, it suffices to show that H ∩ K = 0, andthat H and K generate G.

Since m and k are relatively prime, any element which has both m and k as anexponent must have order 1 (Corollary 2.5.13), so that H ∩ K is indeed the trivialsubgroup.

But if g ∈ G with o(g) = n′, then we may write n′ = m′k′, where m′ divides m andk′ divides k. But then m′ and k′ are relatively prime, so that sm′ + tk′ = 1 for someintegers s and t. But now g = s(m′g)+ t(k′g). Since m′g and k′g have orders k′ and m′,respectively, g is in H +K as desired.

We shall now give a special case of Cauchy’s Theorem, showing that the groups Gp arenontrivial for all primes p dividing the order of G. Cauchy’s Theorem has already beengiven as Problem 24 of Exercises 3.3.23, and will be given again by a different argument,using the case we’re about to give, as Theorem 5.1.1. Both proofs are instructive.

Proposition 4.4.12. (Cauchy’s Theorem for Abelian Groups) Let G be a finite abeliangroup and let p be a prime number dividing the order of G. Then G has an element oforder p.

Proof We argue by induction on |G|, with the case |G| = p a triviality. Let 0 �= g ∈ G.If o(g) is divisible by p, say o(g) = pq, then qg has order p. But if o(g) is not divisibleby p, then |G/〈g〉| is divisible by p. Since |G/〈g〉| < |G|, the induction hypothesis givesan element x ∈ G/〈g〉 of order p. But since x is the image of x ∈ G under the canonicalhomomorphism, Corollary 2.5.19 shows that p must divide the order of x.

This now serves to clarify the situation of Proposition 4.4.11:

Corollary 4.4.13. Let G be a finite abelian group of order n = mk, with (m, k) = 1. LetH ⊂ G be the set of elements in G of exponent m and let K ⊂ G be the set of elementsin G of exponent k. Then H has order m and K has order k.

Proof By Proposition 4.4.12, the order of H cannot be divisible by any prime thatdoesn’t divide m, as then H would have an element whose order does not divide m. Butthen |H| must divide m by Lagrange’s Theorem. Similarly, |K| divides k.

But since G ∼= H ×K, |G| = |H| · |K|. Since n = mk, the result must follow.


As a special case, Corollary 4.4.13 gives a calculation of the order of the p-torsionsubgroup of a finite abelian group for any prime p dividing the order of the group.

Corollary 4.4.14. Let G be a finite abelian group of order n = pr11 . . . prk

k , where p1, . . . , pkare distinct primes. Then the pi-torsion subgroup Gpi

has order prii for 1 ≤ i ≤ k.

The analogue of the preceding statement for nonabelian groups is the second SylowTheorem. Its proof is a little deeper than that of the abelian case. We now give anotheruseful consequence of Corollary 4.4.13.

Corollary 4.4.15. Let G be a finite abelian group and let H be a subgroup. Suppose thatthe orders of H and G/H are relatively prime. Then H is the subgroup of G consistingof all elements of exponent |H|, and G is isomorphic to H ×G/H.

Proof Write n, m and k for the orders of G, H, and G/H, respectively. Then n = mk.Let H ′ ⊂ G be elements of exponent m and let K ⊂ G be the elements of exponent k.We shall show that H = H ′ and that K is isomorphic to G/H.

To see the former, let i : H ⊂ G be the inclusion. Since the elements of H have expo-nent m, we must have i(H) ⊂ H ′. But then i : H → H ′ is an injective homomorphism.Since H and H ′ both have order m, it must be an isomorphism.

Let π : G → G/H be the canonical map. Then kerπ = H. Since |K| and |H| arerelatively prime, we have H ∩ K = 0, and hence π restricts to an injection from K toG/H. But since K and G/H have the same order, π must induce an isomorphism fromK to G/H.

Next, we show that for any prime p that any finite abelian p-group is a direct productof cyclic groups. Together with Proposition 4.4.9, this will give the existence part of theFundamental Theorem:

Theorem 4.4.16. (Fundamental Theorem of Finite Abelian Groups) Every finite abeliangroup is a direct product of cyclic groups of prime power order.

We shall restate it with a uniqueness statement below.

Lemma 4.4.17. Let G be an abelian group of exponent m. Suppose that g ∈ G hasorder m. Let x ∈ G/〈g〉. Then there exists y ∈ G with y = x, such that the order of y inG is equal to the order of x in G/〈g〉.

Proof Let k = o(x) and let n = o(x). Because 〈x〉 is the image of 〈x〉 under thecanonical map π : G→ G/〈g〉 k divides n (Corollary 2.5.19). So n = kl for l ∈ Z.

Note then that kx has order l. But since x has order k, kx = kx = 0, and hencekx ∈ kerπ = 〈g〉. Say kx = sg.

Since g has orderm, the order of sg ism/(m, s). Since kx = sg, this says l = m/(m, s),and hence l · (m, s) = m. Since n = kl is the order of x, it must divide the exponent ofG, which is m. Thus, k divides (m, s). But then s = kt, and hence y = x− tg has orderk.

For the next result, we only use the special case of Lemma 4.4.17 in which the exponentm is a prime power. The reader is invited to re-examine the proof in this simpler case.

Proposition 4.4.18. Let p be a prime. Then any finite abelian p-group is a directproduct of cyclic groups.


Proof By Proposition 4.4.12, |G| is a power of p. We argue by induction on |G|. Theinduction begins with |G| = p, in which case G itself is cyclic.

By Lagrange’s Theorem, |G| is an exponent for G. Let pj be the smallest power ofp which is an exponent for G. Then there must be an element g ∈ G of order pj . LetH = G/〈g〉, with π : G → H the canonical map. Then |H| < |G|, so our inductionhypothesis gives an isomorphism ν : Zpr1 × · · · × Zprk → H.

Let xi = ν ◦ ιi(1), where ιi : Zpri → Zpr1 × · · · × Zprk is the inclusion of the i-thfactor of Zpr1 × · · · ×Zprk . Then xi has order pri . By Lemma 4.4.17, there are elementsyi ∈ G with π(yi) = xi, such that yi has order pri as well.

Define ν : Zpr1 × · · · × Zprk × Zpj → G by ν(m1, . . . ,mk+1) = m1y1 + · · ·+mkyk +mk+1g. Since the orders of the yi and of g are as stated, ν ◦ ιi is a homomorphism for1 ≤ i ≤ k+1 by Proposition 2.5.17. Thus, ν is a homomorphism by Proposition 2.10.10.We claim that ν is an isomorphism.

Suppose that α = (m1, . . . ,mk+1) is in the kernel of ν. Then

0 = π ◦ ν(m1, . . . ,mk+1)= ν(m1, . . . ,mk).

Since ν is an isomorphism, m1 = · · · = mk = 0. But then 0 = ν(α) = mk+1g. Sinceo(g) = pj , mk+1 = 0, and hence α = 0. Thus, ν is injective.

Let x ∈ G, and suppose that ν−1 ◦ π(x) = (m1, . . . ,mk). Then π(x− (m1y1 + · · ·+mkyk)) = 0. Since kerπ = 〈g〉, x− (m1y1 + · · ·+mkyk) = mk+1g for some integer mk+1.But then x = ν(m1, . . . ,mk+1), and hence ν is onto.

This completes the proof of the existence part of the Fundamental Theorem. We shallnow discuss uniqueness. First consider this.

Lemma 4.4.19. Let f : G → H be a homomorphism of abelian groups. Then for eachprime p, f restricts to a homomorphism fp : Gp → Hp.

Proof The order of f(g) must divide the order of g. If g ∈ Gp, then the order of g is apower of p, so the same must be true of f(g). Thus, f(g) ∈ Hp.

For those who understand the language of category theory, this implies that passageto the p-torsion subgroup provides a functor from abelian groups to abelian p-groups.Even better, the next result shows that the isomorphisms μG of Corollary 4.4.10 arenatural transformations. The proof is obtained by chasing elements around the diagram.

Lemma 4.4.20. Let G and H be finite abelian groups and let p1, . . . , pk be the collectionof all distinct primes that divide either |G| or |H|. Let f : G→ H be a homomorphism.Then the following diagram commutes.

Gp1 × · · · ×Gpk G

Hp1 × · · · ×Hpk H

��

fp1×···×fpk

��μG

∼=

��

f

��μH

∼=

Here, μG and μH are the isomorphisms of Corollary 4.4.10.

Note that if p divides |G| but not |H|, then Hp = 0, so that fp is the constanthomomorphism to 0. Similarly, if p divides |H| but not |G|, then Gp = 0, and fp is


the unique homomorphism from 0 to Hp. Note that since μG and μH are isomorphisms,the above diagram shows that f is an isomorphism if and only if fp1 × · · · × fpk

is anisomorphism. We obtain an immediate corollary.

Corollary 4.4.21. Let f : G → H be a homomorphism of finite abelian groups. Thenf is an isomorphism if and only if fp : Gp → Hp is an isomorphism for each prime pdividing either |G| or |H|.

Thus, a uniqueness statement for the Fundamental Theorem will follow if we can provethe uniqueness of the decomposition (from Proposition 4.4.18) of an abelian p-group as aproduct of cyclic groups. Fix a prime p. Then, by ordering the exponents, we can writeany finite abelian p-group as a product Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk.Proposition 4.4.22. Suppose that

ν : Zpr1 × · · · × Zprk Zps1 × · · · × Zpsl ,��∼=

where 1 ≤ r1 ≤ · · · ≤ rk, and 1 ≤ s1 ≤ · · · ≤ sl. Then k = l and ri = si for 1 ≤ i ≤ k.Proof We argue by induction on the order of the group. There is only one way inwhich we can get a group of order p, so the induction starts there.

Write Zp ⊂ Zps for the cyclic subgroup 〈ps−1〉. Note that the elements of Zp are theonly ones in Zps of exponent p.

By Problem 2 of Exercises 2.10.12, the elements of Zpr1 × · · · × Zprk which haveexponent p are precisely those in the product Zp × · · · × Zp of the subgroups of orderp in each factor. Thus, Zpr1 × · · · × Zprk has pk elements of exponent p, all but one ofwhich has order p.

Similarly, Zps1×· · ·×Zpsl has pl elements of exponent p. Since isomorphisms preserveorder, k = l. Moreover, ν restricts to an isomorphism ν : Zp × · · · ×Zp → Zp × · · · ×Zp.Passing to factor groups, we obtain an isomorphism

ν : (Zpr1 × · · · × Zprk )/(Zp × · · · × Zp)→ (Zps1 × · · · × Zpsk )/(Zp × · · · × Zp).

By Problem 2 of Exercises 4.1.26, Zps/Zp ∼= Zps−1 . Thus, Problem 14 of the sameexercises shows that we may identify ν with an isomorphism

ν : Zpr1−1 × · · · × Zprk−1 Zps1−1 × · · · × Zpsk−1 .��∼=

The result now follows from induction on order.

The full statement of the Fundamental Theorem is now immediate:

Theorem 4.4.23. (Fundamental Theorem, with Uniqueness) Any finite abelian groupmay be written uniquely as a direct product of cyclic groups of prime power order. Unique-ness means that given any two such decompositions of a given group, the number of factorsof a given order in the two decompositions must be the same.

Exercises 4.4.24.1. Let G be an abelian group. Show that G/Tors(G) is torsion-free.

‡ 2. Show that a finite abelian group G is cyclic if and only if the p-torsion subgroupGp is cyclic for each prime p dividing the order of G. (Hint : This does not requirethe Fundamental Theorem.)


3. Show that a finite abelian group is cyclic if and only if it has no subgroup of theform Zp × Zp for any prime p. (Hint : Apply the Fundamental Theorem.)

† 4. Use the Fundamental Theorem to show that a finite abelian group is cyclic if andonly if it has at most d elements of exponent d for each d dividing the order of G.

5. Give a direct proof (not using the Fundamental Theorem) of the preceding problem.(Hint : One direction has already been done in Problem 11 of Exercises 2.5.21. Forthe other direction, first show via Problem 2 above that we may assume that thegroup has order pr for some prime p and some r ≥ 1. Then show that there mustbe a generator.)

6. Show that the full converse to Lagrange’s Theorem holds for finite abelian groupsG: if k divides the order of G, then G has a subgroup of order k.

7. Show that any finite abelian group may be written as a product Zn1 × · · · × Znk,

where ni divides ni−1 for 1 < i ≤ k.8. Show that the decomposition of the preceding problem is unique.

4.5 The Automorphisms of a Cyclic Group

We calculate the automorphism group of a cyclic group. Recall that for any group Gand for any g ∈ G whose order divides n, there is a unique homomorphism hg : Zn → Gwith hg(1) = g.

Lemma 4.5.1. For n > 1 and m ∈ Z, the homomorphism hm : Zn → Zn is an auto-morphism if and only if m and n are relatively prime.

Proof The image of hm is the cyclic subgroup generated by m. Since the order of mis n/(m,n), hm is onto if and only if (m,n) = 1. But any surjection from a finite set toitself is an injection.

Thus, we’ve identified which homomorphisms from Zn to Zn are automorphisms, butwe have yet to identify the group structure on Aut(Zn).

Recall that there are two operations defined on Zn, addition and multiplication, andthat multiplication gives Zn the structure of a monoid, with unit element 1. Recall thatthe collection of invertible elements in a monoid M forms a group (which we have denotedInv(M)) under the monoid operation of M .

Definition 4.5.2. For n > 1 we write Z×n for the group of invertible elements in the

multiplicative monoid structure on Zn. Thus, m ∈ Z×n if and only if there is a k ∈ Zn

with km = 1 ∈ Zn.

Z×n is an example of a general construction in rings. The set of elements in a ring A

which have multiplicative inverses forms a group under the operation of multiplication.It is called the group of units of A, and may be denoted A×.

We wish to identify the elements of Z×n . The answer turns out to be quite relevant

to our discussion of automorphisms.

Lemma 4.5.3. Let n > 1 and let m ∈ Z. Then m ∈ Z×n if and only if m and n are

relatively prime.


Proof Suppose that m is invertible in Zn, with inverse k. Since km is congruent to1 mod n, we obtain that km − 1 = sn for some integer s. But then km − sn = 1, andhence (m,n) = 1.

Conversely, if rm+ sn = 1, then r is the multiplicative inverse for m in Zn.

Thus, there is a one-to-one correspondence between the elements of Aut(Zn) and Z×n .

We next show that this correspondence is an isomorphism of groups.

Proposition 4.5.4. Define ν : Z×n → Aut(Zn) by ν(m) = hm, where hm : Zn → Zn is

the homomorphism carrying 1 to m. Then ν is an isomorphism of groups.

Proof It suffices to show that hm ◦ hm′ = hmm′ . But it’s easy to see (e.g., by Corol-lary 2.5.5 and Proposition 2.5.17) that hm(k) = mk, and the result follows.

Since multiplication in Zn is commutative, we obtain the following corollary.

Corollary 4.5.5. For any positive integer n, Aut(Zn) ∼= Z×n is an abelian group.

In contrast, most other automorphism groups, even of abelian groups, are nonabelian.It will require a bit more work to display the group structures on these groups of

units Z×n . It will be easiest to do so in the case that n only has one prime divisor. We

will then be able to put this information together, via methods we are about to discuss.First recall from Section 4.4 that if G is abelian and if Gp is its p-torsion subgroup,

then any homomorphism f : G → G restricts to a homomorphism fp : Gp → Gp (i.e.,fp(g) = f(g) for all g ∈ Gp). Moreover, Corollary 4.4.21 shows that f : G → G is anautomorphism if and only if fp is an automorphism for each p dividing |G|. The nextresult follows easily.

Lemma 4.5.6. Let G be an abelian group and let p1, . . . pk be the primes dividing theorder of G. Let Gpi be the pi-torsion subgroup of G for 1 ≤ i ≤ k. Then there is anisomorphism

ϕ : Aut(G) Aut(Gp1)× · · · ×Aut(Gpk)��∼=

defined by ϕ(f) = (fp1 , . . . , fpk).

Let n = pr11 . . . prk

k with p1, . . . , pk distinct primes. Then the pi-torsion subgroup ofZn is isomorphic to Zpri

i. We obtain the following corollary, which may also be proven

using the Chinese Remainder Theorem, below.

Corollary 4.5.7. With n as above, there is an isomorphism

ρ : Z×n

Z×p

r11× · · · × Z×

prkk

��∼=

given by ρ(m) = (m, . . . ,m).

Proof As the reader may easily check, there is a commutative diagram

Z×n

Z×p

r11× · · · × Z×

prkk

Aut(Zn) Aut(Zpr11

)× · · · ×Aut(Zprkk

)

��ρ

��

ν ∼=

��

∼= ν×···×ν

��ϕ

∼=


Thus, the calculation of the automorphism groups of cyclic groups is reduced to thecalculation of Z×

pr for p a prime and r ≥ 1. We begin by calculating the order of Z×pr .

Lemma 4.5.8. For any prime p and any integer r > 0, the group of units Z×pr has order

pr−1(p− 1).

Proof The elements of Zpr that do not lie in Z×pr are the elements m such that (p,m) �=

1. But since p is prime, this means that p divides m, and hence m ∈ 〈p〉 ⊂ Zpr . But 〈p〉has order pr−1, and hence |Z×

pr | = pr − pr−1 = pr−1(p− 1).

Corollary 4.5.7 now allows us to calculate |Z×n |.

Corollary 4.5.9. Let n = pr11 . . . prk

k , where p1, . . . , pk are distinct primes and the expo-nents ri are all positive. Then

|Aut(Zn)| = |Z×n | = (p1 − 1)pr1−1

1 . . . (pk − 1)prk−1k .

In particular, Z×p has order p − 1. In Corollary 7.3.15, we shall show, using some

elementary field theory, that Z×p is always a cyclic group. However, the proof is theoret-

ical, and doesn’t provide any hint as to what elements generate this group. And indeedthere is no known algorithm to obtain a generator other than trial and error. (Some ofthe deeper aspects of number theory are lurking in the background.) So for the practi-cal purpose of identifying the orders of individual elements of Z×

p , direct calculation isnecessary. The reader with some computer expertise may find it interesting to write aprogram for this purpose.

On the other hand, when r > 1, the orders of the elements of Z×p will be the only

open question in our understanding of Z×pr . We need to understand what happens to Z×

pr

as r varies. Thus, let 1 ≤ s < r. Since ps divides pr, there is a homomorphism, whichwe shall denote by π : Zpr → Zps , with π(1) = 1. In fact, π(m) = m for all m ∈ Z. Wecall π the canonical homomorphism. We have m ∈ Z×

pr if and only if (m, p) = 1, whichis true if and only if m ∈ Z×

ps . The next lemma is now immediate.

Lemma 4.5.10. For 1 ≤ s < r and for p a prime, the canonical homomorphism π :Zpr → Zps induces a surjective homomorphism of multiplicative groups

π : Z×pr → Z×

ps .

At this point, the cases p > 2 and p = 2 diverge. One difference that is immediatelyapparent is that Z×

2 = e, while Z×p is nontrivial for p > 2. Thus, for p > 2 we shall define

K(p, r) to be the kernel of π : Z×pr → Z×

p , while K(2, r) is defined to be the kernel ofπ : Z×

2r → Z×4 (and hence K(2, r) is only defined for r ≥ 2). Note that Z×

4 = {±1} hasorder 2.

Lemma 4.5.10 and the First Noether Theorem now give us the order of K(p, r).

Corollary 4.5.11. For p > 2 and r ≥ 1, K(p, r) has order pr−1. Also, for r ≥ 2,K(2, r) has order 2r−2.

For p > 2, the orders of K(p, r) and Z×p are relatively prime, so Corollary 4.4.15 gives

a splitting of Z×pr .


Corollary 4.5.12. For p > 2 and r ≥ 1, K(p, r) is the p-torsion subgroup of Z×pr , and

we have an isomorphism

Z×pr∼= K(p, r)× Z×

p .

The prime 2 is a little different, in that K(2, r) and Z×4 are both 2-groups. But we

get a product formula regardless.

Lemma 4.5.13. For r ≥ 2, we have an isomorphism

Z×2r∼= K(2, r)× Z×

4 .

Proof LetH = {±1} ⊂ Z×2r . ThenH is a subgroup of Z×

2r , and is carried isomorphicallyonto Z×

4 by π. We shall show that Z×2r is the internal direct product of K(2, r) and H,

i.e., the map μ : H ×K(2, r)→ Z×2r given by μ(h, k) = hk is an isomorphism.

The key is that since −1 is not congruent to 1 mod 4, −1 is not in K(2, r), and henceH ∩ K(2, r) = 1. (The unit element here is denoted 1.) This easily implies that μ isinjective.

But since m · k = mk in both Z2r and Z4, either m or −m is in K(2, r) for eachm ∈ Z×

2r . Thus, μ is onto.

We shall show that K(p, r) is cyclic for all primes p and all r ≥ 2. (Unlike the caseof Z×

p , we shall provide a criterion to determine the order of each element.) The key isto understand the order of the elements 1 + ps. As the group operation is multiplicationmodulo pr, we can get a handle on this if we can calculate the powers (x + y)n of thesum of two integers x and y.

The formula for (x + y)n is given by what’s called the Binomial Theorem, which inturn depends on an understanding of the binomial coefficients.

Definition 4.5.14. Let 0 ≤ k ≤ n ∈ Z. We define the binomial coefficient(nk

)by(

n

k

)=

n!k!(n− k)! .

Here, the factorial numbers are defined as usual: 0! = 1, 1! = 1, and n! is definedinductively by n! = (n− 1)! · n for n > 1.

The next lemma is useful for manipulating the binomial coefficients.

Lemma 4.5.15. The binomial coefficients are all integers. Moreover, if 1 ≤ k < n wehave (

n− 1k − 1

)+

(n− 1k

)=

(n

k

).

Proof An easy inspection of the definition shows that(n0

)=

(nn

)= 1 for all n. But

the only other binomial coefficients are the(nk

)with 1 ≤ k < n. Thus, if the displayed

equation is true, then the binomial coefficients are all integers by induction on n, startingwith the case n = 1, which we’ve just verified by inspection.

The left-hand side of the displayed equation expands to

(n− 1)!(k − 1)!(n− k)! +

(n− 1)!k!(n− k − 1)!

=(n− 1)![k + (n− k)]

k!(n− k)! ,

and the result follows.


Our next result is the Binomial Theorem, which is true, with the same proof, notonly in the integers, but in any commutative ring.

Theorem 4.5.16. (Binomial Theorem) Let x and y be integers, and n ≥ 1 be another.Then

(x+ y)n =n∑k=0

(n

k

)xkyn−k

= yn + nxyn−1 +(n

2

)x2yn−2 + · · ·+ nxn−1y + xn

Proof The second line just gives an expansion of the preceding expression, using the(easy) calculations that

(n1

)=

(nn−1

)= n, in addition to the already known calculations

of(n0

)and

(nn

).

We prove the theorem by induction on n. For n = 1, it just says that x+ y = x+ y.Thus, assume that n > 1 and the theorem is true for n− 1. Then

(x+ y)n = (x+ y)n−1(x+ y) =

(n−1∑k=0

(n− 1k

)xkyn−k−1

)(x+ y).

Applying the distributive law to the last expression and collecting terms, we get

xn + yn +n−1∑k=1

xkyn−k[(n− 1k − 1

)+

(n− 1k

)].

But this is exactly the desired sum, by Lemma 4.5.15.

We wish to calculate the order of 1 + ps in K(p, r). By the definitions, the order willbe the smallest integer m such that (1 + ps)m is congruent to 1 mod pr. Since the orderof K(p, r) is a power of p, m must be also. Thus, we shall study the binomial expansionfor (1+ps)p

k

. For this purpose, we shall need to know the largest power of p that dividesthe binomial coefficients

(pk

j

).

Lemma 4.5.17. Let p be prime and let k ≥ 1. Let 1 ≤ j ≤ pk and write j = pab, with(p, b) = 1. Then (

pk

j

)= pk−ac,

with (p, c) = 1.

Proof After making appropriate cancellations, we have(pk

j

)=pk(pk − 1) . . . (pk − j + 1)

1 · 2 . . . (pab) .

Consider the right-hand side as a rational number and rearrange the terms slightly. Weget the product of pk/pab with the product(

pk − 11

)·(pk − 2

2

). . .

(pk − (j − 1)

j − 1

)But for 1 ≤ i < pk, pk− i and i have the same p-divisibility, so that the only contributionto the p-divisibility of our binomial coefficient comes from pk/pab. But this gives thedesired result.


We shall now calculate the order of the generic element of K(p, r). Note that if p > 2,then x ∈ K(p, r) if and only if x = 1 + y for some y divisible by p. Similarly, x ∈ K(2, r)if and only if x = 1 + y for some y divisible by 4.

Proposition 4.5.18. Let p be a prime and let x = 1 + psa, where (p, a) = 1. If p = 2,assume that s > 1. Then x has order pr−s in K(p, r).

Proof As discussed above, the order of x in K(p, r) is the smallest power, pk, of p forwhich xp

k

is congruent to 1 mod pr. So we want to find the smallest value of k suchthat (psa+ 1)p

k − 1 is divisible by pr. Consider the binomial expansion for (psa+ 1)pk

.We have (psa+ 1)p

k

= 1 + pkpsa plus terms of the form(pk

j

)psjaj with j > 1. But then

(psa+ 1)pk − 1 is equal to ps+ka plus the terms

(pk

j

)psjaj with j > 1.

It suffices to show that if j > 1, then(pk

j

)psjaj is divisible by ps+k+1.

By Lemma 4.5.17, if j = pbc, with (p, c) = 1, then(pk

j

)is divisible by pk−b. Thus,(

pk

j

)psjaj is divisible by psj+k−b = psp

bc+k−b, so it suffices to show that spbc+k−b > s+k.But this is equivalent to showing that s[pbc − 1] > b. If b = 0, then c = j > 1, and

the result holds. Otherwise, b > 0, and we may and shall assume that c = 1.For p > 2, we must show that s[pb − 1] > b for all s ≥ 1. But the general case will

follow from that of s = 1, where we show that pb − 1 > b for all b > 0. For p = 2, ourhypothesis is that s > 1, so it suffices to show that 2b − 1 ≥ b for all b > 0.

For any p, pb − 1 = [(p − 1) + 1]b − 1, so we may use the Binomial Theorem again.We see that pb − 1 is equal to b(p − 1) plus some other non-negative terms. But this isgreater than b if p > 2, and is ≥ b if p = 2, precisely as desired.

Corollary 4.5.19. For p > 2 and r ≥ 1, K(p, r) is the cyclic group (of order pr−1)generated by 1 + p.

For p = 2 and r ≥ 2, K(2, r) is the cyclic group (of order 2r−2) generated by 5.

Proof The orders of the groups in question were calculated in Corollary 4.5.11. ByProposition 4.5.18, the 1 + p has the same order as K(p, r) for p > 2, while 5 has thesame order as K(2, r).

Lemma 4.5.13 now gives an explicit calculation of Z×2r .

Corollary 4.5.20. For r ≥ 2, Z×2r is isomorphic to Z2r−2 × Z2.

In particular, Z×2r is not cyclic for r ≥ 3.

On the other hand, when we have established Corollary 7.3.15, we will know that Z×p

is a cyclic group of order p − 1 for all primes p. We shall state the consequence here,though the reader is advised to read ahead for the proof.

This time we make use of Corollary 4.5.12:

Corollary 4.5.21. Let p be an odd prime and let r ≥ 1. Then Z×pr is a cyclic group of

order (p− 1)pr−1.

Exercises 4.5.22.1. Find generators for Z×

p for all primes p ≤ 17.

2. Find a generator for Z×3r for any r.

3. Find a generator for Z×5r for any r.


4. Find a generator for Z×49.

5. Find all elements of order 2 in Z×2r .

6. Find all elements of order 2 in Z×15, Z×

30, and Z×60.

7. Write a computer program to calculate the order of an element in Z×pr , where the

input is p, r, and the element in question. Be sure it checks that the element is aunit.

8. Write a computer program to calculate the order of an element in Z×n . Be sure it

checks that the element is relatively prime to n.

4.6 Semidirect Products

The simplest way to construct new groups from old is to take direct products. But thereare ways to construct more interesting groups if we can calculate the automorphisms ofone of the groups in question.

Suppose given an action of K on H through automorphisms. We may represent thisby a homomorphism α : K → Aut(H), so that the action of k ∈ K takes h ∈ H toα(k)(h). We shall define a group multiplication on the set H × K, to obtain a newgroup, which we denote H �α K, called the semidirect product of H and K given by α:

Definition 4.6.1. Let α : K → Aut(H) be a homomorphism. By the semidirect productof H and K with respect to α, written H �αK, we mean the set H ×K with the binaryoperation given by setting

(h1, k1) · (h2, k2) = (h1 · α(k1)(h2), k1k2).

Proposition 4.6.2. H �α K, with the above product, is a group. There are homomor-phisms π : H �α K → K and ι : H → H �α K, given by π(h, k) = k, and ι(h) = (h, e).The homomorphism ι is an injection onto the kernel of π. Finally, there is a homomor-phism s : K → H �α K given by s(k) = (e, k). This homomorphism has the propertythat π ◦ s is the identity map of K.

Proof We first show associativity.

((h1, k1) · (h2, k2)) · (h3, k3) = (h1 · α(k1)(h2), k1k2) · (h3, k3)= (h1 · α(k1)(h2) · α(k1k2)(h3), k1k2k3)= (h1 · α(k1)(h2) · α(k1)(α(k2)(h3)), k1k2k3).

Here, the last equality holds because α is a homomorphism, so that α(k1k2) = α(k1) ◦α(k2).

Since α(k1) is an automorphism of H, we have

α(k1)(h2) · α(k1)(α(k2)(h3)) = α(k1)(h2 · α(k2)(h3)),

and hence

((h1, k1) · (h2, k2)) · (h3, k3) = (h1 · α(k1)(h2 · α(k2)(h3)), k1k2k3).

But this is precisely what one gets if one expands out (h1, k1) · ((h2, k2) · (h3, k3)).


The pair (e, e) is clearly an identity element for this product, since α(e) is the identityhomomorphism of H. We claim that the inverse of (h, k) is (α(k−1)(h−1), k−1). Thereader is welcome to verify this fact, as well as the statements about π, ι, and s.

Note that if α(k) is the identity automorphism for all k ∈ K (i.e., α : K → Aut(H)is the trivial homomorphism), then H �α K is just the direct product H ×K.

Notation 4.6.3. In the semidirect product H �α K, we shall customarily identify Hwith ι(H) and identify K with s(K), so that the element (h, k) is written as hk. Thus,H �α K = {hk |h ∈ H, k ∈ K}.Lemma 4.6.4. With respect to the above notations, the product in H �α K is given by

h1k1 · h2k2 = h1α(k1)(h2)k1k2

for h1, h2 ∈ H and k1, k2 ∈ K.In particular, this gives khk−1 = α(k)(h). Thus, H �α K is abelian if and only if H

and K are both abelian and α : K → Aut(H) is the trivial homomorphism.

Proof The multiplication formula is clear. But this gives kh = α(k)(h)k, which in turnimplies the stated conjugation formula.

Thus, when α is nontrivial, the new groupH�αK may have some interesting features.Indeed, this method of constructing groups will give us a wide range of new examples.But it also gives new ways of constructing several of the groups we’ve already seen. Forinstance, as we shall see below, the dihedral group D2n is the semidirect product obtainedfrom an action of Z2 on Zn, and A4 is the semidirect product obtained from an actionof Z3 on Z2 × Z2.

We wish to analyze the semidirect products obtained from actions of cyclic groups onother cyclic groups. Since these are, in general, nonabelian, we shall use multiplicativenotation for our cyclic groups. Thus, we shall begin with a discussion of notation.

Recall first the isomorphism ν : Z×n → Aut(Zn) of Proposition 4.5.4. Here, Z×

n is thegroup of invertible elements with respect to the operation of multiplication in Zn, and,for m ∈ Z×

n , ν(m) is the automorphism defined by ν(m)(l) = ml.In multiplicative notation, we write Zn = 〈b〉 = {e, b, . . . , bn−1}. Thus, the element

of Zn written as l in additive notation becomes bl. In other words, in this notation, wehave an isomorphism ν : Z×

n → Aut(〈b〉), where ν(m) is the automorphism defined byν(m)(bl) = (bml).

Now write Zk = 〈a〉 in multiplicative notation, and recall from Proposition 2.5.17that there is a homomorphism hg : Zk → G with hg(a) = g ∈ G if and only if g hasexponent k in G. Moreover, if g has exponent k, there is a unique such homomorphism,and it is given by the formula hg(aj) = gj .

Thus, suppose given a homomorphism α : Zk = 〈a〉 → Z×n , with α(a) = m. Then m

has exponent k in Z×n , meaning that mk is congruent to 1 mod n. Moreover, α(aj) = mj .

By abuse of notation, we shall write α : Zk → Aut(Zn) for the composite of this αwith the isomorphism ν above. Thus, as an automorphism of Zn = 〈b〉, α(aj) is givenby α(aj)(bi) = bm

ji.

Corollary 4.6.5. Let m have exponent k in Z×n , and let α : Zk → Z×

n be the homomor-phism that takes the generator to m. Then writing Zn = 〈b〉, Zk = 〈a〉, and identifyingZ×n with Aut(Zn), we obtain

Zn �α Zk = {biaj | 0 ≤ i < n, 0 ≤ j < k},


where b has order n, a has order k, and the multiplication is given by

biajbi′aj

′= bi+m

ji′aj+j′.

Moreover, the nk elements {biaj | 0 ≤ i < n, 0 ≤ j < k} are all distinct.

Proof The multiplication formula comes via Lemma 4.6.4 from the formula α(aj)(bi) =bm

ji.

Examples 4.6.6.1. Let α : Z2 → Z×

n be the homomorphism that carries the generator to −1. Thenin Zn �α Z2, with the notations above, we see that b has order n, a has order 2,and aba−1 = b−1. Thus, Problem 6 of Exercises 2.8.5 provides a homomorphismf : D2n → Zn�αZ2 with f(a) = a and f(b) = b. Since a and b generate Zn�αZ2, fis onto, and hence an isomorphism, as the two groups have the same order. Thus,semidirect products provide a construction of the dihedral groups that does notrequire trigonometry.

2. Let α : Z2 → Z×8 be the homomorphism that carries the generator to 3. We obtain

a group of order 16,

Z8 �α Z2 = {biaj | 0 ≤ i < 8, 0 ≤ j < 2},where a has order 2, b has order 8, and the multiplication is given by biajbi

′aj

′=

bi+3ji′aj+j′.

3. Let β : Z2 → Z×8 be the homomorphism that carries the generator to 5. We obtain


Z8 �β Z2 = {biaj | 0 ≤ i < 8, 0 ≤ j < 2},where a has order 2, b has order 8, and the multiplication is given by biajbi

′aj

′=

bi+5ji′aj+j′.




′aj

′=

bi+2ji′aj+j′.




′aj

′=

bi+2ji′aj+j′.

6. Let p be an odd prime. Recall from Proposition 4.5.18 that p+ 1 has order p inZ×p2 . Thus, there is a homomorphism α : Zp → Z×

p2 that carries the generator ofZp to p+ 1. We obtain a group of order p3,

Zp2 �α Zp = {biaj | 0 ≤ i < p2, 0 ≤ j < p},where a has order p, b has order p2, and the multiplication is given by biajbi

′aj

′=

bi+(p+1)ji′aj+j′.


Other than the first example, these groups are all new to us. Indeed, the groups oforder 16 in Examples 2 and 3 turn out to be the lowest-order groups that we had notconstructed previously.

The group of Example 5 turns out to be the smallest odd order group that is non-abelian, while for each odd prime p, the group of Example 6 represents one of the twoisomorphism classes of nonabelian groups of order p3.

We may also construct examples of semidirect products where at least one of the twogroups is not cyclic. For instance, since the groups Z×

n are not generally cyclic, we canobtain a number of groups of the form Zn �α K, where K is not cyclic. We are alsointerested in the case where group “H” in the construction of semidirect products is notcyclic. The simplest case is where H = Zn × Zn.

As above, write Zn × Zn = {bicj | 0 ≤ i, j < n}, where b and c have order n andcommute with each other. Since every element of Zn × Zn has exponent n, Proposi-tions 2.10.6 and 2.5.17 show that there is a homomorphism f : Zn×Zn → Zn×Zn withthe property that f(b) = bc and f(c) = c: explicitly, f(bicj) = bici+j .

Lemma 4.6.7. The homomorphism f just described is an automorphism of Zn×Zn. Ithas order n in the group of automorphisms of Zn × Zn.

Proof For k ≥ 1 the k-fold composite of f with itself is given by fk(c) = c andfk(b) = fk−1(bc) = fk−1(b)fk−1(c) = (fk−1(b))c. An easy induction now shows thatfk(b) = bck. Thus, fk is the identity if and only if k is congruent to 0 mod n.

Since fn is the identity, f is indeed an automorphism, with inverse fn−1.

Thus, if we write Zn = 〈a〉, there is a homomorphism

α : Zn → Aut(Zn × Zn)

with α(a) = f . We obtain a construction which is of special interest when n is prime:

Corollary 4.6.8. There is a group G of order n3 given by G = {bicjak | 0 ≤ i, j, k < n},where a, b, and c all have order n, and b commutes with c, a commutes with c, andaba−1 = bc. Thus,

bicjak · bi′cj′ak′ = bi+i′cj+j

′+ki′ak+k′.

There are, of course, at least as many examples of semidirect products as there aregroups with nontrivial automorphism groups. Thus, we shall content ourselves for nowwith these examples.

In our studies of the dihedral and quaternionic groups, we have made serious use ofour knowledge of the homomorphisms out of these groups. There is a similar result forthe generic semidirect product.

First, we need some discussion. Let ι : H → H �α K and s : K → H �α K be thestructure maps: in the ordered pairs notation, ι(h) = (h, e) and s(k) = (e, k). Thenevery element of H �α K may be written uniquely as ι(h) · s(k) with h ∈ H and k ∈ K.This is, indeed, the basis for the HK notation.

In particular, H �α K is generated by the elements of ι(H) and s(K), so that anyhomomorphism f : H �α K → G is uniquely determined by its restriction to the sub-groups ι(H) and s(K). Since ι and s are homomorphisms, this says that f is uniquelydetermined by the homomorphisms f ◦ ι and f ◦ s.


Thus, we will understand the homomorphisms out of H �α K completely if we cangive necessary and sufficient conditions for a pair of homomorphisms f1 : H → G and f2 :K → G to be obtained as f1 = f ◦ ι and f2 = f ◦s for a homomorphism f : H�αK → G.

Proposition 4.6.9. Let α : K → Aut(H) be a homomorphism. Suppose given homo-morphisms f1 : H → G and f2 : K → G for a group G. Then there is a homomorphismf : H �α K → G with f1 = f ◦ ι and f2 = f ◦ s if and only if f2(k)f1(h)(f2(k))−1 =f1(α(k)(h)) for all h ∈ H and k ∈ K.

Such an f , if it exists, is unique, and is given in the HK notation by f(hk) =f1(h)f2(k).

Proof Suppose first that f exists with f ◦ ι = f1 and f ◦ s = f2. In the HK notationwe have that hk = ι(h)s(k), so that f(hk) = (f ◦ ι)(h) · (f ◦ s)(k) = f1(h)f2(k) for allh ∈ H and k ∈ K.

As shown in Lemma 4.6.4, khk−1 = α(k)(h) in H �α K, for all h ∈ H and k ∈ K.Applying f to both sides, we obtain f2(k)f1(h)(f2(k))−1 = f1(α(k)(h)), as claimed.

Conversely, suppose that f2(k)f1(h)(f2(k))−1 = f1(α(k)(h)) for all h ∈ H and k ∈ K.We define f : H �α K → G by f(hk) = f1(h)f2(k). It suffices to show that f is ahomomorphism. But

f(h1k1)f(h2k2) = f1(h1)f2(k1)f1(h2)f2(k2)= f1(h1)f1(α(k1)(h2))f2(k1)f2(k2)= f1(h1α(k1)(h2))f2(k1k2)= f(h1k1h2k2),

for all h1, h2 ∈ H and k1, k2 ∈ K, where the second equality follows from the formulaf2(k1)f1(h2)(f2(k1))−1 = f1(α(k1)(h2)).

This permits a slight simplification when H and K are cyclic.

Corollary 4.6.10. Let α : Zk → Z×n be a homomorphism, with α(a) = m, where Zk =

〈a〉 in multiplicative notation. Write Zn�αZk in multiplicative notation, with Zn = 〈b〉.Then for any group G, there is a homomorphism f : Zn �α Zk → G with f(a) = x andf(b) = y if and only if the elements x, y ∈ G satisfy the following conditions:

1. x has exponent k

2. y has exponent n

3. xyx−1 = ym.

Proof The conditions are clearly necessary, as a has order k, b has order n, andaba−1 = bm in Zn �α Zk.

Conversely, if the three conditions hold, there are homomorphisms f1 : 〈b〉 → G andf2 : 〈a〉 → G given by f1(bi) = yi and f2(aj) = xj . Thus, by Proposition 4.6.9, it sufficesto show that xjyix−j = ym

ji.Since conjugation by xj is a homomorphism, xjyix−j = (xjyx−j)i, so it suffices by the

law of exponents to show that xjyx−j = ymj

. But since xjyx−j = x(xj−1yx−(j−1))x−1,the desired result follows from the third condition by an easy induction argument.

The above is an important ingredient in calculating the automorphism groups of thesemidirect products of cyclic groups. But it does not give good criteria for showing


when the homomorphism f is an isomorphism. For this it will be convenient to use thelanguage and concepts of extensions, as developed in Section 4.7.

We shall now begin to consider the question of the uniqueness of the semidirectproduct: When are two different semidirect products isomorphic? We will not answerthis question completely, but will study some aspects of it. First, what happens if weswitch to a different homomorphism K → Aut(H)?

Recall from Section 3.6 that G acts on the set of homomorphisms from K to G asfollows: if f : K → G is a homomorphism and g ∈ G, then g · f is the homomorphismdefined by (g · f)(k) = gf(k)g−1 for all k ∈ K. By abuse of notation, we shall alsowrite gfg−1 for g · f . We say that the homomorphisms f and gfg−1 are conjugatehomomorphisms.

So it makes sense to ask whether H �α K and H �α′ K are isomorphic when α andα′ are conjugate homomorphisms of K into Aut(H). Another fairly minor way in whichwe could alter a homomorphism α : K → Aut(H) is by replacing it with α ◦ g, whereg ∈ Aut(K). We wish to study the effect of this sort of change as well.

Since we’re considering two different semidirect products of the same groups, we shalluse different notations for their structure maps. Thus, we write ι : H → H �α K,π : H �α K → K, and s : K → H �α K for the structure maps of H �α K, and writeι′ : H → H �α′ K, π′ : H �α′ K → K, and s′ : K → H �α′ K for the structure maps ofH �α′ K.

Proposition 4.6.11. Let α and α′ be homomorphisms of K into Aut(H). Then thefollowing conditions are equivalent:

1. α is conjugate to α′ ◦ g, with g ∈ Aut(K).

2. There is an isomorphism ϕ : H �α K → H �α′ K such that im(ϕ ◦ ι) = im ι′ andim(ϕ ◦ s) = im s′.

3. There is a commutative diagram

H H �α K K

H H �α′ K K��

f

��ι

��

ϕ

�� s

��

g

��ι′ �� s′

in which the vertical maps are all isomorphisms.

Proof Suppose that the second condition holds. Then ϕ restricts to an isomorphismfrom ι(H) to ι′(H), and hence there exists an isomorphism f : H → H making theleft-hand square of the diagram commute.

Similarly, ϕ restricts to an isomorphism from s(K) to s′(K) and there exists anautomorphism g of K such that the right-hand square of the diagram commutes. Thus,the second condition implies the third.

Suppose then that the third condition holds. Then identifying H and K with theirimages under ι′ and s′, we see that ϕ ◦ ι = f and ϕ ◦ s = g. By Proposition 4.6.9, thissays that g(k)f(h)(g(k))−1 = f(α(k)(h)) for all h ∈ H and k ∈ K. But since f(h) ∈ Hand g(k) ∈ K, we have g(k)f(h)(g(k))−1 = α′(g(k))(f(h)) by the conjugation formulathat holds in H �α′ K.


Putting these two statements together, we see that α′(g(k))(f(h)) = (f ◦ α(k))(h),and hence α′(g(k)) ◦ f = f ◦ α(k) as automorphisms of H. But then as maps of Kinto Aut(H), this just says that α′ ◦ g is the conjugate homomorphism to α obtained byconjugating by f ∈ Aut(H). Thus, the third condition implies the first.

Now suppose that the first condition holds, and we’re given g ∈ Aut(K) such that α′◦gis conjugate to α. Thus, there is an f ∈ Aut(H) such that (α′ ◦g)(k) = f ◦α(k)◦f−1 forall k ∈ K. Reversing the steps above, this implies that g(k)f(h)(g(k))−1 = f(α(k)(h))for all h ∈ H and k ∈ K. Thus, by Proposition 4.6.9, There is a homomorphismϕ : H �α K → H �α′ K given by ϕ(hk) = f(h)g(k).

Clearly, ϕ restricts to an isomorphism, f , from ι(H) to ι′(H) and restricts to anisomorphism, g, from s(K) to s′(K), so it suffices to show that ϕ is an isomorphism.Since the elements of H �α′ K may be written uniquely as an element of H times anelement of K, f(h)g(k) = e if and only if f(h) = e and g(k) = e. Since f and g are bothinjective, so is ϕ. Finally, since f and g are both surjective, every element of H �α′ Kmay be written as a product f(h)g(k), and hence ϕ is onto. Thus, the first conditionimplies the second.

This is far from giving all the ways in which semidirect products can be isomorphic,but we shall find it useful nevertheless, particularly when H is abelian. As a sampleapplication, consider this.

Corollary 4.6.12. Let α, α′ : Z5 → Z×11 be homomorphisms such that neither α nor α′

is the trivial homomorphism. Then the semidirect products Z11 �α Z5 and Z11 �α′ Z5

are isomorphic.

Proof By Corollary 7.3.15, Z×11 is a cyclic group of order 10. Alternatively, we can

verify it by hand: 2k is not congruent to 1 mod 11 for k ≤ 5. Since the order of 2 in Z×11

divides 10, we see that 2 has order 10, and hence generates Z×11.

Thus, Z×11 has a unique subgroup H of order 5. Since the images of α and α′ must

have order 5, they induce isomorphisms of Z5 onto H. Thus, the composite

Z5 H Z5��α′

��α−1

gives an isomorphism, which we’ll denote by g : Z5 → Z5. But then α′ = α ◦ g, andhence Z11 �α Z5

∼= Z11 �α′ Z5 by Proposition 4.6.11.

Using the Sylow Theorems in Chapter 5, we’ll be able to show that every groupof order 55 is a semidirect product of the form Z11 �α Z5. Together with the abovecorollary, this shows that there are exactly two isomorphism classes of groups of order55: the product Z11×Z5 (which is cyclic), and the semidirect product Z11�αZ5 obtainedfrom the homomorphism α : Z5 → Z×

11 that sends the generator of Z5 to 4.

Exercises 4.6.13.1. Show that the subgroup s(K) ⊂ H �αK is normal if and only if α : K → Aut(H)

is the trivial homomorphism.

2. Consider the groups of Examples 4.6.6. For the groups of Examples 2, 3, 4, and 5,find the orders of all the elements not contained in 〈b〉. Deduce that the groups ofExamples 2 and 3 cannot be isomorphic.

3. Show that the group of Example 4 of Examples 4.6.6 contains a subgroup isomor-phic to D10.


4. Let α : Z4 → Z×3 be the homomorphism that carries the generator of Z4 to the

unique nontrivial element of Z×3 . Construct an isomorphism f : Q12 → Z3 �α Z4.

5. Consider the group G = Z8 �β Z2 of Example 3 of Examples 4.6.6. Show thatthe subgroup 〈b〉 is not characteristic in G, by constructing an automorphism of Gwhich carries 〈b〉 outside of itself. (Of course, 〈b〉 is normal in G, as it is the groupbeing acted on in the semidirect product.)

6. Show that if m divides the order of an element x ∈ G, then o(x) = m · o(xm).Deduce that if f : G → G′ is a homomorphism and if x ∈ G, then xo(f(x)) ∈ ker fand o(x) = o(f(x)) · o(xo(f(x))). Deduce that if hk ∈ H �α K, then (hk)o(k) ∈ H,and o(hk) = o(k) · o((hk)o(k)). Thus, the determination of the order of hk dependsonly on calculating the element (hk)o(k) ∈ H and on understanding the orders ofthe elements in H.

† 7. Let x, y ∈ G, such that xyx−1 = ym. Show that (yx)k = y1+m+···+mk−1xk for all

k ≥ 2.

† 8. Let m > 1 and k > 1 be integers. Show that (1+m+ · · ·+mk−1)(m−1) = mk−1,and hence 1 +m+ · · ·+mk−1 = (mk − 1)/(m− 1).

9. Find the orders of the elements in the group of Example 6 of Examples 4.6.6. Showalso that the subgroup 〈b〉 is not characteristic. (Hint : Take a look at the proof ofProposition 4.5.18.)

‡ 10. Show that the group constructed in Corollary 4.6.8 for n = 2 is isomorphic to D8.

11. Consider the groups constructed in Corollary 4.6.8 for n a prime number p. Whenp > 2, show that every element in this group has exponent p. (Why is this differentfrom the case n = 2?) Recall that in Problem 3 of Exercises 2.1.16, it was shownthat any group of exponent 2 must be abelian. As we see here, this property is notshared by primes p > 2.

12. Now consider the groups of Corollary 4.6.8 for general values of n. For which valuesof n does the group in question have exponent n? If the exponent is not equal ton, what is it?

13. In this problem we assume a familiarity with the group of invertible 3× 3 matricesover the ring Zn. Let

G =

⎧⎨⎩⎛⎝ 1 x y

0 1 z0 0 1

⎞⎠∣∣∣∣∣∣ x, y, z ∈ Zn

⎫⎬⎭ .

Show that G forms a group under matrix multiplication. Show, using Proposi-tion 4.6.9, that G is isomorphic to the group constructed in Corollary 4.6.8.

† 14. There are three non-identity elements in Z2×Z2, which must be permuted by anyautomorphism. Show that the automorphism group of Z2 × Z2 is isomorphic tothe group of permutations on these elements, so that Aut(Z2 × Z2) ∼= S3. (Hint :Take a look at the proof of Corollary 3.2.11.)

15. Show that any two nontrivial homomorphisms from Z2 to S3 are conjugate. Deducefrom Problem 14 that if α, α′ : Z2 → Aut(Z2 × Z2) are nontrivial, then (Z2×Z2)�α

Z2∼= (Z2×Z2)�α′Z2. Now deduce from Problem 10 that if α : Z2 → Aut(Z2 × Z2)

is nontrivial, then (Z2 × Z2) �α Z2∼= D8.


16. Show that Aut(Z) ∼= {±1} and write α : Z2 → Aut(Z) for the nontrivial homomor-phism. The semidirect product Z�α Z2 is called the infinite dihedral group. Showthat the dihedral groups D2n are all factor groups of the infinite dihedral group.

17. Let α : Z → Aut(Z) be the unique nontrivial homomorphism. The semidirectproduct Z�αZ is called the Klein bottle group, as it turns out to be isomorphic tothe fundamental group of the Klein bottle. Show that the infinite dihedral groupis a factor group of the Klein bottle group.

18. Let Z[ 12 ] ⊂ Q be the set of rational numbers that may be written as m2k with

m ∈ Z and k ≥ 0. In other words, if placed in lowest terms, these elements havedenominators that are powers of 2. Show that Z[ 12 ] is a subgroup of the additivegroup of rational numbers and that multiplication by 2 induces an automorphismof Z[ 12 ]. Deduce the existence of a homomorphism μ2 : Z → Aut(Z[ 12 ]) such thatμ2(k) is multiplication by 2k for all k ∈ Z.

‡ 19. Let G = Z[ 12 ]�μ2 Z. Let H ⊂ Z[ 12 ] be the standard copy of the integers in Q. Findan element g ∈ G such that gHg−1 is contained in H but not equal to H.3

4.7 Extensions

Here, loosely, we study all groups G that contain a given group H as a normal subgroup,such that the factor group G/H is isomorphic to a particular group K. This is the firststep in trying to classify composition series.

Definitions 4.7.1. Let H and K be groups. An extension of H by K consists of agroup G containing H as a normal subgroup, together with a surjective homomorphismf : G→ K with kernel H.4

Examples 4.7.2.1. Let f : D2n → Z2 be induced by f(a) = 1 and f(b) = 0. Then f is an extension of

Zn = 〈b〉 by Z2.

2. The projection map π : Zn × Z2 → Z2 is an extension of Zn by Z2.

3. The unique nontrivial homomorphism Z2n → Z2 is yet another extension of Zn =〈2〉 by Z2.

4. Let f : Q4n → Z2 be induced by f(a) = 1 and f(b) = 0. Then f is an extension ofZ2n = 〈b〉 by Z2.

5. Let α : K → Aut(H) be a homomorphism and let π : H �α K → K be theprojection. Then π is an extension of H by K.

3We could have used the rational numbers Q here in place of Z[ 12], obtaining a construction that

would be slightly simpler for the beginner. However, by using Z[ 12], we obtain a group that is finitely

generated (say, by the generator of H and the generator of the Z that acts), whereas we would not haveif we’d used Q.

4It is sometimes useful to think of an extension of H by K as a group G, together with an embeddingι : H → G and a surjective homomorphism f : G → K with kernel ι(H). In this language, the entire

sequence Hι−→ G

f−→ K constitutes the extension in question. Note that given such data, we can forman extension in the sense of Definition 4.7.1 by replacing G by the group G′ obtained by appropriatelyrenaming the elements in im ι.


Clearly, for appropriate choices of H and K, there may be a number of nonisomorphicextensions of H by K. Moreover, an understanding of the extensions of one group byanother is a key to understanding groups in general.

Recall from Proposition 4.2.5 that every nontrivial finite group G has a compositionseries, which is a sequence e = H0 ⊂ H1 ⊂ · · · ⊂ Hk = G, such that Hi−1 � Hi, andHi/Hi−1 is a nontrivial simple group for 1 ≤ i ≤ k. But this says that Hi is an extensionof Hi−1 by a nontrivial simple group.

In other words, every finite group may be obtained by a sequence of extensions of lowerorder groups by simple groups. But the finite simple groups are all known. Thus, thenext step in a systematic study of finite group theory is to classify all possible extensionsof one finite group by another.

The search for such a classification is known as the Extension Problem, and constitutesone of the major unsolved problems in mathematics.

Notice that there are various ways in which we could desire to classify the extensionsof H by K. For our purposes here, we’d be most interested in simply determining theisomorphism classes of groups G that occur as extensions of H by K. Such a classificationtakes no account of the precise embedding of H in G, nor of the map f : G → K thatdetermines the extension. Thus, such a classification forgets some of the data that’s partof the definition of an extension.

A classification that would forget less data would be to classify the extensions of Hby K up to the equivalence relation obtained by identifying extensions f : G → K andf ′ : G′ → K if there is a commutative diagram

H G K

H G′ K

��⊂

��

g1

��f

��

g2

��

g3

��⊂ ��f ′

in which the vertical maps are all isomorphisms. An even more rigid equivalence relationis the one given by such diagrams where the maps g1 and g3 are required to be theidentity maps of H and K, respectively.

All three equivalence relations are relevant for particular questions. We shall discussthese issues further.

In any case, we shall not be able to solve all the extension questions that mightinterest us, but we can make considerable progress in special cases. These cases will besufficient to classify all groups of order ≤ 63.

Here is a useful result for obtaining equivalences between extensions.

Lemma 4.7.3. (Five Lemma for Extensions) Let f : G→ K be an extension of H by Kand let f ′ : G′ → K ′ be an extension of H ′ by K ′. Suppose given a commutative diagramof group homomorphisms

H G K

H ′ G′ K ′

��⊂

��

g1

��f

��

g2

��

g3

��⊂ ��f ′

where g1 and g3 are isomorphisms. Then g2 is also an isomorphism.


Proof Let x ∈ ker g2. Then (f ′ ◦ g2)(x) = e, and hence g3(f(x)) = e by commutativityof the diagram. Since g3 is injective, f(x) = e, and hence x ∈ ker f = H. But on H ⊂ G,g2 acts as g1 : H → H ′, and hence x ∈ ker g1. But g1 is injective, so we must have x = e.Thus, ker g2 = e, and hence g2 is injective.

Thus, it suffices to show that each y ∈ G′ lies in the image of g2. Since g3 is surjective,f ′(y) = g3(z′) for some z′ ∈ K. But f is also surjective, so z′ = f(z) for some z ∈ G.Thus, f ′(g2(z)) = g3(f(z)) = g3(z′) = f ′(y), so that g2(z) and y have the same imageunder f ′.

But then y(g2(z))−1 lies in the kernel of f ′, which is H ′. Thus, since g1 is surjective,there is an x ∈ H such that g1(x) = y(g2(z))−1. Thus, y = g1(x)g2(z) = g2(xz), andhence y ∈ im g2 as desired.

The Five Lemma for Extensions is a special case of a more general result which isgiven for abelian groups as the Five Lemma in Lemma 7.7.48. The statement of thegeneral case involves five vertical maps, rather than the three displayed here, hence thename. In both cases, the technique of proof is called a diagram chase.

As noted above, if α : K → Aut(H) is a homomorphism, then the projection mapπ : H �α K → K is an extension of H by K. As a general rule, the semidirect productsare the easiest extensions to analyze, so it is useful to have a criterion for an extensionto be one.

Definition 4.7.4. A section, or splitting, for f : G→ K is a homomorphism s : K → G,such that f ◦ s is the identity map of K. A homomorphism f : G → K that admits asection is said to be a split surjection.

An extension of H by K is called a split extension if f : G → K admits a section.The section s is said to split f .

The homomorphism s : K → H �α K is clearly a section for π. We also have aconverse.

Proposition 4.7.5. Split extensions are semidirect products. Specifically, let f : G→ Kbe a split extension of H by K, with splitting map s : K → G. Then s induces ahomomorphism α : K → Aut(H) via α(k)(h) = s(k)h(s(k))−1. Moreover, there is anisomorphism ϕ : H �α K → G such that the following diagram commutes.

H H �α K K

H G K

� ��

� ��

��⊂ι

��

ϕ

�� s1

� ��

� ��

��⊂ �� s

Here, s1 is the standard section for the semidirect product. Explicitly, ϕ is defined byϕ(hk) = hs(k).

Proof Recall from Lemma 3.7.7 that, since H � G, we have a homomorphism Γ : G→Aut(H), given by setting Γ(g)(h) = ghg−1 for all g ∈ G and h ∈ H. The above specifi-cation for α is just the composite Γ ◦ s : K → Aut(H), and hence is a homomorphism asclaimed.

Recall from Proposition 4.6.9 that if f1 : H → G′ and f2 : K → G′ are homomor-phisms such that f2(k)f1(h)(f2(k))−1 = f1(α(k)(h)) for all h ∈ H and k ∈ K, then there


is a unique homomorphism f : H �α K → G′ such that f ◦ ι = f1 and f ◦ s1 = f2:specifically, f(hk) = f1(h)f2(k).

We apply this with f1 equal to the inclusion of H in G and with f2 equal to the sections : K → G. But then the condition we need to verify to construct our homomorphismis that s(k)h(s(k))−1 = α(k)(h), which holds by the definition of α. Thus, there is ahomomorphism ϕ : H�αK → G with ϕ◦ι equal to the inclusion ofH in G, and ϕ◦s1 = s:specifically, ϕ(hk) = hs(k). But this just says that the stated diagram commutes, so itsuffices to show that ϕ is an isomorphism.

Since ϕ(hk) = hs(k), we obtain that (f ◦ ϕ)(hk) = (f ◦ s)(k) = k. But this says thatf ◦ ϕ = π, where π : H �α K → K is the projection map. Thus, we have an additionalcommutative diagram.

H H �α K K

H G K

� ��

� ��

��⊂ι

��

ϕ

��π

� ��

� ��

��⊂ ��f

But this shows ϕ to be an isomorphism by the Five Lemma for Extensions.

Of course, if α : K → Aut(H) is the trivial homomorphism, then the semidirectproduct H �αK is simply H ×K. Thus, the corollary below is immediate from the waythat α depends on s in the statement of Proposition 4.7.5.

Corollary 4.7.6. Let f : G → K be a split extension of H by K, with splitting maps : K → G. Suppose that s(k) commutes with each element of H for all k ∈ K. ThenG ∼= H ×K.

In particular, if f : G → K is a split surjection and if G is abelian, then G ∼=ker f ×K.

Semidirect products are the easiest extensions to construct, and hence may be viewedas the most basic of the extensions of one group by another. For this reason, it is usefulto be able to recognize whether an extension splits. The easiest case is when the twogroups have orders that are relatively prime.

Proposition 4.7.7. Let H and K be groups whose orders are relatively prime and letf : G→ K be an extension of H by K. Then the extension splits if and only if G has asubgroup of order |K|.Proof If s : K → G is a section of f , then s is injective, so s(K) has the same order asK.

Conversely, suppose that K ′ ⊂ G has order |K| and write f1 : K ′ → K for therestriction of f to K ′. Then ker f1 = K ′∩H. Since the orders of K ′ and H are relativelyprime, K ′ ∩H = e, by Lagrange’s Theorem. Thus, f1 is injective. Since K ′ and K have

the same order, f1 is an isomorphism, and the composite Kf−11−−→ K ′ ⊂ G gives a section

for f .

In fact, under the hypotheses above, G always has a subgroup of order |K|.

Theorem (Schur–Zassenhaus Lemma) Let f : G → K be an extension of H by K,where H and K have relatively prime order. Then G has a subgroup of order |K|.


The proof requires the Sylow Theorems and could be given in Chapter 5, but weshall not do so. In practice, when considering a particular extension, one can generallyconstruct the section by hand.

Let us now consider the issue of classifying the split extensions of one group byanother.

Note that the definition of the homomorphism α : K → Aut(H) constructed inProposition 4.7.5 makes use of the specific section s. Thus, we might wonder whether adifferent homomorphism α would have resulted if we’d started with a different section.Indeed, this can happen when the group H is nonabelian.

Example 4.7.8. Let ε : Sn → {±1} take each permutation to its sign. Then ε is anextension of An by Z2. Moreover, setting s(−1) = τ ∈ Sn determines a section of ε ifand only if τ has order 2 and is an odd permutation.

Since the order of a product of disjoint cycles is the least common multiple of theorders of the individual cycles, we see that τ determines a section of ε if and only if itis the product of an odd number of disjoint 2-cycles. For such a τ , we write sτ for thesection of ε with sτ (−1) = τ . Given the variety of such τ , we see that there are manypossible choices of section for ε.

Write ατ : Z2 → Aut(An) for the homomorphism induced by sτ . By definition,ατ = Γ ◦ sτ , where Γ : Sn → An is the homomorphism induced by conjugation. Thus,ατ (−1) = Γ(τ).

But Problem 10 in Exercises 3.7.14 shows that Γ is injective for n ≥ 4. Thus, if n ≥ 4,then each different choice for a section of ε induces a different homomorphism from Z2

to Aut(An).If τ and τ ′ have the same cycle structure, then they are conjugate in Sn, and hence

sτ and sτ ′ are conjugate as homomorphisms into Sn. But then ατ and ατ ′ are conjugateas homomorphisms into Aut(An).

However, as we shall see in the exercises below, if τ is a 2-cycle and τ ′ is a productof three disjoint 2-cycles, then ατ and ατ ′ are not conjugate as homomorphisms intoAut(An) if n > 6. Thus, in this case, Proposition 4.6.11 would not have predicted thatAn�ατ

Z2 and An�ατ′ Z2 would be isomorphic; nevertheless they are isomorphic, as eachone of them admits an isomorphism to Sn. However, the criterion in Proposition 4.6.11 isthe most general sufficient condition we shall give for the isomorphism of two semidirectproducts.

Thus, the classification of even the split extensions of a nonabelian group offers sub-stantial difficulty. But we shall be mainly concerned here with the extensions of anabelian group, a problem for which we can develop some reasonable tools.

Lemma 4.7.9. Let f : G→ K be an extension of H by K, where H is abelian. For x ∈G, write cx : H → H for the automorphism obtained from conjugation by x. Suppose thatf(x) = f(y). Then cx = cy as automorphisms of H. Moreover, there is a homomorphismαf : K → Aut(H) defined by setting αf (k) equal to cx for any choice of x with f(x) = k.

In particular, if f happens to be a split extension and if s : K → G is a section of f ,then the homomorphism from K to Aut(H) induced by s coincides with the homomor-phism αf above. Thus, if H is abelian, the homomorphism from K to Aut(H) inducedby a section of f is independent of the choice of section.

Proof If f(x) = f(y), then y−1x ∈ ker f = H, so x = yh for some h ∈ H. But thencx = cy ◦ ch. Since H is abelian, conjugation of H by h is the identity map, so cx = cyas claimed.


Thus, there is a well defined function αf : K → Aut(H) obtained by setting αf (k)equal to cx for any x with f(x) = k. But if f(x) = k and f(y) = k′, then f(xy) = kk′,so that αf (kk′) = cxy = cx ◦ cy = αf (k)αf (k′), so αf is a homomorphism.

If s : K → G is a section of f , then the homomorphism α : K → Aut(H) induced bythis section is obtained by setting α(k) = cs(k) : H → H. Since f(s(k)) = k, this justsays that α = αf .

Thus, we may make the following definition.

Definition 4.7.10. Let f : G → K be an extension of H by K, where H is abelian.Then the action of K on H induced by f is the homomorphism αf : K → Aut(H)obtained by setting αf (k) equal to the automorphism obtained by conjugating H by anygiven element of f−1(k).

The existence of the homomorphisms αf gives a valuable method of studying theextensions of an abelian group: fix a homomorphism α : K → Aut(H) and classify theextensions whose induced action is α. Do this for all possible homomorphism α, and lookto see if a given group can be written as an extension in more than one way.

We shall illustrate this process by classifying the groups of order 8.

Lemma 4.7.11. Let f : G→ Z2 be an extension of Z4 by Z2. Suppose that the inducedaction αf : Z2 → Z×

4 carries the generator of Z2 to −1. Then G is isomorphic to eitherD8 or Q8.

Proof Let Z4 = 〈b〉 and Z2 = 〈a〉. For any x ∈ G with f(x) = a, we have xbkx−1 =αf (a)(bk) = b−k for all k. Also, f(x2) = a2 = e, so x2 ∈ ker f = 〈b〉. Thus, x2 = bk forsome k.

Note that x fails to commute with b and b−1, so neither of these can equal x2. Thus,either x2 = e or x2 = b2.

If x2 = e, then x has order 2, and hence there is a section of f , specified by s(a) = x.In this case, G ∼= Z4 �αf

Z2, which, for this value of αf , is isomorphic to the dihedralgroup D8 by the first example of Examples 4.6.6.

The remaining case gives x2 = b2. But then there is a homomorphism h : Q8 → Gwith h(a) = x and h(b) = b by Problem 5 of Exercises 2.9.7. As the image of h con-tains more than half the elements of G, h must be onto, and hence an isomorphism.(Alternatively, we could use the Five Lemma for Extensions to show that h is an isomor-phism.)

Q8, of course, is not a split extension of 〈b〉 by Z2, for two reasons. First, we knowthat every element of Q8 which is not in 〈b〉 has order 4, and hence it is impossible todefine a section from Q8/〈b〉 to Q8. Second, if it were a split extension, then it would beisomorphic to D8, which we know is not so.

We have classified the extensions of Z4 by Z2 corresponding to the homomorphismfrom Z2 to Z×

4 that takes the generator of Z2 to −1. Since Z×4 is isomorphic to Z2, there

is only one other homomorphism from Z2 to Z×4 : the trivial homomorphism. Let us

begin with some general observations about extensions whose induced action is trivial.

Lemma 4.7.12. Let f : G→ K be an extension of H by K, where H is abelian. Thenthe induced homomorphism αf : K → Aut(H) is the trivial homomorphism if and onlyif H ⊂ Z(G).


Proof Suppose that αf is the trivial homomorphism. Then for any x ∈ G, let y = f(x).By the definition of αf , we have xhx−1 = αf (y)(h) for all h ∈ H. But αf is the trivialhomomorphism, so that αf (y) is the identity map of H. Thus xhx−1 = h for all h ∈ H,and hence each h ∈ H commutes with every element of G. Thus, H ⊂ Z(G).

But the converse is immediate: if H ⊂ Z(G), then conjugating H by any element ofG gives the trivial homomorphism, and hence αf (y) is the identity element of Aut(H)for every y ∈ K.

The extensions of H containing H in their centers are important enough to merit aname.

Definition 4.7.13. An extension f : G → K of H by K is a central extension if H ⊂Z(G).

Example 4.7.14. As one may easily see, the center of D8 is 〈b2〉. Moreover, D8/〈b2〉 isisomorphic to Z2 × Z2, with generators a and b. Thus, D8 is a central extension of Z2

by Z2 × Z2.

In particular, a central extension of a cyclic group by an abelian group is not neces-sarily abelian. But a central extension of an abelian group by a cyclic group is abelian:

Lemma 4.7.15. Let f : G → Zk be a central extension of the abelian group H by thecyclic group Zk. Then G is abelian.

Proof Write Zk = 〈a〉, and let x ∈ G with f(x) = a. Let g ∈ G. Then f(g) = ai = f(xi)for some i, and hence gx−i ∈ ker f = H. But then g = hxi for some h ∈ H.

If g′ is any other element of G, we may write g′ = h′xj for h′ ∈ H and j ∈ Z. Butthen gg′ = hxih′xj = hh′xi+j , as h′ ∈ Z(G). But multiplying out g′g gives the sameresult, and hence g and g′ commute.

We now return to the groups of order 8.

Lemma 4.7.16. Let f : G→ Z2 be a central extension of Z4 by Z2. Then G is isomor-phic to either Z8 or Z4 × Z2.

Proof Since Z2 is cyclic, Lemma 4.7.15 shows that G is abelian. So we could justappeal to the Fundamental Theorem of Finite Abelian Groups, which tells us that Z8,Z4 ×Z2, and Z2 ×Z2 ×Z2 are the only abelian groups of order 8. Clearly, the first twoare the only ones that are extensions of Z4.

Alternatively, we can argue directly. Write Z4 = 〈b〉 and Z2 = 〈a〉 and let f(x) = a.Then f(x2) = e, and hence x2 ∈ 〈b〉. Now x has even order, so that o(x) = 2o(x2). Thus,if x2 = b or b−1, then x has order 8, and hence G is cyclic.

If x has order 2, then s(a) = x gives a splitting of f , and hence G is the semidirectproduct obtained from the action of Z2 on Z4. But in this case, the action is trivial, sothat the semidirect product thus obtained is just the direct product.

If x has order 4, then x2 = b2. But then b−1x has order 2, and setting s(a) = b−1xgives a section of f , and hence G ∼= Z4 × Z2 as above.

Once again, we have a nonsplit extension: the unique nontrivial homomorphismf : Z8 → Z2 is a nonsplit extension of Z4 by Z2.

We now classify the groups of order 8.

Proposition 4.7.17. Every group of order 8 is isomorphic to exactly one of the followinggroups: Z8, D8, Q8, Z4 × Z2, and Z2 × Z2 × Z2.


Proof If G has an element of order 4, say o(b) = 4, then 〈b〉 has index two in G, andhence is normal in G by Proposition 4.1.12 (or by the simpler argument of Problem 12of Exercises 3.7.14). But then G is an extension of Z4 by Z2. As Z×

4 = {±1}, there areonly two different homomorphisms from Z2 to Z×

4 , the homomorphism that carries thegenerator of Z2 to −1 and the trivial homomorphism. Thus, if G is an extension of Z4

by Z2, then G is isomorphic to one of Z8, D8, Q8, and Z4 × Z2 by Lemmas 4.7.11 and4.7.16.

If G does not have an element of order 4, then every element of G has exponent 2.But then G is abelian by Problem 3 of Exercises 2.1.16. But the fundamental theoremof finite abelian groups shows Z2 × Z2 × Z2 to be the only abelian group of order 8 andexponent 2.

The reader may easily check that no two of the listed groups are isomorphic.

We shall classify all groups of order p3 for odd primes p as well, but this will requiretheoretical results to be given in the next chapter. We now return to the general theoryof extensions.

Lemma 4.7.18. Let f : G→ K and f ′ : G′ → K be extensions of H by K, where H isabelian. Suppose given a commutative diagram

H G K

H G′ K,

��⊂

��

g1

��f

��

g2

��

g3

��⊂ ��f ′

where g1, g2, and g3 are isomorphisms. Then the actions αf , αf ′ : K → Aut(H) inducedby f and f ′ are related by g1 ◦αf (k)◦g−1

1 = αf ′(g3(k)), so that αf is conjugate to αf ′ ◦g3as a homomorphism from K to Aut(H).

Proof For k ∈ K, let x ∈ G with f(x) = k. Then f ′(g2(x)) = g3(k), so αf ′(g3(k))(g1(h)) =g2(x)g1(h)g2(x)−1 = g2(xhx−1) = g1(αf (k)(h)) for all h ∈ H. Thus, αf ′(g3(k)) ◦ g1 =g1 ◦ αf (k), and the result follows.

Remarks 4.7.19. There is a general method that may be used to classify the extensionsof an abelian group H by a group K up to the following equivalence relation: Theextensions f : G→ K and f ′ : G→ K are equivalent if there is a commutative diagram

H G K

H G′ K.

��⊂ ��f

��

g

��⊂ ��f ′

Here, g is forced to be an isomorphism, by the Five Lemma for Extensions.Notice that Lemma 4.7.18 implies that if f and f ′ are equivalent, then the homomor-

phisms αf and αf ′ are equal. Thus, the method under discussion will give a separateclassification for each different homomorphism α : K → Aut(H). Thus, fixing such anα, we shall write Hα to denote H with this particular action of K.

The method of classification is given using the cohomology of groups, a subtopicof homological algebra. The classification theorem gives a one-to-one correspondence


between the equivalence classes of extensions corresponding to the homomorphism αand the elements of the cohomology group H2(K;Hα).5

Nevertheless, a couple of deficiencies in this approach should be pointed out. First,given an element in H2(K;Hα), it is quite tedious to construct the associated extension.And it comes in a form that is difficult to analyze.

Second, the classification is too rigid for many of the questions of interest: there areextensions whose associated actions are both α which are not equivalent in the abovesense, but are equivalent under weaker hypotheses. Thus, there may be fewer isomor-phism classes of groups G that may be written as extensions of H by K with action αthan there are elements of the cohomology group H2(K;Hα).

Nevertheless, the homological approach can be useful.

It is sometimes useful to be able to go back and forth between the existence of acommutative diagram such as that in Lemma 4.7.18 and the existence of a suitableisomorphism from G to G′.

Lemma 4.7.20. Let f : G → K and f ′ : G′ → K be extensions of H by K, and writeι : H → G and ι′ : H → G′ for the inclusions. Then the following conditions areequivalent.

1. There is a commutative diagram

H G K

H G′ K,

��⊂ι

��

g1

��f

��

g2

��

g3

��⊂ι′

��f ′

where g1, g2, and g3 are isomorphisms.

2. There is an isomorphism g : G→ G′ with g(im ι) = im ι′.

Proof If the first condition holds, then the second one does also, as we may take g = g2.Thus, suppose that the second condition holds, and we are given an isomorphism

g : G→ G′ with g(im ι) = im ι′. Then g induces an isomorphism from im ι to im ι′. Sinceι and ι′ are embeddings, there is an isomorphism g1 : H → H such that ι′ ◦ g1 = g ◦ ι.Thus, if we set g2 = g, then the left square of the displayed diagram commutes.

We must now construct g3, which comes from a couple of applications of the FirstNoether Theorem. First, we have surjective maps f : G → K and f ′ : G′ → K, withkernels given by ker f = H and ker f ′ = H. Thus, the First Noether Theorem givesisomorphisms f : G/H

∼=−→ K and f′

: G′/H∼=−→ K such that the following diagrams

commute.

Gf ��

π ��

��

� K

G/Hf

��

G′ f ′��

π ��

��

� K

G′/Hf′

��

5See, for instance, A Course in Homological Algebra, by P.J. Hilton and U. Stammbach, Springer-Verlag, 1971, for an exposition of this classification.


Since g(im ι) = im ι′, the kernel of the composite Gg−→ G′ π−→ G′/H is precisely H, so

the First Noether Theorem provides an isomorphism k : G/H∼=−→ G′/H that makes the

following diagram commute.

H G G/H K

H G′ G′/H K��

g1 ∼=

��⊂ι

��

g ∼=

��π

��

k ∼=

��f

∼=

��⊂ι′

��π ��f′

∼=

Now set g3 = f′ ◦ k ◦ f−1

.

Note that the group H of Lemma 4.7.20 need not be abelian.We now wish to explore the question of when two different homomorphisms, α, α′ :

K → Aut(H) produce isomorphic semidirect products. In this regard, the above lemmaspermit an improvement of the result in Proposition 4.6.11 in the case that H is abelian.

Proposition 4.7.21. Let H be an abelian group and let α and α′ be homomorphisms ofK into Aut(H). Then α is conjugate to a composite α′ ◦ g with g ∈ Aut(K) if and onlyif there is an isomorphism ϕ : H �α K → H �α′ K such that ϕ(im ι) = im ι′, where ιand ι′ are the canonical inclusions of H in H �α K and H �α′ K, respectively.

Proof If α is conjugate to a composite α′ ◦ g with g ∈ Aut(K), then Proposition 4.6.11provides an isomorphism ϕ : H �α K → H �α′ K with ϕ(im ι) = im ι′. Thus, it sufficesto show the converse.

So suppose given an isomorphism ϕ : H�αK → H�α′ K with ϕ(im ι) = im ι′. ThenLemma 4.7.20 provides a commutative diagram of the sort appearing in the hypothesisof Lemma 4.7.18.

By Lemma 4.7.9, the action map απ : K → Aut(H) induced by π : H �α K → K isprecisely α. A similar result holds for α′. Thus, the output of Lemma 4.7.18 is preciselythat α is conjugate to α′ ◦ g for some g ∈ Aut(K).

We would like to eliminate the hypothesis that ϕ(im ι) = im ι′ from the above, andobtain a necessary and sufficient condition for the semidirect products H �α K andH �α′ K to be isomorphic. We can do this if H and K are finite groups whose ordersare relatively prime.

Corollary 4.7.22. Let H and K be finite groups whose orders are relatively prime, andsuppose that H is abelian. Let α and α′ be two homomorphisms from K to Aut(H).Then the semidirect products H �α K and H �α′ K are isomorphic if and only if α isconjugate to α′ ◦ g for some automorphism g of K.

Proof It suffices to show that any homomorphism from H�αK to H�α′K carries ι(H)into ι′(H), and that the analogous statement holds for homomorphisms in the oppositedirection. But this will follow if we show that in either one of the semidirect products,the only elements of exponent |H| are those in H itself.

If |H| is an exponent for x, then it is also an exponent for f(x) for any homomorphismf . Since any element whose order divides both |H| and |K| must be the identity, we seethat any element of either semidirect product which has exponent |H| must be carriedto the identity element by the projection map (π or π′) into K. Since H is the kernel ofthe projection map, the result follows.


We now give a generalization of Corollary 4.7.22.

Corollary 4.7.23. Let H be an abelian group and let α, α′ : K → Aut(H). Supposethat α has the property that any embedding f : H → H �α K must have image equal toι(H). Then H �αK is isomorphic to H �α′ K if and only if α is conjugate to α′ ◦ g forsome g ∈ Aut(K).

Proof Again, it suffices to show that ϕ(im ι) = im ι′, or, equivalently, that im ι =im(ϕ−1 ◦ ι′), for any isomorphism ϕ : H �α K → H �α′ K. But since ϕ−1 ◦ ι′ is anembedding, this follows from the hypothesis.

Corollary 4.7.23 may be applied, for instance, if ι(H) is the only subgroup of H�αKwhich is isomorphic to H. An example where this occurs is for H = 〈b〉 ⊂ D2n for n ≥ 3.Here, we make use of the fact that every element of D2n outside of 〈b〉 has order 2. Sinceb has order n > 2, every embedding of 〈b〉 in D2n must have image in 〈b〉. Since 〈b〉 isfinite, this forces the image to be equal to 〈b〉.

Finally, let us consider the extensions of a nonabelian group H. As shown in Exam-ple 4.7.8, if f : G→ K is even a split extension of H by K, different choices of splittingsmay define different homomorphisms from K into Aut(H). In particular, we do not havea homomorphism from K to Aut(H) which is associated to the extension itself. But wecan define something a little weaker.

As usual, we write ch : H → H for the automorphism induced by conjugating byh ∈ H: ch(h′) = hh′h−1 for all h′ ∈ H. Recall that an automorphism f ∈ Aut(H) isan inner automorphism if f = ch for some h ∈ H. The inner automorphisms form asubgroup, Inn(H) ⊂ Aut(H).

Lemma 4.7.24. The group of inner automorphisms of a group G is a normal subgroupof Aut(G).

Proof Let x ∈ G and write cx for conjugation by x. Let f ∈ Aut(G). We wish to showthat f ◦ cx ◦ f−1 is an inner automorphism. But

f ◦ cx ◦ f−1(g) = f(xf−1(g)x−1)= f(x)f(f−1(g))f(x)−1,

and hence f ◦ cx ◦ f−1 = cf(x).


Definition 4.7.25. We write Out(G) for the quotient group Aut(G)/Inn(G) and call itthe outer automorphism group of G.

Proposition 4.7.26. Let f : G → K be an extension of H by K. Then f determinesa homomorphism α : K → Out(H). Here, for k ∈ K, α(k) is represented by theautomorphism of H given by conjugation by g, for any g ∈ G with f(g) = k.

In particular, if f is a split extension and s is a splitting of f , then α = π ◦αs, whereαs : K → Aut(H) is the homomorphism induced by s and π is the canonical map fromAut(H) to Out(H) = Aut(H)/Inn(H). Thus, if s and s′ are two different sections of f ,then the homomorphisms αs and αs′ agree in Out(H).


Proof If f(g) = f(g′) = k, then g′ = gh for some h ∈ H. But then conjugating by g′

is the same as conjugating first by h and then by g. But conjugation by h is an innerautomorphism of H, and hence represents the identity in Out(H). Thus, conjugation byg′ represents the same element of Out(H) as conjugation by g, and hence the prescriptionabove gives a well-defined function α : K → Out(H).

It suffices to show that α is a homomorphism. But for k, k′ ∈ K, if f(g) = k andf(g′) = k′, then f(gg′) = kk′. But conjugation by gg′ is the composite of the conjugationsby g and by g′.

We should also note that the above is not always very useful. For instance, for n �= 6,Out(Sn) is the trivial group.

Exercises 4.7.27.† 1. Problem 1 of Exercises 4.2.18 provides a normal subgroup H ⊂ A4 of order 4.

Show that H ∼= Z2 × Z2 and that A4 is a split extension of H by Z3.

2. Show that any nonabelian group G of order 12 which contains a normal subgroup oforder 4 must be isomorphic to A4. (Hint : Find a G-set with 4 elements and analyzethe induced homomorphism from G to S4. Alternatively, show that G is a splitextension of a group of order 4 by Z3, and, using the calculation of Aut(Z2 × Z2)given in Problem 14 of Exercises 4.6.13, show that there is only one such splitextension which is nonabelian.)

† 3. Consider the subgroup H ⊂ An of Problem 1. Show that H is normal in S4 andthat S4 is isomorphic to the semidirect product (Z2 × Z2) �α S3, where α : S3 →Aut(Z2 × Z2) is an isomorphism. (Recall the calculation of Aut(Z2 × Z2) given inProblem 14 of Exercises 4.6.13. In the situation at hand, you may enjoy giving anew proof of this calculation, based on showing that the map α above is injective.)

4. Show that if k is odd, then Q4k is isomorphic to Zk �α Z4 for some α : Z4 →Aut(Zk). Calculate α explicitly.

5. Find all the central extensions of Z2 by Z2 × Z2.

6. Let f : G → Z2 be an extension of Z8 by Z2 and suppose that the inducedhomomorphism αf : Z2 → Z×

8 carries the generator of Z2 to 5. Show that fis a split extension, and hence that G is isomorphic to the group Z8 �β Z2 ofExample 3 in Examples 4.6.6. (Hint : Write Z8 = 〈b〉 and Z2 = 〈a〉 and let x ∈ Gwith f(x) = a. Note that x2 = bk for some k, and use this to calculate the orderof each of the elements bix.)


8 carries the generator of Z2 to 3. Show that f is asplit extension, and hence that G is isomorphic to the group Z8 �αZ2 of Example 2in Examples 4.6.6. (Hint : Argue as in the preceding problem. Here, if x2 = bk,note that the fact that x must commute with x2 places restrictions on which valuesof k can occur.)


8 carries the generator of Z2 to −1. Show that f isisomorphic to either D16 or Q16.

9. Classify those groups of order 16 that contain an element of order 8.


10. Let H be any group and let f : G → Z be an extension of H by Z. Show that fsplits.

11. We show that the inner automorphisms of Sn induced by different sections of thesign homomorphism are not always conjugate in Aut(Sn).

(a) Let f ∈ Aut(G) and let Gf = {g ∈ G | f(g) = g}, the fixed points of G underf . Show that Gf is a subgroup of G. Show also that if f and f ′ are conjugatein Aut(G), then Gf and Gf

′are isomorphic as groups.

(b) Let τ ∈ Sn be a 2-cycle and write Anτ for the fixed points of An under the

automorphism obtained from conjugating by τ . Show that Anτ is isomorphic

to Sn−2.

(c) Let H ⊂ S6 be the subgroup consisting of those elements of S6 that commutewith (1 2)(3 4)(5 6). Show that H is a split extension of Z2 × Z2 × Z2 by S3.

(d) Let τ ′ ∈ Sn be a product of three disjoint 2-cycles. Show that the subgroupAn

τ ′fixed by τ ′, is isomorphic to an index 2 subgroup of H × Sn−6, where

H ⊂ S6 is the group of the preceding part.

Deduce that if n > 6, then the automorphism of An obtained from conjugationby a 2-cycle cannot be conjugate in Aut(An) to the automorphism obtained fromconjugation by the product of three disjoint 2-cycles.

‡ 12. Let α : K → Aut(H) and β : K → Aut(H ′) be homomorphisms. Let f : H → H ′

be a group homomorphism such that f(α(k)(h)) = β(k)(f(h)) for all k ∈ K andh ∈ H (i.e., f is a K-map between the K-sets H and H ′). Define f∗ : H �α K →H ′

�β K by f∗(hk) = f(h)k. Show that f∗ is a homomorphism.

13. Let α : K → Aut(H) be a homomorphism. Suppose given f ∈ Aut(H) thatcommutes with every element of the image of α. Show that the homomorphismf∗ : H �α K → H �α K of Problem 12 is defined and gives an automorphism ofH �α K.

‡ 14. In this problem, we calculate the automorphisms of a dihedral group. The case ofD4 is exceptional, and indeed has been treated in Problem 14 in Exercises 4.6.13.Thus, we shall consider D2n for n ≥ 3.

A key fact is given by either Problem 6 of Exercises 2.8.5 or by Proposition 4.6.10:The homomorphisms from D2n to a group G are in one-to-one correspondence withthe choices of x = f(a) and y = f(b) such that x has exponent 2, y has exponentn, and xyx−1 = y−1.

The reader should first recall from Problem 4 of Exercises 3.7.14 that since n ≥ 3,the subgroup 〈b〉 ⊂ D2n is characteristic. Thus, Lemma 3.7.8 provides a restrictionhomomorphism ρ : Aut(D2n)→ Aut(〈b〉), given by setting ρ(f) equal to its restric-tion f : 〈b〉 → 〈b〉. As usual, by abuse of notation, we shall identify Aut(〈b〉) withZ×n , and write ρ : Aut(D2n)→ Z×

n .

(a) Show that any homomorphism f : D2n → D2n whose restriction to 〈b〉 givesan automorphism of 〈b〉 must have been an automorphism of D2n in the firstplace.

(b) Show that the restriction homomorphism ρ : Aut(D2n) → Z×n is a split sur-

jection (i.e., ρ : Aut(D2n)→ Z×n is a split extension of ker ρ by Z×

n ).


(c) Show that an automorphism f is in the kernel of ρ if and only if f(b) = b andf(a) = bia for some i. Deduce that the kernel of ρ is isomorphic to Zn.

(d) Show that the action of Z×n on Zn = ker ρ induced by the split extension is

the same as the usual action of Z×n = Aut(Zn). Deduce that Aut(D2n) is

isomorphic to the semidirect product Zn�id Aut(Zn), where id is the identitymap of Aut(Zn).

15. Show that every automorphism of S3 is inner.

16. Recall from Problem 8 of Exercises 3.7.14 that if n ≥ 3, the subgroup 〈b〉 ⊂ Q4n

is characteristic. Using this as a starting point, give a calculation of Aut(Q4n),n ≥ 3, by methods analogous to those of Problem 14 above. Note that the lack ofa semidirect product structure on Q4n necessitates greater care in the constructionof a splitting for ρ than was required in the dihedral case. (We shall treat theexceptional case of Aut(Q8) in Problem 9 of Exercises 5.4.9.)

‡ 17. Recall from Problem 3 of Exercises 4.2.18 that An is a characteristic subgroupof Sn for n ≥ 3. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ :Aut(Sn)→ Aut(An).

Let Γ : Sn → Aut(An) be the homomorphism induced by conjugation. By Prob-lem 10 in Exercises 3.7.14, Γ is injective for n ≥ 4. Use this fact, together withProposition 4.6.9, to show that ρ is injective for n ≥ 4.

18. We compute the automorphism groups of A4 and S4.

(a) Let H ⊂ A4 be the subgroup of order 4. Show that H is characteristic inA4. Thus, Lemma 3.7.8 provides a restriction homomorphism ρ : Aut(A4)→Aut(H).

(b) Show that the kernel of ρ is isomorphic to H, and hence to Z2 × Z2.

(c) Show that ρ is a split surjection. Deduce that the homomorphism Γ : S4 →Aut(A4) induced by conjugation is an isomorphism.

(d) Deduce from Problem 17 that the homomorphism Γ : S4 → Aut(S4) inducedby conjugation is an isomorphism.

‡ 19. We consider the automorphism groups of the semidirect products G = Zn �α Zkfor selected homomorphisms α : Zk → Aut(Zn). As usual, we write ι : Zn → Gand s : Zk → G for the standard inclusions and write π : G→ Zk for the standardprojection.

The one assumption we shall insist on is that ι(Zn) be a characteristic subgroup ofG. As the reader may easily verify, this is always the case if n and k are relativelyprime. But we have seen other examples, e.g., D2n = Zn �α Z2 or Example 2 ofExamples 4.6.6, where Zn is characteristic in G, and our arguments will apply tothese cases as well. In particular, the material here generalizes that in Problem 14.

As usual, we use multiplicative notation, with Zn = 〈b〉 and Zk = 〈a〉. As discussedin Section 4.6, we have α(aj)(bi) = bm

ji for an integer m with the property thatmk ≡ 1 mod n. The elements of G may be written uniquely as biaj with 0 ≤ i < nand 0 ≤ j < k, and biajbi

′aj

′= bi+m

ji′aj+j′.

Since 〈b〉 is assumed characteristic in G, we have a restriction homomorphism ρ :Aut(G)→ Aut(〈b〉) ∼= Z×

n .


(a) Let f ∈ Aut(G). Show that there is a unique element f ∈ Aut(〈a〉) such thatthe following diagram commutes:

G G

〈a〉〈a〉��

π

��f

��

π

��f

Show that the passage from f to f gives a homomorphism

η : Aut(G)→ Aut(〈a〉) ∼= Z×k .

(b) Show, via Corollary 4.6.10, that the image of η consists of those f ∈ Aut(Zk)such that the following diagram commutes:

Zk Z×n

Zk Z×n .

��

f

��α

��α

Show that such automorphisms f ∈ Aut(Zk) correspond to elements j ∈ Z×k

such that mj ≡ m mod n. Deduce that if α is injective (i.e., if m has order kin Z×

n ), then η is the trivial homomorphism.

(c) Consider the map (ρ, η) : Aut(G)→ Aut(〈b〉)×Aut(〈a〉) given by (ρ, η)(f) =(ρ(f), η(f)). Show that (ρ, η) induces a split surjection of Aut(G) onto Aut(〈b〉)×(im η).

(d) Show that an automorphism f is in the kernel of (ρ, η) if and only if f(b) = band f(a) = bia, where i is chosen so that bia has order k in G. Show, viaProblems 7 and 8 of Exercises 4.6.13, that bia has order k if and only ifi · (mk − 1)/(m− 1) is divisible by n.Show also that passage from f to bi induces an isomorphism from the kernelof (ρ, η) onto a subgroup H ⊂ 〈b〉.

(e) Let β : Aut(Zn) × (im η) → Aut(H) be the action induced by the exten-sion (ρ, η). Show that β restricts on Aut(Zn) to the restriction map r :Aut(Zn) → Aut(H) induced by the fact that H is characteristic in Zn. (Re-call from Problem 9 of Exercises 3.7.14 that every subgroup of a cyclic groupis characteristic.)Deduce that if α is injective, then Aut(G) ∼= H �r Aut(Zn).

(f) Let β : Aut(Zn)× (im η)→ Aut(H) be as above. Compute the restriction ofβ to im η.

(g) Show that if n is prime, then H = Zn.

(h) Give an example of α, n, and k where H is not all of Zn.

Chapter 5

Sylow Theory, Solvability, andClassification

The main theorems of this chapter, e.g., Cauchy’s Theorem, the Sylow Theorems, andthe characterization of solvable and nilpotent groups, could have been given much earlierin the book. In fact, a proof of Cauchy’s Theorem has appeared as Problem 24 inExercises 3.3.23. Indeed, these theorems depend on very little other than the theory ofG-sets (which it might be advisable to review at this time) and the Noether IsomorphismTheorems.

Our main reason for delaying the presentation of these theorems until now was sothat we could develop enough theory to support their applications. Given our analysisof abelian groups, extensions, and some automorphism groups, we will be able to use theresults in this chapter to classify the groups of most orders less than 64, and, hopefully,to bring the study of finite groups into a useful perspective.

Lagrange’s Theorem tells us that if the group G is finite, then the order of everysubgroup must divide the order of G. But there may be divisors of |G| that do notoccur in this manner. For instance, Problem 2 of Exercises 4.2.18 shows that A4 has nosubgroup of order 6.

In this chapter, we shall give some existence theorems for subgroups of particularorders in a given finite group. The first and most basic such result is Cauchy’s Theorem,which says that if G is a finite group whose order is divisible by a prime p, then Ghas an element (hence a subgroup) of order p. We give this in Section 5.1, along withapplications to classification.

In Section 5.2, we study finite p-groups, which, because of Cauchy’s Theorem, areprecisely the finite groups whose order is a power of the prime p. By a proof similar tothe one given for Cauchy’s Theorem, we show that any p-group has a nontrivial center.We give some applications, including most of the pieces of a classification of the groupsof order p3 for p odd. The last step will be given via the Sylow Theorems in Section 5.3.

Section 5.3 is devoted to the Sylow Theorems, which study the p-subgroups of a finitegroup. Extending the result of Cauchy’s Theorem, the Sylow Theorems show that if pis prime and if |G| = pjm with (p,m) = 1, then G must have a subgroup of order pj .Such a subgroup is the highest order p-group that could occur as a subgroup of G, andis known as a p-Sylow subgroup of G (though that is not the initial definition we shallgive for a p-Sylow subgroup).

134

CHAPTER 5. SYLOW THEORY, SOLVABILITY, AND CLASSIFICATION 135

The Sylow Theorems are actually a good bit stronger than a simple existence prooffor subgroups whose order is the highest power of p dividing the order of G. They alsoshow that any two p-Sylow subgroups of G are conjugate and that the number of p-Sylowsubgroups of G is congruent to 1 mod p. This leads to some useful theoretical resultsregarding p-Sylow subgroups which are normal, as well as to some valuable countingarguments for the classification of groups of particular orders. In particular, Sylowtheory can be useful for finding normal subgroups in a particular group.

The Sylow Theorems give a useful point of view for considering finite groups. Finitep-groups are more tractable than general finite groups, and a useful first step in under-standing a given group is to understand its Sylow subgroups and their normalizers. Formany low order groups, an understanding of the entire structure of the group will followfrom this analysis.

Given the Sylow Theorems and the material on semidirect products, extensions, andthe automorphisms of cyclic groups, we are able to classify the groups of many of theorders less than 64. We give numerous classifications in the text and exercises for Sec-tion 5.3.

Section 5.4 studies the commutator subgroup and determines the largest abelian fac-tor group that a group can have. Then, in Section 5.5, and using the commutatorsubgroups, we study solvable groups. Solvable groups are a generalization of abeliangroups. In essence, in the finite case, they are the groups with enough normal subgroupsto permit good inductions on order. One characterization we shall give of the finite solv-able groups is that they are the finite groups whose composition series have no nonabeliansimple groups in the list of successive simple subquotients.

As expected, the induction arguments available for solvable groups provide for strongerresults than we get for arbitrary groups. In Section 5.6, we prove Hall’s Theorem, whichshows that if G is a finite solvable group and if |G| = mn with (m,n) = 1, then G has asubgroup of order m. This fact turns out to characterize the solvable groups, though wewon’t give the converse here.

The nilpotent groups form a class of groups that sits between the abelian groupsand the solvable groups. They are defined by postulating properties that always hold inp-groups, as we showed in Section 5.2. By the end of Section 5.7, we show that the finitenilpotent groups are nothing other than direct products of finite p-groups over varyingprimes p. In the process we will learn some more about the p-groups themselves. Weshall also see that the full converse of Lagrange’s Theorem holds for nilpotent groups: Anilpotent group G has subgroups of every order that divides the order of G.

We close the chapter with a discussion of matrix groups over commutative rings.Here, we assume an understanding of some of the material in Chapters 7 and 10. Thereader who has not yet studied this material should defer reading Section 5.8 until havingdone so.

One of the reasons for studying matrix groups is that the automorphism group of adirect product, Zkn, of k copies of Zn is isomorphic to Glk(Zn), so by studying the latter,we can obtain classification results for groups containing normal subgroups isomorphicto Zkn. This situation arises for some low order groups.

We also calculate the order of Gln(Fq), where Fq is the finite field with q = pr

elements with p prime. We describe some important subquotient groups, PSln(Fq) ofGln(Fq). Most of these projective special linear groups turn out to be simple, but weshall not show it here.


5.1 Cauchy’s Theorem

We now prove Cauchy’s Theorem. An alternative proof was given as Problem 24 ofExercises 3.3.23.

Theorem 5.1.1. (Cauchy’s Theorem) Let G be a finite group and let p be a primedividing the order of G. Then G has an element of order p.

Proof Note that we’ve already verified the theorem in the case that G is abelian:Cauchy’s Theorem for abelian groups was given as Proposition 4.4.12. We argue byinduction on the order of G. The induction starts with |G| = p, in which case G ∼= Zp,and the result is known to be true.

Recall that the nontrivial conjugacy classes in G are those with more than one elementand that the conjugacy class of x is nontrivial if and only if x �∈ Z(G). Recall also that aset of class representatives for the nontrivial conjugacy classes in G is a set X ⊂ G suchthat

1. X ∩ Z(G) = ∅.2. If y ∈ G is not in Z(G), then y is conjugate to some x ∈ X.

3. If x, x′ ∈ X with x �= x′, then x and x′ are not conjugate in G.

Then the class formula (Corollary 3.6.18), shows that if X ⊂ G is a set of classrepresentatives for the nontrivial conjugacy classes in G, then

|G| = |Z(G)|+∑x∈X

[G : CG(x)],

where CG(x) is the centralizer of x in G.As stated, we already know Cauchy’s Theorem to be true for abelian groups. Thus,

if p divides the order of Z(G), then Z(G) has an element of order p, and hence G doesalso.

Thus, we may assume that p does not divide the order of Z(G). Since p does dividethe order of G, the class formula tells us that there must be some x �∈ Z(G) such thatp does not divide [G : CG(x)]. Since |G| = [G : CG(x)] · |CG(x)|, this says that p mustdivide the order of CG(x).

Since x �∈ Z(G), [G : CG(x)], which is equal to the number of conjugates of x in G,cannot equal 1. Thus, |CG(x)| < |G|, so the induction hypothesis provides an element oforder p in CG(x), and hence also in G.

As a sample application, we classify the groups of order 6.

Corollary 5.1.2. Let G be a group of order 6. Then G is isomorphic to either Z6 orD6.

Proof By Cauchy’s Theorem, G has an element, b, of order 3. But then 〈b〉 has index 2in G, and hence 〈b〉 � G by Proposition 4.1.12 (or by the simpler argument of Problem 12of Exercises 3.7.14). Let π : G→ G/〈b〉 be the canonical map.

A second application of Cauchy’s Theorem given an element a ∈ G of order 2. Butthen a �∈ 〈b〉, and hence π(a) = a is not the identity element of G/〈b〉. Thus, a generatesG/〈b〉. Since a and a both have order 2, setting s(a) = a gives a section of π.


Thus, G is a split extension of Z3 by Z2. Since Aut(Z3) ∼= Z×3 = {±1}, there are

exactly two homomorphisms from Z2 to Aut(Z3): the trivial homomorphism and thehomomorphism that sends the generator of Z2 to −1.

The semidirect product given by the trivial homomorphism is Z3 × Z2, which is Z6,because 2 and 3 are relatively prime (e.g., by Problem 4 of Exercises 2.10.12). Thesemidirect product induced by the nontrivial homomorphism from Z2 to Aut(Z3) wasshown to be isomorphic to D6 in the first example of Examples 4.6.6.

Exercises 5.1.3.1. Show that any group of order 10 is isomorphic to either Z10 or D10.

2. Assuming that Z×p is cyclic for any odd prime p (as shown in Corollary 7.3.15),

show that any group of order 2p is isomorphic to either Z2p or D2p.

3. Recall that Cauchy’s Theorem for abelian groups was available prior to this chapter.Give a proof of the preceding problem that doesn’t use Cauchy’s Theorem fornonabelian groups. (Hint : If G is a group of order 2p, then all of its elements musthave order 1, 2, p, or 2p. If all nonidentity elements have order 2, then G must beabelian, and hence Cauchy’s Theorem for abelian groups gives a contradiction. SoG must have an element, b, of order p. Now study the canonical map G→ G/〈b〉,and show that G must have an element of order 2.)

5.2 p-Groups

Recall that a p-group is a group in which the order of every element is a power of p.Lagrange’s Theorem shows that any group whose order is a power of p must be a p-group. But by Cauchy’s Theorem, if |G| has a prime divisor q �= p, then it must have anelement of order q, and hence it is not a p-group. We obtain the following corollary.

Corollary 5.2.1. A finite group is a p-group if and only if its order is a power of p.

The proof used above for Cauchy’s Theorem may be used, with suitable modifications,to prove some other useful facts. For instance:

Proposition 5.2.2. A finite p-group has a nontrivial center (i.e., Z(G) �= e).

Proof Let X be a set of class representatives for the nontrivial conjugacy classes in Gand consider the class formula:

|G| = |Z(G)|+∑x∈X

[G : CG(x)].

If x �∈ Z(G), then [G : CG(x)], which is equal to the number of conjugates of x in G,cannot equal 1. Thus, since [G : CG(x)] divides |G|, we must have [G : CG(g)] = pr forsome r > 0, and hence p divides [G : CG(g)].

Thus, p must divide every term in the class formula other than |Z(G)|, so it mustdivide |Z(G)| as well. In particular, |Z(G)| �= 1, and hence Z(G) �= e.

We first give an immediate consequence of Proposition 5.2.2 to classification.

Corollary 5.2.3. Every group of order p2 is abelian.


Proof We argue by contradiction. If G is not abelian, then Z(G) �= G. But then if|G| = p2, we must have |Z(G)| = p. But conjugation acts trivially on Z(G), so thatZ(G) � G. Thus, G/Z(G) has order p, and hence is cyclic. But now G must be abelianby Lemma 4.7.15, which gives the desired contradiction.

We now consider some more general corollaries to Proposition 5.2.2.

Corollary 5.2.4. The only finite p-group which is simple is Zp.

Proof Let G be a finite simple p-group. The center of any group is a normal subgroup.Since the center of G is nontrivial and G is simple, G must equal its center. Thus, Gis abelian. But the only abelian simple groups are the cyclic groups of prime order, byLemma 4.2.2.

Recall that a composition series for a group G is a sequence

e = H0 ⊂ H1 ⊂ · · · ⊂ Hj = G

such that Hi−1 � Hi and Hi/Hi−1 is a nontrivial simple group for 1 ≤ i ≤ j. ByProposition 4.2.5, every nontrivial finite group has a composition series.

Since a quotient of a subgroup of a p-group is again a p-group, the simple groupsHi/Hi−1 associated to the composition series for a nontrivial finite p-group must all beZp. But a direct proof of this fact will give us a stronger statement: We can construct acomposition series for a p-group G such that all the subgroups in the series are normalin G itself, and not just normal in the next group in the series.

Corollary 5.2.5. Let G be a finite group of order pj, with p prime and j > 0. Thenthere is a sequence

e = H0 ⊂ H1 ⊂ · · · ⊂ Hj = G,

with Hi � G and Hi/Hi−1∼= Zp for 1 ≤ i ≤ j.

Proof We argue by induction on |G|, with the result being trivial for |G| = p. ByProposition 5.2.2, Z(G) �= e, so Cauchy’s Theorem provides a subgroup H1 ⊂ Z(G) with|H1| = p. But any subgroup of Z(G) is normal in G, so we can form the factor groupG/H1.

Since |G/H1| < |G|, the induction hypothesis provides a sequence

e = K1 ⊂ · · · ⊂ Kj = G/H1,

such that Ki � G/H1 and Ki/Ki−1∼= Zp for 2 ≤ i ≤ j.

Let π : G→ G/H1 be the canonical map and set Hi = π−1(Ki) for 2 ≤ i ≤ j. Sincethe pre-image under a homomorphism of a normal subgroup is always normal, we haveHi � G for all i. Also, since Ki = Hi/H1 for i ≥ 1, the Third Noether Theorem showsthat Hi/Hi−1

∼= Ki/Ki−1∼= Zp for 2 ≤ i ≤ j, and the result follows.

Since the subgroup Hi above has order pi, we see that the full converse of Lagrange’sTheorem holds in p-groups.

Corollary 5.2.6. Let G be a finite p group. Then every divisor of |G| is the order of asubgroup of G.


We shall now begin to try to classify the groups of order p3 for primes p > 2. Weshall not yet succeed entirely, because not all the groups we must consider are p-groups.We must also understand the automorphism groups of the groups of order p2, which inthe case of Zp × Zp will require the Sylow theorems.

As in the case of p = 2, the abelian groups of order p3 are Zp3 , Zp2 × Zp, andZp × Zp × Zp. For the nonabelian groups of order p3, the whole story will depend onwhether there exists an element of order p2.

Proposition 5.2.7. Let G be a nonabelian group of order p3 where p is an odd prime.Suppose that G contains an element of order p2. Then G is isomorphic to the semidirectproduct Zp2 �α Zp, where α : Zp → Z×

p2 is induced by α(1) = p+ 1.1

Proof Let b ∈ G have order p2. Then 〈b〉 has index p in G, so 〈b〉 � G by Proposition4.1.12. Thus, G is an extension of Zp2 by Zp, via f : G → Zp. Since G is not abelian,Lemma 4.7.15 shows that the induced homomorphism α : Zp → Z×

p2 = Aut(Zp2) isnontrivial.

We shall use strongly the fact that Z×p2 is abelian. Because of this, the p-torsion

elements form a subgroup, which, according to Corollary 4.5.12, is K(p, 2), the subgroupconsisting of those units congruent to 1 mod p. Corollary 7.3.15 then shows that K(p, 2)is a cyclic group of order p, generated by 1 + p.

Since K(p, 2) is the p-torsion subgroup of Z×p2 , the image of α : Zp → Z×

p2 must lie inK(p, 2). Since α is nontrivial and Zp is simple, α must be an isomorphism from Zp toK(p, 2). But then p+ 1 = α(x) for some generator x of Zp.

But we may change our identification of Zp to make x the canonical generator of Zp.Thus, we may assume that α carries the canonical generator of Zp to 1 + p. Thus, itsuffices to show that f is a split extension, as then G ∼= Zp2 �α Zp, with α as claimed.

Let us continue to write x for our preferred generator of Zp, so that α(x) = p+ 1.Since α is the action induced by the extension f : G → Zp, Lemma 4.7.9 shows thatghg−1 = h1+p for any h ∈ ker f and any g with f(g) = x. To show that f splits, itsuffices to find g ∈ G of order p such that f(g) = x: The splitting is then defined bysetting s(x) = g.

Let a ∈ G with f(a) = x. Then a cannot have order p3, as G would then be cyclic.Alternatively, if a has order p, then a itself gives the desired splitting of f . Thus, weshall assume that a has order p2.

For simplicity of notation, let m = p + 1, so that aha−1 = hm for any h ∈ ker f .Moreover, since α(aj) = mj , ajha−j = hm

j

for h ∈ ker f . Thus, as shown in Problem 7of Exercises 4.6.13, an easy induction on k shows that

(ha)k = h1+m+···+mk−1ak

for k ≥ 1. Thus, taking k = p, Problem 8 of Exercises 4.6.13 gives

(ha)p = hmp−1m−1 ap.

Recall that m = p+ 1. Then m− 1 = p, while the proof of Proposition 4.5.18 showsthat mp is congruent to 1 + p2 mod p3. But then (mp − 1)/(m − 1) is congruent to pmod p2, so that (ha)p = hpap.

Now ap has order p and lies in 〈b〉 = ker f . Thus, ap = bjp for some j relatively primeto p. But then if we take h = b−j , then the element ha has order p, and gives the desiredsplitting for f .

1Thus, G is the group of Example 6 of Examples 4.6.6.


Thus, there is no analogue at odd primes of Q8: Every nonabelian extension of Zp2by Zp is split when p > 2.

Lemma 5.2.8. Let p be an odd prime and let G be a nonabelian group of order p3 thatdoes not contain an element of order p2. Then G is a split extension of Zp × Zp by Zp.

Proof By Corollary 5.2.5, G has a normal subgroup H of order p2. Since groups oforder p2 are abelian, and since G contains no element of order p2, H must be isomorphicto Zp × Zp. Since H has index p, G is an extension of H by Zp. But all elements of Ghave exponent p, so any lift of the generator of G/H to G provides a splitting for thisextension.

We have constructed an example of a nonabelian group of order p3 in which everyelement has exponent p, in Corollary 4.6.8. We would like to show that every nonabeliansplit extension of Zp ×Zp by Zp is isomorphic to this one. To show this we first need tounderstand the automorphism group of Zp × Zp. We will then need to understand theconjugacy classes of homomorphisms from Zp into this group of automorphisms. Butthe automorphism group in question is nonabelian, so that a complete picture of theconjugacy classes of its subgroups of order p will depend on the Sylow Theorems of Sec-tion 5.3. We will content ourselves here with a calculation of the order of Aut(Zp × Zp).Together with the Sylow theorems, this will be sufficient to complete our classification.

Given a basic knowledge of linear algebra, the following result becomes a little moretransparent, and may be extended to a calculation of the order of the automorphismgroup of a product of n copies of Zp.

Lemma 5.2.9. The automorphism group of Zp × Zp has order (p2 − 1)(p2 − p). Inparticular, this group has order jp, where (j, p) = 1.

Proof Write Zp × Zp = {aibj | 0 ≤ i, j ≤ p}. Of course, a and b commute and eachhas order p. Since every element of Zp × Zp has exponent p, the homomorphisms, f ,from Zp × Zp to itself are in one-to-one correspondence with all possible choices of theelements f(a) and f(b). Since there are p2 choices for f(a) and p2 choices for f(b), thisgives p2 · p2 = p4 different homomorphisms from Zp × Zp to itself. We’d like to knowwhich of these are isomorphisms.

Let f : Zp ×Zp → Zp ×Zp be a homomorphism. Then the image of f has order 1, por p2. It is an isomorphism if and only if the image has order p2. So one way to argueat this point is to count up the homomorphisms whose images have order 1 or p, andsubtract this number from p4.

Since Zp × Zp has exponent p, every e �= x ∈ Zp × Zp is contained in exactly onesubgroup of order p. That subgroup is 〈x〉. Thus, the number of subgroups of order pis the number of nonidentity elements in Zp × Zp divided by the number of nonidentityelements of Zp: we have (p2 − 1)/(p− 1) = p+ 1, subgroups of order p in Zp × Zp.

But if H ⊂ Zp × Zp has order p, then the homomorphisms from Zp × Zp to Hare in one-to-one correspondence with the choices of pairs of elements of H, so thereare p2 different homomorphisms from Zp × Zp to H. Only one of these is the trivialhomomorphism, and hence there are p2 − 1 homomorphisms from Zp × Zp onto H.

Multiplying this by the number of subgroups of order p, we see that there are p3 +p2 − p − 1 homomorphisms from Zp × Zp to itself with an image of order p. Since thetrivial homomorphism is the only homomorphism whose image has order 1, this givesexactly p3 + p2 − p homomorphisms from Zp ×Zp to itself that are not automorphisms.But then there are exactly p4− p3− p2 + p automorphisms of Zp×Zp. But this is easilyseen to factor as (p2 − 1)(p2 − p), as desired.


A perhaps simpler argument is as follows. Choose f(a). Any nonidentity choice willdo, so there are p2− 1 choices for f(a). Now consider the choice of f(b). As long as f(b)is not in 〈f(a)〉, f will be an automorphism. Thus, there are p2 − p possible choices forf(b) once f(a) has been chosen. In particular, there are (p2 − 1)(p2 − p) ways to choosean automorphism f .

Exercises 5.2.10.1. Show that every nonabelian group of order p3 is a central extension of Zp by Zp×Zp.

2. Construct a group of order p4 whose center is isomorphic to Zp2 .

3. Find a group of order 16 whose center has order 2.

4. Construct a group of order 81 whose center has order 3.

5. Let p be an odd prime. Construct a group of order p4 whose center has order p.

6. Show that every group of order pn may be generated by n elements.

5.3 Sylow Subgroups

Cauchy’s Theorem shows that if p divides the order of a finite group G, then G has asubgroup of order p. But what if |G| is divisible by p2? Can we then find a subgroup oforder p2?

We cannot argue here by Cauchy’s Theorem and induction, as there may not be anormal subgroup of order p. (E.g., the group G may be simple.) So we need a newapproach to the question.

Note that there are examples of groups G and numbers n such that |G| is divisibleby n, but G has no subgroup of order n. For instance, we’ve seen in Problem 2 ofExercises 4.2.18 that A4 has no subgroup of order 6.

Definitions 5.3.1. Let G be a finite group and let p be a prime. A p-subgroup of G isa subgroup which is a p-group. A p-Sylow subgroup P of G is a maximal p-subgroup ofG, i.e., P is a p-subgroup of G, and any p-subgroup of G containing P must be equal toP .

Examples 5.3.2. In D6, 〈b〉 has order 3. Since no higher power of 3 divides the orderof D6, 〈b〉 cannot be properly contained in a 3-subgroup of D6. Thus, 〈b〉 is a 3-Sylowsubgroup of D6.

Similarly, 〈a〉, 〈ab〉, and 〈ab2〉 are all 2-subgroups of D6. Since 2 itself is the highest2-power that divides |D6|, each of these is a 2-Sylow subgroup of D6.

Lemma 5.3.3. Let H be a p-subgroup of a finite group G. Then H is contained in ap-Sylow subgroup of G.

Proof Let |G| = pim with p relatively prime to m, and let |H| = pj . If i = j, then His a p-Sylow subgroup of G by Lagrange’s Theorem. Otherwise, we argue by inductionon i− j.

From what we know at this point, H itself might be a p-Sylow subgroup of G evenif j < i. If this is the case, we are done. Otherwise, there is a p-subgroup K of G thatstrictly contains H. But then the index of K in G is smaller than that of H, so byinduction, K is contained in a p-Sylow subgroup.


Lemma 5.3.4. Let G be a finite group and let p be a prime dividing the order of G. LetP be a p-Sylow subgroup of G. Then the index of P in its normalizer is relatively primeto p.

Proof Recall that P � NG(P ). Thus, the order ofNG(P )/P is the index of P inNG(P ).If this order is divisible by p, then Cauchy’s Theorem gives a subgroup H ⊂ NG(P )/Pwith |H| = p. But then if π : NG(P )→ NG(P )/P is the canonical map, then π−1H is ap-subgroup of G that strictly contains P .

Corollary 5.3.5. Let G be a finite group and let p be a prime dividing the order of G.Let P be a p-Sylow subgroup of G. Then any p-subgroup of NG(P ) is contained in P .

Proof Let K ⊂ NG(P ) be a p-group. Then K/(K ∩ P ) is isomorphic to a subgroupof NG(P )/P . Since K is a p-group, so is K/(K ∩ P ). Since |NG(P )/P | is prime to p,K/(K ∩ P ) = e, and hence K ⊂ P .

Lemma 5.3.6. Let P be a p-Sylow subgroup of the finite group G. Then any conjugateof P is also a p-Sylow subgroup of G.

Proof Since conjugation gives an automorphism of G, any conjugate of a p-subgroupof G is again a p-subgroup of G. Now for x ∈ G, suppose that xPx−1 not a p-Sylowsubgroup of G. Then xPx−1 must be properly contained in a p-subgroup P ′ of G. Butthen P is properly contained in x−1P ′x, contradicting the statement that P is a p-Sylowsubgroup of G.

We now give yet another variant of the proof of Cauchy’s Theorem.

Theorem 5.3.7. (First Sylow Theorem) Let G be a finite group and let p be a primedividing the order of G. Then any two p-Sylow subgroups of G are conjugate, and thenumber of p-Sylow subgroups of G is congruent to 1 mod p.

Proof Let P be a p-Sylow subgroup of G and let X ⊂ Sub(G) be the orbit of P underthe G-action given by conjugation. (Recall that Sub(G) is the set of subgroups of G.)By Lemma 5.3.6, each element of X is a p-Sylow subgroup of G.

The G-set X may be considered as a P -set by restricting the action. For P ′ ∈ X, theisotropy subgroup of P ′ under the action of P on X is given by

PP ′ = {x ∈ P |xP ′x−1 = P ′}= P ∩NG(P ′).

In particular, P ′ is fixed by P if and only if P ⊂ NG(P ′). By Corollary 5.3.5, this canhappen if and only if P = P ′, as both P and P ′ are p-Sylow subgroups of G.

Thus, the action of P on X has exactly one trivial orbit: {P}. (Here, an orbit istrivial if it has only one element.) The number of elements in the orbit of P ′ ∈ X underthe P -action on X is [P : P ∩NG(P ′)], which divides |P |. Thus, since P is a p-group, thenumber of elements in any nontrivial orbit is divisible by p (hence congruent to 0 modp). In particular, if Y ⊂ X is a set of orbit representatives for the nontrivial orbits of theaction of P on X (i.e., the intersection of each nontrivial P -orbit in X with Y consistsof exactly one element), then the G-set Counting Formula (Corollary 3.3.15) gives

|X| = |{P}|+∑P ′∈Y

[P : P ∩NG(P ′)].


Since each [P : P ∩NG(P ′)] is congruent to 0 mod p and since |{P}| = 1, we see thatthe number of elements in X is congruent to 1 mod p.

It remains to show that X contains every p-Sylow subgroup of G. Let P ′ be anarbitrary p-Sylow subgroup of G, and consider the action of P ′ on X by conjugation. IfP ′ is not in X, then every orbit of this P ′-action must be nontrivial. But that wouldmake the number of elements in X congruent to 0 mod p, rather than 1 mod p. So wemust have P ′ ∈ X.

Theorem 5.3.8. (Second Sylow Theorem) Let G be a finite group and let p be a primedividing the order of G. Let P be a p-Sylow subgroup of G. Then the index of P in G isprime to p. Thus, if |G| = pim, with p and m relatively prime, then |P | = pi.

Moreover, the number of p-Sylow subgroups in G divides m in addition to being con-gruent to 1 mod p.

Proof The index, m, of P in G is the product of [NG(P ) : P ] and [G : NG(P )]. Theformer is prime to p by Corollary 5.3.5. The latter is equal to the number of conjugatesof P , and hence is 1 mod p by the First Sylow Theorem.

Recall from Corollary 5.2.6 that if p is prime and if P is a finite group of order pi,then P has a subgroup of every order dividing the order of P . Applying this to a p-Sylowsubgroup of a group G, the Second Sylow Theorem produces the following corollary.

Corollary 5.3.9. Let G be a finite group and let q be any prime power that divides theorder of G. Then G has a subgroup of order q.

We now give the final Sylow Theorem.

Theorem 5.3.10. (Third Sylow Theorem) Let G be a finite group and let p be a primedividing the order of G. Let P be a p-Sylow subgroup of G. Then the normalizer of P isits own normalizer: NG(NG(P )) = NG(P ).

Proof We claim that P is a characteristic subgroup of NG(P ). The point here is thatif x ∈ P and if f is any automorphism of NG(P ), then f(x) must be a p-torsion element.But by Corollary 5.3.5, every p-torsion element of NG(P ) must be contained in P .

Now suppose that NG(P ) � H. Then conjugation by an element of H gives anautomorphism of NG(P ). But this must preserve P , so P is normal in H. But thenH ⊂ NG(P ).

We now summarize some important facts implicit in the above discussion.

Corollary 5.3.11. Let G be a finite group and let p be a prime dividing the order of G.Suppose there is a p-Sylow subgroup P of G which is normal in G. Then P is the onlyp-Sylow subgroup of G, and P is the set of all p-torsion elements in G. Indeed, P is acharacteristic subgroup of G.

Conversely, if there is only one p-Sylow subgroup of G, then it is normal in G.

Proof The p-Sylow subgroups of G are all conjugate. But if P � G, then it has noconjugates other than itself. Moreover, since NG(P ) = G, P contains all the p-torsionelements of G by Corollary 5.3.5. And since P is a p-group, all its elements are p-torsion,so that P is, in fact, the set of p-torsion elements of G. But then P is characteristic,since any automorphism preserves the p-torsion elements.

If there is only one p-Sylow subgroup, P , of G, then P has no conjugates other thanitself. Thus, [G : NG(P )] = 1, and hence P � G.


In many cases a normal p-Sylow subgroup is sufficient to unfold the entire structureof a group. The point is that if P is a normal p-Sylow subgroup of G, then |P | and|G/P | are relatively prime. Thus, the argument of Proposition 4.7.7 gives the followingcorollary.

Corollary 5.3.12. Suppose that G has a normal p-Sylow subgroup, P , and that G has asubgroup K of order [G : P ]. Then G is the semidirect product P �α K, where α : K →Aut(P ) is induced by conjugation in G.

According to the Schur–Zassenhaus Lemma, which we will not prove, such a subgroupK always exists. However, if [G : P ] is a prime power, the existence of such a K isautomatic from the second Sylow Theorem.

Corollary 5.3.13. Let G be a group of order prqs where p and q are distinct primes.Suppose that the p-Sylow subgroup, P , of G is normal. Let Q be a q-Sylow subgroup ofG. Then G is isomorphic to the semidirect product P �α Q, where α : Q → Aut(P ) isinduced by conjugation in G.

The Sylow Theorems are extremely powerful for classifying low-order finite groups.We shall give a few of their applications now. First, let us complete the classification ofgroups of order p3.

Proposition 5.3.14. Let p be an odd prime and let G be a nonabelian group of orderp3 that contains no element of order p2. Then G is isomorphic to the group constructedin Corollary 4.6.8 for n = p.

Proof By Lemma 5.2.8, G is a semidirect product (Zp ×Zp) �α Zp for some α : Zp →Aut(Zp × Zp). Moreover, α cannot be the trivial homomorphism, as then G would beabelian by Lemma 4.7.15. But by Lemma 5.2.9, the p-Sylow subgroups of Aut(Zp × Zp)have order p, so α must be an isomorphism onto one of these p-Sylow subgroups.

But the p-Sylow subgroups are all conjugate, and hence an isomorphism onto oneof them is conjugate as a homomorphism into Aut(Zp × Zp) to an isomorphism ontoany of the others. Moreover, if α and α′ are isomorphisms onto the same p-Sylowsubgroup of Aut(Zp × Zp), then because they are isomorphisms, we have α = α′ ◦ g foran automorphism g of Zp. Thus, any two nonabelian semidirect products (Zp×Zp)�αZpand (Zp × Zp) �α′′ Zp are isomorphic by Proposition 4.6.11.

Let us return to the theme of discovering normal Sylow subgroups and using them tosplit the group as a semidirect product.

Corollary 5.3.15. Let G be a group of order prm, where m < p, with p prime. Thenthe p-Sylow subgroup of G is normal.

Proof The number of p-Sylow subgroups of G is congruent to 1 mod p and divides m.Since m < p, there must be only one p-Sylow subgroup, which is thus normal.

When used in conjunction with Corollary 5.3.13, we can get some concrete results:

Corollary 5.3.16. Let G be a group of order prqs, where p and q are distinct primes.Suppose that qs < p. Then G is a semidirect product P �α Q for some α : Q→ Aut(P ),where P and Q are the p-Sylow and q-Sylow subgroups of G, respectively.

This becomes more powerful when used in conjunction with what we know aboutautomorphisms:


Corollary 5.3.17. Every group of order 15 is isomorphic to Z15.

Proof Let |G| = 15. By Corollary 5.3.16, G is a semidirect product Z5 �α Z3 for somehomomorphism α : Z3 → Z×

5 . But Z×5 has order 4, which is relatively prime to 3, and

hence α must be the trivial homomorphism. But then G ∼= Z5×Z3, which is isomorphicto Z15.

Now let us consider a deeper sort of application of the Sylow Theorems. Sometimes,under the assumption that a given Sylow subgroup is nonnormal, we can use the countof the number of its conjugates to find the number of elements of p-power order. Thiswill sometimes permit us to find other subgroups that are normal.

Corollary 5.3.18. Let G be a group of order 12 whose 3-Sylow subgroups are not nor-mal. Then G is isomorphic to A4.

Proof We begin by showing that the 2-Sylow subgroup of G is normal. Let P be a3-Sylow subgroup of G. Then the number of conjugates of P is 1 mod 3 and divides 4.Thus, P has either 1 or 4 conjugates. But we’ve assumed that P is nonnormal, so itmust have 4 conjugates.

Note that if P and P ′ are distinct 3-Sylow subgroups, then P ∩ P ′ = e, since P andP ′ both have order 3. But since each 3-Sylow subgroup has exactly two elements of order3, there must be 8 elements of order 3 in G.

The Second Sylow Theorem, on the other hand, shows that G has a subgroup, H,of order 4. Since there are 8 elements of order 3 in G, every element outside of H musthave order 3. Since the elements of H are all 2-torsion, none of their conjugates may lieoutside of H, and hence H must be normal.

By Corollary 5.3.16, G ∼= H �α Z3 for some α : Z3 → Aut(H). Since there are nonontrivial homomorphisms of Z3 into Z×

4 = {±1} and since G is nonabelian, H must beisomorphic to Z2 × Z2.

As shown in either Problem 14 of Exercises 4.6.13 or Problem 3 of Exercises 4.7.27,the automorphism group of Z2 × Z2 is isomorphic to S3. It’s easy to see that there areonly two nontrivial homomorphisms from Z3 into S3

∼= D6, and that one is obtainedfrom the other by precomposing with an automorphism of Z3. Thus, Proposition 4.6.11shows that there is exactly one isomorphism class of semidirect products (Z2×Z2)�αZ3

which is nonabelian.As shown in Problem 1 of Exercises 4.7.27, A4 is a semidirect product (Z2×Z2)�αZ3,

and the result follows.

Thus, the classification of groups of order 12 depends only on classifying the splitextensions of Z3 by the groups of order 4. We shall leave this to the exercises.

Here is a slightly more complicated application of the Sylow Theorems.

Lemma 5.3.19. Let G be a group of order 30. Then the 5-Sylow subgroup of G isnormal.

Proof We argue by contradiction. Let P5 be a 5-Sylow subgroup of G. Then thenumber of conjugates of P5 is congruent to 1 mod 5 and divides 6. Thus, there must besix conjugates of P5. Since the number of conjugates is the index of the normalizer, wesee that NG(P5) = P5.

Since the 5-Sylow subgroups of G have order 5, any two of them intersect in theidentity element only. Thus, there are 6 · 4 = 24 elements in G of order 5. This leaves


6 elements whose order does not equal 5. We claim now that the 3-Sylow subgroup, P3,must be normal in G.

The number of conjugates of P3 is congruent to 1 mod 3 and divides 10. Thus, ifP3 is not normal, it must have 10 conjugates. But this would give 20 elements of order3, when there cannot be more than 6 elements of order unequal to 5, so that P3 mustindeed be normal.

But then P5 normalizes P3, and hence P5P3 is a subgroup of G. Moreover, the SecondNoether Theorem gives

(P5P3)/P3∼= P5/(P5 ∩ P3).

But since |P5| and |P3| are relatively prime, P5 ∩ P3 = e, and hence P5P3 must haveorder 15.

Thus, P5P3∼= Z15, by Corollary 5.3.17. But then P5P3 normalizes P5, which contra-

dicts the earlier statement that NG(P5) has order 5.

Corollary 5.3.20. Every group of order 30 is a semidirect product Z15 �α Z2 for someα : Z2 → Z×

15.

Proof Since the 5-Sylow subgroup P5 is normal, it is normalized by any 3-Sylow sub-group P3, and hence the product P3P5 is a subgroup of order 15, and hence index 2.Since index 2 subgroups are normal, G is an extension of a group of order 15 by Z2. ByCauchy’s Theorem, this extension must be split. Since there is only one group of order15, the result follows.

We shall classify these semidirect products in Exercises 5.3.22.Often in mathematics the statement of a theorem is insufficient, by itself, to derive

all the applications of its method of proof. This is, in general, a good thing. It meansthat the methods have enough substance to be applied in more varied settings. (Thealternative would be to continually learn or write proofs that one never needed to seeagain, which, in the end, would become boring.) Here is an example of a proof based onthe method of proof of the Sylow Theorems.

Proposition 5.3.21. Let G be a group of order 24 that is not isomorphic to S4. Thenone of its Sylow subgroups is normal.

Proof Suppose that the 3-Sylow subgroups are not normal. The number of 3-Sylowsubgroups is 1 mod 3 and divides 8. Thus, if there is more than one 3-Sylow subgroup,there must be four of them.

Let X be the set of 3-Sylow subgroups of G. Then G acts on X by conjugation, so weget a homomorphism f : G→ S(X) ∼= S4. As we’ve seen in the discussion on G-sets, thekernel of f is the intersection of the isotropy subgroups of the elements of X. Moreover,since the action is that given by conjugation, the isotropy subgroup of H ∈ X is NG(H).Thus,

ker f =⋂H∈X

NG(H).

For H ∈ X, the index of NG(H) is 4, the number of conjugates of H. Thus, the orderof NG(H) is 6. Suppose that K is a different element of X. We claim that the order ofNG(H) ∩NG(K) divides 2.

To see this, note that the order of NG(H) ∩NG(K) cannot be divisible by 3. This isbecause any p-group contained in the normalizer of a p-Sylow subgroup must be contained


in the p-Sylow subgroup itself (Corollary 5.3.5). Since the 3-Sylow subgroups have primeorder here, they cannot intersect unless they are equal. But if the order of NG(H) ∩NG(K) divides 6 and is not divisible by 3, it must divide 2.

In consequence, we see that the order of the kernel of f divides 2. If the kernel hasorder 1, then f is an isomorphism, since G and S4 have the same number of elements.

Thus, we shall assume that ker f has order 2. In this case, the image of f has order12. But by Problem 2 of Exercises 4.2.18, A4 is the only subgroup of S4 of order 12, sowe must have im f = A4.

By Problem 1 of Exercises 4.2.18, the 2-Sylow subgroup, P2, of A4 is normal. Butsince ker f has order 2, f−1P2 has order 8, and must be a 2-Sylow subgroup of G. Asthe pre-image of a normal subgroup, it must be normal, and we’re done.

The results we have at this point are sufficient to classify the groups of most ordersless than 64. Almost all of them are semidirect products. We shall treat many examplesin the exercises below.

But for some of the orders less than 64, we must understand more about the subnormalseries of p-groups. We shall develop the material to do this in the next three sections.Then, we must make a deeper analysis of some automorphism groups. For this, we willneed some information from our analysis of matrix groups over finite fields below.

Exercises 5.3.22.1. Show that the 2-Sylow subgroup of S4 is isomorphic to D8. Is it normal in S4?

2. What is the 2-Sylow subgroup of S6?

3. What is the 2-Sylow subgroup of S8?

4. Let H be any group and write Hn for the product H × · · · ×H of n copies of H.Recall from Problem 21 of Exercises 3.6.24 that Sn acts on Hn through automor-phisms via σ · (h1, . . . , hn) = (hσ−1(1), . . . , hσ−1(n)). For any subgroup K ⊂ Sn weget an induced semidirect product Hn

�α K, called the wreath product of H andK, and denoted H �K.

Write Pr for the 2-Sylow subgroup of S2r .

(a) Show that Pr ∼= Pr−1 � S2 for r ≥ 2.

(b) Suppose that 2r−1 < n < 2r. Show that the 2-Sylow subgroup of Sn isisomorphic to the product of Pr−1 with the 2-Sylow subgroup of Sn−2r−1 .

(c) Express the 2-Sylow subgroup of Sn in terms of the binary expansion of n.

5. Prove the analogue of the preceding problem for primes p > 2.

6. Let p and q be primes, with q < p. Assume, via Corollary 7.3.15, that Z×p is cyclic.

Show that there are at most two groups of order pq. When are there two groups,and when only one?

7. Let p and q be primes. Show that every group of order p2q has a normal Sylowsubgroup.

8. Give a simpler, direct argument for Corollary 5.3.18 by studying the homomorphismΛ : G → S(G/P3) obtained via Proposition 3.3.16 from the standard action of Gon the left cosets of a 3-Sylow subgroup of G.

9. Classify the groups of order 12. How does Q12 fit into this classification?



11. Classify the groups of order 24 whose 2-Sylow subgroup is either cyclic or dihedral.


13. We show here that any group G of order 36 without a normal 3-Sylow subgroupmust have a normal 2-Sylow subgroup. Thus, assume that G has a non-normal3-Sylow subgroup P3. Then the action of G on the set of left cosets G/P3 inducesa homomorphism Λ : G→ S(G/P3) ∼= S4.

(a) Show, using Problem 31 of Exercises 3.7.14, that the kernel of Λ must haveorder 3.

(b) Deduce from Problem 2 of Exercises 4.2.18 that the image of Λ is A4. WriteV ⊂ A4 for the 2-Sylow subgroup of A4 and let H = Λ−1V ⊂ G.

(c) Show that H, as an extension of Z3 by Z2×Z2, must be isomorphic to eitherD12 or to Z3 × Z2 × Z2.

(d) Show that every 2-Sylow subgroup of G must be contained in H. Deduce thatif H ∼= Z3 × Z2 × Z2, then the 2-Sylow subgroup of G is normal.

(e) Show that if H ∼= D12, then G has exactly three 2-Sylow subgroups. Deducethat if P2 is a 2-Sylow subgroup of G, then |NG(P2)| = 12. Deduce that Λinduces an isomorphism from NG(P2) to A4.

(f) Show that the only split extension of Z3 by A4 is the trivial extension Z3×A4.Deduce that H must have been isomorphic to Z3 × Z2 × Z2.

14. Classify those groups of order 40 whose 2-Sylow subgroup is isomorphic to one ofthe following:

(a) Z8

(b) D8

(c) Z4 × Z2.

15. Show that any group of order 42 is a split extension of Z7 by a group of order 6.Classify all such groups.

16. Alternatively, show that any group of order 42 is a split extension of a group oforder 21 by Z2. Classify these, and relate this classification to the preceding one.(To do this, you should calculate the automorphism group of the nonabelian groupof order 21. See Problem 19 of Exercises 4.7.27.)


18. Show that any group G of order 60 whose 5-Sylow subgroups are nonnormal mustbe simple. (Hint : First calculate the orders of the normalizers of the 5-Sylowand the 3-Sylow subgroups of G and then go through all possible orders for thesubgroups of G and show that no subgroup of that order can be normal. In somecases, it is possible to show that no subgroup of the stated order can exist at all.The classifications above of the groups whose order properly divides 60 will beuseful.)


19. Show that any simple groupG of order 60 is isomorphic to A5. (Hint : You definitelywant to use the information derived in the solution of the previous exercise. Thekey is finding a subgroup of index 5. It may help to calculate the order of thecentralizer of an element of order 2 in G.)


21. Show that every group G of order 90 has a normal 5-Sylow subgroup. Deduce thatG has a subgroup of index two.

22. Classify the groups of order 90 whose 3-Sylow subgroup is cyclic.

5.4 Commutator Subgroups and Abelianization

The commutator of two group elements, a and b, may be thought of as a measurementof how far a and b are from commuting with each other. The commutators of all pairs ofelements in a group generate a useful normal subgroup called the commutator subgroupof G, and denoted [G,G]. We shall see that G/[G,G] is the largest abelian quotientgroup of G.

Definition 5.4.1. Let a and b be elements of the group G. The commutator of a andb, written [a, b], is defined by

[a, b] = aba−1b−1.

We’ve already used the next lemma repeatedly.

Lemma 5.4.2. The elements a and b commute if and only if [a, b] = e.

The next lemma may be seen by inspection.

Lemma 5.4.3. For a, b ∈ G, [a, b]−1 = [b, a].

Definition 5.4.4. The commutator subgroup of G, written [G,G], is the subgroup gen-erated by all the commutators of the elements of G.

Thus, an element of G is in [G,G] if and only if it is a product of commutators. Note,however, that not every element of [G,G] is actually a commutator itself.

Lemma 5.4.5. Let G be any group. Then the commutator subgroup is a characteristicsubgroup of G, and hence is normal.

Proof Let f be an automorphism of G. Then clearly, f([a, b]) = [f(a), f(b)]. Thus,since f is a homomorphism, it must take a product of commutators to a product ofcommutators.

Proposition 5.4.6. The quotient group G/[G,G] is abelian. In addition, any homomor-phism f : G→ H with H abelian factors through the canonical map π : G→ G/[G,G].

Proof In G/[G,G], we have ab(a)−1(b)−1 = aba−1b−1 = e, so that each pair of elementsa and b of G/[G,G] commutes.

Let f : G → H with H abelian. To see that f factors through the canonical mapto G/[G,G], it suffices to show that the commutator subgroup [G,G] is contained in thekernel of f . But f([a, b]) = [f(a), f(b)], and the latter is the identity element of H, sincef(a) and f(b) commute. But then the generators of [G,G] are contained in the kernel off , and hence [G,G] must be also.


Definition 5.4.7. We write Gab = G/[G,G], and call it the abelianization of G.

Corollary 5.4.8. Every subgroup of G that contains [G,G] is normal. Moreover, ifH � G, then H contains [G,G] if and only if G/H is abelian.

Proof If [G,G] ⊂ H, then H = π−1π(H), where π is the canonical map from G toGab. But every subgroup of the abelian group Gab is normal, and hence H is normal asthe pre-image of a normal subgroup.

If G/H is abelian, then the canonical map from G to G/H factors through Gab. Butthis happens if and only if [G,G] ⊂ H. Conversely, if [G,G] ⊂ H, then G/H is a quotientgroup of Gab by the third Noether Theorem, and hence is abelian.

Exercises 5.4.9.† 1. Show that a group G is abelian, if and only if [G,G] = e.

† 2. Show that if G is nonabelian and simple, then [G,G] = G, and Gab = e.

3. Show that for nonabelian groups of order p3, p prime, the commutator subgroupand center coincide.

4. What is the commutator subgroup of D2n? What is D2nab?

5. What is the commutator subgroup of Q4n? What is Q4nab?

6. What is the commutator subgroup of A4? What is A4ab?

7. What is the commutator subgroup of S4? What is S4ab?

8. What is the commutator subgroup of Sn for n ≥ 5? What is Snab?

‡ 9. We calculate the automorphism group of Q8.

(a) Show that any automorphism f of Q8 induces an automorphism f of Q8ab.

Show that passage from f to f induces a surjective homomorphism η : Aut(Q8)→Aut(Q8

ab).

(b) Show η is a split extension whose kernel is isomorphic to Z2 × Z2.

(c) Deduce that Aut(Q8) is isomorphic to S4.

10. Classify the groups of order 24 whose 2-Sylow subgroups are quaternionic. Showthat there is a unique such group whose 3-Sylow subgroups are not normal. Weshall meet this group again later as the matrix group Sl2(Z3).

5.5 Solvable Groups

The most tractable groups to analyze are the abelian groups. Every subgroup is normal,and hence lots of induction arguments are available. This is amply illustrated by theclassification of finite abelian groups given in Section 4.4.

At the other extreme, we have the nonabelian simple groups. With no nontrivialproper normal subgroups available, it is difficult to get our hands on the structure of thegroup.

The solvable groups give a generalization of abelian groups that share some of the niceproperties found in the abelian setting. In particular, knowing that a group is solvablewill open up a collection of tools for revealing its structure.


Solvability has long been known to be an important idea. For instance, solvabilityplays a key role in Galois’ famous proof that there does not exist a formula based on theoperations of arithmetic and n-th roots for finding solutions of polynomials of degree 5or higher. We shall give an argument for this in Section 11.11.

Thus, it is of interest to determine whether a given group is solvable. Of particularimportance in this regard is the famous Feit–Thompson Theorem:

Theorem (Feit–Thompson Theorem) Every group of odd order is solvable.

The Feit–Thompson Theorem is considered to be the first major breakthrough alongthe way to the classification of the finite simple groups. Its proof is quite lengthy and iswell beyond the scope of this book.2 But the fact that odd order groups are solvable isan important one to keep in mind for intuition’s sake.

We shall need a more compact notation for commutator subgroups, as we shall bediscussing commutator subgroups of commutator subgroups, etc.

Definition 5.5.1. For i ≥ 0, defineG(i) inductively byG(0) = G andG(i) = [G(i−1), G(i−1)]for i ≥ 1. We call G(i) the i-th derived subgroup of G.

Thus, G(1) = [G,G]. Given this, let us make the following definition.

Definition 5.5.2. A group G is solvable if G(k) = e for some k ≥ 0.

The virtue of this definition is that it is easy to work with. We shall give an equivalentdefinition below that may be more intuitive. In the meantime, we can give a quickworkup of some basic properties. The next lemma is immediate from Problems 1 and 2of Exercises 5.4.9.

Lemma 5.5.3. Every abelian group is solvable. A simple group, on the other hand, issolvable if and only if it is abelian.

Lemma 5.5.4. Let f : G→ H be a surjective homomorphism. Then f(G(i)) = H(i) forall i ≥ 0.

Proof We shall show that f([G,G]) = [H,H]. This then shows that f : [G,G]→ [H,H]is surjective, and the result follows by induction on k.

But since f([a, b]) = [f(a), f(b)], f([G,G]) ⊂ [H,H]. And since f is surjective, thegenerators of [H,H] must lie in f([G,G]).

Corollary 5.5.5. A quotient group of a solvable group is solvable.

Proof If G(k) = e, so is H(k) for any quotient group H of G.

Lemma 5.5.6. Let H be a subgroup of G. Then H(i) ⊂ G(i) for all i ≥ 0.

Proof Once again, by induction, it suffices to treat the case of i = 1. But since H ⊂ G,if h, k ∈ H, then [h, k] ∈ [G,G].

Of course, H(i) need not equal G(i) ∩H.

Corollary 5.5.7. A subgroup of a solvable group is solvable.2The proof takes up the entirety of a 255-page journal issue. See W. Feit and J. Thompson, Solvability

of groups of odd order, Pacific J. Math. 13 (1963), 775–1029.


Recall that a subnormal series for G consists of a sequence of subgroups

e = Gk � Gk−1 � · · · � G0 = G.

Here, each Gi+1 is normal in Gi, but need not be normal in G if i > 1. (Note that we’vereversed the order of the indices from that used previously.)

Definition 5.5.8. A subnormal series is abelian if the quotient groups Gi/Gi+1 areabelian for 0 ≤ i < k.

Proposition 5.5.9. A group G is solvable if and only if it has an abelian subnormalseries.

Proof Suppose that G is solvable, with G(k) = e. Consider the sequence

e = G(k) � G(k−1) � · · · � G(0).

Since G(i+1) = [G(i), G(i)], the quotient group Gi/Gi+1 is just (G(i))ab

, which is certainlyabelian, so that the sequence above is an abelian subnormal series.

Conversely, given an abelian subnormal series

e = Gk � Gk−1 � · · · � G0 = G,

we claim that G(i) ⊂ Gi for all i, which then forces G(k) = e. Clearly, G(0) = G0, so as-sume inductively that G(i−1) ⊂ Gi−1. But then G(i) ⊂ [Gi−1, Gi−1] by Lemma 5.5.6, so itsuffices to see that [Gi−1, Gi−1] ⊂ Gi. But this follows immediately from Corollary 5.4.8,since Gi−1/Gi is abelian.

The above characterization of solvable groups can be quite useful.

Corollary 5.5.10. Let H be a normal subgroup of G. Then G is solvable if and only ifboth H and G/H are solvable.

Proof Suppose that G is solvable. Then H is solvable by Corollary 5.5.7, and G/H issolvable by Corollary 5.5.5.

Conversely, suppose that H and G/H are solvable. Let e = Hk � · · · � H0 = Hbe an abelian subnormal series for H and let e = Kl � · · · � K0 = G/H be an abeliansubnormal series for G/H. Then if π : G → G/H is the canonical map, we have anabelian subnormal series

e = Hk � · · · � H0 = π−1Kl � π−1Kl−1 � · · · � π−1K0 = G

for G.

Also, the following corollary is immediate from Corollary 5.2.5.

Corollary 5.5.11. Let p be a prime. Then any finite p-group is solvable.

Note that other than this last result, we have made no use of finiteness in the dis-cussion of solvability. Recall that a composition series for a group G is a subnormalseries

e = Gk � Gk−1 � · · · � G0 = G

where the subquotient groups Gi/Gi+1 of G are all nontrivial simple groups. By Propo-sition 4.2.5, every nontrivial finite group has a composition series. The Jordan–HolderTheorem shows that any two composition series for a group G have the same length, andgive the same collection of subquotient groups Gk−1/Gk, . . . , G/G1, though possibly ina different order. However, for economy of means, we shall not use it in the followingproof.


Proposition 5.5.12. A finite group is solvable if and only if it has a composition seriesin which the simple subquotient groups are all cyclic of prime order. If G is solvable,then every composition series for G has that form.

Proof A composition series whose simple subquotient groups are cyclic is certainly anabelian subnormal series, so any group that has such a composition series is solvable byProposition 5.5.9.

Conversely, suppose that G has a composition series in which one of the quotientsGi/Gi+1 is not cyclic of prime order. Since the only abelian simple groups have primeorder (Lemma 4.2.2), Gi/Gi+1 must be a nonabelian simple group. But ifG were solvable,the subgroup Gi would be also. And this would force the quotient Gi/Gi+1 to be solvableas well, contradicting Lemma 5.5.3.

Exercises 5.5.13.1. Show that Sn is not solvable if n ≥ 5.

2. Show that A5 is the only group of order ≤ 60 that is not solvable.

3. Show that the Feit–Thompson Theorem is equivalent to the statement that everynonabelian simple group has even order.

5.6 Hall’s Theorem

There is a generalization of Sylow theory that works in solvable groups. It is due toPhilip Hall.

Definition 5.6.1. Let G be a finite group. A subgroup H ⊂ G is a Hall subgroup if |H|and [G : H] are relatively prime.

Our purpose in this section is to prove the following theorem, which shows that afinite solvable group G has Hall subgroups of all possible orders n dividing |G| such that(n, |G|/n) = 1.

Theorem 5.6.2. (Hall’s Theorem) Let G be a finite solvable group, and let |G| = nkwith (n, k) = 1. Then G has a subgroup of order n. Moreover, any two subgroups oforder n are conjugate.

In fact, Hall also showed a converse to this. We shall state it here, but its proof isbeyond the scope of this book.

Theorem (The Hall Converse) Let G be a finite group with |G| = pr11 . . . prk

k withp1, . . . , pk distinct primes. Suppose that G has a Hall subgroup of order |G|/pri

i for eachi = 1, . . . , k. Then G is solvable.

Thus, the existence of Hall subgroups of every order n dividing G with (n, |G|/n) = 1is equivalent to solvability.

There is an interesting special case of the Hall Converse. Suppose that |G| = prqs

with p and q prime. Then a p-Sylow subgroup of G is a Hall subgroup of order |G|/qsand a q-Sylow subgroup of G is a Hall subgroup of order |G|/pr. The Hall Converseproduces an immediate corollary.

Corollary (Burnside’s Theorem) Let G be a finite group whose order has only twoprime divisors. Then G is solvable.


In fact, Burnside’s Theorem is one of the ingredients in the proof of the Hall Converse,and, like the converse itself, is beyond the scope of this book. We shall now concentrateon proving Hall’s Theorem.

Lemma 5.6.3. Let G be a finite group and let H be a Hall subgroup of G of order n.Let K � G with |K| = m. Then H ∩ K is a Hall subgroup of K of order (m,n) andHK/K is a Hall subgroup of G/K of order n/(m,n).

Proof Write |G| = nk. Since H is a Hall subgroup, this gives (n, k) = 1. Writem = n′k′ with n′ dividing n and k′ dividing k. Thus, n′ = (m,n).

Of course, |H ∩K| divides (m,n) = n′, so the order of H/(H ∩ K) is divisible byn/(m,n) = n/n′.

The Noether Isomorphism Theorem gives H/(H ∩ K) ∼= HK/K, so |H/(H ∩K)|divides |G/K| = (n/n′) · (k/k′). But also, |H/(H ∩K)| divides |H| = n. Since n and kare relatively prime, |H/(H ∩K)| must divide n/n′.

Thus, |H/(H ∩K)| = n/(m,n), which forces |H ∩K| = (m,n).

Recall that a “minimal subgroup” of G was defined to be a nontrivial subgroupH ⊂ G such that H has no nontrivial proper subgroups. Thus, H is minimal among thenontrivial subgroups of G. We define minimal normal subgroups by analogy.

Definition 5.6.4. We say that H is a minimal normal subgroup of G if e �= H � G andif no nontrivial proper subgroup of H is normal in G.

If G is finite, then the existence of minimal normal subgroups in G is easy. Justchoose the nontrivial normal subgroup of the lowest order.

Lemma 5.6.5. Let G be a finite solvable group and let H be a minimal normal subgroupof G. Then H is an abelian group of exponent p for some prime p.

Proof The commutator subgroup [H,H] is a characteristic subgroup of H. SinceH � G, this gives [H,H] � G by Corollary 3.7.9.

Thus, since H is a minimal normal subgroup of G, either [H,H] = H or [H,H] = e.Suppose that [H,H] = H. Then the derived subgroups H(i) = H for all i > 0. Butsolvable groups are by definition those for whichH(i) = e for some i. SinceH is nontrivial,it thus cannot be solvable, contradicting the fact that a subgroup of a solvable group issolvable (Corollary 5.5.7).

Thus, [H,H] = e, and hence H is abelian. Let p be a prime dividing |H| and let Hp bethe p-torsion subgroup of G. Then Hp is a characteristic subgroup of H (Lemma 4.4.19),and hence Hp � G. Since Hp is nontrivial by Cauchy’s Theorem, we see that H mustbe an abelian p-group for some prime p. But then the elements of exponent p form anontrivial characteristic subgroup of H, so H must have exponent p as well.

We are now ready to prove Hall’s Theorem.

Proof of Theorem 5.6.2 We wish to show that for any divisor n of |G| such that(n, |G|/n) = 1 there are Hall subgroups of G of order n and that any two such subgroupsare conjugate.

We argue by induction on |G|. The proof breaks up into two parts. First, we assumethat G contains a nontrivial normal subgroup, K, whose order is not divisible by k.

Write |K| = n′k′ with n′ dividing n and k′ dividing k. The induction hypothesisgives a Hall subgroup H/K of G/K of order n/n′. Thus, H ⊂ G has order nk′. Since


k′ < k, the induction hypothesis shows that there’s a Hall subgroup H ⊂ H of order n.This gives the desired subgroup of G.

Suppose now that H and H ′ are two subgroups of G of order n. By Lemma 5.6.3,HK/K and H ′K/K are Hall subgroups of order n/n′ in G/K, and hence are conjugateby the induction hypothesis. But if x conjugates HK/K to H ′K/K, then x conjugatesHK to H ′K. But then x conjugates H to a Hall subgroup of H ′K of order n. Andinduction shows that xHx−1 is conjugate to H ′ in H ′K.

Thus, we are left with the case where the order of every nontrivial normal subgroupof G is divisible by k. Let P � G be a minimal normal subgroup. Then Lemma 5.6.5,shows that P is an abelian group of exponent p for a prime number, p. Since |P | = pr isdivisible by k and since (n, k) = 1, we must have k = pr. Thus, P is a p-Sylow subgroupof G. Since P � G, there is only one subgroup of G of order k = pr, and hence P is theunique minimal normal subgroup of G.

The quotient group G/P is solvable, so it has a minimal normal subgroup Q/P ,whose order is qs for some prime number q. Let Q be a q-Sylow subgroup of Q. Thenthe canonical map π : G→ G/P restricts to an isomorphismQ ∼= Q/P (Corollary 5.3.12),so Q = QP .

We claim that NG(Q) is the desired subgroup of G of order n. To show this, it sufficesto show that [G : NG(Q)] = pr.

To see this, first note that, since Q is normal in G, every conjugate of Q in G mustlie in Q. Thus, every conjugate of Q in G is a q-Sylow subgroup of Q, and is conjugateto Q in Q by the First Sylow Theorem. So Q has the same number of conjugates in Qas in G, and hence

[G : NG(Q)] = [Q : NQ(Q)].

We claim now that NQ(Q) = Q. Since [Q : Q] = pr, this will complete the proof that|NG(Q)| = n.

Since P is the p-Sylow subgroup of Q and is normal there, it contains every p-torsionelement of Q. Since Q ⊂ NQ(Q), the only way the two could differ is if |NQ(Q)| isdivisible by p. Thus, it suffices to show that NQ(Q) ∩ P = e.

Note first that if x ∈ P normalizes Q and if y ∈ Q, then xyx−1y−1 ∈ P ∩ Q = e.Thus, x commutes with every element of Q. Since P is abelian and Q = QP , we see thatNQ(Q) ∩ P ⊂ Z(Q), so it suffices to show that Z(Q) = e.

Recall that Q/P � G/P , and hence Q � G. Since the center of a group is acharacteristic subgroup, we see that Z(Q) � G. Suppose that Z(Q) is nontrivial. SinceP is the unique minimal normal subgroup of G, this implies that P ⊂ Z(Q). But then Pnormalizes Q. Since Q = QP , we must have NQ(Q) = Q, so that Q � Q. But Q is theq-Sylow subgroup of Q, so this implies that Q is characteristic in Q and hence normalin G, contradicting the assumption that G has no normal subgroups whose order is notdivisible by k. Thus, Z(Q) = e, and hence NG(Q) is a Hall subgroup of G of order n, asdesired.

It now suffices to show that if H is another subgroup of G of order n, then H isconjugate to NG(Q). Since |H| = |G/P | is prime to |P |, the canonical map inducesan isomorphism H → G/P . Thus, H ∩Q = Q′ maps isomorphically to Q/P under thecanonical map, and hence is a q-Sylow subgroup of Q. Thus Q and Q′ are conjugate in Q,hence in G. But then NG(Q) and NG(Q′) are conjugate as well. Now Q′ = (H∩Q) � H,so H ⊂ NG(Q′). Since the two groups have the same order, the result follows.


5.7 Nilpotent Groups

Nilpotent groups form a class of groups that lie between the solvable and the abeliangroups. They form a generalization of the properties found among the finite p-groups.The fact that a finite p-group, P , has a nontrivial center is a very useful one, and itdemonstrates that finite p-groups are, in fact, better behaved than the general run ofsolvable groups. And since P/Z(P ) is again a p-group, the center of P/Z(P ) must benontrivial as well. Thus, induction arguments are possible. This forms the basic ideathat we will generalize to define nilpotency.

In the end, we shall see that the finite nilpotent groups are nothing more than carte-sian products of p-groups. But the abstractions we will develop along the way turn outto be useful for our understanding of the p-groups, if nothing else.

Definition 5.7.1. Let G be a group. Define the subgroups Zi(G) inductively as fol-lows. Z0(G) = e and Z1(G) = Z(G). Suppose given Zi−1(G), which is assumed tobe normal in G, and let π : G → G/Zi−1(G) be the canonical map. Then Zi(G) =π−1(Z(G/Zi−1(G))).

Of course, as the pre-image of a normal subgroup of G/Zi−1(G), Zi(G) � G, andthe induction continues.

Definition 5.7.2. A group G is nilpotent if Zk(G) = G for some k ≥ 0.

Just so the abstraction doesn’t get away from us here, we prove the following lemma.

Lemma 5.7.3. Finite p-groups are nilpotent.

Proof Suppose that Zi−1(G) �= G. Then G/Zi−1(G) is a nontrivial p-group, andmust have a nontrivial center. Thus, Zi(G) is strictly larger than Zi−1(G). But then if|G| = pj , we must have Zk(G) = G for some k ≤ j.

Since the quotient group Zi(G)/Zi−1(G) is the center of G/Zi−1(G), and henceabelian, the sequence

e = Z0(G) � · · · � Zk(G) = G

is an abelian subnormal series when G is nilpotent.

Proposition 5.7.4. Nilpotent groups are solvable.

The next result gives a useful property of nilpotent groups, which we have not yetverified for p-groups. (We dealt with a special case in Proposition 4.1.12 by a different,but related, argument.)

Proposition 5.7.5. Let H be a subgroup of the nilpotent group G, with H �= G. ThenH cannot be its own normalizer in G.

Proof Let H ⊂ G be its own normalizer. We shall show that H must equal G. SinceG is nilpotent, it suffices to show that Zi(G) ⊂ H for all i.

We show this by induction on i, with the case i = 0 immediate. Suppose thatZi−1(G) ⊂ H, and let π : G → G/Zi−1(G) be the canonical map. Since Zi−1(G) ⊂ H,we haveH = π−1(π(H)). This is easily seen to imply thatNG(H) = π−1(NG/Zi−1(G)(π(H))),so π(H) must be its own normalizer in G/Zi−1(G).

Since Zi(G) = π−1(Z(G/Zi−1(G))), it suffices to show that Z(G/Zi−1(G)) ⊂ π(H).In other words, it suffices to assume that i = 1, and show that Z(G) ⊂ H. But theelements of Z(G) normalize every subgroup of G, so that Z(G) ⊂ NG(H) = H, and theresult follows.


Corollary 5.7.6. Every subgroup of a finite nilpotent group is part of a subnormal series.

Proof Let H be the subgroup and G the group. If H is not already normal in G, thenits normalizer at least strictly contains it, so that the index of the normalizer is strictlysmaller than the index of H. Thus, the result follows by induction on the index.

We can use this in our efforts to classify the low order groups.

Corollary 5.7.7. Let G be a nonabelian group of order 16 that does not contain anelement of order 8. Then G has an index 2 subgroup H such that H has an element oforder 4.

Proof If every element of G has order 2, then G is abelian by Problem 3 of Exer-cises 2.1.16. Thus, G must have an element x of order 4. If 〈x〉 is not normal in G,then we may take H to be its normalizer. Otherwise, take H to be the pre-image of asubgroup of order 2 in G/〈x〉.

Thus, we shall be able to classify the groups of order 16 without first understandingthe structure of Aut(Z2 × Z2 × Z2). This is useful, because Aut(Z2 × Z2 × Z2) turnsout to be a simple group of order 168.

Lemma 5.7.8. A product H ×K is nilpotent if and only if both H and K are nilpotent.

Proof This will follow once we show that Zi(H ×K) = Zi(H)× Zi(K) for all i ≥ 0.We show this by induction on i, with the case i = 0 immediate.

Suppose Zi−1(H ×K) = Zi−1(H)×Zi−1(K). Then the product of canonical maps

π × π : H ×K → (H/Zi−1(H)

)× (K/Zi−1(K)

)has kernel Zi−1(H ×K). Thus, it suffices to show that

Z((H/Zi−1(H)

)× (K/Zi−1(K)

)) = Z(H/Zi−1(H))×Z(K/Zi−1(K)).

But it is easy to see that for any groups H ′ and K ′, Z(H ′ ×K ′) = Z(H ′)×Z(K ′).

But now we can characterize finite nilpotent groups.

Proposition 5.7.9. A finite group is nilpotent if and only if it is the internal directproduct of its Sylow subgroups, meaning that if p1, . . . , pk are the primes dividing |G|,then the pi-Sylow subgroup, Gpi , of G is normal for all i, and G is the internal directproduct of Gp1 , . . . , Gpk

.

Proof If G is isomorphic to Gp1 × · · · × Gpk, then G is nilpotent by Lemmas 5.7.3

and 5.7.8. Thus, we prove the converse. Note that for abelian groups, the converse isgiven as Proposition 4.4.9.

Let G be nilpotent and let Gpi be a pi-Sylow subgroup of G. Then the third SylowTheorem shows that the normalizer of Gpi

is its own normalizer in G. But since G isnilpotent, Proposition 5.7.5 shows that the normalizer of Gpi must be G itself, so thatGpi � G. We shall show that this is sufficient to obtain the stated product formula.

If 1 ≤ i, j ≤ k and if i �= j, let x ∈ Gpiand let y ∈ Gpj

. Consider the commutator[x, y] = xyx−1y−1. Since Gpj

is normal in G, xyx−1 ∈ Gpj, and hence the commutator

[x, y] is also. But the normality of Gpi shows that yx−1y−1 and hence also [x, y] are inGpi

as well. But because the identity is the only element that has prime power order formore than one prime, Gpi

∩Gpj= e, and hence [x, y] = e.


Thus, the elements of Gpi commute with the elements of Gpj whenever i �= j, and wehave a homomorphism μ : Gp1 × · · · × Gpk

→ G given by μ(g1, . . . , gk) = g1 · · · gk. Weclaim that μ is an isomorphism. By the second Sylow Theorem, Gp1 × · · · ×Gpk

has thesame order as G, so it suffices to show that μ is injective.

Suppose that μ(g1, . . . , gk) = e. We wish to show that gi = e for all i. Assume bycontradiction that i is the smallest index for which gi �= e. We cannot have i = k, asthen gi = e as well. Thus, g−1

i = gi+1 · · · gk. But the order of (e, . . . , e, gi+1, · · · , gk) inGp1 ×· · ·×Gpk

is the least common multiple of the orders of gi+1, . . . , gk (by Problem 2of Exercises 2.10.12), and hence the order of its image, under μ is divisible only by theprimes pi+1, . . . , pk. But the image in question happens to be g−1

i , which is a pi-torsionelement, and we have our desired contradiction.

We obtain a converse to Lagrange’s Theorem for nilpotent groups.

Corollary 5.7.10. Let G be a finite nilpotent group. Then every divisor of the order ofG is the order of a subgroup of G.

Proof Write |G| = pr11 . . . prk

k where p1, . . . , pk are distinct primes. Then any divisor of|G| has the form m = ps11 . . . psk

k with 0 ≤ si ≤ ri for i = 1, . . . , k.Let Gpi be the pi-Sylow subgroup of G. Then there is a subgroup Hi of Gpi of

order psii by Corollary 5.2.6. By Proposition 5.7.9, G has a subgroup isomorphic to

H1 × · · · ×Hk.

Exercises 5.7.11.1. Give an example of a solvable group that’s not nilpotent.

2. Classify the groups of order 16 containing a copy of D8.

3. Classify the groups of order 16 containing a copy of Q8.

4. Classify the groups of order 16 containing a copy of Z4 × Z2.

5. Which groups appear in more than one of the three preceding problems? Which ofthem contain an element of order 8?

6. Classify the groups of order 24 whose 2-Sylow subgroups are isomorphic to Z4×Z2.

5.8 Matrix Groups

Perhaps the most important missing piece at this point in being able to classify loworder groups is an understanding of the automorphism groups of the Sylow subgroupsthat arise in them. So far, we know how to calculate the automorphisms only of cyclicgroups, and groups such as dihedral and quaternionic groups, whose automorphisms arecalculated in Exercises 4.7.27 and 5.4.9.

Here, we shall investigate the automorphisms of groups of the form Zn × · · · × Zn,with special attention paid to the case where n is prime. There are a number of groupsof order < 64 which have Sylow subgroups of this form.

Unlike the preceding material, we shall need to make use of some theory we have notyet developed: basic ring theory and linear algebra. The reader who has not yet beenexposed to this material should come back to this section after reading the appropriatechapters.


Of course, for any ring A, the cartesian product A × · · · × A of n copies of A is theunderlying abelian group of the free A-module of rank n. Consonant with this, we shallwrite

A× · · · ×A︸︷︷︸n times

= An.

Linear algebra gives techniques of studying the A-module homomorphisms from An

to itself. Recall that if N is an A-module, then EndA(N) denotes the set of A-modulehomomorphisms from N to itself, while AutA(N) denotes the group (under the operationof composition of homomorphisms) of A-module isomorphisms from N to itself. Recallalso that an A-module homomorphism is an isomorphism if and only if it is bijective.Since every A-module isomorphism is an isomorphism of abelian groups, forgetting theA-module structure gives an injective group homomorphism

ϕ : AutA(N)→ Aut(N).

For A = Zn, Corollary 7.7.16 shows that every group homomorphism between Zn-modules is a Zn-module homomorphism. Thus, the next lemma is immediate.

Lemma 5.8.1. For all k ≥ 1, forgetting the Zn-module structure gives an isomorphismϕ : AutZn

(Zkn)∼=−→ Aut(Zkn).

Recall that the group of invertible n×n matrices over a ring A is denoted by Gln(A).Here, the group structure comes from matrix multiplication. Corollary 7.10.9 shows thatallowing matrices to act on the left on column vectors gives a group isomorphism fromGln(A) to AutA(An).3 Thus, for any A, allowing matrices to act on the left on columnvectors gives an injective ring homomorphism

ιn : Gln(A)→ Aut(An).

For A = Zn, Lemma 5.8.1 allows us to do even better.

Corollary 5.8.2. The standard left action of matrices on column vectors induces a groupisomorphism

ιk : Glk(Zn)→ Aut(Zkn).

In particular, the study of matrix groups will give us calculations of automorphismgroups of groups that we care about. Recall that the transpose, M t, of an n× n matrixM = (aij) is the matrix whose ij-th entry is aji.

Lemma 5.8.3. For any commutative ring A, there is a group automorphism α : Gln(A)→Gln(A) defined by α(M) = (M t)−1.

Proof It is easy to check that for any pair of matrices M1,M2 ∈ Mn(A), we have(M1M2)t = M t

2Mt1. This is enough to show that the transpose of an invertible matrix

is invertible, with (M t)−1 = (M−1)t. Since inversion of matrices also reverses order, theresult follows.

3Here, if A is not commutative, then this applies only to the standard right module structure on An.


For finite rings A, the set of n × n matrices is finite, and hence so is Gln(A). (If Ais infinite, it is not hard to show that Gln(A) is infinite for n ≥ 2.) The next result willassist in calculating the order of Gln(A) when A is finite.

Proposition 5.8.4. Let A be a commutative ring. Consider the standard action ofGln(A) on columns, and let en be the n-th canonical basis vector. Then the isotropysubgroup of en under this action is isomorphic to An−1

�ιn−1 Gln−1(A).

Proof The isotropy subgroup, Gln(A)en, of en consists of those invertible matrices

whose last column is en. Thus, we are concerned with the invertible matrices of the form(M 0X 1

), where M is (n−1)×(n−1), X is a 1×(n−1) row matrix, 0 is the (n−1)×1

column matrix whose coordinates are all 0, and the 1 in the lower corner is the 1 × 1matrix 1. Note that M must lie in Gln−1(A), as otherwise the first n− 1 rows of the fullmatrix would be linearly dependent.

Conversely, if M is any element of Gln−1(A) and X is any 1 × (n − 1) row matrix,(M 0X 1

)is invertible, with inverse

(M−1 0−XM−1 1

), and hence lies in the isotropy

group Gln(A)en.

Thus, if α is the automorphism of Gln(A) replacing a matrix with the inverse of its

transpose, then H = α(Gln(A)en) is the set of all matrices of the form

(M X0 1

)with

M ∈ Gln−1(A) and X ∈ An−1, where we identify An−1 with the set of (n−1)×1 columnmatrices. Of course, Gln(A)en

is isomorphic to H, so we may study the latter.Note that the elements of H multiply according to the formula(

M X0 1

)(M ′ Y0 1

)=

(MM ′ X +MY

0 1

).

Thus, there is an isomorphism f : An−1�ιn−1Gln−1(A)→ H, via f(X,M) =

(M X0 1

).

We can apply this to calculate the order of the general linear group of a finite field.It is shown in Section 11.3 that the order of any finite field must have the form q = pr

for p prime, and that for any such q, there is a unique field, which we shall denote byFq, of order q. Of course, Fp = Zp.

Corollary 5.8.5. Let Fq be the field of order q = pr. Then for n ≥ 1,

|Gln(Fq)| = (qn − 1)(qn − q) . . . (qn − qn−1).

Proof We argue by induction on n. For n = 1, Gl1(Fq) ∼= F×q has order q− 1, and the

assertion holds.Suppose the result is true for n− 1. Then the formula for the size of an orbit gives

|Gln(Fq)| = |Gln(Fq) · en| · |Fn−1q �ιn−1 Gln−1(Fq)|

= |Gln(Fq) · en| · qn−1 · (qn−1 − 1) . . . (qn−1 − qn−2)= |Gln(Fq) · en| · (qn − q) . . . (qn − qn−1).

Thus, it suffices to show that the orbit of en under the action of Gln(Fq) has qn − 1elements, and hence that for any nonzero element v ∈ Fnq , there is an M ∈ Gln(Fq) withMen = v.


But this follows immediately from elementary linear algebra: v is linearly independentover Fq, so there is a basis of Fnq that contains v.

Matrix groups over finite fields have another important use in the theory of finitegroups. They are a source of finite simple groups. Recall that if A is a commutative ring,then the determinant induces a surjective group homomorphism

det : Gln(A)→ A×

for n ≥ 1. The special linear group Sln(A) is defined to be the kernel of det. Since amatrix is invertible if and only if its determinant is a unit, we see that Sln(A) consistsof all n× n matrices over A of determinant 1.

Recall that aIn is the matrix whose diagonal entries are all equal to a and whoseoff-diagonal entries are all 0. It’s an easy exercise to show that det(aIn) = an.

Definition 5.8.6. Let A be a commutative ring and define H ⊂ Sln(A) by

H = {aIn | a ∈ A with an = 1}.

Then the projective special linear group is defined as PSln(A) = Sln(A)/H.

It turns out that the subgroup H above is the center of Sln(A).

Theorem Let Fq be the field with q = pr elements. The projective special linear groupPSln(Fq) is simple for all q if n ≥ 3 and is simple for q > 3 if n = 2.

The proof is accessible to the techniques of this volume, but we shall not give it. It issomewhat analogous to the proof that An is simple for n ≥ 5. There, the cycle structuregave us a handle on the conjugacy classes. Here, the conjugacy classes are determinedby the rational canonical form.

Exercises 5.8.7.1. Give a new proof that Aut(Z2

2) = Gl2(Z2) is isomorphic to S3.

2. Show that Gl3(Z2) has a subgroup of index 7 that is isomorphic to S4.

3. What are the normalizers of all the Sylow subgroups of Gl3(Z2)?

4. Show that Gl3(Z2) is simple.

5. Complete the classification of the groups of order 24.

6. Let q = pr with p prime. Show that one of the p-Sylow subgroups of Gln(Fq) is theset of matrices with 1’s on the diagonal and 0’s below it (i.e., the upper triangularmatrices with 1’s on the diagonal).

7. Show that the p-Sylow subgroups of Gl3(Zp) are isomorphic to the group con-structed in Corollary 4.6.8 for n = p.

8. Write Tn ⊂ Gln(Fq) for the subgroup of upper triangular matrices with 1’s on thediagonal. Then Tn acts on Fnq through automorphisms by an action ι′n obtained byrestricting the action of the full general linear group. Show that Tn is isomorphicto Fn−1

q �ι′n−1Tn−1.

9. Show that D8 embeds in Gl2(A) whenever A has characteristic �= 2.


10. Show that any element of order 4 in Z24 is part of a Z4-basis. Calculate the order

of Gl2(Z4).

11. Classify the groups of order 48 whose 2-Sylow subgroup is Z4 × Z4.

12. What are the orders of Sln(Fq) and PSln(Fq)?

13. Show that the 2-Sylow subgroup of Sl2(Z3) is isomorphic to Q8 and is normal inSl2(Z3). (Hint : Look at the elements of Sl2(Z3) of trace 0.)

14. Deduce that the 2-Sylow subgroup of Gl2(Z3) is a group of order 16 containingboth D8 and Q8.


16. What is the 2-Sylow subgroup of Sl2(Z5)?

17. What is the 2-Sylow subgroup of Sl2(Z4)?

18. Show that the subgroup H of Definition 5.8.6 is the center of Sln(A).

Chapter 6

Categories in Group Theory

Category theory, a.k.a. abstract nonsense, is a branch of mathematics that attempts toidentify phenomena that crop up in many different types of mathematical structures.One may think of it as the study of those things that happen for reasons which arepurely formal. For this reason, the facts that hold for categorical reasons are often notthe deepest ones in a given field. But categorical patterns repeat in so many contextsthat they form a collection of truths that may be instantly recognized and used.

What category theory gives us is a language by which to describe patterns common todifferent areas of mathematics. Verifying that these patterns hold in a given context cantake some work. For instance, we will give some concepts regarding commonly occurringuniversal properties. In each category of interest, one must make a construction, oftenunique to that category, to show that objects satisfying the universal property actuallyexist. Once that construction has been made, one can revert to formal arguments inusing it.

Categorical language is well suited to describing the overview of the relationshipsbetween objects in a given context. For instance, classification results may be expressedin categorical terms. For reasons like this, some of the more important mathematicalquestions can be phrased in terms of categorical language. Formal study of categorytheory will not generally be useful in solving them, but categorical thinking can sometimessuggest a viable approach.

In the first two sections, we give the most general categorical concepts: categoriesand functors. The next three sections introduce some important kinds of universal prop-erties: initial and terminal objects, products, coproducts, pushouts, and pullbacks. Wegive constructions realizing these universal properties in various categories of interest,particularly in the categories of sets, groups, and abelian groups. The constructions re-alizing these universal properties in these three categories can be quite different. Thus,the reader can see that a general categorical notion can behave quite differently in differ-ent categories. The insight coming from the categorical concept tells us nothing aboutwhether or how that concept might be realized in a particular category.

In Section 6.6, we define the notion of a free object on a set in a given categoryand construct the free groups and the free abelian groups. Then in Section 6.7, we givethe technique of describing a group by generators and relations. This technique can bevaluable in studying infinite groups, and arises frequently in topology, as the fundamentalgroup of a CW complex with one vertex is expressed naturally in this form.

We then define direct and inverse limits, giving constructions for them in variouscategories. Direct limits will be used in the construction of algebraic closures of fields.

163

CHAPTER 6. CATEGORIES IN GROUP THEORY 164

Inverse limits are used to construct the ring of p-adic integers, Zp.Then, generalizing free functors, we define adjoint functors, giving examples and

showing their utility for deducing general results. We then close with a description ofcolimits and limits indexed on a partially ordered set or a small category. This materialhas applications to sheaves and to various of the constructions in algebraic geometry andhomotopy theory, and generalizes many of the universal properties described earlier inthe chapter.

Because of the number of concepts and constructions here that may be new to thereader, Sections 6.2, 6.3, 6.4, 6.5, and 6.8 have been divided into subsections, each withits own set of exercises.

6.1 Categories

The notion of a category is very abstract and formal, and is best understood by goingback and forth between the definition and some examples. The point is that the word“object” has to be taken as a black box at first, as there are categories with manydifferent sorts of objects. The objects in a category are the things you want to study.So the category of sets has sets as objects, the category of groups has groups as objects,etc.

The morphisms in a category determine the ways in which the objects are related.For instance, when studying sets, one generally takes the functions between two sets asthe morphisms. With this in mind, here is the abstraction.

Definition 6.1.1. A category, C, consists of a collection of objects, ob(C), togetherwith morphisms between them. Specifically, if X and Y are objects of C (written X,Y ∈ob(C)) there is a set of morphisms from X to Y , denoted MorC(X,Y ). Morphisms maybe composed, just as functions or homomorphisms may be composed: If Z is anotherobject of C, then for any f ∈ MorC(Y,Z) and any g ∈ MorC(X,Y ) there is a compositemorphism f ◦ g ∈ MorC(X,Z).

This composition rule is required to satisfy two properties. First, there’s an associa-tivity law: if h ∈ MorC(W,X), then for f and g as above, we have f ◦ (g ◦h) = (f ◦ g)◦has elements of MorC(W,Z). Second, each object has an identity morphism. We write1X ∈ MorC(X,X) for the identity of X. The defining property of the identity morphismsis that g ◦ 1X = 1Y ◦ g = g for all g ∈ MorC(X,Y ).

Example 6.1.2. One of the most ubiquitous categories is the category of sets and func-tions, which we denote by Sets. The objects of Sets consist of all possible sets. Themorphisms between X and Y in Sets are all the functions from X to Y . The identityfunction is the usual one, and the properties of composition of functions are easily seento satisfy the requirements for this to be a category.

A very large number of the categories we will look at can be thought of as subcate-gories of the category of sets and functions:

Definition 6.1.3. A category D is a subcategory of C if ob(D) is a subcollection ofob(C), MorD(X,Y ) is a subset of MorC(X,Y ) for all objects X and Y of D, and thelaws of composition for the two categories coincide, as do the identity morphisms.

It may come as a surprise that not all categories are equivalent (in a formal sensewe shall give below, but which includes the intuitive notion of two categories being


isomorphic) to subcategories of Sets.1 But the structure of Sets does provide a verygood intuition for what a category is.

Of course, one can define categories from now until doomsday, and very few of theones chosen randomly will be mathematically interesting. In particular, if one wants tostudy something like groups, one should not take the morphisms between two groups tobe all functions between their underlying sets. The morphisms in a given category shouldconvey information about the basic structure of the objects being considered. Withoutthat, a study of the category will not shed light on the mathematics one wants to do.

Examples 6.1.4.1. The category in which we’ve been working for the most part here is the category Gp,

of groups and homomorphisms. Thus, the objects are all groups and the morphismsbetween two groups are the group homomorphisms between them.

2. An interesting subcategory of Gp is the category of abelian groups and homomor-phisms, which we denote by Ab. Thus, the objects of Ab are all abelian groups,while the morphisms between two abelian groups are the group homomorphismsbetween them.

The inclusion of Ab in Gp is an example of what’s called a full subcategory:

Definition 6.1.5. A subcategory D of C is a full subcategory if for any objects X,Y ∈ob(D), the inclusion MorD(X,Y ) ⊂ MorC(X,Y ) is bijective.

It is sometimes useful to restrict the morphisms in a category, rather than the ob-jects. Thus, we can form a subcategory of Gp whose objects are all groups, but whosemorphisms are the injective homomorphisms. Note that in many cases here, the set ofmorphisms between two objects is the null set. But that doesn’t cause any problems asfar as the definition of a category is concerned.

It is sometimes useful to make new categories out of old ones.

Definition 6.1.6. Let C and D be categories. The product category C ×D is definedas follows. The objects of C × D are the ordered pairs (X,Y ) with X ∈ ob(C) andY ∈ ob(D). The morphisms of C ×D are given by

MorC×D((X,Y ), (X ′, Y ′)) = MorC(X,X ′)×MorD(Y, Y ′).

Here, the latter product is the cartesian product of sets.We write f×g : (X,Y )→ (X ′, Y ′) for the morphism corresponding to f ∈ MorC(X,X ′)

and g ∈ MorD(Y, Y ′), and define composition by

(f × g) ◦ (f ′ × g′) = (f ◦ f ′)× (g ◦ g′)

for f ′ × g′ : (X ′, Y ′)→ (X ′′, Y ′′) in C ×D.

One of the most important ideas about the relationships between objects in a categoryis that of isomorphism:

Definition 6.1.7. An isomorphism in a category C is a morphism f : X → Y in Cwhich admits an inverse. Here, an inverse for f is a morphism g : Y → X such thatf ◦ g = 1Y and g ◦ f = 1X .

1One example of a category not equivalent to a subcategory of Sets is the homotopy category en-countered in topology.


Clearly, a morphism in Sets is an isomorphism in Sets if and only if it is a bijection.Also, a morphism in Gp (or Ab) is an isomorphism in Gp (resp. Ab) if and only if it is agroup isomorphism in the sense we have been studying.

As in the case of finite group theory, one of the most important problems in studyinga given category is being able to classify the objects.

Definitions 6.1.8. The isomorphism class containing a given object consists of the col-lection of all objects isomorphic to it. Clearly, any two isomorphism classes having anobject in common are equal.

A classification of the objects that have a given property consists of a determinationof all the individual isomorphism classes with that property.

One example of a classification is the determination of a set of representatives of theisomorphism classes with a given property. This means a set S = {Xi | i ∈ I} of objectssuch that every isomorphism class in C having the given property contains exactly oneof the objects in S.

Not every collection of isomorphism classes has a set of representatives. For instance,the isomorphism classes of torsion-free abelian groups do not form a set.

Even if there is a classification of the objects with a given property, it can be difficultto determine which isomorphism class contains a given object. In particular, given twoobjects in a given category, there may not be any algorithmic procedure to determinewhether the objects are isomorphic.

For instance, think about the classifications we’ve seen of the groups in different or-ders. At first glance, it can be quite difficult to decide whether two groups are isomorphic.And the classifications for different orders can involve very different criteria.

The development of criteria to determine whether two objects are isomorphic is oneof the most important problems in studying a given category. It should be noted thatgeneral category theoretic techniques are rarely useful for solving this problem. However,the notion of functors, to be studied in Section 6.2, can be useful for showing that twoobjects are not isomorphic.

There are partial forms of invertibility which can be useful.

Definitions 6.1.9. A left inverse for a morphism f : X → Y in C is a morphismg : Y → X in C such that g ◦ f = 1X . A left inverse is often called a retraction, and Xis often called a retract of Y in C.

Similarly, a right inverse for f : X → Y in C is a morphism g : Y → X in C suchthat f ◦ g = 1Y . A right inverse for f is often called a section of f .

We shall also consider categories of monoids. We write Mon for the category whoseobjects are all monoids and whose morphisms are the homomorphisms of monoids. Wewrite Amon for the full subcategory whose objects are the abelian monoids.

Exercises 6.1.10.1. Show that if f : X → Y is an isomorphism in C, then its inverse is unique.

2. Show that every injection in Sets admits a left inverse, and that every surjectionin Sets admits a right inverse. Are these inverses unique?

3. Does every injective group homomorphism have a left inverse in Gp? Does everysurjective group homomorphism have a right inverse in Gp?

4. Let g be a left inverse for f : X → Y in C. Suppose that g admits a left inversealso. Show then that f is an isomorphism with inverse g.


5. Suppose that f : G → K is a morphism in Gp that admits a right inverse. Showthen that G is a split extension of ker f by K.

6. Suppose that i : H → G is a morphism in Gp that admits a left inverse. Show thati must be injective. Show that if i(H) � G, then G ∼= H ×G/i(H). What can yousay if i(H) is not normal in G?

7. Let G be a group and let G-sets be the category whose objects are all G-setsand whose morphisms are all G-maps. Show that the isomorphisms in G-sets areprecisely the G-isomorphisms.

8. Let C be any category and let X ∈ ob(C). Show that MorC(X,X) is a monoidunder composition of morphisms. We call it the monoid of endomorphisms of X,and write MorC(X,X) = EndC(X). Show that this monoid is a group if and onlyif every morphism from X to itself is an isomorphism in C.

9. Let M be a monoid. Show that there is a category which has one object, O,such that the monoid of endomorphisms of O is M . Thus, monoids are essentiallyequivalent to categories with one object.

6.2 Functors

Just as we can learn about a group by studying the homomorphisms out of it, we canlearn about a category by studying its relationship to other categories. But to getinformation back, we shall need to define ways to pass from one category to another thatpreserve structure. We study covariant functors here, and study contravariant functorsin a subsection below.

Definition 6.2.1. Let C and D be categories. A functor, F : C → D assigns to each ob-ject X ∈ ob(C) an object F (X) ∈ ob(D) and assigns to each morphism f ∈ MorC(X,Y )a morphism F (f) ∈ MorD(F (X), F (Y )), such that

F (1X) = 1F (X)

for all X ∈ ob(C), and

F (f) ◦ F (g) = F (f ◦ g)

for each pair of morphisms f and g in C for which the composite f ◦ g is defined.

Functors are often used in mathematics to detect when objects of a given categoryare not isomorphic.

Lemma 6.2.2. Let F : C → D be a functor and suppose that f : X → Y is an isomor-phism in C. Then F (f) : F (X) → F (Y ) is an isomorphism in D. Thus, if Z and Ware objects of C, and if F (Z) and F (W ) are not isomorphic in D, then Z and W mustnot be isomorphic in C.

Proof Let g be the inverse of f . Then F (g) ◦ F (f) = F (g ◦ f) = F (1X) = 1F (X).Similarly, F (f) ◦ F (g) = 1F (Y ). Thus, F (g) is an inverse for F (f) in D.


Indeed, in a lot of ways, the preceding lemma is the main reason for looking atfunctors, to the point where the definition of functor is tailored to make it true. Theutility of this approach lies in the fact that we can often find functors to categories thatare much more manageable than the one we started in.

Examples 6.2.3.1. Let p be a prime. Then for an abelian group G, the set Gp of p-torsion elements of G

is a subgroup of G. As shown in Lemma 4.4.19, if f : G→ H is a homomorphismwith H abelian, then f restricts to a homomorphism fp : Gp → Hp. Clearly,gp ◦ fp = (g ◦ f)p for any homomorphism g : H → K with K abelian, and (1G)p =1(Gp), so there is a functor Fp : Ab → Ab given by setting Fp(G) = Gp for allabelian groups G, and setting Fp(f) = fp for f : G → H a homomorphism ofabelian groups.

Notice that while Fp(f) may be an isomorphism even if f is not, Corollary 4.4.21shows that if f : G→ H is a homomorphism between finite abelian groups, then fis an isomorphism if and only if Fp(f) is an isomorphism for each prime p dividingeither |G| or |H|.Thus, on the full subcategory of Ab whose objects are the finite abelian groups,the collection of functors Fp, for p prime, together detect whether a morphism isan isomorphism.

2. Let C be any category and let X be an object of C. We define a functor F : C →Sets as follows. For an object Y of C, we set F (Y ) = MorC(X,Y ). If f : Y → Zis a morphism from Y to Z in C, we define F (f) : MorC(X,Y )→ MorC(X,Z) bysetting F (f)(g) = f ◦ g for all g ∈ MorC(X,Y ). (It is often customary to write f∗instead of F (f).) The reader may easily check that F is a functor.

3. Define φ : Gp → Sets by setting φ(G) = G, the underlying set of G. For ahomomorphism f : G → H, we set φ(f) = f . We call φ the forgetful functor thatforgets the group structure on G.

More generally, any functor that simply forgets part of the structure of a given objectis called a forgetful functor. Thus, there is also a forgetful functor from Gp to Mon.

Definition 6.2.4. A functor C × C → D is often called a bifunctor on C, and may bethought of as a functor of two variables.

Example 6.2.5. There is a bifunctor F : Sets × Sets → Sets obtained by settingF (X,Y ) = X × Y and setting F (f × g) = f × g, for f ∈ MorSets(X,X ′) and g ∈MorSets(Y, Y ′). Here, the right-hand function f×g is the usual mapping f×g : X×Y →X ′ × Y ′ given by (f × g)(x, y) = (f(x), g(y)).

Exercises 6.2.6.1. Show that functors preserve left inverses and right inverses.

2. Show that there is a functor F : Gp → Gp such that F (G) is the commutatorsubgroup of G for every group G. Here, if f : G → G′ is a group homomorphism,then F (f) is obtained by restricting f to the commutator subgroup.

3. Give an example of groups H and G such that H is a retract of G, but Z(H) is nota retract of Z(G). Deduce that there is no functor on Gp that takes each group toits center.


4. Show that there is a functor F : Gp → Ab such that F (G) = Gab for all groupsG, and if f : G → G′ is a group homomorphism, then F (f) = fab, the uniquehomomorphism such that the following diagram commutes.

G G′

Gab G′ab��

πG

��f

��

πG′

��fab

Here, for any group H, we write πH for the canonical map from H to Hab =H/[H,H].

5. Show that the inclusion of a subcategory is a functor. In particular the inclusionof a category in itself is called the identity functor.

6. Let φ : Gp → Sets be the forgetful functor. Suppose that φ(f) is an isomor-phism. Is f then an isomorphism? Alternatively, suppose that φ(G) and φ(G′) areisomorphic in Sets. Are G and G′ then isomorphic in Gp?

7. Show that passage from a monoid M to its group of invertible elements Inv(M)gives a functor from Mon to Gp (or from Amon to Ab).

8. LetG be a group and letG be the category with one object, O, such that the monoidof endomorphisms of O is G. Show that there is a one-to-one correspondencebetween G-sets and the functors from G into Sets.

9. Let M and M ′ be monoids and let M and M ′ be categories with one object whosemonoids of endomorphisms are M and M ′, respectively. Show that there is a one-to-one correspondence between the monoid homomorphisms from M to M ′ andthe functors from M to M ′.

10. Let C and D be categories.

(a) Show that there are functors πC : C ×D → C and πD : C ×D → D obtainedby setting πC(X,Y ) = X, πD(X,Y ) = Y , πC(f × g) = f and πD(f × g) = g.

(b) For an object Y of D, show that there is a functor ιY : C → C ×D obtainedby setting ιY (X) = X × Y and ιY (f) = f × 1Y .

Contravariant Functors

Now we come to the fact that there’s more than one kind of functor. The functors we’veconsidered so far are called covariant functors. The other kind of functors are calledcontravariant. They behave very much like covariant ones, except that they reverse thedirection of morphisms.

Definition 6.2.7. A contravariant functor F : C → D associates to each X ∈ ob(C)an object F (X) ∈ ob(D) and to each morphism f ∈ MorC(X,Y ) a morphism F (f) ∈MorD(F (Y ), F (X)), such that

F (1X) = 1F (X)


for all X ∈ ob(C), and

F (f) ◦ F (g) = F (g ◦ f)

for each pair of morphisms f and g in C for which the composite g ◦ f is defined.

Contravariant functors may be viewed as covariant functors out of a different category.

Definition 6.2.8. Let C be a category. The opposite category, Cop, is defined by setting

ob(Cop) = ob(C) and setting MorCop(X,Y ) = MorC(Y,X)

for all X,Y ∈ ob(C). Here, if f : X → Y is a morphism in C, we write fop : Y → X forthe associated morphism in Cop. We define composition in Cop by setting fop ◦ gop =(g ◦ f)op, and set the identity morphism of X ∈ ob(Cop) equal to (1X)op.

The point here is that any contravariant functor on C may be viewed as a covariantfunctor on Cop. The reader may supply the details:

Proposition 6.2.9. For any category C, there is a contravariant functor I : C → Cop

defined by setting I(X) = X for X ∈ ob(C) and setting I(f) = fop for f ∈ MorC(X,Y ).If F : C → D is a contravariant functor, then there is a covariant functor F op :

Cop → D defined by setting F op(X) = F (X) for X ∈ ob(C) = ob(Cop) and settingF op(fop) = F (f) for f ∈ MorC(X,Y ). Thus, F itself factors as the composite

CI−→ Cop F op

−−→ D.

Exercises 6.2.10.1. Let C be any category and let Y be an object of C. For X ∈ ob(C), let F (X) =

MorC(X,Y ). For f ∈ MorC(X,X ′), define F (f) : MorC(X ′, Y )→ MorC(X,Y ) byF (f)(g) = g ◦ f . (It is often customary to write f∗ instead of F (f).) Show that Fdefines a contravariant functor from C to Sets.

2. Let C be a category. Show that there is a functor Hom : Cop × C → Sets, givenby setting Hom(X,Y ) = MorC(X,Y ) for each pair of objects X,Y of C. Here, iff : X ′ → X and g : Y → Y ′ are morphisms in C, we define

Hom(fop, g) : MorC(X,Y )→ MorC(X ′, Y ′)

by Hom(fop, g)(h) = g ◦ h ◦ f for all h ∈ MorC(X,Y ).

3. Show that a contravariant functor takes isomorphisms to isomorphisms.

4. Show that a contravariant functor takes morphisms that admit left inverses tomorphisms that admit right inverses and vice versa.

5. LetG be a group and letG be the category with one object, O, such that the monoidof endomorphisms of O is G. Show that there is a one-to-one correspondencebetween right G-sets and the contravariant functors from G into Sets.


6.3 Universal Mapping Properties: Products and Co-products

An object, or the relationship between a family of objects connected by morphisms, issometimes uniquely determined by what morphisms exist either in or out of the objector family. This phenomenon is known as a universal mapping property, and is perhapsbest understood through examples. We treat initial and terminal objects here, and thengive subsections for products, coproducts, and free products.

The simplest examples of universal mapping properties are initial and terminal objectsfor a category:

Definitions 6.3.1. Suppose that X ∈ ob(C) has the property that MorC(X,Y ) hasexactly one element for each Y ∈ ob(C). Then X is called an initial object for C.

Alternatively, if MorC(Y,X) has exactly one element for each Y ∈ ob(C), then X iscalled a terminal object for C.

If X is both an initial object and a terminal object for C, we call X a zero object forC.

One example that may not be immediately obvious is that the null set is an initialobject for Sets. The point here is that a function is determined by where it sends theelements of its domain. Since the null set has no elements, there is exactly one way tospecify where its elements go.

Exercises 6.3.2.1. Show that any two initial objects for C must be isomorphic.

2. Show that any two terminal objects for C must be isomorphic.

3. Show that a set with one element is a terminal object for Sets.

4. Show that the trivial group e is a zero object in both Gp and in Ab.

5. Let G be a group. Show that the G-set with one element is a terminal object inG-sets.

6. Let G be any nontrivial group. Show that G-sets does not have an initial object.

Products

Other universal mapping properties express relationships between more than one ob-ject. These properties often recur in many different categories, and are given namesappropriate to that fact. The one we’ve seen the most of is products:

Definition 6.3.3. Let X,Y ∈ ob(C). A product for X and Y in C consists of anobject, Z ∈ ob(C), together with two morphisms, πX : Z → X and πY : Z → Y , calledprojections. The universal property that makes these a product is as follows: For anyobject W ∈ ob(C) and any two morphisms f : W → X and g : W → Y , there is a unique


morphism (f, g) : W → Z such that the following diagram commutes:

W

Z Y

X,��

��

��

��

��

f

��

��

�

h

��g

��πY

��πX

Here, h = (f, g).

Thus, the projection maps are part of the definition of a product, and not just theobject Z itself.

Examples 6.3.4.1. The cartesian product, together with the usual projections, is the product in Sets:

a function into the product is determined by its component functions, which areprecisely the composites of the original function with the projection maps. Indeed,a complete argument showing that the cartesian product is the product in thecategory of sets is given in Proposition 1.6.5.

2. In Gp, the direct product has been shown in Lemma 2.10.5 to be the categoricalproduct, and this remains true in Ab.

In Gp and Ab, the product has some extra features (like the inclusion maps) that donot occur in Sets or more general categories.

The universal mapping property that defines the product has at most one solution (upto isomorphism) for a given pair of objects X and Y . This is because of the uniqueness ofthe map from W to Z that makes the diagram commute in the definition of the product:

Proposition 6.3.5. Suppose given objects and morphisms

XπX←−− Z πY−−→ Y and X

π′X←−− Z ′ π′

Y−−→ Y

in C such that both (Z, πX , πY ) and (Z ′, π′X , π

′Y ) satisfy the universal mapping property

for the product of X and Y in C. Let f = (πX , πY ) be the unique morphism from Z toZ ′ which makes the following diagram commute. (Here, we use the universal mapping

property of Xπ′

X←−− Z ′ π′Y−−→ Y .)

Z

Z ′ Y

X��

��

��

��

��

πX

��

��

�

f

��

πY

��π′

Y

��π′

X

Then f is an isomorphism.


Proof By the universal property of X πX←−− ZπY−−→ Y , there is a unique morphism

g : Z ′ → Z such that πX ◦ g = π′X and πY ◦ g = π′

Y .Now g◦f : Z → Z is the unique morphism such that πX◦g◦f = πX and πY ◦g◦f = πY .

Thus, g ◦ f = 1Z .Similarly, f ◦ g = 1Z′ , and the result follows.

Exercises 6.3.6.1. What is the product in G-sets? Is the product of two orbits an orbit? What are

its isotropy groups?

2. Suppose that for each pair (X,Y ) of objects of C we are given a product for X andY . Show that passage from a pair of objects to their product gives a functor fromC × C to C.

Coproducts

It is frequent in categorical mathematics to ask what happens to a definition if you turnall the arrows around. This is called dualization. The dual of a product is called acoproduct.

Definition 6.3.7. Let X,Y ∈ ob(C). A coproduct for X and Y in C is an object Ztogether with morphisms ιX : X → Z and ιY : Y → Z, called the natural or canonicalinclusions, with the following universal property: For any pair of morphisms f : X →Wand g : Y → W , there is a unique morphism 〈f, g〉 : Z → W such that the followingdiagram commutes:

Y

X Z

W,

��ιY

��

��

��

��

�

g��ιX

��

f��

��

��h

Here, h = 〈f, g〉.Examples 6.3.8.

1. Proposition 2.10.6, when applied strictly to abelian groups and their homomor-phisms, shows that the direct product, together with the canonical inclusions,

Gι1−→ G×H ι2←− H,

is the coproduct in the category Ab of abelian groups.

2. The coproduct in Sets is called the disjoint union. Here, the disjoint union of Xand Y is denoted X

∐Y . Its elements are the union of those in X and those in Y ,

where the elements from X are not allowed to coincide with any of the elements ofY . Formally, this may be accomplished as follows. Set theory tells us that there’ssome set, S, containing both X and Y . The disjoint union of X and Y is definedto be the subset of S × {1, 2} consisting of the elements (x, 1) with x ∈ X and theelements (y, 2) with y ∈ Y . We then have ιX(x) = (x, 1) and ιY (y) = (y, 2).


Now if f : X → W and g : Y → W are functions, we define 〈f, g〉 : X∐Y → W

by setting 〈f, g〉(x, 1) = f(x) and setting 〈f, g〉(y, 2) = g(y). Since ιX and ιYare disjoint and have X

∐Y as their union, this is well defined and satisfies the

universal property for the coproduct.

Thus, unlike the case of the product, the forgetful functor from Ab to Sets doesnot preserve coproducts. As we shall see presently, the forgetful functors Ab → Gpand Gp → Sets also fail to preserve coproducts. Indeed, the coproducts in these threecategories are given by completely distinct constructions.

Exercises 6.3.9.1. What is the coproduct in G-sets?

2. Show that the direct product is not the coproduct in Gp, even if the two groupsare abelian.

3. Prove the analogue of Proposition 6.3.5 for coproducts, showing that any two co-products for a pair of objects are isomorphic via a morphism that makes the ap-propriate diagram commute.

Free Products

The coproduct in Gp is called the free product. We shall give the construction now. LetH and K be groups, and write X = H

∐K for the disjoint union of the sets H and K.

By a word in X we mean a formal product x1 � x2 � · · · � xk, with xi ∈ X for all i. (Wewrite � for the formal product operation to distinguish it from the multiplications in Hand K.) The empty word is the word with no elements.

The elements in the free product H ∗K will not be the words themselves, but ratherequivalence classes of words. The equivalence relation we shall use is derived from aprocess called reduction of words. Words may be reduced as follows.

First, if xi and xi+1 are both in the image of ιH : H → X, then x1 � · · · � xk reducesto x1 � · · ·� (xixi+1)� · · ·�xk, where xixi+1 is the product of xi and xi+1 in H. There is asimilar reduction if xi and xi+1 are both in the image of ιK : K → X. Finally, if xi is theidentity element of either H or K, then x1�· · ·�xk reduces to x1�· · ·�xi−1�xi+1�· · ·�xk.In particular, for either e ∈ H or e ∈ K, then the singleton word e reduces to the emptyword.

We shall refer to the reductions above as elementary reductions, and write w ↘e w′ orw′ ↙e w if the word w′ is obtained from the word w by one of these elementary reductions.More generally, given a sequence w0 ↘e w1 ↘e . . .↘e wn of elementary reductions, we saythat w0 reduces to wn, and write w0 ↘ wn (or wn ↙ w0). We shall also declare thatw ↘ w, so that ↘ is reflexive.

The equivalence relation we shall use is the equivalence relation generated by theprocess of reduction, in the sense of Section 1.5. This means that the words w and w′

are equivalent, written w ∼ w′, if there is a sequence w = w0, . . . , wn = w′ of words suchthat for i = 0, . . . , n− 1, either wi ↘ wi+1 or wi ↙ wi+1.

Note that some words (e.g., h � h′ � eK � k, where h, h′ ∈ H, k ∈ K, and eK is theidentity of K) can be reduced in more than one way. (Also, as in the case of wordscontaining an adjacent pair such as eK � k above, two different elementary reductionoperations can produce exactly the same word as a result. However, both cases arerepresented by the same symbols, w ↘e w′.) An examination of words which admit morethan one reduction shows the following important fact.


Lemma 6.3.10. Suppose given two different elementary reduction operations on a wordw, giving w ↘e w1 and w ↘e w2. Then either w1 = w2 or there is a word w3 withw1 ↘e w3 ↙e w2.

It is convenient to be able to represent an element of the free product by what’sknown as a reduced word:

Definition 6.3.11. We say that a word x1 � · · · � xk is reduced if xi �= e for 1 ≤ i ≤ kand if no adjacent pair xi, xi+1 comes from the same group (H or K). We also declarethe empty word to be reduced.

The next lemma is fundamental for the understanding of free products.

Lemma 6.3.12. Let w be a word in X = H∐K. Then there is a unique reduced word,

w′ such that w ↘ w′.

Proof We argue by induction on the number of letters in w. If w is reduced, then itcannot be reduced further, and we take w′ = w.

Otherwise, there is an elementary reduction from w to a word v. By induction, thereis a unique reduced word w′ with v ↘ w′. But then w ↘ w′. For the uniquenessstatement, suppose that w′′ is reduced and that w ↘ w′′, say, via w ↘e w1 ↘e . . .↘e w′′.If w1 = v, then w′ = w′′ by the induction hypothesis. Otherwise, by Lemma 6.3.10, thereis a word w with v ↘e w ↙e w1. But if w ↘ w′′′ with w′′′ reduced, then the uniquenessstatement of the induction hypothesis gives w′ = w′′′ = w′′.

Recall that ∼ is the equivalence relation generated by reduction.

Proposition 6.3.13. Every equivalence class under ∼ contains exactly one reduced word.If w is a reduced word, then every word in the equivalence class of w under ∼ reduces tow.

Proof Suppose that w ∼ w′. Then there is a sequence w = w0, . . . , wn = w′ such thatfor each i = 0, . . . , n− 1, either wi ↘ wi+1 or wi ↙ wi+1. In either case, Lemma 6.3.12tells us that the unique reduced word w′′ to which wi reduces is equal to the uniquereduced word to which wi+1 reduces. In particular, each wi reduces to w′′, and there canbe at most one reduced word in such a sequence.

We define the elements of the free product H ∗ K to be the equivalence classes ofwords under ∼. Thus, the elements of H ∗K are in one-to-one correspondence with thereduced words. We define a binary operation on the words themselves by concatenation:

(x1 � · · · � xk) · (y1 � · · · � yl) = x1 � · · · � xk � y1 � · · · � yl.In word notation, we write w · w′ for the product in this sense of the words w and w′.But clearly, if w reduces to w1 and w′ reduces to w′

1, then w · w′ reduces to w1 · w′1.

Thus, concatenation induces a well defined binary operation on the equivalence classesof words.

Lemma 6.3.14. Concatenation of words induces a group structure on the free productH ∗K.

Proof Concatenation is clearly an associative operation on the set of words themselves,with the empty word as an identity element. Since it preserves equivalence classes, thefree product inherits a monoid structure from the set of words themselves. But theinverse of x1 � · · · � xk in H ∗K is given by x−1

k � · · · � x−11 .


The inclusions of H and K in X induce group homomorphisms from H and K toH ∗K, which we’ll denote by ιH and ιK .

Proposition 6.3.15. The free product H ∗K, together with the homomorphisms ιH andιK is the coproduct of H and K in the category of groups.

Proof Given group homomorphisms f : H → G and g : K → G, we define 〈f, g〉 on Xto be f on H ⊂ X and to be g on K ⊂ X. Given a word x1 � · · · � xk in X, we may thendefine 〈f, g〉(x1 � · · · � xk) to be (〈f, g〉(x1)) . . . (〈f, g〉(xk)). Stipulating that 〈f, g〉 carriesthe empty word to the identity element of G, we obtain a monoid homomorphism fromthe monoid of words in X into G. Since two words differing by an elementary reductionare carried to the same element of G (since f and g are homomorphisms), passage toequivalence classes of words gives a group homomorphism from H ∗K into G.

Clearly, 〈f, g〉 ◦ ιH = f and 〈f, g〉 ◦ ιK = g. Since the images of ιH and ιK generateH ∗K, 〈f, g〉 is the unique homomorphism with this property.

Free products can be somewhat intractable, especially in comparison to the groups wehave considered previously. But they are important in understanding groups in general.For instance, finitely generated free groups, which we shall study in Section 6.6, areiterated free products of copies of the integers. We shall see that any finitely generatedgroup is a factor group of such a free group, and this opens up a new way of looking atgroups, called generators and relations, which we shall study in Section 6.7.

A concrete example of free products “in nature” is PSl2(Z), which turns out to beisomorphic to Z2 ∗ Z3.

It should be noted that not all categories have products, coproducts, or other of theconstructions we shall give.

Exercises 6.3.16.1. Show that if H and K are both nontrivial, then H ∗K is infinite.

2. Show that if H and K are both nontrivial, then H ∗K is nonabelian even if H andK are both abelian.

3. Show that the homomorphisms ιH : H → H ∗K and ιK : K → H ∗K are injective.

4. What is H ∗ e?5. Show that the same constructions that give the product and coproduct in Gp give

a product and coproduct in Mon.

6. Show that the same constructions that give the product and coproduct in Ab givea product and coproduct in Amon.

6.4 Pushouts and Pullbacks

We now give a generalization of products and coproducts. These are universal mappingproperties not just of pairs of objects, but of particular collections of objects and mor-phisms. We treat pullbacks now, and treat pushouts in a separate subsection. In furthersubsections, we will construct pushouts in Gp, Ab, and Sets.


Definition 6.4.1. Suppose given a pair of morphisms into Z, f : X → Z and g : Y → Z.This gives us the following diagram:

X

Y Z��

f

��g

.

A pullback for this diagram consists of the following data: We have an object W andmorphisms g : W → X and f : W → Y such that

1.

W X

Y Z

��g

��f

��f

��g

commutes.

2. For every commutative diagram

W ′ X

Y Z

��g′

��f ′

��f

��g

there is a unique morphism h : W ′ →W such that the following diagram commutes.

W ′

W X

Y Z��

��

��

��

��

f ′

��

��

��

h

��g′

��f

��g

��f

��g

In other words, the pullback is a universal completion of the original 3-object diagramto a commutative square. When the context is totally unambiguous, one often refers tothe object W as the pullback, but the proper usage is that “pullback” refers to the wholesquare.

Exercises 6.4.2.1. Show that pullbacks in Sets, Gp, and Ab may all be constructed as follows. Given

a diagram

X

Y Z��

f

��g

,


let Y ×Z X ⊂ Y ×X be given by

Y ×Z X = {(y, x) | g(y) = f(x)}.

Then the pullback of the original diagram is given by

Y ×Z X X

Y Z

��g

��

f

��

f

��g

,

where f(y, x) = y and g(y, x) = x. The construction Y ×Z X is sometimes calledthe fibre product of Y and X over Z.

2. Show that pullbacks, when they exist, are unique.

3. Show that if C has a terminal object, ∗, then the pullback of the unique diagram

X

Y ∗��

��

is the product of X and Y (i.e., show that the pullback of this diagram satisfies thesame universal property as the product). Thus, pullbacks do generalize products.

4. Suppose given a pullback diagram

W X

Y Z

��g

��f

��f

��g

in either Gp or Ab. Show that ker f ∼= ker f and that ker g ∼= ker g.

5. In either Gp or Ab, show that the pullback of

X

e Y��

f

��

is ker f .

Pushouts

Pushout is the notion dual to pullback. In Sets and Gp, pushouts are more delicate andharder to construct than pullbacks, but they can be very important.


Definition 6.4.3. The pushout of a diagram

X Y

Z

��f

��g

is a commutative diagram

X Y

Z W

��f

��g

��g

��f

which is universal in the sense that if

X Y

Z W ′

��f

��

g

��g′

��f ′

is another commutative diagram, then there is a unique morphism h : W → W ′ suchthat the following diagram commutes.

X Y

Z W

W ′

��f

��

g

��g

��

��

��

��

��

g′��f

��

f ′��

��

��h

As in the case of pullbacks, it is customary to refer to the object W as the pushoutas well.

Exercises 6.4.4.1. Show that pushouts, when they exist, are unique.

2. Show that if C has an initial object, ∗, then the pushout of the unique diagram

∗ Y

Z

��

��

is the coproduct of Y and Z.


3. Suppose given a pushout diagram

X Y

Z W

��f

��g

��g

��f

in C, where f is an isomorphism. Deduce that f is also an isomorphism.

4. Show, using the universal mapping property, that in Sets, Gp, or Ab, the pushoutof a product diagram

X × Y Y

X

��πY

��

πX

is the terminal object. (In Sets, we must assume that neither X nor Y is the nullset.)

Pushouts in Gp and Ab

Pushouts are closely related to coproducts, and are often constructed using them. Wefirst give the construction of pushouts in Ab.

Proposition 6.4.5. The pushout in Ab of a diagram

A B

C

��f

��g

is the diagram

A B

C (C ×B)/D,

��f

��g

��ιB

��ιC

where D = im(g,−f) = {(g(a),−f(a)) | a ∈ A} ⊂ C × B, and where the maps ι are thecomposites of the standard inclusions of B and C in the product with the canonical mapfrom C ×B to its factor group.

Proof First, note that factoring out by D makes the square commute. Now given acommutative square

A B

C G

��f

��g

��g′

��f ′


in Ab, there is a unique homomorphism h from the coproduct C × B into G such thath ◦ ιB = g′ and h ◦ ιC = f ′. But then h((g(a),−f(a)) = f ′g(a) − g′f(a) = 0 bycommutativity of the diagram. Thus, h factors through the factor group of C ×B by D.But factorizations of homomorphisms through factor groups are unique.

In order to understand pushouts in the category of groups, we will need the followingconcept.

Definition 6.4.6. Let G be a group and let S be any subset of G. We write 〈〈S〉〉 ⊂ Gfor the subgroup generated by T = {gxg−1 | g ∈ G, x ∈ S}, the set of all conjugatesof the elements of S. We call 〈〈S〉〉 the normal subgroup generated by S. As usual, ifS = {g1, . . . , gn}, we write 〈〈S〉〉 = 〈〈g1, . . . , gn〉〉.

From a different viewpoint, if S is contained in the subgroup H � G and if thesubgroup 〈〈S〉〉 = 〈T 〉 above equals H, we say that H is normally generated by S in G.

Lemma 6.4.7. Let S be a subset of the group G. Then 〈〈S〉〉 is the smallest normalsubgroup of G that contains S.

Proof Clearly, any normal subgroup containing S must contain T , and hence mustcontain the subgroup generated by T . Thus, it suffices to show that 〈〈S〉〉 is normal inG.

A generic element of 〈〈S〉〉 is a product of powers of elements of T . Since any conjugateof an element of T is also in T , the result follows.

The pushout in Gp is called the free product with amalgamation. The constructionproceeds precisely as in the abelian case. We give it as an exercise.

We shall see below that every group may be presented in terms of “generators andrelations,” which is an example of a free product with amalgamation.

Exercises 6.4.8.1. Show that the pushout in Gp of the diagram

G H

K

��f

��g

is the diagram

G H

K (K ∗H)/〈〈S〉〉,

��f

��

g

��ιH

��ιK

where S = {g(x) · f(x)−1 |x ∈ G}, and where the maps ι are the composites of thestandard inclusions of H and K in the free product with the canonical map fromK ∗H to its factor group.


2. What is the pushout in Gp of a diagram of the following form?

H G

e

��f

��

What would the pushout be in Ab if the diagram were also in Ab?

3. Suppose given a pushout diagram

A B

C D

��f

��g

��g

��f

in Ab.

(a) If f is injective, show that f is also.

(b) Show that f is surjective if and only if f is.

4. Suppose the diagram of the previous problem is a pushout square in Gp ratherthan Ab.

(a) If f is surjective, show that f is also.

(b) Show that f may be surjective when f is not.

Pushouts in Sets

In Sets, the pushout is again a quotient of the coproduct. This time we proceed byplacing an equivalence relation on the coproduct and passing to the set of equivalenceclasses. Thus, suppose given a diagram

X Y

Z

��f

��g

in Sets. Our equivalence relation on the coproduct Y∐Z will be the equivalence relation,

∼, generated by the following notion of basic equivalence: There is a basic equivalencebetween ιY (y) with ιZ(z) in Y

∐Z if there is an element x ∈ X with f(x) = y and

g(x) = z. In this case, we write either ιY (y) � ιZ(z) or ιZ(z) � ιY (y), as the relationin question is symmetric.

Since ∼ is the equivalence relation generated by �, we see that two elements v, v′ ∈Y

∐Z are equivalent under ∼ if and only if there is a sequence

v = v0 � v1 � . . . � vn = v′.

We define the pushout object P to be the set of equivalence classes in Y∐Z under

∼. We then have a function π : Y∐Z → P that takes each element to its equivalence

class.


Exercises 6.4.9.1. Show that the pushout in sets of the above diagram is

X Y

Z P,

��f

��g

��ιY

��ιZ

where the functions ι are the composites of the standard inclusions into the co-product with the quotient function π.

2. If f is injective, show that ιZ is also. (In this case, P is often denoted Z ∪g Y .)

3. If f is surjective, show that ιZ is also.

6.5 Infinite Products and Coproducts

We’ve discussed products and coproducts of pairs of objects in a category. This issufficient, by induction, to provide products or coproducts for any finite collection ofobjects. But in some cases, as we shall see in our discussion of rings and modules, it isnecessary to make use of products or coproducts of infinite families of elements.

We define infinite products and coproducts here and construct the products. We giveconstructions of infinite coproducts in a separate subsection.

We shall consider sets of objects {Xi | i ∈ I}. In other words, I is a set, called theindexing set, and Xi is an object of our category for each i ∈ I.Definitions 6.5.1. The product of a set {Xi | i ∈ I} of objects of C is an object X,together with morphisms πi : X → Xi, called projections, with the following universalproperty. If fi : Y → Xi is a morphism in C for each i ∈ I, then there is a uniquemorphism f : Y → X such that πi ◦ f = fi for all i ∈ I.

The coproduct of a set {Xi | i ∈ I} of objects of C is an object X, together withmorphisms ιi : Xi → X for each i ∈ I, called the natural inclusions, with the followinguniversal property. If fi : Xi → Y is a morphism in C for each i ∈ I, then there is aunique morphism f : X → Y such that f ◦ ιi = fi for all i ∈ I.

Once again, the products in Sets, Gp, and Ab come from the same construction. Weshall see in our discussion of adjoint functors in Section 6.9 that this is no coincidence.The construction of products in Sets is as follows. The elements of the product

∏i∈I Xi

are collections (xi | i ∈ I) such that xi ∈ Xi for all i ∈ I. In other words, the element(xi | i ∈ I) consists of a choice, for each i ∈ I, of an element of Xi. We call (xi | i ∈ I) atuple or an I-tuple.

Another way of thinking about the elements of a product is as follows. Let S be aset that contains each Xi. (The existence of such a set is one of the standard axioms ofset theory.) Then an I-tuple in

∏i∈I Xi is a function from I to S whose value on i ∈ I

must lie in Xi. From this viewpoint, we can see that the collection of all such I-tuples,∏i∈I Xi, is a set. The projection πi :

∏i∈I Xi → Xi takes (xi | i ∈ I) to xi.

The act of choosing elements xi ∈ Xi for each i ∈ I suggests that the Axiom of Choiceis involved here. This is indeed the case.

Axiom of Choice (second form) Let {Xi | i ∈ I} be a family of nonempty sets. Thenthe product

∏i∈I Xi is nonempty.


We shall discuss the relationship between the two given forms of the Axiom of Choicein Exercises 6.5.11.

The following argument does not make use of the Axiom of Choice. The output, if onedoes not accept that axiom, is that if

∏i∈I Xi = ∅, then ∅ is the product of {Xi | i ∈ I}

in Sets. In particular, any infinite family of sets has a product in Sets.

Proposition 6.5.2. The product∏i∈I Xi, together with its projection maps, is the prod-

uct of {Xi | i ∈ I} in Sets.

Proof Suppose given a collection of functions fi : Y → Xi for each i ∈ I. Definef : Y → ∏

i∈I Xi by f(y) = (fi(y) | i ∈ I). Then πi ◦ f = fi for all i ∈ I. But twoelements of the product are equal if their coordinates are all equal, so f is the uniquefunction with this property.

Exercises 6.5.3.1. Show that the category of finite groups and homomorphisms does not have infinite

products.

2. Show that if∏i∈I Xi is the product of {Xi | i ∈ I} in an arbitrary category C, then

for each Y ∈ ob(C) there is an isomorphism of sets between MorC(Y,∏i∈I Xi)

and∏i∈I MorC(Y,Xi), where the latter product is that in Sets. (In fact, this

isomorphism is natural, in the sense of natural transformations, Section 6.9.)

3. Suppose given objects {Gi | i ∈ I} in either Gp or Ab. Define a binary operationon

∏i∈I Gi by (xi | i ∈ I) · (yi | i ∈ I) = (xiyi | i ∈ I). Show that

∏i∈I Gi, with this

multiplication, is the product of {Gi | i ∈ I} in the category in question.

Constructions of Infinite Coproducts

The infinite coproduct in Sets is the infinite disjoint union.

Definition 6.5.4. Let {Xi | i ∈ I} be a family of sets. We define∐i∈I Xi as follows.

Let S be a set that contains each Xi. Then∐i∈I Xi is the subset of S × I consisting of

the pairs (x, i) such that x ∈ Xi. The inclusion ιi : Xi →∐i∈I Xi takes x ∈ Xi to the

pair (x, i).

Proposition 6.5.5. The disjoint union, together with the structure maps ιi, is the co-product in the category of sets. Thus, given a family {Xi | i ∈ I} of sets, together withfunctions fi : Xi → Y for all i ∈ I, there is a unique function f :

∐i∈I Xi → Y such

that f ◦ ιi = fi for all i.

Proof Given the functions fi : Xi → Y , define f :∐i∈I Xi → Y by f(x, i) = fi(x) for

each pair (x, i) ∈∐i∈I Xi. Since (x, i) = ιi(x), f satisfies the requirement that f ◦ ιi = fi

for all i, and is the unique function that does so.

The infinite coproduct in Gp is called the infinite free product. It is constructed byprecise analogy to the free product of two groups. Thus, given a collection {Gi | i ∈ I} ofgroups, we set X =

∐i∈I Gi, the disjoint union of the sets Gi. The infinite free product∗i∈I Gi is obtained by a passage to equivalence classes from an equivalence relation, ∼,

on the set of all finite words in X.Here, as in the case of the free product of two groups, the equivalence relation ∼ is

the equivalence relation generated by a notion of elementary reductions. The elementaryreductions are defined as follows.


Definition 6.5.6. Let X =∐i∈I Gi and let x1 � · · · � xn be a word in X.

1. If xi is the identity in the group Gj that contains it, then

x1 � · · · � xn ↘e x1 � · · · � xi−1 � xi+1 � · · · � xn.

2. If xi and xi+1 both lie in the same group Gj ⊂ X, then

x1 � · · · � xn ↘e x1 � · · · � (xixi+1) � · · · � xn.

As in the case of the free product of two groups,∗i∈I Gi is a group under the operationinduced by concatenation of words. The canonical map ιi : Gi → ∗i∈I Gi takes g ∈ Gito the equivalence class of the singleton word g. The proof of the following propositionproceeds precisely as did that of Proposition 6.3.15.

Proposition 6.5.7. Let {Gi | i ∈ I} be any family of groups. Then the infinite freeproduct ∗i∈I Gi, together with the canonical maps ιi : Gi →∗i∈I Gi, is the coproduct of{Gi | i ∈ I} in the category of groups.

Intuition might now suggest that the infinite coproduct in Ab would coincide withthe infinite product. However, this is not the case (mainly because we cannot add aninfinite number of group elements). Rather, the coproduct is a subgroup of the infiniteproduct, called the direct sum.

Definition 6.5.8. Let {Ai | i ∈ I} be a family of abelian groups. We define the directsum

⊕i∈I Ai to be the subset of

∏i∈I Ai consisting of the tuples (ai | i ∈ I) for which

ai = 0 for all but finitely many i. We define ιi : Ai →⊕

i∈I Ai as follows: For a ∈ Ai,the coordinates of ιi(a) are all 0 except for the i-th coordinate, which is a.

We shall customarily write∑i∈I ai for the I-tuple (ai | i ∈ I) ∈

⊕i∈I Ai. Note that

the sum∑i∈I ai is actually finite, as all but finitely many of the ai are 0.

Clearly,⊕

i∈I Ai is a subgroup of the product∏i∈I Ai, and the maps ιi : Ai →⊕

i∈I Ai are homomorphisms. The next lemma is immediate from the definitions.

Lemma 6.5.9. Let {Ai | i ∈ I} be a family of abelian groups. Then in the direct sum⊕i∈I Ai, we have ∑

i∈Iai =

∑i∈I

ιi(ai),

where the sum on the right is the sum, with respect to the group operation in⊕

i∈I Ai,of the finitely many terms ιi(ai) for which ai �= 0.

Proposition 6.5.10. Let {Ai | i ∈ I} be a family of abelian groups. Then the directsum,

⊕i∈I Ai, together with the structure maps ιi : Ai →

⊕i∈I Ai, is the coproduct in

the category of abelian groups.In other words, if B is an abelian group and if fi : Ai → B is a homomorphism for

each i ∈ I, then there is a unique homomorphism f :⊕

i∈I Ai → B such that f ◦ ιi = fifor all i ∈ I.

Explicitly, the homomorphism f is given by

f

(∑i∈I

ai

)=

∑i∈I

fi(ai),

where the right-hand side is the sum in B of the finite number of terms fi(ai) for whichai �= 0.


Proof That f must satisfy the stated formula is immediate from Lemma 6.5.9 and therequirement that f ◦ ιi = fi for all i ∈ I. Moreover, the definition of ιi shows that iff :

⊕i∈I → B is defined by the stated formula, then f ◦ ιi = fi for all i, as required.

Thus, it suffices to show that setting f(∑

i∈I ai)

=∑i∈I fi(ai) defines a homo-

morphism from⊕

i∈I Ai to B. But that is immediate from the fact that each fi is ahomomorphism.

Coproducts are so important in the study of abelian groups that it is often customaryto refer to the finite products as direct sums. Thus, A1⊕· · ·⊕Ak is an alternative notationfor A1 × · · · ×Ak.Exercises 6.5.11.

1. Suppose given a family {Xi | i ∈ I} of sets. Define π :∐i∈I Xi → I by setting π ◦ ιi

equal to the constant map from Xi to i, for each i ∈ I. Show that the elements of∏i∈I Xi are in one-to-one correspondence with the functions s : I →∐

i∈I Xi suchthat π ◦ s = 1I . Deduce that the two forms of the Axiom of Choice are equivalent.

2. Show that the direct sum⊕

i∈I Ai is the subgroup of the product∏i∈I Ai generated

by⋃i∈I ιi(Ai).

3. What are the infinite coproducts in Mon and Amon?

4. For any category C, show that if X is the coproduct of {Xi | i ∈ I}, then for any ob-ject Y , there is an isomorphism of sets between MorC(X,Y ) and

∏i∈I MorC(Xi, Y ).

6.6 Free Functors

Free functors are actually a special case of the adjoint functors discussed in Section 6.9.For simplicity, and because they constitute a very important special case, we give aseparate treatment of them.

We give an illustration first. Let X be a set. Since abelian groups are sets withadditional structure, we can consider the functions f : X → A, where A is an abeliangroup. The free abelian group onX is an abelian group, FA(X), together with a functionηX : X → FA(X) which is universal in the sense that if f : X → A is any function fromX to an abelian group, then there is a unique group homomorphism f : FA(X) → Asuch that the following diagram commutes.

Xf ��

ηX ��

��

�� A

FA(X)f

��

As usual, we shall refer to FA(X) as the free abelian group on X, even though thatname more properly applies to the function ηX .

We shall give a construction of free abelian groups below. The following propositionshows how useful they can be.

Proposition 6.6.1. Let f : A→ FA(X) be an extension of the abelian group B = ker fby the free abelian group FA(X). Suppose that A is also abelian. Then the extensionsplits, and A is isomorphic to B × FA(X).


Proof We shall see below that the function ηX : X → FA(X) is always injective. Forsimplicity, we shall make use of this fact, and treat X as a subset of FA(X).

Group extensions are always onto, so that X is contained in the image of f . Thus,for each x ∈ X there is an element g(x) ∈ A such that f(g(x)) = x. But this defines afunction g : X → A. Since A is abelian, the universal property of free abelian groupsprovides a homomorphism g : FA(X) → A such that g(x) = g(x) for all x ∈ X. Weclaim that g is a section of f .

Thus, we wish to show that f ◦ g is the identity map of FA(X). But the universalproperty shows that homomorphisms out of FA(X) are determined by their restrictionto X. And f ◦ g(x) = f(g(x)) = x. Thus, f ◦ g has the same effect on X as the identitymap, so that f ◦ g is the identity as desired.

Thus, the extension splits. Since A is abelian, FA(X) acts trivially on B = ker f ,and hence A is the product of B and FA(X).

We define free functors in general as follows. (More general definitions are possible.)

Definition 6.6.2. Suppose given a category C with a forgetful functor to Sets. (In otherwords, the objects of C are sets with additional structure, and passage from the objectsto the underlying sets and from morphisms to the underlying functions provides a functorφ : C → Sets.) Then a free C-object on a set X consists of an object F (X) ∈ ob(C)together with a function ηX : X → F (X) from X to the underlying set of F (X) with theuniversal property that if f : X → A is any function from X to a C-object, then there isa unique C-morphism f : F (X)→ A such that the following diagram commutes.

Xf ��

ηX ��

��

� A

F (X)f

��

The free objects we’re most interested in will be very easy to construct. The key isthe following lemma.

Lemma 6.6.3. Let C be a category with a forgetful functor to Sets. Suppose given afamily of sets {Xi | i ∈ I} such that

1. Each set Xi has a free C-object ηi : Xi → F (Xi).

2. The collection {F (Xi) | i ∈ I} has a coproduct in C.

Write⊕

i∈I F (Xi) for the coproduct in C of the F (Xi) and write η :∐i∈I Xi →⊕

i∈I F (Xi) for the unique function that makes the following diagram commute for eachi.

Xi F (Xi)

∐i∈I

Xi

⊕i∈I

F (Xi)

��ηi

��

ιi

��

ιi

��η

Here, the maps ιi are the canonical maps for the coproducts, in Sets on the left and inC on the right.

Then η is the free C-object on∐i∈I Xi.


Proof Suppose given a function f :∐i∈I Xi → A, with A and object of C. For i ∈ I,

write fi : Xi → A for the composite

Xiιi−→

∐i∈I

Xif−→ A.

Then since ηi is the free C-object on Xi, there is a unique C-morphism f i : F (Xi)→ Asuch that f i ◦ ηi = fi.

Now apply the universal property of the coproduct in C, obtaining the unique mor-phism f from

⊕i∈I F (Xi) to A such that f ◦ ιi = f i.

An easy diagram chase shows that f ◦ η = f , and uniqueness follows by compositionof the uniqueness property of the coproduct in C and the uniqueness properties of thefree C-objects ηi.

For any set X, we have X =∐x∈X{x}. Thus, if C has arbitrary coproducts, it will

have arbitrary free objects, provided that a one-point set has a free object. But we haveseen free objects for the one-point set already in Gp and Ab: if G is a group and if g ∈ G,there is a unique homomorphism f : Z → G with f(1) = g. Thus, if η : {x} → Z takesx to 1, then η is the free object on {x} in both Gp and Ab. (One-point sets are the onlysets for which free groups are abelian, and therefore are the only sets for which the freegroups and the free abelian groups coincide.)

Proposition 6.6.4. A group is a free abelian group if and only if it is a direct sum ofcopies of Z.

A group is a free group if and only if it is a free product of copies of Z.

Proof By the preceding lemma,⊕

x∈X Z is the free abelian group on X, while ∗x∈X Zis the free group on X.

But given a direct sum or free product of copies of Z, just let X be the indexing setfor the sum or free product, and this says that our sum or free product is free in theappropriate category.

We can turn free objects into free functors as follows. The proof follows from theuniversal property and is left to the reader.

Proposition 6.6.5. Let C be a category with a forgetful functor to Sets. Suppose thatfor every set X we are given a free C-object ηX : X → F (X). Then for any functionf : X → Y there is a unique C-morphism F (f) : F (X) → F (Y ) such that F (f) ◦ ιX =ιY ◦ f .

The correspondence that takes X to F (X) and takes f to F (f) is a functor F :Sets→ C.

In practice, we identify X with its image under ιX in any of the categories underdiscussion. Thus, the free group on X is given by ∗x∈X〈x〉, where each x ∈ X hasinfinite order, and the generic element may be written xn1

1 . . . xnk

k where x1, . . . , xk ∈ Xare not necessarily distinct, and n1, . . . , nk ∈ Z.

There is a similar additive notation for the free abelian group on X. Its elements maybe written in the form n1x1 + · · ·+ nkxk, where x1, . . . , xk ∈ X and n1, . . . , nk ∈ Z. Inthis case, because the group is abelian, we may collect terms, and assume that x1, . . . , xkare distinct.

Exercises 6.6.6.


1. Show that if f : G→ F (X) is an extension of a group H by the free group on X,then f is a split extension.

2. Show that there exist group extensions f : G → FA(X) which cannot be split,where FA(X) is a free abelian group.

3. Let f : F (X) → G be a homomorphism, with F (X) the free group on X. Showthat f is onto if and only if G is generated by {f(x) |x ∈ X}.

4. Let X be a set and let F (X) and FA(X) be the free group and the free abeliangroup on X, respectively. Let f : F (X) → FA(X) be the unique homomorphismsuch that the following diagram commutes.

X

F (X) FA(X)��

��

��η

��

��

�η

��f

Here, the maps η are the canonical inclusions of X in the free group and free abeliangroup, respectively.

Show that f factors uniquely through an isomorphism f : F (X)ab ∼= FA(X) fromthe abelianization of F (X) onto FA(X).

5. Let f : FA(X)→ A be a homomorphism, with A an abelian group and with FA(X)the free abelian group on X. Show that f is onto if and only if G is generated by{f(x) |x ∈ X}.

6. What is the free G-set on a set X?

7. What is the free monoid on a set X?

8. What is the free abelian monoid on a set X?

6.7 Generators and Relations

Let G be a group and suppose that the set X ⊂ G generates G. Then there is a uniquehomomorphism f : F (X) → G from the free group on X whose restriction to X is itsinclusion in G. By Problem 3 of Exercises 6.6.6, f is surjective.

In this way, we see that every group is a factor group of a free group. But that,by itself, isn’t very useful information. We need to be able to say something about thekernel, K, of f : F (X)→ G before we can get any useful information out of it.

Free groups are very complicated objects, and we shall not attempt to say anythingabout the case where X is infinite. Thus, we shall assume that G is finitely generated,by the set X = {x1, . . . , xk}. We shall write F (X) = F (x1, . . . , xk).

It happens that subgroups of free groups are always free groups.2 But even if thefree group is finitely generated, if it is free on more than one generator, then it turns outto have many subgroups that are not finitely generated. But some infinitely generatedsubgroups of a free group turn out to be normally generated by a finite set.

2Our favorite proof of this uses the topology of covering spaces of graphs. See Section 3.8 of E.H.Spanier’s Algebraic Topology, McGraw–Hill, 1966.


Definitions 6.7.1. A finite presentation for a group G consists of a finite subset X ={x1, . . . , xk} ⊂ G which generates G, together with a finite subset R = {R1, . . . , Rl} ofthe free group F (x1, . . . , xk) on x1, . . . , xk, such that the kernel of the homomorphism

f : F (x1, . . . , xk)→ G

that’s induced by {x1, . . . , xn} ⊂ G is normally generated by R in F (x1, . . . , xk).Given the above presentation for G, we say that G has generators X and relations

R. In general, given variables x1, . . . , xn and elements R1, . . . , Rm of the free groupF (x1, . . . , xn), we write 〈x1, . . . , xn |R1, . . . , Rm〉 for the group with generators x1, . . . , xnand relations R1, . . . , Rm. Thus,

〈x1, . . . , xn |R1, . . . , Rm〉 = F (x1, . . . , xn)/〈〈R1, . . . , Rm〉〉.If a group G admits a finite presentation, then we say that G is finitely presented.

Note that a finite presentation of G is the construction of a pushout diagram

F (r1, . . . , rm) F (x1, . . . , xn)

e G.

��

��

It turns out that not every finitely generated group is finitely presented. Thus, sub-groups of finitely generated free groups can be very complicated indeed.

One can show, via the theory of covering spaces, that the commutator subgroupof the free group, F (x, y), on two letters is infinitely generated. But recall that theabelianization, F (x, y)ab, is naturally isomorphic to FA(x, y), the free abelian group ofx and y. Thus, the next example shows that the commutator subgroup of F (x, y) isnormally generated by a single element, xyx−1y−1.

Example 6.7.2. We show here that the free abelian group on two letters has the pre-sentation

FA(x, y) = 〈x, y |xyx−1y−1〉.First note that since FA(x, y) is abelian, xyx−1y−1 is in the kernel of the map π :

F (x, y) → FA(x, y) which takes x to x and y to y. Thus, there is an induced mapπ : G→ FA(x, y), where G = 〈x, y |xyx−1y−1〉 = F (x, y)/〈〈xyx−1y−1〉〉.

Note that x and y commute in G. Thus, since x and y generate G, G is abelian.Thus, the universal property of the free abelian group provides a homomorphism ϕ :FA(x, y)→ G with ϕ(x) = x and ϕ(y) = y.

Now notice that the composites ϕ ◦ π and π ◦ ϕ restrict to the identity map ongenerators for the two groups. Thus, ϕ and π are inverse isomorphisms.

The hard part in showing that a given group H is given by a particular presentationis generally in producing a homomorphism from H to the stated quotient of the freegroup on the generators. In the above example, the homomorphism is obtained via theuniversal property for the free abelian group.

Exercises 6.7.3.1. Give a finite presentation for the free abelian group on n generators.


2. Show that Zn = 〈x |xn〉.3. Show that D2n = 〈a, b | a2, bn, aba−1b〉.4. Show that Q4n = 〈a, b | b2n, a2bn, aba−1b〉.5. Find a presentation for A4.

6. Show that 〈x, y |x2, y2, (xy)3〉 is a presentation for S3.

7. Show that 〈x, y |x2, y2, (xy)n〉 is a presentation for D2n for each n ≥ 2.

8. Show that 〈x, y, z |x2, y2, z2, (xy)3, (yz)3, xzx−1z−1〉 is a presentation for S4, viasetting x = (1 2), y = (2 3), and z = (3 4).

9. Let G = 〈x, y |xn, ym, xyx−1yk〉 for integers n,m, k with m,n > 0. Show that G isfinite and that 〈y〉 � G. Do x and y necessarily have order n and m, respectively?

10. Let G = 〈x, y |x4, y4, (xy)4〉. Show that Q8 and Z4 × Z4 are both factor groupsof G. What extra relations, when added to those of G, will give Q8? What extrarelations, when added to those of G, will give Z4 × Z4? Do any of the relationsoriginally in G become redundant when we do this?

11. Show that 〈x, y |xyx−1y−2〉 is a presentation for the group Z[ 12 ]�μ2Z of Problem 19of Exercises 4.6.13.

12. Given finite presentations for H and K and a homomorphism α : K → Aut(H),derive a finite presentation for H �α K.

6.8 Direct and Inverse Limits

We consider yet another pair of examples of universal properties that replace one diagramwith another. We treat inverse limits here and treat direct limits in a separate subsection.

Inverse limits are related to pullbacks. We suppose given objects Xi for i ≥ 0 andmorphisms pi : Xi → Xi−1 for i ≥ 1. (The indices i here are integers.) We call this dataan inverse system. Pictorially, this gives

· · · X3 X2 X1 X0.�� p3 ��p2 ��p1

The inverse limit of this system can be thought of as a universal object to tack on at−∞. What this means is that the inverse limit is a universal diagram of the followingsort.

X

· · · X1 X0.��

��

��

�� p1

Definition 6.8.1. Suppose given an inverse system, consisting of objects Xi for i ≥ 0and morphisms pi : Xi → Xi−1 for i ≥ 1. An inverse limit for this system consistsof an object X = lim←−Xi, together with morphisms πi : X → Xi for i ≥ 0 such thatpi ◦ πi = πi−1 for i ≥ 1, and the following universal property is satisfied:

For each object Y and each collection of morphisms fi : Y → Xi such that pi ◦ fi =fi−1 for all i ≥ 1, there is a unique morphism f : Y → X such that πi ◦ f = fi for alli ≥ 0.


We give the construction of inverse limits in Sets as follows. Suppose given an inversesystem of sets:

· · · X3 X2 X1 X0.�� p3 ��p2 ��p1

Let X be the subset of the infinite product∏i≥0Xi consisting of those tuples (xi | i ≥

0) with the property that pi(xi) = xi−1 for all i ≥ 1, and let πi : X → Xi be therestriction to X of the i-th projection map of the product (which we shall also write asπi :

∏i≥0Xi → Xi).

Proposition 6.8.2. Using the notations just given, the set X, together with the mapsπi : X → Xi, constitutes the inverse limit of the stated inverse system in Sets.

Proof Suppose given functions fi : Y → Xi, i ≥ 0, such that pi ◦ fi = fi−1 for i ≥ 1.Then the universal property of products gives a function f : Y →∏

i≥0Xi with πi◦f = fifor all i. As tuples, this says f(y) = (fi(y) | i ≥ 0) for all y ∈ Y . But since pi ◦ fi = fi−1,we have f(y) ∈ lim←−Xi for all y ∈ Y . In other words, f takes values in the inverse limit.

By construction, πi ◦ f = fi. Moreover, f is the unique map with this propertybecause X is a subset of the product, and a map into the product is determined by itscoordinate functions.

We now give a useful way to interpret the construction of inverse limits. We shall write(. . . , xi, . . . , x1, x0) for an element of the product

∏i≥0Xi, reversing the usual order.

Then there is a “shift” map s :∏i≥0Xi →

∏i≥0Xi given by s(. . . , xi, . . . , x1, x0) =

(. . . , pi+1(xi+1), . . . , p2(x2), p1(x1)). It is easy to see that the inverse limit lim←−Xi is theset of all x ∈ ∏

i≥0Xi such that s(x) = x. In other words, the inverse limit is the set offixed points of the shift map on

∏i≥0Xi.

As was the case for products and pullbacks, the construction of inverse limit thatworks in Sets also produces the inverse limit in most of the other categories we’ve beenconsidering.

Proposition 6.8.3. Suppose given an inverse system

· · · G3 G2 G1 G0�� h3 ��h2 ��h1

in the category of groups, and let G ⊂∏i≥0Gi be the inverse limit of the induced inverse

system of sets (i.e., G = {(gi | i ≥ 0) |hi(gi) = gi−1 for all i ≥ 0}). Then G is a subgroupof the direct product of the Gi, and with this multiplication, together with the restrictionsto G of the projection maps πi :

∏i≥0Gi → Gi, forms the inverse limit of this system in

the category of groups.In addition, if the groups Gi are abelian, so is G, and in this case, G, together with

the projection maps is the inverse limit of the given system in the category of abeliangroups.

Proof That G is a subgroup of the direct product follows from the fact that the hiare group homomorphisms. Since the direct product is the product in the category ofgroups, G, together with the projection maps, satisfies the universal property for theinverse limit by exactly the same proof as that given for the category of sets. A similarargument works for abelian groups.


We shall see in Chapter 7 that the same construction gives the inverse limit in thecategory of rings. We shall study one such example, the p-adic integers, Zp, in somedepth. Zp is the inverse limit of the unique system of ring homomorphisms

· · · Zp3 Zp2 Zp.��

Exercises 6.8.4.1. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets is

surjective, then so is each function πi : lim←−Xi → Xi.

2. Show that if each of the functions pi : Xi → Xi−1 in an inverse system of sets isbijective, then so is each function πi : lim←−Xi → Xi.

3. Show that the construction given for Sets, Gp, and Ab also produces the inverselimit in G-sets, Mon, and Amon.

4. Show that if X = lim←−Xi is the inverse limit for an inverse system consisting ofobjects Xi, i ≥ 0, and morphisms pi : Xi → Xi−1 in a category C, then for anyobject Y of C, MorC(Y,X) is the inverse limit in Sets of the MorC(Y,Xi).

Direct Limits

Direct limit is the concept dual to inverse limit:

Definition 6.8.5. A direct system in C consists of objects Xi for i ≥ 0, together withmorphisms ji : Xi → Xi+1 for i ≥ 0. Pictorially, we get

X0 X1 X2 X3 · · ·��j0 ��j1 ��j2 ��

A direct limit for this system consists of an object X = lim−→Xi, together with morphismsιi : Xi → X for i ≥ 0, such that ιi+1 ◦ ji = ιi for i ≥ 0 and such that the followinguniversal property is satisfied. Given an object Y and morphisms fi : Xi → Y such thatfi+1◦ji = fi for all i ≥ 0, there is a unique morphism f : lim−→Xi → Y such that f ◦ιi = fifor all i ≥ 0.

Surprisingly (since the direct limit is related to coproducts and pushouts in the sameway that the inverse limit is related to products and pullbacks), the construction of thedirect limit is the same in all the categories we’ve been discussing. The intuition is thatif the morphisms ji are all injective, then the direct limit is the union of the Xi. (SeeProblem 6 of Exercises 6.8.8.) If the ji are not injective, we need to work a little harderto construct the direct limit.

Assume that we’re working in Sets. We shall construct lim−→Xi by passage to equiva-lence classes from an equivalence relation on the disjoint union

∐i≥0Xi.

Here, we use the equivalence relation generated by the following notion of elementaryequivalence: if x ∈ Xi, we set x equivalent to ji(x) ∈ Xi+1. We write x ∼e ji(x). Wethen declare two elements x, x′ ∈ ∐

i≥0Xi to be equivalent, written x ∼ x′, if there isa sequence x = x0, . . . , xn = x′ such that for i = 0, . . . , n − 1, either xi ∼e xi+1 orxi+1 ∼e xi.

We define lim−→Xi to be the set of equivalence classes under ∼, and write π :∐i≥0Xi →

lim−→Xi for the quotient map. The structure maps ιi : Xi → lim−→Xi are defined to be the


composites

Xi

∐i≥0

Xi lim−→Xi,��ιi ��π

where ιi : Xi →∐i≥0Xi is the natural inclusion of Xi in the coproduct.

Proposition 6.8.6. The above construction gives the direct limit in Sets.

Proof Given functions fi : Xi → Y with fi+1 ◦ ji = fi for all i ≥ 0, there is a uniquefunction f from the coproduct

∐i≥0Xi to Y such that f ◦ ιi = fi for all i, where ιi is

the natural inclusion of Xi in the coproduct. Write ≈ for the equivalence relation on∐i≥0Xi given by setting x ≈ x′ if f(x) = f(x′). Since fi+1 ◦ ji = fi for all i ≥ 0, we see

that x ∼e x′ implies that x ≈ x′.By Proposition 1.5.3, if x ∼ x′, then f(x) must equal f(x′). Thus, f factors (neces-

sarily uniquely, since π is onto) through a function f : lim−→Xi → Y . But now f clearlyhas the desired properties.

Example 6.8.7. Let p be a prime. We shall construct a group by taking direct limitsof cyclic p-groups. Specifically, note the equivalence class of p mod pi has order pi−1 inthe additive group Zpi . Thus, there is a group homomorphism ji−1 : Zpi−1 → Zpi whichtakes 1 to p: explicitly, ji−1(n) = pn. Since p has order pi−1, ji−1 is injective.

Clearly, the maps {ji | i ≥ 1} form a direct system in Ab. We write Tp for the directlimit of this system, which by Problem 4 below is an abelian group.

Exercises 6.8.8.1. For any direct system of sets, show that any element of the direct limit lies in the

image of ιi for some i.

2. In Sets, show that if x, y ∈ Xi are equivalent in lim−→Xi, then there exists k ≥ 0such that ji+k ◦ · · · ◦ ji(x) = ji+k ◦ · · · ◦ ji(y).

3. Suppose given a direct system in Gp. Show that the direct limit of the underlyingsets has a unique group structure that makes it the direct limit of this system inGp.

† 4. Repeat the preceding problem in Ab, G-sets, Mon, and Amon.

5. In Sets, suppose there is a set U such that Xi ⊂ U for all i ≥ 0. Suppose also thateach composite

Xiji−→ Xi+1 ⊂ U

coincides with the inclusion Xi ⊂ U . Show that we may identify the direct limitlim−→Xi with the union

⋃i≥0Xi ⊂ U .

† 6. More generally, suppose given a direct system of sets such that each function ji :Xi → Xi+1 is injective. Show that the structure maps ιi : Xi → lim−→Xi are alsoinjective. In addition, show that lim−→Xi is the union of the images of the Xi.

7. Let Tp, as defined above, be the direct limit of the cyclic groups Zpi and let ιi :Zpi → Tp be the structure map for the direct limit. By Problem 6, ιi is an injection.Show that Tp is a p-torsion group and that the image of ιi the subgroup consistingof those elements of exponent pi.


8. Let p be a prime and let Z[ 1p ] ⊂ Q be the subgroup of the rational numbersconsisting of all quotients of the form m

pi , where m ∈ Z and i ≥ 0. Show that Z[ 1p ]is isomorphic to the direct limit of the system

Zp→ Z

p→ Z→ . . .

where each group in the direct system is Z and each map is multiplication by p.

9. Show that the torsion group Tp above is isomorphic to the factor group Z[ 1p ]/Z.

10. Show that any countable abelian torsion group is a direct limit of finite abeliangroups.

11. Show that if G is finite, then a countable G-set is a direct limit of finite G-sets.

12. Show that if X = lim−→Xi is the direct limit for a direct system consisting of objectsXi, i ≥ 0, and morphisms ji : Xi → Xi+1 in a category C, then for any object Yof C, MorC(X,Y ) is the inverse limit in Sets of the MorC(Xi, Y ).

6.9 Natural Transformations and Adjoints

In our study of semidirect products, we found that it was interesting to be able todetermine when two homomorphisms are related in certain ways (e.g., conjugation).Similarly, it is often of interest to relate the effect of two different functors.

Definition 6.9.1. Let C and D be categories, and let F and G be two functors from Cto D. A natural transformation η : F → G consists of a D-morphism ηX : F (X)→ G(X)for each X ∈ ob(C) such that for each f ∈ MorC(X,Y ), the following diagram commutes.

F (X) F (Y )

G(X) G(Y )��

ηX

��F (f)

��

ηY

��G(f)

We say that η is a natural isomorphism if ηX is an isomorphism in D for each objectX of C.

With straightforward modifications, we obtain the definitions of natural transforma-tions or isomorphisms between two contravariant functors.

The value of naturality is not immediately apparent. However, it turns out to be astrong tool in certain kinds of mathematical argument.

On the theoretical level, naturality permits us to define equivalence of categories.

Definition 6.9.2. An equivalence of categories between C and D consists of functorsF : C → D and G : D → C, together with natural isomorphisms between G ◦ F and theidentity functor of C, and between F ◦G and the identity functor of D.

One place where equivalence of categories turns out to be useful is in the notion ofskeletal subcategories.


Definition 6.9.3. Let D be a subcategory of C, and suppose that the inclusion functori : D → C is an equivalence of categories. Then we say that D is a skeleton (or skeletalsubcategory) of C.

We shall illustrate a nice feature of skeletal subcategories presently. First, we haveanother definition.

Definition 6.9.4. A small category is a category whose objects form a set.

For instance, consider the category of finite groups. It is not small. In fact, thecollection of all objects that are isomorphic to the trivial group is too big to be a set.But we can recover most of the benefits of smallness if we can construct a skeleton whichis small.

Recall from Corollary 2.6.6 that there are countably many isomorphism classes offinite groups. In particular, the isomorphism classes of finite groups form a set. (Thisdoes not happen in Gp, as there is no bound on the cardinality of the underlying sets ofthe objects in Gp.)

Proposition 6.9.5. Let D be a full subcategory of C, and suppose that we’re given, foreach X ∈ ob(C) an isomorphism ηX : X ∼= F (X) for some object F (X) of D. Then Dis a skeleton for C.

Proof We claim that there is a unique way to make F into a functor from C to D insuch a way that η is a natural transformation. But if this is the case, we’re done, sinceη then provides a natural isomorphism from the identity functor of C to i ◦F , as well asfrom the identity functor of D to F ◦ i, where i is the inclusion of D in C.

Because the morphisms ηX are all isomorphisms, it is easy to give the functorialityof F : For f : X → Y in C, we take F (f) = ηY ◦ f ◦ η−1

X . This immediately gives thenaturality diagram for η, and the functoriality of F follows by cancelling inverses.

However, the fact that we have to choose an isomorphism ηX : X ∼= F (X) for eachobject X of C can be problematical from a set-theoretic point of view, as it requiresan extension from sets to classes of the Axiom of Choice. Thus, it is not clear that thecategory of finite groups has a small skeleton. For any group G, we could use Cayley’sTheorem to identify G with a subgroup of S(G), the group of symmetries on G, butthere’s no natural ordering of the elements of G, and hence no natural identification ofS(G) with S|G|.

However, this need not be a serious issue. Any real mathematical question that wecared about would probably either concern isomorphism classes of objects or concerninformation relating to particular sorts of diagrams of objects. The latter questions cangenerally be phrased in terms of functors from a small category into the category onewishes to study. Thus, for the questions we care about, the next result will show thatwe do not really need a small skeleton.

The proof of the next proposition is based on the method of proof of Proposition 6.9.5.It is left to the reader.

Proposition 6.9.6. Let D be a full subcategory of C such that D contains an object inevery isomorphism class of objects of C. Let E be any small category. Then any functorfrom E into C is naturally isomorphic to a functor that takes value in D.

Also, since D is full in C, any two functors from E into D that are naturally isomor-phic as functors into C are also naturally isomorphic as functors into D.


Thus, the category of finite subgroups of S∞, which is small, contains all the infor-mation of interest with regard to the category of finite groups.

Just as group isomorphisms are special examples of homomorphisms, equivalences ofcategories are special examples of what are called adjoint functors.

Definitions 6.9.7. We say that a pair of functors R : C → D and L : D → C is anadjoint pair if for each X ∈ ob(D) and Y ∈ ob(C), there is an isomorphism (of sets)ΦX,Y : MorC(L(X), Y ) ∼= MorD(X,R(Y )) which is natural in the following sense: givenmorphisms f1 : X ′ → X in D and f2 : Y → Y ′ in C, then the following diagramcommutes.

MorC(L(X), Y ) MorD(X,R(Y ))

MorC(L(X ′), Y ′) MorD(X ′, R(Y ′))

��ΦX,Y

��L(f1)∗ ◦ f2∗

��f∗1 ◦R(f2)∗

��ΦX′,Y ′

Here, g∗(h) = g ◦ h = h∗(g).If R and L form an adjoint pair, we call R a right adjoint for L and call L a left

adjoint for R. The pair itself can be called an adjunction.

Another way of expressing the naturality property for adjoint functors is as follows.Fixing X ∈ ob(D), then filling in the blanks provides two functors, MorC(L(X),−)and MorD(X,R(−)), from C to Sets, and ΦX,− provides a natural isomorphism be-tween them. At the same time, for each Y ∈ ob(C), we have contravariant functorsMorC(L(−), Y ) and MorD(−, R(Y )) from D to Sets, and Φ−,Y provides a natural iso-morphism between them.

It turns out that we’ve already seen several examples of adjoint functors. For instance,if C comes equipped with a forgetful functor to Sets and if ηX : X → F (X) is a freeC-object for each set X, then the universal property of free objects says precisely that

ηX∗ : MorC(F (X), Y ) ∼= MorSets(X,Y )

for each object Y of C. The right adjoint here is the forgetful functor to Sets, whichdoesn’t really require notation. Naturality in the variable Y is immediate, while natu-rality in the first variable is shown in Proposition 6.6.5.

We’ve seen some other examples as well.

Exercises 6.9.8.1. Show that abelianization is a left adjoint to the inclusion of Ab in Gp.

2. LetH be a subgroup of the groupG. Then there is a forgetful functor fromG-sets toH-sets. Show that this forgetful functor has a left adjoint, given by L(X) = G×HXfor any H-set X.

3. Let H be a subgroup of the group G and let R : G-sets→ Sets be given by R(X) =XH , the H-fixed point set of X. Define L : Sets→ G-sets by L(X) = (G/H)×X.Show that these form a pair of adjoint functors.

4. Show that passage from a monoid to its group of invertible elements is a rightadjoint to the forgetful functor from groups to monoids.


5. Suppose given an adjunction ΦX,Y : MorC(L(X), Y ) ∼= MorD(X,R(Y )). For X ∈ob(D), let ηX : X → R(L(X)) be given by ηX = ΦX,L(X)(1L(X)). Show that η is anatural transformation from the identity functor to R ◦ L. It is customarily calledthe unit of the adjunction. Show also that this unit determines the isomorphism Φas follows: for f ∈ MorC(L(X), Y ), ΦX,Y (f) = R(f) ◦ ηX .

6. Suppose given an adjunction ΦX,Y : MorC(L(X), Y ) ∼= MorD(X,R(Y )). For Y ∈ob(C), let εY : L(R(Y )) → Y be given by εY = Φ−1

R(Y ),Y (1R(Y )). Show that εgives a natural transformation from L ◦ R to the identity functor. We call it thecounit of the adjunction. Show that it determines the isomorphism Φ as follows:for g ∈ MorD(X,R(Y )), Φ−1

X,Y (g) = εY ◦ L(g).

7. Show that left adjoints preserve coproducts, pushouts, and direct limits. Thismeans, for instance, that if we apply a left adjoint to a pushout diagram, theresulting diagram is also a pushout diagram.

8. Show that right adjoints preserve products, pullbacks, and inverse limits.

The last two problems should explain some of the results in the preceding sections,as well as predict some others that we haven’t proven yet. That, indeed is one of themain points in favor of category theory. And in some cases, knowing about adjoints canhelp explain why a theorem is true. For instance, it may be instructive to reconsider thenotion of adjoint when one looks at the Frobenius Reciprocity Theorem in representationtheory.

6.10 General Limits and Colimits

Products, coproducts, pushouts, pullbacks, direct and inverse limits are examples of acouple of much more general types of construction. In all these cases, we’re given somekind of diagram in C, and we’re asked to extend it to a larger diagram that satisfies auniversal property.

The key to generalizing is in what constitutes a diagram.

Definition 6.10.1. Let I be a small category and let C be an arbitrary category. Byan I-diagram in C, we mean a functor X : I → C. We shall also refer to X as a diagramin C with indexing category I.

We’ll show in Exercises 6.10.7 how to interpret the universal constructions we’vealready seen in terms of functors.

By analogy with the constructions we’ve seen, the limit of an I-diagram in C willbe defined to be a diagram defined on a larger indexing category that satisfies a certainuniversal property.

Definition 6.10.2. For a category I, let I+ be the category defined as follows. Theobjects of I+ consist of those in I, together with one new object, denoted ∗. Themorphisms in I+ between objects that come from I are the same as those in I (i.e., I isa full subcategory of I+). In addition, we declare that MorI+(∗, i) has a single elementfor all objects i of I+, while MorI+(i, ∗) = ∅ if i �= ∗.

There is, of course, a unique composition law which makes this a category.Clearly, ∗ is an initial object for I+. Indeed, I+ is the result of adjoining an initial

object to I. (If I already had an initial object, that object will no longer be initial inI+.)


Definition 6.10.3. Let I be a small category and let X : I → C be a functor. A limitfor X, denoted lim←−X, is a functor X : I+ → C whose restriction to I ⊂ I+ is X, andwhich is universal in the following sense: if Y : I+ → C is another such extension ofX to I+, then there is a unique natural transformation η : Y → X such that ηi is theidentity map of Xi whenever i is an object of I.

Thus, the limit of an I-diagram is an I+-diagram.Colimits are defined in terms of the category obtained by adjoining a terminal object

to I.

Definition 6.10.4. For a category I, let I+ be the category defined as follows. Theobjects of I+ consist of those in I, together with one new object, denoted ∗. Themorphisms in I+ between objects that come from I are the same as those in I (i.e.,I is a full subcategory of I+). In addition, we declare that MorI+(i, ∗) has a singleelement for all objects i of I+, while MorI+(∗, i) = ∅ if i �= ∗.

Colimits are obtained as follows.

Definition 6.10.5. Let I be a small category and let X : I → C be a functor. A colimitfor X, denoted lim−→X, is a functor X : I+ → C whose restriction to I ⊂ I+ is X, andwhich is universal in the following sense: if Y : I+ → C is another such extension of X toI+, then there is a unique natural transformation η : X → Y such that ηi is the identitymap of Xi whenever i is an object of I.

Limit and colimit constructions indexed by arbitrary small categories can be quitecomplicated. It is often desirable to restrict attention to diagrams indexed by the categoryassociated to a partially ordered set. (Recall that a partial ordering on a set I is a relation,generally denoted ≤ by default, that is reflexive, transitive, and antisymmetric.)

Definition 6.10.6. Let I be a partially ordered set. The category, I, associated to I isthe category whose objects are the elements of I, and whose morphisms are defined asfollows: for i, j ∈ I, MorI(i, j) is empty unless i ≤ j, in which case it has one element,which we denote by i ≤ j.

I is a category, as transitivity gives the composition of morphisms, while reflexivitygives the existence of identity morphisms. Since I is a set, I is small.

Exercises 6.10.7.1. Show that a small category I comes from a partially ordered set if and only if the

following properties hold.

(a) For each i, j ∈ ob(I), MorI(i, j) has at most one element.

(b) If i �= j and if MorI(i, j) �= ∅, then MorI(j, i) = ∅.2. Let I be a set. Then we may consider it to be a partially ordered set in whichi ≤ j if and only if i = j. Let I be the induced category and let C be an arbitrarycategory. Note that an I-diagram, X, in C is simply a family {X(i)|i ∈ I} of C-objects indexed by I. Show that specifying a limit for X is equivalent to specifyinga product for {X(i)|i ∈ I} and that specifying a colimit for X is equivalent tospecifying a coproduct for {X(i)|i ∈ I}.


3. Let I be a partially ordered set with three elements, i, j and k, where i ≤ j, i ≤ k,and there are no other nonidentity morphisms in the induced category I. Showthat ordinary diagrams in a category C which have the form

• •

•

��

��

are in one-to-one correspondence with I-diagrams in C. Under this correspondence,show that specifying a colimit for a functor X : I → C is equivalent to specifyinga pushout for the associated ordinary diagram in C.

4. Give the dual of the preceding exercise.

5. Let N be the non-negative integers with the usual ordering, and with associatedcategory N . Show that direct systems in a category C are in one-to-one corre-spondence with functors from N into C. Under this correspondence, show thatspecifying a colimit for an N -diagram is equivalent to specifying a direct limit forthe associated direct system.

6. Give the dual of the preceding exercise.

7. Let I be a small category and let C be any category. Show that there is a category,which we denote CI , whose objects are the I-diagrams in C and whose morphismsare the natural transformations between them.

8. Suppose we’re given a construction that produces a colimit for every I-diagram inC. Show that we get an induced functor from CI to CI+ .

9. Suppose we’re given a construction that produces a limit for every I-diagram in C.Show that we get an induced functor from CI to CI

+.

10. Show that left adjoints preserve colimits of I-diagrams.

11. Show that right adjoints preserve limits of I-diagrams.

Chapter 7

Rings and Modules

The study of rings and their modules is vastly more complicated than that of groups. Inthis chapter, long as it is, we shall do little more than set the stage for this study. Weshall give some examples that we shall study in greater depth in later chapters, and shallgive a number of basic definitions, constructions, and questions for further study.

Section 7.1 defines concepts such as zero-divisors, unit groups, and algebras overcommutative rings. Examples are given from matrix rings, the rings Zn, the complexnumbers, C, and its subrings, the division ring H of quaternions, and the p-adic integers,Zp.

Section 7.2 defines left, right, and two-sided ideals. Quotient rings are defined, andtheir universal property is given. Principal ideals and generators for ideals are discussed.Operations on ideals are considered, and their use is illustrated via the Chinese Remain-der Theorem.

Section 7.3 develops polynomials in one and several variables. Evaluation maps arestudied, with the usual applications to the theory of roots. Finitely generated algebrasare discussed. The cyclotomic extensions Z[ζn] and Q[ζn] are featured as examples.

Section 7.4 studies the symmetric polynomials, showing that they form a polynomialalgebra on the elementary symmetric functions. Applications are given to the theory ofdiscriminants.

Section 7.5 develops group rings and monoid rings. It is shown that polynomials forman example of the latter. Augmentation ideals are studied, and are generalized to thenotion of augmented algebras. The relationship between Z[Zn] and Z[ζn] is studied inthe exercises.

Then, in Section 7.6, the theory of prime and maximal ideals is developed for commu-tative rings. Krull dimension is introduced. So are the ascending and descending chainconditions for ideals, in the context of the discussion of maximality principles for familiesof ideals.

Section 7.7 introduces modules. Cyclic modules, generating sets, annihilators, andinternal and external operations on modules are studied. The theories of free modulesand of exact sequences are developed. The section closes with a discussion of ringswithout identity.

Using the techniques developed for studying modules, the ascending and descendingchain conditions are studied for modules and for left and right ideals in Section 7.8. Therelationship between chain conditions and extensions of modules is studied, as is therelationship between the Noetherian property and finite generation. The section closeswith the Hilbert Basis Theorem and its application to finitely generated algebras over a

201

CHAPTER 7. RINGS AND MODULES 202

Noetherian ring.Then, Section 7.9 classifies the vector spaces over a division ring, and shows that the

finitely generated free modules over a commutative ring have a well defined rank.Section 7.10 gives the relationship between matrix rings and the endomorphism rings

of finitely generated free modules.We close the chapter with a development of rings and modules of fractions. Local rings

and localization are studied. The field of fractions of an integral domain is constructed,showing that a ring is an integral domain if and only if it is a subring of a field.

7.1 Rings

We give some basic definitions and examples of rings. We then treat the complex num-bers, the quaternions, and the p-adic integers in subsections.

We shall follow the tradition that insists that a ring should have a multiplicativeidentity. We give a brief discussion of rings without identity in the last subsection ofSection 7.7.

Definitions 7.1.1. A ring A is a set with two binary operations, addition and multi-plication.1 Addition gives A the structure of an abelian group, with identity element0. Multiplication provides A with a (not necessarily abelian) monoid structure, withidentity element 1. The two operations interrelate via the distributive laws:

a(b+ c) = ab+ ac and (b+ c)a = ba+ ca

for all a, b, c ∈ A.If the multiplication does satisfy the commutative law, we call A a commutative ring.Let A and B be rings. Then a ring homomorphism f : A → B from A to B is a

function such that f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b) for all a, b ∈ A, andf(1) = 1. As usual, a ring isomorphism is a ring homomorphism that is bijective.

A subring of a ring A is a subgroup B of the additive group of A, such that B isclosed under multiplication and contains the multiplicative identity element 1.

Finally, the invertible elements in the multiplicative monoid of a ring A are called theunits of A. They form a group under multiplication, denoted A×.

Let f : A → B be a ring homomorphism. Then f gives a group homomorphismbetween the additive groups of A and B, so we can talk about its kernel. Note that thekernel of f cannot be a subring of A unless 1 = 0 in B, because f(1) = 1. Nevertheless,ker f is more than just a subgroup of A. It is a two-sided ideal, a concept we shall studybelow.

Images are a different story:

Lemma 7.1.2. Let f : A → B be a ring homomorphism. Then the image of f is asubring of B.

The next example is the exception, rather than the rule.

Example 7.1.3. There is a ring with only one element. We denote the element by 0 andset 0 + 0 = 0 and 0 · 0 = 0. In other words, both addition and multiplication are givenby the unique binary operation on a one-element set. The element 0 serves as identityfor both operations. Thus, with the multiplicative identity set equal to 0, the propertiesof a ring are satisfied.

We call this ring the 0 ring and denote it by 0.1The use of A for a ring derives from anneau, the French word for ring.


The next lemma shows that the 0 ring is the only ring in which 0 = 1.

Lemma 7.1.4. Let A be a ring. Then 0 · a = a · 0 = 0 for all a ∈ A.

Proof We have

0a = (0 + 0)a = 0a+ 0a.

But then subtracting 0a from both sides gives 0 = 0a. The case of a · 0 is similar.

The distributive laws now give a useful fact.

Corollary 7.1.5. Let A be a ring and let −1 ∈ A be the additive inverse of 1. Then forany a ∈ A, (−1) · a = a · (−1) is the additive inverse of a.

The next result is an important generalization of the preceding ones. The proof isleft to the reader.

Lemma 7.1.6. Let A be a ring. For m ∈ Z, write m for the m-th power of 1 in theadditive group of A. (Thus, if m is positive, m is the sum of m copies of 1 in A.) Thenfor any a ∈ A, m · a = a ·m is the m-th power of a in the additive group of A.

Examples 7.1.7. The most obvious examples of rings are the integers, rational num-bers, and real numbers, Z, Q, and R, with the usual addition and multiplication. Thelatter two examples turn out to have some interesting subrings, but Z does not: Since 1generates Z as an abelian group, the only subring of Z is Z itself.

We’ve also seen (in Proposition 2.4.7) that multiplication in Zn induces a commutativering structure on Zn such that the canonical map π : Z→ Zn is a ring homomorphism.(Since π is surjective, this is the only ring structure that makes π a ring homomorphism.)

The integers play a special role in ring theory:

Lemma 7.1.8. Let A be a ring. Then there is a unique ring homomorphism from Z toA. Moreover, the image of each m ∈ Z under this homomorphism commutes2 with everyelement of A.

Proof Ring homomorphisms are required to preserve the multiplicative identity ele-ments. But recall from Proposition 2.5.6 that for any group G, and any element x ∈ G,there is a unique group homomorphism from Z to G that carries 1 to x. Thus, there isat most one ring homomorphism from Z to A: as a homomorphism of additive groups,it must be the homomorphism f : Z → A for which f(1) = 1. It suffices to show thatthis group homomorphism is also a ring homomorphism for any ring A, and that f(m)commutes with every element of A.

Note that since f is a homomorphism of additive groups and since f(1) = 1, f(n)must be the n-th power of 1 in the additive group structure on A, for all n ∈ Z. Thus,Lemma 7.1.6 says that for each n ∈ Z, f(n) commutes with every element of a and thatf(n) · a is the n-th power of a in the additive group structure on A.

Thus, f(m)f(n) is the m-th power of the n-th power of 1 with respect to the additivegroup operation in A. So f is a ring homomorphism by the power laws for a group.

2When we say that two elements of a ring commute, we mean this with respect to the operation ofmultiplication. The point is that the addition operation is abelian so that two elements always commuteadditively.


In categorical language, this implies that Z is an initial object for the category ofrings. Clearly, 0 is a terminal object.

We now give some examples of unit groups.

Examples 7.1.9.1. It is easy to see, e.g., via the order inequalities for multiplication, that Z× = {±1}.2. In Lemma 4.5.3, it is shown that m ∈ Z×

n if and only if (m,n) = 1. Then, inCorollary 4.5.7, it is shown that if n = pr11 . . . prk

k , where p1, . . . , pk are distinctprimes, then

Z×n∼= Z×

pr11× · · · × Z×

prkk

.

Finally, the unit groups Z×pr are explicitly calculated in Corollaries 4.5.20 and 4.5.21.

We now give some language for describing properties of rings and their elements.

Definitions 7.1.10. Suppose we have two elements a, b in a ring A, neither of which is0, such that ab = 0. Then a and b are said to be zero-divisors in A. More specifically, ifA is noncommutative, a is a left zero-divisor and b is a right zero-divisor.

We say that an element a ∈ A is nilpotent if an = 0 for some integer n. Note thatany nonzero nilpotent element is a zero-divisor.

Generalizing the obvious facts about the integers, we say that an integral domain, ordomain, for short, is a nonzero commutative ring with no zero-divisors.

A nonzero ring is called a division ring if every nonzero element in it is a unit. If thering is also commutative, we call it a field.

The next lemma connects some of these definitions.

Lemma 7.1.11. A division ring has no zero-divisors. Thus, every field is an integraldomain.

Proof Let A be a division ring and let a, b ∈ A such that a �= 0 and ab = 0. It sufficesto show that b = 0.

Because A is a division ring and a �= 0, a has a multiplicative inverse, a−1. But then0 = a−1 · 0 = a−1 · ab = (a−1a)b = b.

Of course, Z is an example of an integral domain, and Q and R are examples of fields.The following lemma could have been left as an exercise, but we shall make sufficient

use of it that a proof has been given. The reader may wish to investigate alternativeproofs.

Lemma 7.1.12. Let n > 1. Then Zn is an integral domain if and only if n is prime.Moreover, for any prime number p, Zp is a field.

Proof Suppose that n is not prime. Then we can write n = kl, where k and l are bothgreater than 1. But then 1 < k, l < n, so k and l are nonzero in Zn. But k · l = kl = 0,and hence k and l are zero-divisors in Zn.

On the other hand, if p is prime, then each nonzero element of Zp is shown to be aunit in Lemma 4.5.3.

In a noncommutative ring, it is useful to keep track of the elements that commutewith all other elements of the ring.


Definition 7.1.13. Let A be a ring. Then the center of A is the set

{a ∈ A | ab = ba for all b ∈ A}consisting of those elements of a that commute with every element of A.

Lemma 7.1.14. Let A be a ring. Then the center of A is a subring of A.

Our first examples of noncommutative rings come from matrices.

Definitions 7.1.15. Let A be a ring (not necessarily commutative). We write Mn(A)for the collection of n × n matrices with coefficients in A. Given elements aij ∈ A with1 ≤ i ≤ n and 1 ≤ j ≤ n, we write (aij) ∈Mn(A) for the matrix whose ij-th entry is aij .

We define addition in Mn(A) so that the ij-th entry of (aij) + (bij) is aij + bij anddefine multiplication so that the ij-th entry of (aij) · (bij) is

∑nk=1 aikbkj . Note that the

order of the factors in the last expression is important if A is noncommutative.We write 0 for the matrix whose entries are all 0, and write In for the matrix whose

off-diagonal entries are all 0 and whose diagonal entries are all 1, e.g., I3 =(

1 0 00 1 00 0 1

).

More generally, for a ∈ A, we write aIn for the matrix whose diagonal entries are allequal to a and whose off-diagonal entries are all 0. We write ι : A→Mn(A) for the mapgiven by ι(a) = aIn for all a ∈ A.

We summarize the most basic properties of matrix rings:

Lemma 7.1.16. Let A be a ring. Then Mn(A) is a ring under the operations of matrixaddition and matrix multiplication, and the map ι : A→Mn(A) is a ring homomorphism.

Moreover, the elements ι(a) = aIn behave as follows: if M = (aij) ∈ Mn(A), thenaIn ·M is the matrix whose ij-th entry is a · aij for all i, j, while M · aIn is the matrixwhose ij-th entry is aij · a for all i, j.

Thus, if B is the center of A, then ι(B) is contained in the center of Mn(A).

The units in Mn(A) are those matrices M for which there exists an M ′ ∈ Mn(A)such that M ′M = MM ′ = In. Note that this precisely says that M is invertible in theusual sense from linear algebra.

Definitions 7.1.17. Let A be a ring. We write Gln(A) for Mn(A)×, the group ofinvertible n× n matrices. We call it the n-th general linear group of A.

The reader should be warned that if A is not commutative, then the usual sort ofdeterminant theory fails for matrices over A, and hence invertibility cannot be detectedby the familiar methods.

When A is commutative, on the other hand, there is a good theory of determinants,which we shall develop in Chapter 10. We shall show that a matrix in Mn(A) is invertibleif and only if its determinant is in A×.

Another example relates to a generalization of matrix rings.

Definition 7.1.18. Let G be an abelian group and let EndZ(G) be the set of all grouphomomorphisms from G to itself. We call it the endomorphism ring of G.

Here, the ring operations are defined as follows. For f, g ∈ EndZ(G), we let f + gbe the homomorphism given by (f + g)(x) = f(x) + g(x), where additive notation isused for the group operation in G. The product, f · g, of f, g ∈ EndZ(G) is simply thecomposition f ◦ g.

As usual, the justification for this definition is left to the reader:


Lemma 7.1.19. Under these operations, EndZ(G) is a ring, with additive identity thetrivial homomorphism, and multiplicative identity the identity map.

We shall generalize this when we study A-modules. If M is an A-module, then theA-module homomorphisms from M to itself will be shown to form a subring, EndA(M),of EndZ(M). If A is commutative, and if An is the free A-module of rank n, we shallsee that EndA(An) is isomorphic as a ring to Mn(A). Thus, endomorphism rings dogeneralize matrix rings when A is commutative.

A useful concept in ring theory is that of an algebra over a commutative ring A.

Definitions 7.1.20. An algebra over a commutative ring A consists of a ring B togetherwith a ring homomorphism ν : A→ B such that the image of ν is contained in the centerof B. We call ν the structure map for B as an A-algebra. If B is a commutative ring,we call it a commutative A-algebra.

If B and C are A-algebras, via ν : A → B and μ : A → C, then an A-algebrahomomorphism from B to C is a ring homomorphism f : B → C such that the followingdiagram commutes.

B

A

C��f

��ν

��μ

Examples 7.1.21.1. Let B be any ring and let A be the center of B. Then the inclusion of A in B

makes B an A-algebra.

2. Let B be any ring and let A be the center of B. Let ι : B → Mn(B) be the ringhomomorphism given by ι(b) = bIn for b ∈ B. Then Lemma 7.1.16 shows that ι(A)is contained in the center of Mn(B).3 Thus, Mn(B) is an A-algebra via ι ◦ i, wherei : A ⊂ B is the inclusion, and ι : B →Mn(B) is an A-algebra map.

We now show that the notions of rings and of Z-algebras are equivalent.

Lemma 7.1.22. Every ring A has a unique structure as a Z-algebra, and every ringhomomorphism f : A→ B is a Z-algebra homomorphism.

Proof The fact that a ring has a unique Z-algebra structure is just a restatement ofLemma 7.1.8: For any ring A, there is a unique ring homomorphism from Z to A, andits image lies in the center of A.

Now if ν : Z → A and μ : Z → B are the unique ring homomorphisms and iff : A → B is any ring homomorphism, we must have f ◦ ν = μ. Thus, f is a Z-algebrahomomorphism.

As in the case of groups, the direct product gives the simplest method of obtainingnew rings from old.

Definition 7.1.23. Let A and B be rings. Then the direct product A × B is the ringwhose additive group is the direct product of the additive groups of A and B and whosemultiplication is given coordinate-wise. In other words, the underlying set of A × B isthe cartesian product of A and B, and the operations are given by

(a, b) + (c, d) = (a+ c, b+ d) and (a, b) · (c, d) = (ac, bd).3Indeed, Problem 10 of Exercises 7.1.27 shows that ι(A) is equal to the center of Mn(B).


The reader should supply the justification that A×B is a ring:

Lemma 7.1.24. Let A and B be rings. Then the direct product A×B is a ring and theprojection maps from A×B to A and B are both ring homomorphisms.

Moreover, if C is a ring and if f : C → A × B is a map, say f(c) = (f1(c), f2(c)),then f is a ring homomorphism if and only if f1 : C → A and f2 : C → B are ringhomomorphisms.

In category theoretic terms, Lemma 7.1.24 just says that the direct product is theproduct in the category of rings.

The finite direct products A1 × · · · × An are defined analogously, and satisfy a sim-ilar universal property. Indeed, the product of an infinite family of rings is a ring viacoordinate-wise multiplication:

Definition 7.1.25. Suppose given a family {Ai | i ∈ I} of rings. We write∏i∈I Ai for

the ring structure on the product of the Ai whose operations are coordinate-wise additionand multiplication. Thus,

(ai | i ∈ I) + (bi | i ∈ I) = (ai + bi | i ∈ I) and (ai | i ∈ I) · (bi | i ∈ I) = (aibi | i ∈ I).

Again, the reader should justify the definition:

Lemma 7.1.26. Let {Ai | i ∈ I} be a family of rings. Then∏i∈I Ai is a ring under

coordinate-wise addition and multiplication. Here 0 is the I-tuple whose coordinatesare all 0, and 1 is the I-tuple whose coordinates are all 1. The projection maps πj :∏i∈I Ai → Aj are ring homomorphisms for all j ∈ I.If B is a ring and if f : B → ∏

i∈I Ai is a map, say f(b) = (fi(b) | i ∈ I) for eachb ∈ B, then f is a ring homomorphism if and only if fi is a ring homomorphism for alli ∈ I.Exercises 7.1.27.

1. Give the proof of Lemma 7.1.6.

2. Show that every subring of a field is an integral domain.4

3. We say that n ∈ Z is square free if it is not divisible by the square of any primenumber. Show that Zn contains no nonzero nilpotent elements if and only if n issquare free.

4. Suppose given a ring structure on the additive group Zn. Show that the multiplica-tive identity element must generate Zn as an additive group. Deduce that any ringstructure on Zn is isomorphic to the usual one. (Hint : Let e be the multiplicativeidentity for a given ring structure on Zn. Show that e fails to generate Zn addi-tively if and only if there is an integer d > 1 such that d divides n and such thate is the additive d-th power of some element x ∈ Zn. Now show that no such ringstructure could exist.)

5. Let f : A → B be a ring homomorphism. Show that f restricts to a grouphomomorphism f : A× → B×. (Note then that the passage from rings to theirunit groups gives a functor from rings to groups.)

6. Give the proof of Lemma 7.1.16.4We shall see in Section 7.11 that a ring is an integral domain if and only if it is a subring of a field.


7. Let A and B be rings. Show that Mn(A×B) is isomorphic to Mn(A)×Mn(B).

8. Show that Mn(A) contains nonzero nilpotent elements, and hence zero-divisors, forany nonzero ring A, provided that n ≥ 2.

9. Let A and B be subrings of a ring C. Let

R ={(

a c0 b

) ∣∣∣∣ a ∈ A, b ∈ B and c ∈ C}.

Show that R is a subring of M2(C).

† 10. Let B be the center of A. Show that ι : A → Mn(A) restricts to an isomorphismfrom B onto the center of Mn(A).

11. Let f : A→ B be a ring homomorphism. Show that there is a ring homomorphismf∗ : Mn(A)→Mn(B) defined by setting f∗((aij)) equal to the matrix whose ij-thentry is f(aij).


13. Show that the unit group of EndZ(G) is the automorphism group Aut(G) of G.

14. Let G be the direct sum of a countably infinite number of copies of Z. Find anelement of EndZ(G) which has a left inverse, but is not a unit.

15. Show that the center of a direct product A×B is the direct product of the centersof A and B.

16. Show that the group of units of A×B is A× ×B×.

‡ 17. Suppose given ring homomorphisms f : A → C and g : B → C. Recall that thepullback of f and g is the diagram

A×C B B

A C

��f

��g

��

g

��f

where A ×C B = {(a, b) ∈ A × B | f(a) = g(b)}, and f and g are defined byf(a, b) = b and g(a, b) = a.

(a) Show that A×C B is a subring of A×B and that f and g are ring homomor-phisms.

(b) Suppose given a ring R and ring homomorphisms f ′ : R→ B and g′ : R→ Asuch that f ◦ g′ = g ◦ f ′. Show that there is a unique ring homomorphismh : R→ A×C B such that f ◦h = f ′ and g ◦h = g′. Show that h must satisfyh(r) = (g′(r), f ′(r)) for all r ∈ R.

(c) Show that the natural map (A×C B)× → A× ×C× B× is an isomorphism.

18. Show that a direct product A × B of commutative rings is an integral domain ifand only if one of A and B is an integral domain and the other is the zero ring.


19. Show that a direct product A×B of rings is a division ring if and only if one of Aand B is a division ring and the other is the zero ring.



22. Let A be a ring. Show that Mm(A) × Mn(A) is isomorphic to the subring of

Mm+n(A) consisting of matrices of the form(M 00 N

), where M is an m ×m

matrix, N is an n×n matrix, and the 0’s are zero-matrices of the appropriate size.

The Complex Numbers

The complex numbers, C, give yet another example of a field. We review the definitionand basic properties of C.

Definitions 7.1.28. The elements of the complex numbers are formal sums a+ bi, witha, b ∈ R. This presentation is unique in the sense that if a, b, c, d ∈ R, then a+bi = c+diif and only if a = c and b = d. Thus, the elements of C are in one-to-one correspondencewith ordered pairs (a, b) of real numbers. So we may identify C with the plane R2, whichgives us a geometric picture of the complex numbers.

We identify a ∈ R with the complex number a+ 0i. We say that a complex numberis in R if it has this form.

Similarly, we write i for the complex number 0 + 1i, and refer to the numbers ai =0 + ai, with a ∈ R as the pure imaginary numbers.

If z = a+ bi and w = c+ di are complex numbers, with a, b, c, d ∈ R, we define theirsum by z+w = (a+ c)+(b+d)i, and define their product by zw = (ac− bd)+(ad+ bc)i.

For z = a+ bi with a, b ∈ R, we define the complex conjugate, z, of z by z = a− bi.Again, there are verifications left to the reader.

Lemma 7.1.29. The above operations of addition and multiplication give C the struc-ture of a commutative ring. Moreover, the standard inclusion R ⊂ C is a ring homomor-phism, and hence the additive and multiplicative identities of C are given by 0 = 0 + 0iand 1 = 1 + 0i, respectively.

The function C → C that carries each z ∈ C to its complex conjugate, z, is a ringhomomorphism.

For z = a+ bi with a, b ∈ R, we have the equality

zz = a2 + b2 ∈ R ⊂ C.

In particular, zz ≥ 0 for all z, and zz = 0 if and only if z = 0.

We can now find a multiplicative inverse for every nonzero element of C.

Corollary 7.1.30. The complex numbers, C, is a field. Explicitly, if 0 �= z ∈ C, then zis a unit in C with inverse

z−1 =1zz· z.

Here, 1/(zz) is the inverse of the nonzero real number zz.


Proof Since zz �= 0, the result follows from the facts that C is commutative and thatR is a subring of C.

The notion of planar distance from analytic geometry may be derived from the com-plex product zz.

Definition 7.1.31. We define the absolute value | | : C→ R by

|z| =√zz =

√a2 + b2

for z = a+ bi with a, b ∈ R.

A very useful tool in studying the complex numbers is the complex exponential func-tion.

Definition 7.1.32. Let x ∈ R. We define the complex exponential eix ∈ C by

eix = cosx+ i sinx.

More generally, if z = x+ iy ∈ C, with x, y ∈ R, we define ez ∈ C by

ez = exeiy = ex cos y + iex sin y,

where ex is the ordinary real exponential function applied to x.

We shall be primarily concerned with the exponentials eix with x ∈ R.

Lemma 7.1.33. Let x, y ∈ R, then ei(x+y) = eix · eiy for x, y ∈ R, where the producton the right is complex multiplication. Moreover, ei·0 = 1. Thus, there is a grouphomomorphism exp : R→ C× defined by exp(x) = eix.

Proof Complex multiplication gives

eix · eiy = (cosx+ i sinx) · (cos y + i sin y)= (cosx cos y − sinx sin y) + i(cosx sin y + sinx cos y)= cos(x+ y) + i sin(x+ y),

by the formulæ for the sine and cosine of a sum. But the last line is precisely ei(x+y).From the known values of the sine and cosine of 0, we see that ei·0 = 1, as claimed,

so that setting exp(x) = eix gives a monoid homomorphism from R to the multiplicativemonoid of C. But any monoid homomorphism from a group to a monoid must take valuein the group of invertible elements of the monoid, which in this case is C×.

The exponential allows us to define some important elements of C, the standardprimitive n-th roots of unity.

Definition 7.1.34. Let n ≥ 1. The standard primitive n-th root of unity, ζn ∈ C, isdefined by

ζn = ei·2π/n = cos(

2πn

)+ i sin

(2πn

).

The fact that 2π is the smallest positive real number whose sine is 0 and whose cosineis 1 gives us the following lemma.


Lemma 7.1.35. Let n ≥ 1. Then ζn has order n in C×.

There is a noncommutative division ring, called the quaternions, H, that is given bya construction similar to that of the complex numbers. We shall treat it in the next setof exercises. Both C and H are examples of what are called Clifford algebras over R.We shall treat the general case of Clifford algebras in Chapter 9.

We shall not prove it here, but C and H are the only division rings that are algebrasover R in such a way that the induced real vector space structure is finite dimensional.

Exercises 7.1.36.1. Write Z2 = 〈T 〉 = {1, T}. Show that Z2 acts on C via T · z = z for all z ∈ C.

Show that the fixed point set CZ2 is equal to R.

2. Write Zn = 〈T 〉 = {1, T, . . . , Tn−1} for n ≥ 2. Show that Zn acts on C viaT k · z = ζkn · z for all k, where ζn = ei·2π/n is the standard primitive n-th root ofunity in C. Show that 1+ζn+ · · ·+ζn−1

n is a fixed point under this action. Deducethat 1 + ζn + · · ·+ ζn−1

n = 0.

3. Show that |zw| = |z| · |w| for all z, w ∈ C, that |z| ≥ 0 for all z ∈ C, and that|z| = 0 if and only if z = 0.

4. Show that |z + w| ≤ |z|+ |w| for all z, w ∈ C.

5. Show that i is a square root of −1 in C.

6. Let S1 = {z ∈ C | |z| = 1}. Show that S1 is a subgroup of C×.

7. Show that eix ∈ S1 for all x ∈ R, and hence exp gives a homomorphism from Rto S1. What is the kernel of exp?

8. Show that exp : R→ S1 is onto. Deduce that S1 is isomorphic as a group to R/Z.

9. Show that there is a homomorphism expC : C→ C× given by expC(z) = ez.

10. Show that eiπ = −1 ∈ C.

11. Show that ζ1 = 1, ζ2 = −1, and ζ4 = i.

12. Let n = mk with m and k positive integers. Show that ζkn = ζm.

13. Show that if n = 2m with m a positive integer, then ζk+mn = −ζkn.

14. Let m > 0 be an odd integer. Show that the subgroup of C× generated by −ζmcoincides with the subgroup generated by ζ2m.

15. Define φ : C → M2(R) by φ(a + bi) =(a −bb a

)when a, b ∈ R. Show that φ is an

injective ring homomorphism. Show that φ(S1) = SO(2) and that φ(eiθ) = Rθ,the matrix that rotates R2 through the angle θ, for all θ ∈ R.

16. We construct the ring of quaternions. The quaternionic groups Q4n, which weanalyzed in the material on group theory, will be seen to embed naturally in thegroup of units of the quaternions.

An alternative name for the quaternions is the Hamiltonians. Because of this,and the fact that Q is already taken, we shall write H for the quaternions. Theelements of H are formal sums a + bi + cj + dk, where a, b, c, and d are real


numbers. Here, a + bi + cj + dk = a′ + b′i + c′j + d′k if and only if a = a′,b = b′, c = c′, and d = d′, and hence we may identify H with the set R4 of 4-tuples (a, b, c, d) of real numbers. We define the addition in H accordingly, so that(a+ bi+ cj+ dk) + (a′ + b′i+ c′j+ d′k) = (a+ a′) + (b+ b′)i+ (c+ c′)j+ (d+ d′)k.

We write 1 = 1 + 0i + 0j + 0k, i = 0 + 1i + 0j + 0k, j = 0 + 0i + 1j + 0k andk = 0 + 0i+ 0j + 1k. For a ∈ R, we write a = a · 1 = a+ 0i+ 0j + 0k. Thus, weidentify the real multiples of 1 with R.

The multiplication in H is defined so that 1 is the identity element, i2 = j2 = k2 =−1, ij = −ji = k, jk = −kj = i and ki = −ik = j. Assembling this information,we see that if α = a+ bi+ cj + dk and α′ = a′ + b′i+ c′j + d′k, then

αα′ = (aa′ − bb′ − cc′ − dd′) + (ab′ + ba′ + cd′ − dc′)i+ (ac′ + ca′ + db′ − bd′)j + (ad′ + da′ + bc′ − cb′)k.

Unlike the complex numbers, we see that multiplication in H is noncommutative.

(a) Show that H is a ring and that the real numbers R ⊂ H form a subring.(b) Show that the center of H is R, so that H is an R-algebra.(c) Show that if we set x equal to either i, j, or k, then the set of all elements of

H of the form a + bx with a, b ∈ R forms a subring of H that is isomorphicto C.

(d) Define conjugation in H as follows: for α = a + bi + cj + dk, we set α =a− bi− cj − dk. Show that the following equalities hold for all α, α′ ∈ H.

i. α+ α′ = α+ α′

ii. α · α′ = α′ · αiii. 1 = 1.iv. α = α.(According to definitions we shall give below, the first three of the aboveproperties say that conjugation is an antiautomorphism of H, and hence H isa self-opposite ring. This is useful in the study of matrices over H.)

(e) Show that α and α commute for all α ∈ H, and that if α = a+ bi+ cj + dk,then αα is the real number a2 + b2 + c2 + d2.

(f) Define the absolute value | | : H→ R by

|α| =√αα =

√a2 + b2 + c2 + d2

for α = a+ bi+ cj+ dk with a, b, c, d ∈ R. Show that |α · α′| = |α| · |α′| for allα, α′ ∈ H, that |α| ≥ 0 for all α ∈ H, and that |α| = 0 if and only if α = 0.

(g) Show that every nonzero element α ∈ H is a unit in H (and hence H is adivision ring). Here, if α �= 0, show that

α−1 =1αα

α.

(h) Show that if we set b = cos(π/n) + i sin(π/n) and set a = j, then the sub-group of H× generated by a and b is isomorphic to Q4n, the quaternionicgroup of order 4n. (In particular, note that Q8 is isomorphic to the subgroup{±1,±i,±j,±k} of H×. When studying Q8 alone, rather than the quater-nionic groups in general, it is customary to use the notation ±1,±i,±j,±kfor its elements.)


(i) Let S3 = {α ∈ H | |α| = 1}. Show that S3 is a subgroup of H× containingthe above copies of Q4n. Show also that S1 embeds as a subgroup of S3.

17. Let S = {α = ai+ bj + ck ∈ H | a, b, c ∈ R and |α| = 1}.5 Show that α2 = −1 forall α ∈ S. Deduce that H has infinitely many subrings isomorphic to C.

The p-adic Integers

We construct the p-adic integers, Zp. For k ≥ 2, let εk : Zpk → Zpk−1 be the unique ringhomomorphism (in fact, the unique group homomorphism that carries 1 to 1) from Zpk

to Zpk−1 . The p-adic integers are the inverse limit of the maps εk.To avoid a dependence on category theory here, we shall give the construction explic-

itly. Zp is a subring of the infinite product∏k≥1 Zpk . It is customary to write the tuples

in the infinite product in the reverse order from the usual one. Thus, a typical elementlooks like (. . . ,mk, . . . ,m2,m1), where each mk lies in Zpk . Under this convention, wehave

Zp = {(. . . ,mk, . . . ,m2,m1) | εk(mk) = mk−1 for all k ≥ 2}.Recall that the ring operations in the infinite product are defined via coordinate-wise

addition and multiplication, so the fact that the εk are ring homomorphisms verifies theassertion above:

Lemma 7.1.37. Zp is a subring of∏k≥1 Zpk .

The additive and multiplicative identities, of course, are 0 = (. . . , 0, . . . , 0, 0) and1 = (. . . , 1, . . . , 1, 1), respectively. The next lemma is elementary, but important.

Lemma 7.1.38. Let πk : Zp → Zpk be induced by the projection onto the k-th factor.Then πk is a ring homomorphism for all k ≥ 1.

The p-adic integers are important in number theory. Despite the fact that they aremade up from quotient rings of Z, the next lemma will show that the p-adic integersthemselves form a ring of characteristic 0. (See Definition 7.2.12.)

Lemma 7.1.39. The unique ring homomorphism ι : Z→ Zp is an embedding.

Proof We have ι(m) = (. . . ,m, . . . ,m,m), so m ∈ ker ι if and only if m is divisible bypk for all k ≥ 0.

The structure of the unit group is one of the interesting features of Zp. Recall that asplitting map, or section, for a homomorphism f : G→ K is a homomorphism s : K → Gsuch that f ◦ s is the identity map of K. A map f that admits a section is called a splitsurjection. Recall from Corollary 4.7.6 that if f : G → K is a split surjection with Gabelian, then G ∼= ker f ×K.

Proposition 7.1.40. An element a = (. . . ,mk, . . . ,m2,m1) of Zp is a unit if and onlyif m1 is a unit in Zp. The map

π1 : Z×p → Z×

p

is a split surjection.5Geometrically, S is a 2-dimensional subsphere of S3.


Proof Let a = (. . . ,mk, . . . ,m2,m1) ∈ Zp. Since the multiplication in Zp is coordinate-wise, if a is a unit in Zp, then each mk must be a unit in Zpk .

Conversely, suppose that m1 is a unit. According to Lemma 4.5.3, an element m isa unit in Zpk if and only if (m, p) = 1. In particular, m is a unit in Zpk if and only ifεk(m) = m is a unit in Zpk−1 . Thus, if m1 is a unit, then each mk must be a unit in Zpk .

Let nk be the inverse of mk in Zpk . Since each εk is a ring homomorphism, we musthave εk(nk) = nk−1. Thus, setting b = (. . . , nk, . . . , n2, n1) specifies an element of Zpthat is easily seen to be a multiplicative inverse for a.

It remains to construct a splitting map for π1 : Z×p → Z×

p . For p = 2, there’s nothingto show, as Z×

2 is the trivial group. For p > 2, we consider the proof of Corollary 4.5.12.Let ηk : Zpk → Zp be the unique ring homomorphism between these rings. Then thekernel of the induced map ηk : Z×

pk → Z×p has order relatively prime to |Z×

p | = p − 1.Thus, the proof of Corollary 4.4.15 shows that if K is the subgroup of Z×

pk consisting

of those elements of exponent p − 1, then ηk : K∼=−→ Z×

p . But this says that there is aunique section sk : Z×

p → Z×pk of ηk.

But then εk ◦ sk is a section for ηk−1 : Z×pk−1 → Z×

p , so εk ◦ sk = sk−1. Thus, setting

s(m) = (. . . , sk(m), . . . , s2(m),m)

gives a well defined function s : Z×p → Zp, that is easily seen to give a section for

π1 : Z×p → Z×

p .

Exercises 7.1.41.† 1. Show that Zp is an integral domain. (Hint : Suppose that ab = 0 in Zp, with

a = (. . . ,mk, . . . ) and b = (. . . , nk, . . . ). Suppose that mk �= 0 in Zpk . What doesthis say about the p-divisibility of mk+l and nk+l?)

2. Let p be an odd prime. Show that the kernel of π1 : Z×p → Z×

p is isomorphic to the

additive group of Zp. Deduce that the torsion subgroup of Z×p maps isomorphically

to Z×p under π1.

3. Show that the kernel of π2 : Z×2 → Z×

4 is isomorphic to the additive group of Z2.

Deduce that the torsion subgroup of Z×2 maps isomorphically to Z×

4 under π2.

† 4. Let a be in the kernel of the additive homomorphism π1 : Zp → Zp. Show thata = pb for some b ∈ Zp. (Hint : Consider the diagram

Zpk Zpk+1

Zpk−1 Zpk

��⊂

��

εk

��

εk+1

��⊂

where the inclusion maps send 1 to p.)

† 5. Let a = (. . . ,mk, . . . ,m2,m1) ∈ Zp. Suppose that mk = 0 in Zpk for all k ≤ r and

that mr+1 �= 0 in Zpr+1 . Show that a = pru, where u ∈ Z×p .


6. Let fk : Ak → Ak−1 be a ring homomorphism for k ≥ 2. Give an isomorphism(lim←−Ak

)× ∼= lim←−(A×k ).

7. Give Zpk the discrete topology for all k. Show that Zp is a closed subspace of theproduct (Tikhonov) topology on

∏k≥1 Zpk , and hence is compact by the Tikhonov

Theorem. Show that ι(Z) is dense in this topology on Zp. For this reason, Zp maybe thought of as a completion of Z in the topology that Z inherits from Zp.

7.2 Ideals

The ideals in a ring are a very important part of its structure. We give the most basicresults on ideals in this section, and shall continue to study them throughout the materialon rings.

Definitions 7.2.1. Let A be a ring. A left ideal of A is a subgroup a ⊂ A of the additivegroup of A with the additional property that if x ∈ a, then for any a ∈ A, we have ax ∈ a.(In other words, a is closed on the operation of multiplying its elements on the left byany element of A.)

From the other side, a right ideal of A is a subgroup a ⊂ A of the additive group ofA with the additional property that if x ∈ a, then for any a ∈ A, we have xa ∈ a.

A two-sided ideal of A is a subgroup of the additive group of A which is simultaneouslya left ideal and a right ideal.

If A is commutative, all ideals are two-sided, and we refer to them simply as ideals.

Examples 7.2.2.1. Let A be a ring. Then A is a two-sided ideal of itself.

2. Let A be a ring and let 0 ⊂ A be the trivial subgroup. Since a · 0 = 0 · a = 0 for alla ∈ A, 0 is a two-sided ideal of A.

Another set of examples comes from the principal ideals.

Definition 7.2.3. Let A be a ring and let x ∈ A. The principal left ideal, Ax, generatedby x is defined by

Ax = {ax | a ∈ A}.Similarly, the principal right ideal generated by x is xA = {xa | a ∈ A}.If A is commutative, then Ax = xA, and we call it simply the principal ideal generated

by x, and denote it by (x).

Principal ideals play a role in ideal theory similar to the role of cyclic subgroups ingroup theory:

Lemma 7.2.4. Let A be a ring and let x ∈ A. Then Ax and xA are the smallest leftand right ideals, respectively, that contain x.

Some commutative rings have no ideals other than the principal ones.

Definitions 7.2.5. A principal ideal ring is a commutative ring in which every ideal isprincipal.

A principal ideal domain, or P.I.D., is an integral domain in which every ideal isprincipal.


Indeed, the integers are better behaved than the general integral domain:

Lemma 7.2.6. Every subgroup of Z is a principal ideal. In particular, Z is a P.I.D.

Proof We know (Proposition 2.3.2) that the subgroups of Z all have the form 〈n〉 ={nk | k ∈ Z} for some n ∈ Z. But 〈n〉 is precisely (n), the principal ideal generated byn.

We shall see below that if K is a field,6 then the polynomial ring K[X] is a principalideal domain.

Principal ideals are useful in recognizing division rings and fields.

Lemma 7.2.7. A nonzero ring A is a division ring if and only if the only left ideals ofA are 0 and A.

Proof Let A be a division ring and let a be a nonzero left ideal of A. Thus, there is anx ∈ a with x �= 0. But then x is a unit in A. Now 1 = x−1x ∈ a, as a is a left ideal. Butthen a = a · 1 ∈ a for the same reason, for all a ∈ A. Thus, a = A.

Conversely, suppose that A is a ring with no left ideals other than 0 and A. Let0 �= x ∈ A. Then Ax is a nonzero left ideal, and hence Ax = A. Thus, there is anelement y ∈ A with yx = 1. Thus, every nonzero element in the multiplicative monoidof A has a left inverse. As in Problem 13 of Exercises 2.1.16, it is easy to see that everynonzero element of A is a unit.

Two-sided ideals play a role in ring theory somewhat analogous to that played bynormal subgroups in the theory of groups.

Proposition 7.2.8. Let a be any subgroup of the additive group of a ring A, and letπ : A → A/a be the canonical map. Then A/a has a ring structure that makes π a ringhomomorphism if and only if a is a two-sided ideal of A. Such a ring structure, if itexists, is unique.

Proof Suppose given a ring structure on A/a such that π is a ring homomorphism.Then writing a ∈ A/a for π(a), we must have a · b = ab, so the ring structure is indeedunique. Moreover, for x ∈ a, 1 = 1 + x. Thus, for a ∈ A, a = a · 1 = a · 1 + x = a+ ax.In other words, a + ax represents the same element of A/a as a does. But this saysa+ ax− a ∈ a, and hence ax ∈ a for a ∈ A and x ∈ a.

Thus, a has to be a left ideal. But a similar argument, obtained by multiplying 1and 1 + x on the right by a, shows that a must also be a right ideal, so that a is in facttwo-sided.

For the converse, suppose that a is a two-sided ideal. We wish to show that themultiplication a · b = ab gives a well defined binary operation on A/a. This will besufficient to complete the proof, as then π is a homomorphism with respect to thisoperation, so that the verifications that A/a is a ring under this multiplication willfollow from the fact that A is a ring.

Thus, for a, b ∈ A and x, y ∈ a, we must show that ab and (a+x)(b+y) represent thesame element of A/a, and hence that (a+ x)(b+ y)− ab ∈ a. But (a+ x)(b+ y)− ab =ay + xb+ xy, which is in a because a is a two-sided ideal.

The rings A/a satisfy a universal property similar to that of factor groups in grouptheory. We call them quotient rings of A.

6The use of K for a field derives from the German Korper , which means field in the mathematicalsense. Amusingly, the German word for a farmer’s field is a cognate of ours: Feld.


Proposition 7.2.9. Let f : A→ B be a ring homomorphism. Then ker f is a two-sidedideal of A. If a is another two-sided ideal of A, then a ⊂ ker f if and only if there is aring homomorphism f : A/a → B which makes the following diagram commute, whereπ : A→ A/a is the canonical map:

Af ��

π ��

��

� B

A/af

��

Moreover, if a ⊂ ker f , then the homomorphism f : A/a → B making the diagramcommute is unique.

When a = ker f , f induces a ring isomorphism between A/a and the image of f .

Proof By Proposition 4.1.7, there is a unique homomorphism of additive groups, f :A/a → B making the diagram commute if and only if a ⊂ ker f . And if a = ker f , thishomomorphism f gives a group isomorphism onto the image of f by the First NoetherIsomorphism Theorem.

Thus, it suffices to show that ker f is a two-sided ideal and that f is a ring homomor-phism whenever a is a two-sided ideal contained in ker f . We leave these verifications tothe reader.

Since π : Z → Zn is a surjective ring homomorphism and has kernel (n), we obtainthe following corollary.

Corollary 7.2.10. There is an isomorphism of rings between Zn and the quotient ringZ/(n).

The next corollary is another consequence of Proposition 7.2.9.

Corollary 7.2.11. Let f : D → B be a ring homomorphism, where D is a division ringand B is any nonzero ring. Then f is injective.

Proof The kernel of f is a two-sided ideal, and hence must be either 0 or D. As B isnonzero, ker f �= D. So ker f = 0, and f is injective.

Recall that for any ring B, there is a unique ring homomorphism f : Z → B. Wemay obtain useful information about B by applying Proposition 7.2.9 with a = ker f .

Definition 7.2.12. Let A be a ring and let f : Z→ A be the unique ring homomorphismfrom Z to A. Let ker f = (n), with n ≥ 0. Then we say that A has characteristic n. Inparticular, A has characteristic 0 if and only if f is injective.

Examples 7.2.13. Since the kernel of the canonical map π : Z → Zn is (n), Zn hascharacteristic n. On the other hand, Z is a subring of Q, R, C, and H, so these havecharacteristic 0.

Proposition 7.2.14. Let n > 0. Then a ring A has characteristic n if and only if it hasa subring isomorphic as a ring to Zn. If A does have characteristic n, then the subringin question is the image of the unique ring homomorphism f : Z→ A.


Proof If A has characteristic n with n > 0, then the image of f is isomorphic to Z/(n)by Proposition 7.2.9. Conversely, suppose that A has a subring B that is isomorphic asa ring to Zn. Then if f ′ : Z → B is the unique ring homomorphism, we may identifyf ′ with the canonical map π : Z → Z/(n). In particular, the kernel of f ′ is (n). But ifi : B → A is the inclusion, then i ◦ f ′ is the unique ring homomorphism from Z to A.Since i is injective, the kernel of i ◦ f ′, whose generator is by definition the characteristicof A, is precisely the kernel of f ′.

Let K be a field. Then the image of the unique ring homomorphism f : Z → K, asa subring of K, must be an integral domain. Thus, Lemma 7.1.12 shows that if f is notinjective (i.e., if the image of f is finite), then f must have image Zp for some prime p.

Corollary 7.2.15. Let K be a field of nonzero characteristic. Then the characteristicof K is a prime number.

We can generalize the notion of a principal ideal as follows.

Definitions 7.2.16. Let A be a ring and let S be any subset of A. Then the left idealgenerated by S, written AS, is given as follows:

AS = {a1x1 + · · ·+ akxk | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}.

Similarly, the right ideal generated by S is given by

SA = {x1a1 + · · ·+ xkak | k ≥ 1, ai ∈ A and xi ∈ S for 1 ≤ i ≤ k}.

When A is commutative, of course, AS = SA, and we denote it by (S). If S is finite,say S = {x1, . . . , xn}, we write (S) = (x1, . . . , xn).

For A noncommutative, the two-sided ideal generated by S is

ASA = {a1x1b1 + · · ·+ akxkbk | k ≥ 1, ai, bi ∈ A and xi ∈ S for 1 ≤ i ≤ k}.

The next lemma is immediate from the definitions.

Lemma 7.2.17. Let A be a ring and let S be a subset of the center of A. Then theleft, right, and two-sided ideals generated by A all coincide, i.e., AS = SA = ASA. Inparticular, if A is commutative, then ASA = (S).

Warning: In the case of left ideals, we may apply the distributive law to reduce thesums a1x1 + · · ·+ akxk until the elements xi are all distinct. Thus, for instance, A{x} =Ax = {ax | a ∈ A}, the principal left ideal generated by x. The analogous result holds forthe right ideals. But in the case of two-sided ideals in noncommutative rings, we cannotmake this kind of reduction in general. Thus, in A{x}A, the principal two-sided idealgenerated by x, we must consider all elements of the form a1xb1 + · · ·+akxbk with k ≥ 1and with ai, bi ∈ A for 1 ≤ i ≤ k.Lemma 7.2.18. Let A be a ring and let S ⊂ A. Then AS, SA, and ASA are thesmallest left, right, and two-sided ideals, respectively, that contain S.

This now gives the following universal property.


Corollary 7.2.19. Let A be a ring and let S ⊂ A, and let a ⊂ A be the two-sided idealgenerated by S. Let f : A→ B be a ring homomorphism whose kernel contains S. Thenthere is a unique homomorphism f : A/a→ B such that the following diagram commutes

Af ��

π ��

��

� B

A/af

��

,

where π : A→ A/a is the canonical map.

Proof The kernel of f is a two-sided ideal of A containing S, so we must have a ⊂ ker fby Lemma 7.2.18. The result now follows from Proposition 7.2.9.

When a ring homomorphism f : B → A is surjective, we can use information aboutA and about ker f to deduce information about B.

An important topic in ideal theory is a discussion of the various ways in which wecan obtain new ideals from old. Recall first (from Lemma 4.1.18) that if G is a groupand if H and K are subgroups of G with K ⊂ NG(H), then KH = {kh | k ∈ K, h ∈ H}is a subgroup of G. Thus, if G is abelian, written in additive notation, and if H and Kare any subgroups of G, we have a subgroup H +K given by

H +K = {h+ k |h ∈ H, k ∈ K}.

The next lemma is elementary.

Lemma 7.2.20. Let a and b be left (resp. right) ideals in the ring A. Then the subgroupsa + b and a ∩ b are also left (resp. right) ideals of A.

We have already seen an important use of sums of ideals. Indeed, the greatest commondivisor of two integers, m and n, was defined to be the generator of the ideal (m) + (n).

We can generalize the notion of the ideals generated by a set.

Definitions 7.2.21. Let A be any ring. Let a be a left ideal of A, let b be a right idealof A, and let S be any subset of A. We write

aS = {a1x1 + · · ·+ akxk | k ≥ 1, ai ∈ a and xi ∈ S for 1 ≤ i ≤ k}Sb = {x1b1 + · · ·+ xkbk | k ≥ 1, xi ∈ S and bi ∈ b for 1 ≤ i ≤ k}

Thus, under either of the above definitions,

ab = {a1b1 + · · ·+ akbk | k ≥ 1, ai ∈ a and bi ∈ b for 1 ≤ i ≤ k}.

These constructions behave as follows.

Lemma 7.2.22. Let A be any ring. Let a be a left ideal of A, let b be a right ideal ofA, and let S be any subset of A. Then aS is a left ideal of A and Sb is a right ideal ofA. Combining these two facts, we see that ab is a two-sided ideal of A.

Finally, if a and b are both two-sided ideals of A, then we have an inclusion of two-sided ideals ab ⊂ a ∩ b.


Proof We leave the assertions of the first paragraph to the reader. For the finalstatement, consider an element of the form a1b1 + · · ·+ akbk with ai ∈ a and bi ∈ b forall i. Since b is a left ideal, each term aibi lies in b. Since a is a right ideal, each termaibi lies in a. So their sum lies in a ∩ b.

We shall explore the effects of taking products and intersections of ideals in Z in thenext set of exercises.

The following use of sums and intersections can be very useful in the study of rings,as well as for the module theory we shall introduce in Section 7.7.

Proposition 7.2.23. (Chinese Remainder Theorem) Suppose given two-sided ideals a1, . . . , akof the ring A such that ai + aj = A whenever i �= j. Let

f : A→ A/a1 × · · · ×A/akbe given by f(a) = (a, . . . , a). Then f induces a ring isomorphism

f : A/b∼=−→ A/a1 × · · · ×A/ak,

where b =(⋂k

i=1 ai

).

Proof By the obvious extension of Lemma 7.1.24 to k-fold products, f is a ring homo-morphism. Thus, it suffices to show that f is surjective, with kernel

⋂ki=1 ai.

The latter statement is immediate, as the kernel of a map into a product is alwaysthe intersection of the kernels of the associated maps into the factors. Thus, we needonly show that f is surjective.

But the product A/a1 × · · · × A/ak is generated as an abelian group by the imagesof the A/ai under the canonical inclusions. Thus, it suffices to show that these imagesare contained in the image of f . We show this for i = 1.

Now, we claim that it suffices to show that (1, 0, . . . , 0) is in the image of f . Tosee this, suppose that f(x) = (1, 0, . . . , 0). Since f is a ring homomorphism f(ax) =(a, . . . , a) ·(1, 0, . . . , 0) = (a, 0, . . . , 0) for all a ∈ A, and hence A/a1×0×· · ·×0 is indeedcontained in the image of f .

We now make use of the fact that a1 + ai = A for all i �= 1. Thus, for 2 ≤ i ≤ k,there are elements ai ∈ a1 and bi ∈ ai, with ai + bi = 1. Then if πi : A → A/ai is thecanonical map for 1 ≤ i ≤ k, we have π1(bi) = 1 and πi(bi) = 0 for 2 ≤ i ≤ k.

Let b = b2 · · · bk. Since the canonical maps πi are ring homomorphisms, π1(b) = 1,while πi(b) = 0 for 2 ≤ i ≤ k. But this just says that f(b) = (1, 0, . . . , 0) as desired.

Exercises 7.2.24.1. Let A and B be rings. Show that the left ideals of A×B all have the form a× b,

where a is a left ideal of A and b is a left ideal of B.

2. Let A be an integral domain and let a, b ∈ A. Show that (a) = (b) if and only ifthere is a unit x ∈ A× such that ax = b.

3. Let A be a ring. Show that there is a ring homomorphism from Zn to A if andonly if the characteristic of A divides n. Show that this ring homomorphism, if itexists, is unique.

4. Let f : A→ B be a ring homomorphism. Show that f−1(a) is a left (resp. right ortwo-sided) ideal of A for each left (resp. right or two-sided) ideal a of B.


† 5. Let f : A → B be a surjective ring homomorphism. Show that f(a) is a left(resp. right or two-sided) ideal of B whenever a is a left (resp. right or two-sided)ideal of A. Deduce that there is a one-to-one correspondence between the ideals ofB and the ideals of A containing ker f .

6. Show that every subgroup of Zn is a principal ideal, and hence that Zn is a principalideal ring.

7. Find a ring homomorphism f : A → B that is not surjective and an ideal a ⊂ Asuch that f(a) is not an ideal of B.

8. Let a ⊂ A be a left ideal. Show that a = A if and only if a contains an elementwhich has a left inverse.

9. Show that a principal left ideal Ax = A if and only if x has a left inverse.

10. A division ring has been shown to have no two-sided ideals other than 0 and thering itself. Show that the paucity of two-sided ideals does not characterize divisionrings: Show that if K is a field and n ≥ 2, then Mn(K) has no two-sided idealsother than 0 and itself. (Hint : The cleanest way to show this uses linear algebra.The reader may wish to come back to this problem after reading Sections 7.9 and7.10.)

† 11. Repeat the preceding problem, replacing K by a division ring D.

† 12. Let A be a ring. Define the opposite ring Aop as follows. As an abelian group, Aop

is just A. For a ∈ A, we write a for the associated element of Aop. We define themultiplication in Aop by a · b = ba. Show that Aop is a ring. Show that the leftideals of A are in one-to-one correspondence with the right ideals of Aop, and thatthe right ideals of A are in one-to-one correspondence with the left ideals of Aop.

‡ 13. Let A be a ring. An anti-automorphism of A is an automorphism f : A→ A of theadditive group of A, such that f(ab) = f(b)f(a) for all a, b ∈ A and f(1) = 1. Showthat A admits an anti-automorphism if and only if A and Aop are isomorphic.

14. We say that a ring is self-opposite if it admits an anti-automorphism. Show thatcommutative rings are self-opposite.

15. Let A be a self-opposite ring. Show that Mn(A) is self-opposite for all n ≥ 1.(Hint : See Problem 5 of Exercises 7.10.23.)

16. Let m,n ∈ Z. Calculate the generator of the product ideal (m)(n).

17. Let m,n ∈ Z. Calculate the generator of (m) ∩ (n).

18. Let n = pr11 . . . prk

k , where p1, . . . , pk are distinct primes. Show that Zn is isomorphicas a ring to the direct product Zpr1

1× · · · × Zprk

k.

19. Use the preceding problem to give a quick proof that if n = pr11 . . . prk

k , then Aut(Zn)is isomorphic to Aut(Zpr1

1)× · · · ×Aut(Zprk

k).

20. Suppose given an infinite family {ai | i ∈ I} of left ideals of A. Define the sum ofthis infinite family by∑

i∈Iai = {ai1 + · · ·+ aik | k ≥ 1, i1, . . . , ik ∈ I, and aij ∈ aij for 1 ≤ j ≤ k}.


(a) Show that∑i∈I ai is the smallest left ideal that contains each ai.

(b) If S is any subset of A, show that AS =∑x∈S Ax.

7.3 Polynomials

One of the most important constructions in ring theory is that of polynomial rings. Westudy the polynomials in one variable now and study the polynomials in several variablesin a subsection (page 228).

Definitions 7.3.1. Let A be a ring (not necessarily commutative) and let X be avariable. We define a ring A[X] as follows. The elements of A[X] are formal sums∑ni=0 aiX

i = anXn+ · · ·+ a1X + a0, where n is a nonnegative integer and ai ∈ A for all

i. We often use functional notation for these sums, writing, say, f(X) =∑ni=0 aiX

i forthe element above. If an �= 0, we say that f(X) has degree n and that an is its leadingcoefficient. Otherwise, if m is the largest integer for which am �= 0, we identify f(X) withthe polynomial

∑mi=0 aiX

i = amXm+ · · ·+a1X+a0, and hence it has degree m. (In the

case of 0 = 0X0, there are no possible identifications with lower degree polynomials, andwe say that 0 has degree −∞.) We say that a polynomial f(X) is monic if its leadingcoefficient is 1.

As is implicit in the notations above, we write either a or aX0 for the degree 0polynomial whose 0-th coefficient is a. We write ι : A → A[X] for the map given byι(a) = aX0 for all a ∈ A.

We also write Xn for 1Xn and write X for X1.Note that two polynomials are equal if and only if their coefficients are equal. (Here,

if f(X) =∑ni=0 aiX

i, then the coefficients of the Xm for m > n are all implicitly 0.)Addition of polynomials is obtained by adding the coefficients: If f(X) =

∑ni=0 aiX

i

and g(X) =∑ni=0 biX

i (where n is the larger of the degrees of f and g), then

f(X) + g(X) =n∑i=0

(ai + bi)Xi.

Multiplication of polynomials is given as follows: if f(X) =∑ni=0 aiX

i and g(X) =∑mi=0 biX

i, then their product is given by

f(X)g(X) =n+m∑i=0

⎛⎝ i∑j=0

ajbi−j

⎞⎠Xi.

Note that if A is noncommutative, then the order of the products aibi−j is important.Finally, we give a special name to some of the simplest polynomials: We shall refer to

a polynomial of the form aXi for some a ∈ A and i ≥ 0 as a monomial. Note that everypolynomial is a sum of monomials, so that the monomials generate A[X] as an abeliangroup.

We will occasionally use other letters, such as T or Y for the variable of a polynomialring. We shall refer to the variable as an indeterminate element over the ring A.

Again, the reader should justify the definitions:

Lemma 7.3.2. Let A be a ring. Then A[X] is a ring under addition and multiplicationof polynomials, and ι : A→ A[X] is a ring homomorphism.

The element X ∈ A[X] commutes with every element of ι(A), and if A is commuta-tive, so is A[X].


Note that since every polynomial is a sum of monomials, the distributive law showsthat the multiplication in A[X] is determined by the rule aXi · bXj = abXi+j . In partic-ular, we see that if A is a subring of the real numbers R, then the multiplication formulais exactly the one we get by multiplying polynomial functions over R. So polynomialrings are familiar objects after all.

Polynomial rings have a very important universal property.

Proposition 7.3.3. Let g : A→ B be a ring homomorphism and let b ∈ B be an elementthat commutes with every element of g(A). Then there is a unique ring homomorphismgb : A[X]→ B such that gb(X) = b and the following diagram commutes.

Ag ��

ι ��

��

� B

A[X]gb

��

Moreover, gb satisfies the formula

gb

(n∑i=0

aiXi

)=

n∑i=0

g(ai)bi.

In this formula, we adopt the convention that b0 = 1 for any b.

Proof For the uniqueness statement, note that any ring homomorphism gb that carriesX to b and makes the diagram commute must satisfy the stated formula:

gb

(n∑i=0

aiXi

)=

n∑i=0

gb(ι(ai))(gb(X))i

=n∑i=0

g(ai)bi.

Also, the stated formula does make the diagram commute and carry X to b. Thus,it suffices to show that gb, as defined by the formula, gives a ring homomorphism.

That it is a homomorphism of additive groups follows from the distributive law in B.And gb(1) = ι(1) = 1, so it suffices to show that gb(f1(X)f2(X)) = gb(f1(X))gb(f2(X))for all f1(X), f2(X) ∈ A[X].

Since the monomials generate A[X] additively, distributivity shows that it is sufficientto check this for the case that f1(X) and f2(X) are monomials. Thus, we must checkthat for a, a′ ∈ A and i, j ≥ 0, we have g(aa′)bi+j = g(a)big(a′)bj . Since b commuteswith any element of g(A), so does bi by induction. Thus, g(a)big(a′)bj = g(a)g(a′)bibj .That bibj = bi+j follows from the power laws for monoids (Problem 5 of Exercises 2.2.18).Thus, the result follows from the fact that g is a ring homomorphism.

If A is commutative, then so is A[X], and A[X] is an A-algebra via ι : A→ A[X]. Assuch, it satisfies an important universal property.

Corollary 7.3.4. Let A be a commutative ring and let B be an A-algebra, via ν : A→ B.Then for each b ∈ B, there is a unique A-algebra homomorphism εb : A[X] → B withthe property that εb(X) = b. It is given by the formula

εb(n∑i=0

aiXi) =

n∑i=0

ν(ai)bi,


where we set b0 = 1 for any b.

Proof Since ν : A → B is the structure map for B as an A-algebra, each elementof ν(A) commutes with each b ∈ B. Thus, Proposition 7.3.3 provides a unique ringhomomorphism νb : A[X] → B with νb ◦ ι = ν (i.e., νb is an A-algebra map) andνb(X) = b. So we set εb = νb. The formula comes from Proposition 7.3.3.

Definitions 7.3.5. Let B be an A-algebra and let b ∈ B. We shall refer to the uniqueA-algebra homomorphism, εb : A[X] → B, that carries X to b as the evaluation mapobtained by evaluating X at b. In consonance with this language, for f(X) ∈ A[X], wesometimes write εb(f(X)) = f(b), as it is obtained by evaluating f(X) at X = b.

We shall write A[b] for the image of εb. We call it the A-subalgebra of B generatedby b, or the algebra obtained by adjoining b to A.

Note that since εb : A[X]→ A[b] is onto, A[b] must be commutative even if B is not.

Examples 7.3.6. Recall that every ring is a Z-algebra in a unique way. Consider thecomplex numbers, C, as a Z-algebra, and form the Z-subalgebra, Z[i], obtained byadjoining i to Z. The ring Z[i] is known as the Gaussian integers.

Note that i has order 4 in the unit group C×. Thus, the powers of i consist of i, −1,−i, and 1. Collecting terms, we see that every element of Z[i] has the form m+ni, withm,n ∈ Z, and hence

Z[i] = {m+ ni ∈ C |m,n ∈ Z}.As the addition and multiplication operations are inherited from those in C, we have(m+ ni) + (k+ li) = (m+ k) + (n+ l)i and (m+ ni) · (k+ li) = (mk− nl) + (ml+ nk)ifor m,n, k, l ∈ Z.

Note also that since C is a Q-algebra, the same arguments may be used if we replaceZ by Q: Adjoining i to Q, we obtain

Q[i] = {a+ bi ∈ C | a, b ∈ Q}.Recall from Lemma 7.1.35 that for n ≥ 1, the complex number ζn = ei2π/n has order

n in C×. Also, ζ4 = i. Thus, the next example is a generalization of the previous one.

Examples 7.3.7. The non-negative powers of ζn are 1, ζn, . . . , ζn−1n . Thus, if A = Z or

Q, the elements of A[ζn] may all be written as sums a0 + a1ζn + · · · + an−1ζn−1n with

a0, . . . , an−1 ∈ A. We shall refer to Z[ζn] as the cyclotomic integers obtained by adjoininga primitive n-th root of unity and to Q[ζn] as the cyclotomic extension of Q obtained byadjoining a primitive n-th root of unity.

As we saw in the case of ζ4 = i, elements ofA[ζn], withA = Z or Q, do not have uniquerepresentations as sums of the form a0 + a1ζn + · · · + an−1ζ

n−1n with a0, . . . , an−1 ∈ A.

We shall defer the determination of a unique representation for these elements untilSection 8.8.

We shall show in Proposition 8.2.5 that the rings Q[ζn] are all subfields of C.

Remarks 7.3.8. Let A be a commutative ring. Note that if A is a subring of the centerof B and if we view B as an A-algebra whose structure map is the inclusion of A in B,then for b ∈ B, we have

εb(n∑i=0

aiXi) =

n∑i=0

aibi, and hence

A[b] = {n∑i=0

aibi |n ≥ 0 and ai ∈ A for 0 ≤ i ≤ n}.


We now give an important property of A[b].

Lemma 7.3.9. Let B be an A-algebra and let b ∈ B. Then A[b] is the smallest A-subalgebra of B that contains b.

Proof An A-subalgebra of B is any subring of B containing ν(A), where ν : A → Bis the structure map of B as an A-algebra. Thus, if C is an A-subalgebra of B thatcontains b, it must contain every sum

∑ni=0 ν(ai)b

i, and hence contains A[b].Since A[b] is an A-subalgebra of B containing b, the result follows.

We shall now compute the kernels of the evaluation maps εa : A[X] → A, for a ∈ Aand give some applications. The next result is a version of the Euclidean algorithm.

Proposition 7.3.10. Let A be any ring and let g(X) =∑ni=0 aiX

i be a polynomialwhose leading coefficient, an, is a unit in A. Then for any polynomial f(X) ∈ A[X], wemay find polynomials q(X), r(X) ∈ A[X] such that f(X) = q(X)g(X) + r(X) and suchthat the degree of r(X) is strictly less than that of g(X).

Proof We argue by induction on the degree of f(X). If f(X) has degree less than n,then we may simply take q(X) = 0 and take r(X) = f(X), and the statement is satisfied.

Thus, suppose that f(X) =∑mi=0 biX

i with bm �= 0 and that the result is true forall polynomials of degree less than m. In particular, we may assume that m ≥ n. Butthen (bma−1

n Xm−n)g(X) is a polynomial of degree m whose leading coefficient is thesame as that of f(X), and hence f(X)− (bma−1

n Xm−n)g(X) has degree less than m. Byinduction, we may write f(X) − (bma−1

n Xm−n)g(X) = q1(X)g(X) + r(X), where r(X)has degree less than n. But then we may take q(X) = q1(X) + (bma−1

n Xm−n) and wehave f(X) = q(X)g(X) + r(X), as desired.

As the reader may recall from calculus, the simplest application for division of poly-nomials is to the theory of roots.

Definition 7.3.11. Let A be a commutative ring and let f(X) ∈ A[X]. Let B be anA-algebra. We say that b ∈ B is a root of f if f(X) lies in the kernel of the evaluationmap εb : A[X]→ B obtained by evaluating X at b.

In particular, a ∈ A is a root of f if and only if f(a) = 0.

Corollary 7.3.12. Let A be a commutative ring and let a ∈ A. Then the kernel of theevaluation map εa : A[X] → A is (X − a), the principal ideal generated by X − a. Inparticular, a ∈ A is a root of f(X) ∈ A[X] if and only if f(X) = (X − a)q(X) for someq(X) ∈ A[X].

Proof Since X−a has degree 1 and leading coefficient 1, we may write any polynomialf(X) ∈ A[X] as f(X) = q(X)(X−a)+r(X) where r(X) has degree ≤ 0. In other words,f(X) = q(X)(X − a) + r, where r ∈ A.

But the evaluation map is a ring homomorphism, so that f(a) = q(a)(a− a) + r = r.Thus, f(X) ∈ ker εa if and only if r = 0. Thus, f(X) ∈ ker εa if and only if f(X) =q(X)(X − a) for some q(X) ∈ A[X]. But f(X) = q(X)(X − a) just says that f(X) is inthe principal ideal generated by X − a.

Corollary 7.3.13. Let A be an integral domain and let f(X) ∈ A[X] be a polynomialof degree n. Then f has at most n distinct roots in A.


Proof Suppose that a ∈ A is a root of f . Then f(X) = (X − a)g(X), where g(X) hasdegree n− 1. Thus, by induction, there are at most n− 1 roots of g. Thus, it suffices toshow that if b �= a is a root of f , then b is a root of g.

But 0 = f(b) = (b−a)g(b). Since b−a �= 0 and since A is an integral domain, b mustbe a root of g.

We obtain a very useful consequence for field theory.

Corollary 7.3.14. Let K be a field. Then any finite subgroup of K× is cyclic.

Proof Recall from Problem 4 of Exercises 4.4.24 that a finite abelian group is cyclic ifand only if it has at most d elements of exponent d for each d which divides its order.We claim that K× itself has at most d elements of exponent d for any integer d.

To see this, note that 1 is the identity element of K×, so that if a ∈ K× has exponentd, we have ad = 1. But this says that a is a root of the polynomial Xd− 1. Since Xd− 1has degree d, it can have at most d distinct roots, and hence K× can have at most ddistinct elements of exponent d.

As promised during our discussion of group theory, we obtain the following corollary.

Corollary 7.3.15. Let K be a finite field. Then K× is a cyclic group.

Corollary 7.3.15 is a useful consequence of the determination of the kernel of theevaluation map εa : A[X] → A. Later on, we shall derive useful consequences fromthe determinations of the kernels of the evaluation maps εb : A[X] → B for particularA-algebras B, and elements b ∈ B. Thus, Corollary 7.3.12 should be thought of as a firststep in an important study.

Exercises 7.3.16.1. Give the proof of Lemma 7.3.2.

2. Suppose that A is noncommutative. What are the elements of the center of A[X]?

3. Let f(X), g(X) ∈ A[X]. Show that the degree of f(X)+ g(X) is less than or equalto the larger of the degrees of f and g.

4. Let f(X), g(X) ∈ A[X]. Show that the degree of f(X) · g(X) is less than or equalto the sum of the degrees of f and g.

5. Show that A[X] is an integral domain if and only if A is.

6. Show that the units of A[X] are the image under ι of the units of A.

7. Let

M =(

1 10 1

)∈M2(R).

Consider the evaluation map εM : R[X] → M2(R). Show that a polynomialf(X) ∈ R[X] lies in the kernel of εM if and only if f(1) = 0 and f ′(1) = 0, wheref ′(X) is the derivative of f in the sense of calculus.


8. Let

M =(a 10 a

)∈M2(R)

for some real number a. Consider the evaluation map εM : R[X]→M2(R). Showthat a polynomial f(X) ∈ R[X] lies in the kernel of εM if and only if f(a) = 0 andf ′(a) = 0, where f ′(X) is the derivative of f in the sense of calculus.

9. Let p be a prime. What are the elements of Z[1/p] ⊂ Q?

10. What are the elements of A[X2] ⊂ A[X]?

11. What are the elements of A[Xn] ⊂ A[X]?

12. What are the elements of Z[2X] ⊂ Z[X]?

13. Show that C = R[i].

‡ 14. Show that Q[i] is a subfield of C.

15. What are the elements of Z[√

2] ⊂ R?

16. What are the elements of Q[√

2] ⊂ R?

17. Let m be a positive, odd integer. Show that Z[ζm] = Z[ζ2m] as a subring of C.

18. Show that every element of Z[ζ8] may be written in the form a + bζ8 + cζ28 + dζ3

8

with a, b, c, d ∈ Z. Write out the formula for the product (a + bζ8 + cζ28 + dζ3

8 ) ·(a′ + b′ζ8 + c′ζ2

8 + d′ζ38 ).

19. Let A be a ring and let 1 �= a ∈ A. Show that

ak − 1 = (a− 1)(1 + a+ · · ·+ ak−1)

for k ≥ 1.

‡ 20. Cyclotomic units Let m and k be relatively prime to n. Show that the element(ζmn − 1)/(ζkn − 1) ∈ Q[ζn] actually lies in Z[ζn]. Deduce that (ζmn − 1)/(ζkn − 1) isa unit in Z[ζn].

More generally, if (m,n) = (k, n) �= n, show that (ζmn − 1)/(ζkn − 1) is a unit inZ[ζn]. The elements (ζmn − 1)/(ζkn − 1) are often called the cyclotomic units.

21. Let n > 3 be odd. Show that the cyclotomic unit (ζ2n − 1)/(ζn − 1) has infinite

order in Z[ζn]×.

22. If (m,n) = 1 and m �≡ ±1 mod n, show that the cyclotomic unit (ζmn − 1)/(ζn− 1)has infinite order in Z[ζn]×. (Hint : If 0 < φ < π/2, show that

ei(θ−φ) + ei(θ+φ) = reiθ

for some positive real number r. Show also that r > 1 if φ < π/3.)

23. Let n be an even number greater than 6. Show that Z[ζn]× is infinite.

24. Show that in the category of rings, Z[X] is the free object on a one-element set.


25. Show that every element of Z6 is a root of X3 − X ∈ Z6[X]. Deduce that thehypothesis that A be an integral domain in Corollary 7.3.13 is necessary.

26. Show that X2 + 1 has infinitely many roots in the quaternion ring, H. (The issuehere is not zero-divisors, as H is a division ring. The point here is that evaluatingpolynomials at an element α ∈ H does not give a ring homomorphism from H[X]to H unless α lies in the center of H.)

27. Show that Xp −X has infinitely many roots in∏∞i=1 Zp.

28. Let p1, . . . , pk be distinct primes and let m be the least common multiple of p1 −1, . . . , pk − 1. Show that every element of Zn is a root of Xm+1 − X, wheren = p1 . . . pk.

29. Let n > 1 be an integer and let (n,X) be the ideal of Z[X] generated by n and X.Show that Z[X]/(n,X) is isomorphic to Zn.

30. Let B be any ring and let A be the center of B. Let a ∈ A. Show that there is aunique homomorphism εa : B[X]→ B such that εa(X) = a and εa ◦ ι = 1B , whereι : B → B[X] is the standard inclusion. Show that the kernel of εa is the two-sidedideal of B[X] generated by X − a.

31. Let A be a ring and X a variable. We define the ring of formal power series, A[[X]],with coefficients in A.

The elements of A[[X]] are formal sums∑∞i=0 aiX

i, with ai ∈ A for all i ≥ 0. Weshall refer to them as formal power series with coefficients in A. Unlike the case ofpolynomial rings, we do not assume that only finitely many of the coefficients arenonzero; indeed, all of the coefficients may be nonzero.

Here, the sum and product operations are given by

∞∑i=0

aiXi +

∞∑i=0

biXi =

∞∑i=0

(ai + bi)Xi, and

( ∞∑i=0

aiXi

)·( ∞∑i=0

biXi

)=

∞∑i=0

⎛⎝ i∑j=0

ajbi−j

⎞⎠Xi.

We have an inclusion map η : A[X] → A[[X]], that identifies a polynomial, f(X),of degree n with the formal power series whose coefficients of index ≤ n agree withthose of f , and whose higher coefficients are all 0.

(a) Show that A[[X]] is a ring and that η is a ring homomorphism.

(b) Show that the polynomial 1 − X is a unit in A[[X]] (but not in A[X]) withinverse

∑∞i=0X

i = 1 +X +X2 + · · ·+Xn + · · · .(c) Let A be commutative. Show that a formal power series

∑∞i=0 aiX

i is a unitin A[[X]] if and only if a0 ∈ A×.

Polynomials in Several Variables

We can also take polynomials in more than one variable. When considering polynomialsin several variables, the most common case, which is the default if nothing more is said,is to assume that the variables commute with each other. In fact, we shall assume that


this is the case whenever we use the word polynomial, but some people would explicitlycall these “polynomials in several commuting variables.”

Definitions 7.3.17. Suppose given n variables X1, . . . , Xn. A primitive7 monomial inthese variables is a product Xi1

1 · · ·Xinn , with ij ≥ 0 for 1 ≤ j ≤ n. We abbreviate

Xi11 · · ·Xin

n = XI for I = (i1, . . . , in) and write

I = {(i1, . . . , in) | i1, . . . , in ∈ Z, i1, . . . , in ≥ 0}for the set of all n-tuples of non-negative integers. (We call the elements of I multi-indices.)

For any ring A, we define a monomial inX1, . . . , Xn with coefficient in A to be a prod-uct aXi1

1 · · ·Xinn where a ∈ A and Xi1

1 · · ·Xinn is a primitive monomial in X1, . . . , Xn.

A polynomial in X1, . . . , Xn with coefficients in A is a finite sum of monomials inX1, . . . , Xn with coefficients in A.

The elements of the polynomial ring A[X1, . . . , Xn] are the polynomials inX1, . . . , Xn

with coefficients in A. Since there’s no preferred order for the multi-indices of the mono-mials, we shall write the generic polynomial in the form

f(X1, . . . , Xn) =∑I∈I

aIXI ,

where all but finitely many of the coefficients aI ∈ A are 0. So f is the sum of the finitelymany monomials whose coefficients are nonzero.

Addition of polynomials is obtained as in the one-variable case, by adding the co-efficients of the monomials: if f(X1, . . . , Xn) =

∑I∈I aIX

I and g(X1, . . . , Xn) =∑I∈I bIX

I , then

f(X1, . . . , Xn) + g(X1, . . . , Xn) =∑I∈I

(aI + bI)XI .

The multiplication in A[X1, . . . , Xn] is set up to have the following effect on mono-mials: If I = (i1, . . . , in) and J = (j1, . . . , jn), then aXI · bXJ = abXI+J , whereI + J = (i1 + j1, . . . , in + jn). For generic polynomials, this is expressed as follows:If f(X1, . . . , Xn) =

∑I∈I aIX

I and g(X1, . . . , Xn) =∑I∈I bIX

I , then

f(X1, . . . , Xn) · g(X1, . . . , Xn) =∑I∈I

⎛⎝∑J≤I

aJbI−J

⎞⎠XI .

Here, for I = (i1, . . . , in) and J = (j1, . . . , jn) in I, J ≤ I means that jk ≤ ik for allindices k (so that ≤ is only a partial order on I), and we define I −J to be the expectedmulti-index (i1 − j1, . . . , in − jn).

We write 0 for the multi-index whose coordinates are all 0, and identify a ∈ Awith a · X0. We write ι : A → A[X1, . . . , Xn] for the map that takes each a ∈ A toa = aX0 ∈ A[X1, . . . , Xn].

We identify each primitive monomial XI with 1XI and identify each variable Xi withthe primitive monomial X(0,...,0,1,0,...,0), where the 1 occurs in the i-th place.

When n = 2, we shall customarily write A[X,Y ] in place of A[X1, X2].We require the usual justifications.

7The use of the word primitive here has no connection whatever to the notion of a primitive elementin a Hopf algebra.


Lemma 7.3.18. Suppose given a ring A and variables X1, . . . , Xn. Then the polynomi-als A[X1, . . . , Xn] form a ring under the above addition and multiplication formulæ, andι : A→ A[X1, . . . , Xn] is a ring homomorphism.

Each variable Xi commutes with every element of ι(A), and the variables Xi and Xj

commute with each other for all i, j. If A is commutative, so is A[X1, . . . , Xn]. Moregenerally, a polynomial f(X1, . . . , Xn) lies in the center of A[X1, . . . , Xn] if and only ifits coefficients lie in the center of A.

Polynomials in several variables satisfy universal properties similar to those of poly-nomials in one variable. We leave the proofs to Exercises 7.3.27.

Proposition 7.3.19. Let g : A → B be a ring homomorphism and suppose given ele-ments b1, . . . , bn ∈ B such that the following two conditions hold.

1. Each bi commutes with every element of g(A).

2. For i, j ∈ {1, . . . , n}, bi commutes with bj.

Then there is a unique ring homomorphism gb1,...,bn: A[X1, . . . , Xn] → B such that

gb1,...,bn(Xi) = bi for 1 ≤ i ≤ n and the following diagram commutes.

Ag ��

ι �� B

A[X1, . . . , Xn]gb1,...,bn

��

Moreover, gb1,...,bnsatisfies the formula

gb1,...,bn

(∑I∈I

aIXI

)=

∑I∈I

g(aI)bI .

Here, if I = (i1, . . . , in), we set bI = bi11 . . . binn , where b0i is set equal to 1 for all i.

As in the case of a single variable, Proposition 7.3.19 admits a useful restatement inthe context of algebras over a commutative ring.

Corollary 7.3.20. Let A be a commutative ring and let B be an A-algebra, via ν : A→B. Suppose given b1, . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}.Then there is a unique A-algebra homomorphism εb1,...,bn : A[X1, . . . , Xn]→ B with theproperty that εb(Xi) = bi for 1 ≤ i ≤ n. It is given by the formula

εb1,...,bn

(∑I∈I

aIXI

)=

∑I∈I

ν(aI)bI .

Here, if I = (i1, . . . , in), we set bI = bi11 . . . binn , where b0i is set equal to 1 for all i.

We now codify some terms.

Definitions 7.3.21. Let A be a commutative ring. Let B be an A-algebra and letb1, . . . , bn ∈ B such that bi commutes with bj whenever i, j ∈ {1, . . . , n}. We shall referto the A-algebra homomorphism εb1,...,bn

as the evaluation map obtained by evaluatingXi at bi for 1 ≤ i ≤ n. We write A[b1, . . . , bn] for the image of εb1,...,bn

. We callA[b1, . . . , bn] the A-subalgebra of B generated by b1, . . . , bn.

Suppose now that B is a commutative A-algebra. We say that B is finitely generatedas an A-algebra (or has finite type as anA-algebra) ifB = A[b1, . . . , bn] for some collectionof elements b1, . . . , bn ∈ B.


A special case of the evaluation maps is of fundamental importance in algebraicgeometry.

Proposition 7.3.22. Let A be a commutative ring and let a1, . . . , an ∈ A. Let

a = (X1 − a1, . . . , Xn − an),the ideal of A[X1, . . . , Xn] generated by X1 − a1, . . . , Xn − an. Then a is the kernel ofthe evaluation map εa1,...,an

: A[X1, . . . , Xn]→ A.

Proof Since each Xi−ai lies in the kernel of εa1,...,an , Corollary 7.2.19 shows that thereis a unique ring homomorphism εa1,...,an : A[X1, . . . , Xn]/a→ A such that the followingdiagram commutes.

A[X1, . . . , Xn]εa1,...,an ��

π�� A

A[X1, . . . , Xn]/aεa1,...,an

��

Here, π : A[X1, . . . , Xn] → A[X1, . . . , Xn]/a is the canonical map. It suffices to showthat εa1,...,an is an isomorphism.

Let ι = π ◦ ι : A → A[X1, . . . , Xn]/a, where ι is the A-algebra structure map ofA[X1, . . . , Xn]. Then

εa1,...,an ◦ ι = εa1,...,an◦ π ◦ ι = εa1,...,an

◦ ι = 1A,

where the last equality comes from the fact that εa1,...,anis an A-algebra homomorphism.

Thus, it suffices to show that ι is onto.Since every element of A[X1, . . . , Xn] is a sum of monomials and since π is onto, it

suffices to show that π(aXi11 . . . Xin

n ) lies in the image of ι for all a ∈ A and all n-tuples(i1, . . . , in) of nonnegative integers. Since π(Xi − ai) = 0, we have π(Xi) = ι(ai) for alli. Thus, π(aXi1

1 . . . Xinn ) = ι(aai11 . . . ainn ), and the result follows.

The next result could have been left as an exercise in universal properties, but it issufficiently important that we give it here.

Proposition 7.3.23. Let X1, . . . , Xn be indeterminates over the ring A. Then there isa unique ring homomorphism

α : (A[X1, . . . , Xn−1])[Xn]→ A[X1, . . . , Xn]

that carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions ofA in the two sides. Moreover, α is an isomorphism whose inverse is the unique ringhomomorphism

β : A[X1, . . . , Xn]→ A[X1, . . . , Xn−1][Xn]

which carries Xi to Xi for 1 ≤ i ≤ n, and commutes with the natural inclusions of A inthe two sides.

Thus, A[X1, . . . , Xn] may be considered to be the polynomial ring in the variable Xn

with coefficients in A[X1, . . . , Xn−1]. In particular, every polynomial f(X1, . . . , Xn) inn variables over A may be written uniquely in the form

f(X1, . . . , Xn) = fk(X1, . . . , Xn−1)Xkn + · · ·+ f0(X1, . . . , Xn−1)

for some k ≥ 0, where fi(X1, . . . , Xn−1) ∈ A[X1, . . . , Xn−1] for all i.


Proof Here, the natural inclusion of A in A[X1, . . . , Xn−1][Xn] is the composite

Aι−→ A[X1, . . . , Xn−1]

ι′−→ A[X1, . . . , Xn−1][Xn],

where ι is the standard inclusion of A in the polynomial ring in n− 1 variables and ι′ isthe standard inclusion of A[X1, . . . , Xn−1] in its polynomial ring in the variable Xn.

Notice that the variables Xi lie in the centers of each of the rings under consideration.Thus, the existence and uniqueness of β is immediate from Proposition 7.3.19. For α,note that Proposition 7.3.19 provides a unique homomorphism α0 : A[X1, . . . , Xn−1] →A[X1, . . . , Xn] that carries each Xi to Xi and commutes with the natural inclusions ofA in both sides. But now the universal property of a polynomial ring in a single variableallows us to extend α0 to a map α carrying Xn to Xn.

For the uniqueness of α, we claim that any two ring homomorphisms f, g : A[X1, . . . , Xn−1][Xn]→B that agree on A and agree on each of the variables Xi must be equal. To see this,note first that f and g must agree on A[X1, . . . , Xn−1] by the uniqueness property ofProposition 7.3.19. But then, since f and g also agree on Xn, they must be equal by theuniqueness property of Proposition 7.3.3.

But now notice that each of the composites α ◦ β and β ◦ α carries each Xi to Xi

and commutes with the standard inclusions of A. Thus, α ◦ β is the identity by theuniqueness property of Proposition 7.3.19, while β ◦ α is the identity by the uniquenessproperty derived in the preceding paragraph.

If A is commutative and f(X1, . . . , Xn) ∈ A[X1, . . . , Xn], then there is a functionf : An → A which takes (a1, . . . , an) to f(a1, . . . , an), the effect of evaluating f at(a1, . . . , an). Indeed, much of our intuition about polynomials will have come from cal-culus, where polynomials are treated as functions.

Of course, if A is finite, then there are only finitely many functions from An to A,but infinitely many polynomials in A[X1, . . . , Xn]. Thus, in this case, a polynomial isnot determined by its effect as a function on An.

Finiteness is not the only difficulty in trying to treat polynomials as functions. Forinstance, if A =

∏∞i=1 Z2, then the polynomial X2 is easily seen to induce the identity

map of A, which is also induced by X.This suggests that if we wish to interpret polynomials as functions, then we might

think about restricting attention to integral domains with infinitely many elements. First,we shall formalize the situation.

Definition 7.3.24. Let X be any set and let A be a ring. The ring of functions fromX to A, written Map(X,A), is the ring whose elements are the functions from X to Aand whose operations are given by

(f + g)(x) = f(x) + g(x) and (f · g)(x) = f(x)g(x)

for all f, g ∈ Map(X,A) and x ∈ X. The additive and multiplicative identities are theconstant functions to 0 and 1, respectively.8

The next lemma follows from the fact that the evaluation maps are ring homomor-phisms.

8If X is a compact topological space, then the ring C(X) of continuous functions from X to C playsan important role in the theory of C∗ algebras. Of course, C(X) is a subring of Map(X,C).


Lemma 7.3.25. Let A be a commutative ring and let φ : A[X1, . . . , Xn]→ Map(An, A)be the map which takes each polynomial f to the function obtained by evaluating f onthe elements of An. Then φ is a ring homomorphism.

Thus, two polynomials, f and g, induce the same function from An to A if and onlyif f − g ∈ kerφ.

Thus, the elements of A[X1, . . . , Xn] are determined by their effect as functions ifand only if kerφ = 0. The next proposition gives a slightly stronger result when A is aninfinite domain.

Proposition 7.3.26. Let S be an infinite subset of the integral domain A. Suppose givena polynomial f(X1, . . . , Xn) in n variables over A with the property that f(a1, . . . , an) = 0for all a1, . . . , an ∈ S. Then f = 0.

Thus, an element of A[X1, . . . , Xn] is determined by its effect on S × · · · × S ⊂ An

for any infinite subset S of A.

Proof We argue by induction on n, with the result being true for n = 1, as f wouldhave infinitely many roots.

Suppose, inductively, that the result is true for polynomials of n − 1 variables, andlet a ∈ S. Then evaluating Xn at a gives a polynomial f(X1, . . . , Xn−1, a) in n − 1variables over A, which vanishes on all (n− 1)-tuples, a1, . . . , an−1 of elements of S. Bythe inductive assumption, f(X1, . . . , Xn−1, a) = 0.

Write B = A[X1, . . . , Xn−1]. By Proposition 7.3.23, we may identify A[X1, . . . , Xn]with B[Xn]. Under this identification, f(X1, . . . , Xn−1, a) is the image of f under theevaluation map

εa : B[Xn]→ B

obtained by evaluating Xn at a. Thus, when f is thought of as an element of B[Xn],each a ∈ S is a root of f . Since B is a domain and since f has infinitely many roots inB, f = 0.


2. Give the proof of Proposition 7.3.19.

3. Give the proof of Corollary 7.3.20.

‡ 4. Let B be an A-algebra and let b1, . . . , bn ∈ B such that bi commutes with bjwhenever i, j ∈ {1, . . . , n}. Show that A[b1, . . . , bn] is the smallest A-subalgebra ofB containing b1, . . . , bn.

5. Let A be a commutative ring. What are the elements of A[X2, X3] ⊂ A[X]?

6. Let A be a commutative ring. What are the elements of A[X3, X4] ⊂ A[X]?

7. Let A be a commutative ring and let X and Y be indeterminates. Consider theA-algebra structure on A[X]×A[Y ] given by the structure map

AΔ−→ A×A ι×ι−−→ A[X]×A[Y ]

where Δ is the diagonal map Δ(a) = (a, a) and the maps ι are the standardstructure maps for the polynomial rings A[X] and A[Y ]. What are the elements ofthe A-subalgebra of A[X]×A[Y ] generated by (X, 0) and (0, Y )?


8. Show that Z[1/2, 1/3] = Z[1/6] as subrings of Q. (Hint : By Problem 4, it sufficesto show that 1/2, 1/3 ∈ Z[1/6] and that 1/6 ∈ Z[1/2, 1/3].)

9. Show that Q is not finitely generated as a Z-algebra.

10. Let m and n be relatively prime positive integers. Show that Z[ζm, ζn] = Z[ζmn]as subrings of C. Here, ζk = ei·

2πk is the standard primitive k-th root of unity.

11. Let A be a commutative ring. Show that A[X1, . . . , Xn] is the free commutativeA-algebra on the set {X1, . . . , Xn}.

7.4 Symmetry of Polynomials

We consider some issues relating to permutations of the variables in a polynomial ring.Throughout this section, A is a commutative ring.

Suppose given n variables, X1, . . . , Xn. Then the symmetric group on n letters, Sn,acts on the set of variables in the obvious way: σ ·Xi = Xσ(i). By Corollary 7.3.20, theinduced map

σ : {X1, . . . , Xn} → {X1, . . . , Xn} ⊂ A[X1, . . . , Xn]

extends uniquely to an A-algebra homomorphism

σ∗ : A[X1, . . . , Xn]→ A[X1, . . . , Xn]via σ∗(f(X1, . . . , Xn)) = f(Xσ(1), . . . , Xσ(n)).

Since an A-algebra homomorphism out of A[X1, . . . , Xn] is determined by its effect onthe variables, σ ·f = σ∗(f) is easily seen to give an action of Sn on A[X1, . . . , Xn] throughA-algebra homomorphisms. We first consider the fixed points of this action.

Definition 7.4.1. We write A[X1, . . . , Xn]Sn for the fixed points of the action of Sn onA[X1, . . . , Xn]. Thus, a polynomial f(X1, . . . , Xn) is in A[X1, . . . , Xn]Sn if and only ifσ∗(f) = f for all σ ∈ Sn. We shall refer to A[X1, . . . , Xn]Sn as the symmetric polynomialsin n variables over A.

The reader may easily check that the symmetric polynomials form an A-subalgebraof A[X1, . . . , Xn]. We shall show that it is a polynomial ring on certain symmetricpolynomials. We shall argue by an induction in which the next lemma will play a part.

Lemma 7.4.2. Let π : A[X1, . . . , Xn] → A[X1, . . . , Xn−1] be the A-algebra map thattakes Xi to Xi if i < n and takes Xn to 0. Then π carries the symmetric polynomialsin n variables into the symmetric polynomials in n− 1 variables, inducing an A-algebrahomomorphism

π : A[X1, . . . , Xn]Sn → A[X1, . . . , Xn−1]Sn−1 .

Proof Recall from Proposition 7.3.23 that A[X1, . . . , Xn] may be identified with thepolynomial ring on A[X1, . . . , Xn−1] in the variable Xn. Thus, any nonzero element ofA[X1, . . . , Xn] may be written uniquely in the form

f(X1, . . . , Xn) =m∑i=0

fi(X1, . . . , Xn−1)Xin


for some m ≥ 0, with f0, . . . , fm ∈ A[X1, . . . , Xn−1], and fm �= 0.Now Sn−1 acts as permutations on the first n − 1 variables of the polynomials in n

variables, and if σ ∈ Sn−1, then

σ∗(f) =m∑i=0

σ∗(fi)Xin.

But by the uniqueness of the representation of polynomials in n variables as polynomialsin Xn with coefficients in A[X1, . . . , Xn−1] we see that if f ∈ A[X1, . . . , Xn]Sn , then eachfi must be fixed by each σ ∈ Sn−1. Thus, if f is a symmetric polynomial in n variables,then each fi is a symmetric polynomial in n− 1 variables.

But since π(Xn) = 0, we see that π(f) = f0, and the result follows.

We now define the elements we will show to be the polynomial generators of thesymmetric polynomials.

Definition 7.4.3. The elementary symmetric polynomials, or elementary symmetricfunctions, on n variables are defined by

si(X1, . . . , Xn) =

⎧⎪⎨⎪⎩1 if i = 0∑

j1<···<jiXj1 . . . Xji if 1 ≤ i ≤ n0 if i > n.

More generally, if B is a commutative ring and if b1, . . . , bn ∈ B, then we writesi(b1, . . . , bn) ∈ B for the element obtained by evaluating si(X1, . . . , Xn) at b1, . . . , bn.Note that we may do this because the elementary symmetric functions have integercoefficients.

Thus, for instance, s1(X1, . . . , Xn) = X1 + · · · + Xn, or at the other extreme,sn(X1, . . . , Xn) = X1 · · ·Xn.

Note that the summands of si(X1, . . . , Xn) are the products of i distinct variables,and that each such product occurs exactly once. Thus, if σ ∈ Sn, then σ∗ permutes thesummands of each si(X1, . . . , Xn), and hence the elementary symmetric functions areindeed symmetric polynomials.

The elementary symmetric functions are ubiquitous in mathematics. One of thereasons for this is that they give the formula for computing the coefficients of a polynomialwith a predetermined collection of roots.

Lemma 7.4.4. Suppose given n elements, a1, . . . , an ∈ A, not necessarily distinct. Thenthe following equality holds in A[X].

(X − a1) . . . (X − an) = Xn − s1Xn−1 + · · ·+ (−1)nsn,

where si = si(a1, . . . , an) for i = 1, . . . , n.

Proof In the expansion of (X − a1) . . . (X − an) from the distributive law, the termsof degree k in X are those obtained by choosing k of the factors from which to choosean X, leaving n− k factors of the form (−ai). Adding these up, we get⎛⎝ ∑

j1<···<jn−k

(−aj1) . . . (−ajn−k)

⎞⎠ ·Xk = (−1)n−ksn−k(a1, . . . , an) ·Xk

for the monomial of degree k in the resulting polynomial in X.


The reader may easily verify the following lemma.

Lemma 7.4.5. Let π : A[X1, . . . , Xn]Sn → A[X1, . . . , Xn−1]Sn−1 be obtained by settingXn = 0, as above. Then

π(si(X1, . . . , Xn)) ={si(X1, . . . , Xn−1) if 0 ≤ i < n

0 if i ≥ n

There are various definitions of degree for a polynomial in several variables. Thenotion of total degree is one of them.

Definition 7.4.6. The total degree of a monomial aXr11 . . . Xrn

n is r1 + · · · + rn. Thetotal degree of a polynomial f ∈ A[X1, . . . , Xn] is the maximum of the total degrees ofthe nonzero monomials in f .

We are now ready to show that A[X1, . . . , Xn]Sn is a polynomial algebra on thegenerators s1(X1, . . . , Xn), . . . , sn(X1, . . . , Xn).

Proposition 7.4.7. Let αn : A[Y1, . . . , Yn] → A[X1, . . . , Xn]Sn be the A-algebra homo-morphism that carries Yi to si(X1, . . . , Xn), where Y1, . . . , Yn are indeterminates over A.Then α is an isomorphism.

Proof We argue by induction on n. Since s1(X) = X, the result is trivial for n = 1.Thus, we shall assume, inductively, that αn−1 is an isomorphism.

We first show that αn is surjective. Suppose given a symmetric polynomial f(X1, . . . , Xn).We shall argue by induction on the total degree of f that f is in the image of α. Theresult is trivial in total degree 0.

Since αn−1 is surjective, there is a polynomial, g, of n− 1 variables such that

π(f) = g(s1(X1, . . . , Xn−1), . . . , sn−1(X1, . . . , Xn−1))

Thus, our calculation of the effect of π on the elementary symmetric functions showsthat if we set

h(X1, . . . , Xn) = f(X1, . . . , Xn)− g(s1(X1, . . . , Xn), . . . , sn−1(X1, . . . , Xn)),

then h lies in the kernel of π.Thus, if we write h(X1, . . . , Xn) =

∑mi=0 hi(X1, . . . , Xn−1)Xi

n, we see that h0 =π(h) = 0, and hence h(X1, . . . , Xn) = Xn · k(X1, . . . , Xn) for some polynomial k.

Let πi : A[X1, . . . , Xn]→ A[X1, . . . , Xn−1] be the A-algebra map obtained by setting

πi(Xj) =

⎧⎨⎩ Xj if j < i0 if j = i

Xj−1 if j > i,

and write πi : A[X1, . . . , Xn]Sn → A[X1, . . . , Xn−1]Sn−1 for the induced map. Thus,π = πn. Since h is in the kernel of π, symmetry shows that it must also be in the kernelof πi for 1 ≤ i ≤ n− 1. Since h(X1, . . . , Xn) = Xn · k(X1, . . . , Xn), we have

0 = πn−1(h(X1, . . . , Xn)) = πn−1(Xn)πn−1(k(X1, . . . , Xn)).

Since πn−1(Xn) = Xn−1 is not a zero divisor in A[X1, . . . , Xn−1], k must lie in the kernelof πn−1, and hence k(X1, . . . , Xn) = Xn−1 · p(X1, . . . , Xn) for a suitable polynomial p,by the argument above.


An easy induction now shows that h(X1, . . . , Xn) = X1 · · ·Xn · q(X1, . . . , Xn) for asuitable polynomial q. Since both h and X1 · · ·Xn = sn(X1, . . . , Xn) are symmetric, andsince sn(X1, . . . , Xn) is not a zero divisor, we see that q must be a symmetric polynomial.

Note that by the known properties of π, h must have total degree less than or equalto that of f , while q has total degree less than that of h. Thus, q(X1, . . . , Xn) is in theimage of αn by induction. But then f must also be in the image of αn.

To see that αn is injective, we argue by contradiction. Let f(Y1, . . . , Yn) be anelement of the kernel of αn with the lowest possible total degree. Write f(Y1, . . . , Yn) =∑mi=0 fi(Y1, . . . , Yn−1)Y in. Note that

0 = (π ◦ αn)(f) = f0(s1(X1, . . . , Xn−1), . . . , sn−1(X1, . . . , Xn−1)).

Since αn−1 is injective, this forces f0(Y1, . . . , Yn−1) to be 0. But then

f(Y1, . . . , Yn) = Yn · g(Y1, . . . , Yn), and hence αn(f) = sn(X1, . . . , Xn) · αn(g).Thus, g lies in the kernel of αn. Since its total degree is less than that of f , this contradictsthe minimality of the degree of f .

Another important topic regarding symmetries of polynomials concerns discriminants.

Proposition 7.4.8. Consider the following element of A[X1, . . . , Xn]:

δ(X1, . . . , Xn) =∏i<j

(Xj −Xi).

Then for σ ∈ Sn, we have σ∗(δ) = ε(σ) · δ, where ε(σ) is the sign of σ.

Proof We may may identify the set of indices i < j in the above product with theset, S, of all two-element subsets of {1, . . . , n}. Of course, Sn acts on S, via σ({i, j}) ={σ(i), σ(j)}. Using this action to re-index the factors of δ, we see that

δ(X1, . . . , Xn) =∏i<j

(Xmax(σ(i),σ(j)) −Xmin(σ(i),σ(j))).

For i < j, σ∗(Xj −Xi) = Xσ(j) −Xσ(i) = ±(Xmax(σ(i),σ(j)) −Xmin(σ(i),σ(j))), where thesign is positive if σ preserves the order of i and j and is negative otherwise. Thus,

σ∗(δ) = (−1)kδ,

where k is the number of pairs i < j for which σ(j) < σ(i).Define ψ : Sn → ±1 by ψ(σ) = (−1)k where (−1)k is defined by the equation

σ∗(δ) = (−1)kδ. Then it follows immediately from the fact that the homomorphisms σ∗define an action of Sn on A[X1, . . . , Xn] that ψ is a homomorphism.

It suffices to show that ψ = ε. Since Sn is generated by transpositions, it suffices toshow that if τ is a transposition, then ψ(τ) = −1.

Thus, let τ = (i j), with i < j. Then the pairs on which τ reverses order consist ofthe pairs i < k and k < j, where i < k < j, together with the pair i < j itself. Thesigns associated with i < k < j cancel each other out, leaving a single −1 from the pairi < j.

Definition 7.4.9. The discriminant, Δ(X1, . . . , Xn), is defined by Δ = δ2, where δ =∏i<j(Xj −Xi), as above.


Since each σ∗(δ) = ±δ, we obtain the following corollary.

Corollary 7.4.10. The discriminant, Δ, is a symmetric polynomial in the variablesX1, . . . , Xn. Thus, there is a unique polynomial Fn(X1, . . . , Xn), such that

Δ(X1, . . . , Xn) = Fn(s1(X1, . . . , Xn), . . . , sn(X1, . . . , Xn)).

Example 7.4.11. The discriminant on two variables is given by

Δ(X,Y ) = (Y −X)2 = X2 + Y 2 − 2s2(X,Y ).

Now note that

(s1(X,Y ))2 = (X + Y )2 = X2 + Y 2 + 2s2(X,Y ).

Thus, Δ = s21 − 4s2, and hence F2(X,Y ) = X2 − 4Y .

We shall calculate F3(X,Y, Z) in the next set of exercises. It takes considerably morework than F2. Since the polynomials Fn(X1, . . . , Xn) have integer coefficients, one maysave considerable time by using a computer algebra package to calculate them.

Recall from Lemma 7.4.4 that if a polynomial f(X) factors as a product of monicpolynomials of degree 1, then the coefficients of f are, up to sign, the elementary sym-metric functions on the roots.

Definition 7.4.12. Let f(X) = a0Xn + a1X

n−1 + · · ·+ an be a polynomial of degree nin A[X] and let Fn(X1, . . . , Xn) be the polynomial determined in Corollary 7.4.10. Thenthe discriminant, Δ(f), of f is the following element of A:

Δ(f) = Fn(−a1/a0, a2/a0, . . . , (−1)nan/a0).

Of course, we don’t know much at this point about the polynomials Fn, so the follow-ing corollary expresses the discriminant in a more familiar setting. It follows immediatelyfrom Lemma 7.4.4 and Corollary 7.4.10.

Corollary 7.4.13. Let f(X) be a monic polynomial in A[X], and suppose that f factorsas

f(X) = (X − b1) . . . (X − bn)

in B[X], where B is a commutative ring that contains A as a subring. Then the discrim-inant, Δ(f), may be calculated by

Δ(f) = Δ(b1, . . . , bn)

=∏i<j

(bj − bi)2.

Exercises 7.4.14.1. Write X2

1 + · · · + X2n as a polynomial in the elementary symmetric functions on

X1, . . . , Xn.


2. Show that πi : A[X1, . . . , Xn]Sn → A[X1, . . . , Xn−1]Sn−1 is injective when restrictedto the A-submodule consisting of the symmetric polynomials of total degree lessthan n.

3. Let f(X) = aX2 +bX+c be a quadratic in A[X]. Show that Δ(f) = (b2−4ac)/a2.Deduce that a quadratic f(X) ∈ R[X] has a root in R if and only if Δ(f) ≥ 0.

4. Calculate the polynomial F3(X,Y, Z) that’s used to calculate the discriminant ofa cubic.

† 5. Show that if f(X) = X3 + pX + q is a monic cubic with no X2 term, then Δ(f) =−4p3 − 27q2.

7.5 Group Rings and Monoid Rings

Polynomial rings are a special case of a more general construction called monoid rings.

Definitions 7.5.1. Let M be a monoid, and write the product operation of M in mul-tiplicative notation, even if M is abelian. We define A[M ], the monoid ring of M overA, as follows. The elements of A[M ] are formal sums

∑m∈M amm, such that the coef-

ficient am is zero for all but finitely many m in M . (We abbreviate this by saying thatam = 0 almost everywhere.) As in the case of polynomials, addition is given by addingcoefficients: (

∑m∈M amm) + (

∑m∈M bmm) =

∑m∈M (am + bm)m.

The multiplication is defined as follows.( ∑m∈M

amm

)( ∑m∈M

bmm

)=

∑m∈M

⎛⎝ ∑{(x,y) | xy=m}

axby

⎞⎠m.

The last sum ranges over all ordered pairs (x, y) of elements in M whose product is m.In the generic monoid, that collection of ordered pairs can be difficult to determine. Butin the examples we are concerned with, there will be no difficulty.

We identify each m ∈ M with the element of A[M ] whose m-th coefficient is 1 andwhose other coefficients are all 0. Thus, we have a natural inclusion M ⊂ A[M ].

Similarly, we identify a ∈ A with the element of A[M ] whose m-th coefficient is 0for all m �= 1, and whose coefficient of 1 is equal to a. Here, we write 1 for the identityelement of M . We write ι : A→ A[M ] for the map that gives this identification.

In the case where the monoid is a group G, we call A[G] the group ring (or groupalgebra, if A is commutative) of G over A.

Example 7.5.2. Write Z2 = 〈T 〉 = {1, T}. Then for any ring A,

A[Z2] = {a+ bT | a, b ∈ A}.

Here, if a, b, c, d ∈ A, we have (a+bT )+(c+dT ) = (a+c)+(b+d)T and (a+bT )·(c+dT ) =(ac+ bd) + (ad+ bc)T .

Lemma 7.5.3. Let A be a ring and let M be a monoid. Then A[M ] is a ring under theabove operations, and ι : A→ A[M ] is a ring homomorphism.

The inclusion M ⊂ A[M ] is a monoid homomorphism from M to the multiplicativemonoid of A[M ], and each m ∈M commutes with every element of ι(A).

If A is commutative, then A[M ] is an A-algebra, via ι : A→ A[M ].


Proof Everything here is straightforward, with the possible exception of the associa-tivity of multiplication in A[M ]. We shall treat that, and leave the rest to the reader.

Thus, suppose given α, β, γ ∈ A[M ], with

α =∑m∈M

amm, β =∑m∈M

bmm and γ =∑m∈M

cmm.

Then

αβ =∑m∈M

dmm, where dm =∑

{(x,y) | xy=m}axby.

Thus,

(αβ)γ =∑m∈M

⎛⎝ ∑{(w,z) |wz=m}

dwcz

⎞⎠m

=∑m∈M

⎛⎝ ∑{(w,z) |wz=m}

⎛⎝ ∑{(x,y) | xy=w}

axby

⎞⎠ cz

⎞⎠m

=∑m∈M

⎛⎝ ∑{(x,y,z) | xyz=m}

axbycz

⎞⎠m.

Here, the last equality comes from distributing the terms cz through the sums∑

{(x,y) | xy=w} axby.For each ordered triple (x, y, z) with xyz = m, there is exactly one summand axbycz thusobtained.

As the reader may now check, the expansion of α(βγ) produces exactly the sameresult.

Monoid rings have an important universal property.

Proposition 7.5.4. Let f : A → B be a ring homomorphism. Let M be a monoid andlet g : M → B be a homomorphism from M to the multiplicative monoid of B. Supposethat for each m ∈M , g(m) commutes with every element of f(A). Then there is a uniquering homomorphism h : A[M ]→ B such that the following diagram commutes:

A A[M ] M

B

��ι

��

��

��

f��

h

�� ⊃

��

��

��

g

Explicitly, we have

h

( ∑m∈M

amm

)=

∑m∈M

f(am)g(m).

Proof The element∑m∈M amm is, in fact, a finite sum of terms amm. Moreover, if

j : M ⊂ A[M ] is the inclusion, then amm = ι(am)j(m). Thus, if h : A[M ]→ B is a ring


homomorphism making the diagram commute, then we must have h(∑

m∈M amm)

=∑m∈M f(am)g(m), and hence h is unique as claimed.Thus, it suffices to show that if f and g satisfy the stated hypotheses, then h, as

defined by the formula above, is a ring homomorphism. By the distributive law inB and the fact that f is a homomorphism of abelian groups, we see that h is also ahomomorphism of abelian groups. Also, h(1) = f(1)g(1) = 1, so it suffices to show thath(αβ) = h(α)h(β) for all α, β ∈ A[M ].

By the distributive laws in both A[M ] and B, it suffices to show that h(am · a′m′) =h(am)h(a′m′) for all a, a′ ∈ A and m,m′ ∈ M . But h(am · a′m′) = h(aa′mm′) =f(aa′)g(mm′) = f(a)f(a′)g(m)g(m′), while h(am)h(a′m′) = f(a)g(m)f(a′)g(m′), sothe result follows from the fact that f(a′) commutes with g(m).

In the case of algebras over a commutative ring, the statement becomes simpler.

Corollary 7.5.5. Let A be a commutative ring and let B be an A-algebra. Let M be amonoid and let g : M → B be a monoid homomorphism from M to the multiplicativemonoid of B. Then there is a unique A-algebra homomorphism h : A[M ]→ B such thatthe composite M ⊂ A[M ] h−→ B is equal to g. Explicitly,

h

( ∑m∈M

amm

)=

∑m∈M

ν(am)g(m),

where ν : A→ B is the structure map of B as an A-algebra.

Proof If B is an A-algebra with structure map ν, then ν(A) is contained in the centerof B, and hence every element of ν(A) commutes with every element of g(M). Thus,Proposition 7.5.4 provides a unique ring homomorphism h, given by the displayed for-mula, such that the diagram in the statement of Proposition 7.5.4 commutes, with freplaced by ν. But the left-hand triangle of the diagram commutes if and only if h isan A-algebra homomorphism, while the right-hand triangle commutes if and only if thecomposite M ⊂ A[M ] h−→ B is equal to g.

Note that if G is a group, then a monoid homomorphism g : G → B from G to themultiplicative monoid of B always takes value in B×, the group of invertible elementsin the multiplicative monoid of B. Thus, g restricts to give a group homomorphismg : G→ B×. We obtain the following corollary.

Corollary 7.5.6. Let A be a commutative ring and let B be an A-algebra. Let G bea group and let f : G → B× be a homomorphism. Then there is a unique A-algebrahomomorphism h : A[G] → B such that the composite G ⊂ A[G] h−→ B is equal to f .Explicitly,

h

⎛⎝∑g∈G

agg

⎞⎠ =∑g∈G

ν(ag)f(g),

where ν : A→ B is the structure map of B as an A-algebra.

There is an important homomorphism from a group ring A[G] to A.

Definitions 7.5.7. Let A be a ring and let G be a group. Then the standard aug-mentation of the group ring A[G] is the unique ring homomorphism (which exists by


Proposition 7.5.4) ε : A[G] → A with the property that ε(g) = 1 for all g ∈ G and suchthat ε◦ι = 1A, where ι is the standard inclusion of A in A[G]. Explicitly, Proposition 7.5.4gives

ε

⎛⎝∑g∈G

agg

⎞⎠ =∑g∈G

ag.

We define the augmentation ideal, I(G) (or IA(G), if the ring A is allowed to vary),to be the kernel of ε.

Thus, we can obtain information about A[G] from information about A and informa-tion about I(G). We shall study augmentation ideals in the next set of exercises.

In the case that A is commutative, augmentation ideals admit a nice generalizationto the study of certain A-algebras.

Definition 7.5.8. Let A be a commutative ring. An augmented A-algebra consists ofan A-algebra, B, together with an A-algebra homomorphism ε : B → A. Here, since A isgiven the A-algebra structure map 1A, this simply means that ε is a ring homomorphismsuch that ε ◦ ν = 1A, where ν is that A-algebra structure map of B.

The ideal of the augmentation ε is its kernel.

Examples 7.5.9.1. Let A be a commutative ring and let a ∈ A. Then the evaluation map εa : A[X]→A is an augmentation. By Corollary 7.3.12, the ideal of this augmentation is theprincipal ideal (X − a).

2. Similarly, ifA is commutative and a1, . . . , an ∈ A, then the evaluation map εa1,...,an :A[X1, . . . , Xn]→ A is an augmentation. Here, Proposition 7.3.22 tells us the aug-mentation ideal is (X1 − a1, . . . , Xn − an).

3. Let M be any monoid and let A be a commutative ring. Then the trivial monoidhomomorphism from M to the multiplicative monoid of A induces an augmentationmap ε : A[M ]→ A via ε(

∑m∈M amm) =

∑m∈M am.

Exercises 7.5.10.1. Complete the proof of Lemma 7.5.3.

2. Show that in the group ring of a group G over A, we may write∑{(x,y) | xy=g}

axby =∑x∈G

axbx−1g.

3. Let Z2 = 〈T 〉 = {1, T}. For any ring A, show that (1 − T )(1 + T ) = 0 in A[Z2],and hence A[Z2] has zero-divisors.

4. Show that A[Zn] has zero-divisors for any ring A and any integer n > 1. Deducethat if A is any ring and if G is a group with torsion elements of order > 1, thenA[G] has zero-divisors.

5. Let A be a ring and let G be a finite group. Let Σ ∈ A[G] be the sum of all theelements of G: Σ =

∑g∈G g. Show that for any x ∈ A[G], xΣ = (ι ◦ ε(x))Σ, where

ε : A[G] → A is the standard augmentation and ι : A → A[G] is the standardinclusion.


‡ 6. Let ζn = ei·2π/n ∈ C be the standard primitive n-th root of unity. Let A be anysubring of C and let Zn = 〈T 〉 = {1, T, . . . , Tn−1}. Show that there is a uniqueA-algebra homomorphism f : A[Zn]→ A[ζn] such that f(T ) = ζn. Show that f isonto.

‡ 7. Let Zn = 〈T 〉 = {1, T, . . . , Tn−1}. Let f : Z[Zn]→ Z[ζn] be the homomorphism ofProblem 6.

(a) Show that Σ = 1 + T + · · · + Tn−1 lies in the kernel of f , and hence there’san induced, surjective ring homomorphism

f : Z[Zn]/(Σ)→ Z[ζn],

where (Σ) is the principal ideal generated by Σ.

(b) Suppose given integers m, k which are relatively prime to n. Let l > 0 be aninteger such that m ≡ kl mod n, and let

umk = 1 + T k + · · ·+ (T k)l−1,

where the right-hand side is interpreted as lying in Z[Zn]/(Σ).Show that if we chose a different integer l with m ≡ kl mod n, then the aboveformula gives rise to precisely the same element of Z[Zn]/(Σ).Show that umk is a unit in Z[Zn]/(Σ), and that f(umk) is the cyclotomic unit(ζmn − 1)/(ζkn − 1) from Problem 20 of Exercises 7.3.16.

8. Show that there is a commutative diagram of ring homomorphisms

Z[Zn] Z[Zn]/(Σ)

Z Zn

��π

��ε

��ε

��π

where ε is the standard augmentation, the maps π are the canonical maps to therespective quotient rings, and Σ = 1 + T + · · ·+ Tn−1 as in Problem 7.

(a) Show that the above diagram induces an isomorphism

h : Z[Zn]∼=−→ Z×Zn (Z[Zn]/(Σ)) ,

via h(α) = (ε(α), π(α)) for α ∈ Z[Zn]. (See Problem 17 of Exercises 7.1.27.)For this reason, we call the diagram a pullback square.Deduce that there is a pullback square:

Z[Zn]× (Z[Zn]/(Σ))×

Z× Z×n

��π

��

ε

��

ε

��π


(b) Let m1, . . . ,ms, k1, . . . , ks be relatively prime to n and let r1, . . . , rs > 0. Letu = ur1m1k1

. . . urs

msks, the unit of Z[Zn]/(Σ) constructed in Problem 7. Show

that u lifts to a unit of Z[Zn] if and only if mr11 . . .mrs

s ≡ ±kr11 . . . krss mod n.

9. Let X = {Xk | k ≥ 0}. Show that the operation Xk · X l = Xk+l turns X intoa monoid isomorphic to the monoid, N, of non-negative integers under addition.Show that the monoid ring A[X] is isomorphic to the polynomial ring A[X] for anyring A.

10. Let X be as in the previous exercise and let Xn = X× · · · ×X︸︷︷︸n times

be the direct

product of X with itself n times. For any ring A, show that the polynomial ringA[X1, . . . , Xn] is isomorphic to the monoid ring A[Xn].

‡ 11. Let A be a ring and let f : M → M ′ be a monoid homomorphism. Show thatthere is a ring homomorphism f∗ : A[M ] → A[M ′] defined by f∗(

∑m∈M amm) =∑

m′∈M ′(∑m∈f−1(m′) am)m′.

‡ 12. Show that if f : A→ B is a ring homomorphism and M is a monoid, then there isa ring homomorphism f∗ : A[M ]→ B[M ] defined by

f∗(∑m∈M

amm) =∑m∈M

f(am)m.

13. Let G be a group and let A be a commutative ring. Let α =∑g∈G agg be an

element of A[G]. Show that α is in the center of A[G] if and only if axgx−1 = agfor all x, g ∈ G. The reader who has studied module theory should now show thatthe center of A[G] is free as an A-module, and give a basis for it.

14. Show that if A is a ring and if M1 and M2 are monoids, then there is an iso-morphism between (A[M1])[M2] and A[M1 ×M2]. (Hint : One could either arguedirectly, or via universal mapping properties. The latter method is quicker, andless cumbersome notationally.)

15. Let A and B be rings and let M be a monoid. Show that (A×B)[M ] is isomorphicto A[M ]×B[M ].

† 16. Let A be a ring and let a be the principal two-sided ideal in the polynomial ringA[X] generated by Xn − 1. Show that the quotient ring A[X]/a is isomorphic tothe group ring A[Zn] of Zn over A. (Hint : There is a direct proof, that uses moduletheory, and there is an easier one using universal mapping properties.)

17. Let a be a two-sided ideal of A and let M be a monoid. Let π : A → A/a be thecanonical map and let π∗ : A[M ]→ (A/a)[M ] be the ring homomorphism given byπ∗(

∑m∈M amm) =

∑m∈M π(am)m (see Problem 12). Show that π∗ is a surjection

whose kernel is the two-sided ideal of A[M ] generated by a. (Here, we identify awith its image under ι : A→ A[M ].)

18. Let A be any ring and let Z2 = 〈T 〉 = {1, T}. Determine precisely which elementsa+bT of A[Z2], a, b ∈ A, lie in the augmentation ideal IA(Z2). Deduce that IA(Z2)is the principal two-sided ideal generated by T − 1.


19. Let A be a commutative ring and let G be a group. Let S ⊂ G be a generating setfor G (i.e., G = 〈S〉 as in Definitions 2.2.11). Show that the augmentation idealIA(G) is the two-sided ideal of A[G] generated by {x− 1 |x ∈ S}. (Hint : Considerthe proof of Proposition 7.3.22.)

20. Let f : G → K be a split surjection of groups. Find a set of generators (asan ideal) for the kernel of the ring homomorphism f∗ : A[G] → A[K] given byf∗(

∑g∈G agg) =

∑k∈K(

∑g∈f−1k ag)k, where A is any commutative ring. (This

map was shown to be a ring homomorphism in Problem 11.)

21. Show that a group ring A[G] over a commutative ring A is self-opposite.

22. Show that a group ring A[G] over a self-opposite ring A is self-opposite.

23. Show that if A is commutative, then the passage from monoids M to the monoidrings A[M ] gives a left adjoint to the forgetful functor from A-algebras to theirmultiplicative monoids.

24. Let A be commutative. Construct the free A-algebra on an arbitrary set X.

25. Let A be commutative. Construct the free commutative A-algebra on an arbitraryset X. (So far, we’ve only handled the case where X is finite.)

26. Show that if A is commutative, then the passage from groups G to the group ringsA[G] provides a left adjoint to the functor that takes an A-algebra to its group ofunits.

7.6 Ideals in Commutative Rings

We learned a lot about groups by studying their factor groups. Similarly, quotient ringswill be useful for studying rings. The drawback here, of course, is that while the kernelof a surjective group homomorphism is again a group, the kernel of a surjective ringhomomorphism is only an ideal and not a ring.

Thus, there is no good analogue for a composition series for a ring. However, we canmimic the first step by which the composition series for a finite group was constructed.In that step, we constructed a surjective homomorphism from our finite group to a simplegroup.

The analogue for rings of a simple group is a ring with no two-sided ideals other than0 and the ring itself, and hence has no useful quotient rings.

Definition 7.6.1. A nonzero ring is Jacobson simple if it has no two-sided ideals otherthan 0 and the ring itself.

Problem 11 of Exercises 7.2.24 shows that if D is a division ring, then Mn(D) isa Jacobson simple ring for all n ≥ 1. (We shall give a slicker argument for this inChapter 12.) There are other examples of noncommutative Jacobson simple rings morecomplicated than these. But in the commutative case, Jacobson simple rings are as niceas we could ask. The next corollary is immediate from Lemma 7.2.7.

Corollary 7.6.2. A commutative ring is Jacobson simple if and only if it is a field.

We wish to study those quotient rings A/a that are not the 0 ring, and hence, we areconcerned with those ideals a that are not equal to A. We shall give them a name.


Definition 7.6.3. An ideal a of A is proper if a ⊂ A is a proper inclusion (i.e., if a �= A).

We now introduce two important kinds of proper ideals.

Definitions 7.6.4. Let A be a commutative ring. A proper ideal, p, of A is prime ifab ∈ p implies that either a or b must be in p.

A proper ideal, m, of A is maximal if the only proper ideal containing m is m itself.

Examples 7.6.5. Let A be a commutative ring. Then it is immediate from the defini-tions that the ideal 0 of A is prime if and only if A is an integral domain.

Note that the ideal 0 is maximal if and only if A has no proper ideals other than 0.But we’ve just seen in Corollary 7.6.2 that this occurs if and only if A is a field.

These two examples give a characterization of prime and maximal ideals in terms ofthe quotient rings they determine.

Lemma 7.6.6. Let A be a commutative ring. Then a proper ideal p of A is prime ifand only if the quotient ring A/p is an integral domain. Also, a proper ideal m of A ismaximal if and only if the quotient ring A/m is a field. Thus, every maximal ideal isprime.

Proof Let p be a prime ideal of A. We must show that A/p has no zero-divisors. Thus,let a and b be elements of A/p such that a · b = 0. But this says that ab = 0, and henceab ∈ p. But then one of a or b is in p, and hence that element goes to 0 in A/p.

Conversely, if A/p is an integral domain and if ab ∈ p, then ab = 0, so that one of aand b must be 0 in A/p. But then one of a and b must be in p.

For the case of maximal ideals, the correspondence between the ideals in a quotientring A/a and the ideals in A containing a (Problem 5 of Exercises 7.2.24) shows that aproper ideal m of A is maximal if and only if the 0 ideal is maximal in A/m. As shownin the preceding example, this occurs if and only if A/m is a field.

Examples 7.6.7.1. For any prime p, the ideal (p) ⊂ Z is maximal, as Zp is a field. In fact, Lemma 7.1.12

shows that if n > 1, then (n) is a prime ideal of Z if and only if n is a prime number.Thus, every nonzero prime ideal of Z is maximal.

2. Let K be a field and let a ∈ K. Then the evaluation map εa : K[X] → Kis surjective, so its kernel is a maximal ideal. By Corollary 7.3.12, ker εa is theprincipal ideal generated by X − a.

3. Let π1 : Zp → Zp be the ring homomorphism from the p-adic integers onto Zpthat’s induced by the projection map. (See Lemma 7.1.38.) Then kerπ1, which byProblem 4 of Exercises 7.1.41 is the principal ideal generated by p, is maximal.

The next lemma is an easy, but important observation.

Lemma 7.6.8. Let f : A→ B be a ring homomorphism between commutative rings andlet p be a prime ideal of B. Then f−1(p) is a prime ideal of A.

Note that the analogue of the preceding lemma for maximal ideals is false: Leti : Z→ Q be the inclusion. Then 0 is a maximal ideal of Q, while f−1(0) = 0, which isprime in Z, but not maximal.

Prime ideals satisfy an important primality principle with respect to the ideal productof Definitions 7.2.21. This would have made a good exercise, but its uses are ubiquitousenough that we give it here.


Lemma 7.6.9. Let p be a prime ideal in the commutative ring A and let a and b beideals with ab ⊂ p. Then at least one of a and b must be contained in p.

Proof If ab ⊂ p and a �⊂ p, let a ∈ a with a �∈ p. Then ab ∈ p for all b ∈ b. Since a �∈ p,we must have b ∈ p for all b ∈ b.

By Lemma 7.2.22, ab ⊂ a ∩ b for any ideals a, b of a commutative ring.

Corollary 7.6.10. Let p be a prime ideal in the commutative ring A and let a and b beideals such that a ∩ b ⊂ p. Then at least one of a and b must be contained in p.

Prime ideals may be used to define a notion of dimension for a commutative ring.

Definition 7.6.11. Let A be a commutative ring. We say that the Krull dimension ofA is ≤ n if whenever we have a sequence

p0 ⊂ p1 ⊂ · · · ⊂ pk

of proper inclusions (i.e., inclusions that are not onto) of prime ideals of A, we havek ≤ n. Thus, the Krull dimension of A is finite if there is a maximum length for suchsequences of proper inclusions of prime ideals in A, and if

p0 ⊂ p1 ⊂ · · · ⊂ pn

is the longest such sequence that can occur in A, then the Krull dimension is n.

Examples 7.6.12.1. Since a field has only one prime ideal, 0, every field has Krull dimension 0.

2. The first example in Examples 7.6.7 shows that the longest chains of proper inclu-sions of prime ideals in Z are those of the form 0 ⊂ (p), for p prime. Thus, Z hasKrull dimension 1.

Let us return to the analogy with group theory. As noted above, we’ve made use ofthe fact that any nontrivial finite group has a nontrivial simple group as a factor group.By analogy, we would like to show that any commutative ring has a Jacobson simple ringas a quotient ring. Since the only commutative rings that are Jacobson simple are fields,this is equivalent to showing that every commutative ring has a maximal ideal.

To show this, we shall make use of Zorn’s Lemma, which is equivalent to the Axiomof Choice. The reader can find the proof of this equivalence in any good book on settheory.9

First we shall need some setup. Recall that a partially ordered set is a set X togetherwith a relation, ≤, which is reflexive (x ≤ x for all x ∈ X), transitive (if x ≤ y andy ≤ z, then x ≤ z), and antisymmetric (if x ≤ y and y ≤ x, then x = y). In a partiallyordered set, it is not required that an arbitrary pair of elements in X be comparable bythe relation. In other words, there may exist pairs x, y such that neither x ≤ y nor y ≤ xis true.

On the other hand, a partially ordered set in which any pair of elements is comparableis called a totally ordered set.

Let S be a subset of a partially ordered set. We say that x ∈ X is an upper boundfor S if s ≤ x for all s ∈ S. Note that the upper bound x need not lie in S.

Finally, a maximal element of X is an element y such that if x ∈ X with y ≤ x, thenx and y must be equal.

9There’s a good exposition of Zorn’s Lemma and the Well Ordering Principle in Topology, by JamesDugundji, Allyn and Bacon, 1966.


Lemma 7.6.13. (Zorn’s Lemma) Let X be a nonempty partially ordered set such thatevery totally ordered subset of X has an upper bound in X. Then X has at least onemaximal element.

Corollary 7.6.14. Let A be a commutative ring. Then every proper ideal of A is con-tained in a maximal ideal.

Proof Let a be a proper ideal of A, and let X be the set of all proper ideals of A thatcontain a. Put a partial ordering on X by b ≤ b′ if b ⊂ b′. Then a maximal element ofX with respect to this ordering is precisely a maximal ideal containing a.

Since a ∈ X, X is nonempty, so it suffices to show that any totally ordered subset ofX has an upper bound in X. Let S be a totally ordered subset of X. We claim that theunion

⋃b∈S b of the ideals in S is such an upper bound.

Of course, unions of ideals are not generally ideals, but if we’re dealing with a totallyordered set of ideals, this problem disappears: For x, y ∈ ⋃

b∈S b, we can find b, b′ ∈ Ssuch that x ∈ b and y ∈ b′. But since any two elements of S are comparable, eitherb ≤ b′ or b′ ≤ b. But in either case, x + y is in the larger of the two, as is ax for anya ∈ A. But then x + y and ax are in

⋃b∈S b, and hence the union is an ideal. Clearly,

it is an upper bound for the ideals in S, so it suffices to show that it is proper. But if1 ∈ ⋃

b∈S b, then 1 must be an element of some b ∈ S. But this is impossible, since theideals in X are all proper.

If we make extra assumptions on our ring, we can obtain an even stronger sort ofmaximality principle for ideals.

Definitions 7.6.15. Let A be a commutative ring. We say that A is Noetherian, or hasthe ascending chain condition (a.c.c.) if any ascending sequence of inclusions of ideals ofA is eventually constant. In other words, given a sequence

a1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · ·there is some n ≥ 1 such that an = ak for all k ≥ n.

Alternatively, we say that A is Artinian, or has the descending chain condition (d.c.c.)if any decreasing sequence of ideals is eventually constant. This means that given asequence

a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · ·there is some n ≥ 1 such that an = ak for all k ≥ n.

It turns out that the Artinian condition (in a commutative ring) implies that the ringis Noetherian. The Artinian condition is a very strong one, and fails to hold in manyrings of interest. For instance, as we shall see in the exercises below, the only Artinianintegral domains are fields.

The descending chain condition for left or right ideals turns out to be extremelyuseful for certain topics in noncommutative ring theory. For instance, descending chainconditions are important in the study of group rings K[G], where K is a field and G isa finite group.

Here is the strengthened maximality principle that holds in Noetherian rings.

Proposition 7.6.16. Let A be a Noetherian ring and let X be the partially ordered setof ideals in A (ordered by inclusion). Let S be any nonempty subset of X. Then S, withthe partial ordering inherited from X, has maximal elements.


Proof We argue by contradiction. Suppose that S has no maximal elements, and leta1 be an element of S. Since a1 is not a maximal element of S, we can find a properinclusion a1 ⊂ a2, with a2 in S.

Since no element of S is maximal, induction now shows that we may choose a sequence

a1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · ·

of elements of S, such that each inclusion ak−1 ⊂ ak is proper.10 But this contradicts thehypothesis that A is Noetherian, and hence S must have maximal elements after all.

We shall study Noetherian rings in much greater detail in the context of moduletheory.

Exercises 7.6.17.1. Let f : A→ B be a surjective ring homomorphism. Show that there is a one-to-one

correspondence between the prime ideals of B and those primes of A that containker f . Show also that there is a one-to-one correspondence between the maximalideals of B and those maximal ideals of A containing ker f .

2. Let n > 1. Show that every prime ideal of Zn is maximal. Given the primedecomposition of n, list all the maximal ideals, m, of Zn and calculate the quotientfields Zn/m.

3. Show that an integral domain has Krull dimension 0 if and only if it is a field.

4. Let p be a prime ideal in the commutative ring A and let A[X] · p be the idealgenerated by p in A[X]. Show that A[X] · p is a prime ideal of A[X].

5. Let A be an integral domain which is not a field. Show that the principal ideal(X − a) of A[X] (here, a ∈ A) is a prime ideal that is not maximal. Deduce thatA[X] has Krull dimension ≥ 2.

6. Let p be a prime and let n be any integer. Show that the ideal (p,X − n) of Z[X],generated by p and X − n, is maximal.

7. Let f : Q[X]→ Q[i] ⊂ C be the unique Q-algebra map that carries X to i. Recallfrom Problem 14 of Exercises 7.3.16 that Q[i] is a field. Show that ker f = (X2+1),and hence (X2 + 1) is a maximal ideal in Q[X].

8. Let Z4 = 〈T 〉 = {1, T, T 2, T 3}. Recall from Problem 6 of Exercises 7.5.10 that thereis a unique, surjective, Q-algebra homomorphism f : Q[Z4]→ Q[i] with f(T ) = i.Show that ker f = (T 2 + 1), and hence (T 2 + 1) is a maximal ideal in Q[Z4].

9. Let X1, . . . , Xn be indeterminates over a field K. Let a1 ∈ K and for 2 ≤ i ≤n, let fi(X) ∈ K[X1, . . . , Xi−1] ⊂ K[X1, . . . , Xn]. Show that (X1 − a1, X2 −f2(X), . . . , Xn − fn(X)) is a maximal ideal of K[X1, . . . , Xn].

10. Let X1, . . . , Xn be indeterminates over a field K. Find a chain p0 ⊂ · · · ⊂ pn ofproper inclusions of prime ideals of K[X1, . . . , Xn]. Deduce that K[X1, . . . , Xn]has Krull dimension ≥ n.

10Note that the infinite induction here makes use of the Axiom of Choice. Thus, while our conclusionis stronger than that of the Zorn’s Lemma argument for the existence of maximal ideals, the Noetherianproperty has not permitted us to dispense with the use of the choice axiom.


11. Let A and B be commutative rings. What are the prime ideals of A×B?

Suppose that A and B have finite Krull dimension. Calculate the Krull dimensionof A×B in terms of the dimensions of A and B.

12. Show that Z is Noetherian but not Artinian.

13. Show that Zn is both Noetherian and Artinian.

14. Let A be an Artinian integral domain and let a be a nonzero element of A. Showthat the descending chain condition on

(a) ⊃ (a2) ⊃ · · · ⊃ (an) ⊃ · · ·

implies that a is a unit. Deduce that A must be a field. Deduce that a finite domainis a field.

15. Let A be an Artinian commutative ring. Show that every prime ideal of A ismaximal.

7.7 Modules

Modules play a role in ring theory analogous to the role of G-sets in understandinggroups. Among other things, module theory will enable us to obtain a better grasp ofthe theory of ideals, and hence of the structure of the rings themselves.

We give some general concepts and constructions here, and study free modules andexact sequences in subsections. We then close the section with a subsection on the notionof rings without identity elements.

Definitions 7.7.1. Let A be a ring. A left A-module is an abelian group M togetherwith an operation that assigns to each a ∈ A and m ∈ M an element am ∈ M subjectto the following rules:

a(m+m′) = am+ am′

(a+ a′)m = am+ a′m1 ·m = m,

a(a′m) = (aa′)m,

for all a, a′ ∈ A and all m,m′ ∈M .If M is an abelian group, then an operation A×M →M making M an A-module is

called an A-module structure on M .A submodule of M is a subgroup M ′ ⊂ M of the additive group of M , such that

am ∈ M ′ for all a ∈ A and m ∈ M ′ (i.e., M ′ is closed under multiplication by elementsof A).

An A-module homomorphism between the left A-modules M and N is a homomor-phism f : M → N of abelian groups such that f(am) = af(m) for all a ∈ A and allm ∈M .

Right modules and their homomorphisms are defined by reversing the order of thea’s and m’s in the above definitions.


Note that if K is a field, then a K-module is nothing other than a vector space over K,and a K-module homomorphism between vector spaces is nothing other than a K-linearmap.

Another case where we understand what it means to be a module is that of Z-modules.The proof of the next lemma is similar to that of Lemma 7.1.6, which we left to the reader,so we shall leave this to the reader as well.

Lemma 7.7.2. Every abelian group G is a Z-module. Moreover, the Z-module structureon G is unique: for n ∈ Z and g ∈ G, ng is the n-th power of g in the group structureof G. (Thus, if n > 0, ng = g + · · ·+ g, the sum of n copies of g.) Finally, every grouphomomorphism between abelian groups is a Z-module homomorphism.

When A is commutative, any left A-module may be made into a right A-module bysetting m · a = am. This process is clearly reversible, so if A is commutative, then thereis a one-to-one correspondence between the left and right A-modules.

Not only does the above correspondence fail for noncommutative rings, but the the-ories of left modules and of right modules can be quite different for some rings.

Example 7.7.3. Let A be a ring. Then A is a left A-module via the structure mapa · a′ = aa′. Indeed, the equalities that demonstrate that this is an A-module structureare all part of the definition of a ring.

The submodules of A under this structure are precisely the left ideals of A.Note that A is also a right A-module, via a′ · a = a′a. The submodules under this

right module structure are the right ideals of A.

In particular, module theory will help us study ideals, and hence will provide impor-tant information about the ring A.

Here is another source of A-modules.

Example 7.7.4. Let f : A → B be a ring homomorphism. Then B is a left A-modulevia a · b = f(a)b. Also, B is a right A-module via b · a = bf(a).

For simplicity, we shall restrict attention to left modules for the rest of this section.Analogues of all the results will be true for right modules. Thus, for the duration, “A-module” will mean left A-module. We shall, however, use the word “left” to emphasizecertain one-sided properties.

Given the modules from the examples above, we can construct yet more via factorgroups.

Lemma 7.7.5. Let N be a submodule of the A-module M . Then the factor group M/Nis an A-module via a·m = am. Moreover, if f : M →M ′ is an A-module homomorphism,then there is a factorization

Mf ��

π ��

��

� M ′

M/Nf

��

(i.e., an A-module homomorphism f making the diagram commute) if and only if N ⊂ker f . The factorization f , if it exists, is unique.

Finally, the kernel of an A-module homomorphism is always a submodule, and hencewe obtain an isomorphism of A-modules f : M/(ker f)

∼=−→ im f from any A-modulehomomorphism f : M →M ′.


We now analyze the A-module homomorphisms out of A.

Lemma 7.7.6. Let M be a left A-module. Then for each m ∈ M there is a uniqueA-module homomorphism fm : A→M with fm(1) = m.

Proof We must have fm(a) = fm(a · 1) = afm(1) = am, so uniqueness follows, and itsuffices that this formula defines an A-module homomorphism. But that follows imme-diately from the definition of module.

In category theoretic terms, this says that A is the free A-module on one generator.There’s another way to express the result which introduces a useful piece of notation.

Definition 7.7.7. Let M and N be A-modules. We write HomA(M,N) for the set ofA-module homomorphisms from M to N .11

Lemma 7.7.8. Let M and N be A-modules. Then HomA(M,N) is an abelian groupunder the operation that sets (f + g)(m) = f(m) + g(m) for f, g ∈ HomA(M,N).

Under certain circumstances (e.g., if A is commutative), HomA(M,N) is naturally anA-module. We shall discuss the details in Section 9.7. We now give an easy consequenceof Lemma 7.7.6.

Corollary 7.7.9. Let M be an A-module. Then there is an isomorphism of abeliangroups, ε : HomA(A,M)

∼=−→M , given by ε(f) = f(1).

ε is an example of what’s called an evaluation map, as it is obtained by evaluatingthe homomorphisms at 1. As shown in Proposition 9.7.4, ε is an A-module isomorphismunder the appropriate conventions on the group of homomorphisms.

We return to the study of the homomorphisms from A to M .

Definitions 7.7.10. Let m ∈ M and let fm : A → M be the unique A-module homo-morphism that carries 1 to m. Then we write Am for the image of fm and write Ann(m)(or AnnA(m), if there’s more than one ring over which M is a module) for the kernel.Thus,

Am = {am | a ∈ A} andAnn(m) = {a ∈ A | am = 0}.

We shall refer to Ann(m) as the annihilator of m, and say that a kills (or annihilates) mif am = 0.

In general, we say that N is a cyclic module if N = An for some n ∈ N .

Corollary 7.7.11. Let M be a left A-module and let m ∈ M . Then Ann(m) is a leftideal of A, and the cyclic module Am is isomorphic to A/Ann(m).

There is also the notion of the annihilator of a module.

Definition 7.7.12. Let M be an A-module. Then the annihilator of M , written asAnn(M) or AnnA(M), is given by

Ann(M) = {a ∈ A | am = 0 for all m ∈M}=

⋂m∈M

Ann(m).

11We use the same notation for the set of A-module homomorphisms between a pair of right A-modules.This can, at times, require some specificity of language.


Unlike the case of the annihilator of an element, we have the following.

Lemma 7.7.13. The annihilator of an A-module is a two-sided ideal of A.

Warning: If A is not commutative, then Ann(Am) may not be equal to Ann(m). Inaddition, Ann(m) may not be a two-sided ideal, while Ann(Am) always is.

Definitions 7.7.14. Let f : A → B be a ring homomorphism. Let M be a B-module.The A-module structure on M induced by f is given by the structure map a ·m = f(a)mfor a ∈ A and m ∈M .

Let N be an A-module. We say that a B-module structure on N is compatible withthe original A-module structure if the A-module structure induced by f from the B-module structure is the same as the one we started with (i.e., f(a) ·m = am for all a ∈ Aand m ∈M).

We wish to study these relationships in the case where f is the canonical map π :A→ A/a, where a is a two-sided ideal of A. Here, if M is an A/a module, then we shallimplicitly consider M to be an A-module under the structure induced by π.

Proposition 7.7.15. Let a be a two-sided ideal in the ring A. Then an A-module Madmits an A/a-module structure compatible with the given A-module structure if and onlyif a ⊂ Ann(M). Such an A/a-module structure is unique if it exists. Finally, if M and Nare A/a-modules, then every A-module homomorphism from M to N is an A/a-modulehomomorphism as well.

Proof If M is an A/a-module, then, in the induced A-module structure, if x ∈ a andm ∈M , then xm = π(x)m = 0, because π(x) = 0. Thus, a ⊂ Ann(M).

Conversely, if a ⊂ Ann(M), given a ∈ A, x ∈ a, and m ∈ M , (a + x)m = am.Thus, setting a ·m = am gives a well defined correspondence from A/a×M to M . Andthis correspondence is easily seen to satisfy the conditions required for an A/a-modulestructure on M .

Two A/a-module structures that induce the same A-module structure must be equal,because π : A→ A/a is surjective.

Finally, given an A-module homomorphism f : M → N between A/a-modules, wehave f(am) = f(am) = af(m) = af(m), since the A-module structures on both M andN are induced by the given A/a-module structures.

For A = Z, this gives a characterization of Zn-modules.

Corollary 7.7.16. An abelian group M admits a Zn-module structure if and only ifM has exponent n as a group. The Zn-module structure on M , if it exists, is unique.Finally, if M and N are Zn-modules, then every group homomorphism f : M → N is aZn-module homomorphism.

Proof We use the fact that Zn = Z/(n). We know that every abelian group M hasa unique Z-module structure, in which k · m is the k-th power of m in the abeliangroup structure on M for all k ∈ Z and m ∈ M . Thus, k annihilates M if and only ifM has exponent k. But since (n) is the smallest ideal of Z containing n, we see thatn annihilates M if and only if (n) ⊂ Ann(M). The result now follows directly fromProposition 7.7.15.

Now we consider operations on submodules. The constructions that work are similarto those that work for ideals.


Definitions 7.7.17. Let N1 and N2 be submodules of M and let a be a left ideal of A.We define N1 +N2 and aN1 by

N1 +N2 = {n+m |n ∈ N1,m ∈ N2}aN1 = {a1n1 + · · ·+ aknk | k ≥ 1, ai ∈ a, and ni ∈ N1 for i = 1, . . . , k}

Lemma 7.7.18. Let N1 and N2 be submodules of the A-module M and let a be a leftideal of A. Then N1 +N2, aN1, and N1 ∩N2 are all submodules of M .

The modules aM have the following virtue.

Lemma 7.7.19. Let M be an A-module and let a be a two-sided ideal of M . ThenM/aM is an A/a-module. Moreover, if N is any A/a-module and if f : M → N is anA-module homomorphism, then there is a unique factorization

Mf ��

π �� N

M/aMf

��

where f : M/aM → N is an A/a-module homomorphism.

Proof Clearly, a ⊂ Ann(M/aM), and hence M/aM is an A/a-module by Proposi-tion 7.7.15. If f : M → N is an A-module homomorphism, with N an A/a-module,then a ⊂ Ann(N), and hence aM must lie in ker f . Thus, there is a unique A-modulehomomorphism f : M/aM → N making the diagram commute, and f is an A/a-modulehomomorphism by Proposition 7.7.15.

In categorical terms, this says that the passage from M to M/aM is a left adjoint tothe forgetful functor from A/a-modules to A-modules.

We can also consider external operations on modules. Recall that the direct sumM1 ⊕ · · · ⊕Mk of a finite collection of abelian groups is an alternative notation for thedirect product.

Definition 7.7.20. Let M1, . . . ,Mk be A-modules. By the direct sum, M1 ⊕ · · · ⊕Mk,of M1, . . . ,Mk, we mean their direct sum as abelian groups, endowed with the A-modulestructure obtained by setting a(m1, . . . ,mk) = (am1, . . . , amk), for each a ∈ A and eachk-tuple (m1, . . . ,mk) ∈M1 ⊕ · · · ⊕Mk.

Note that the canonical inclusions ιi : Mi → M1 ⊕ · · · ⊕Mk are A-module maps, asare the projection maps πi : M1 ⊕ · · · ⊕Mk → Mi. The proof of the following lemma isidentical to that of its analogue for abelian groups (see Proposition 2.10.6).

Lemma 7.7.21. Let fi : Mi → N be A-module homomorphisms for 1 ≤ i ≤ k. Thenthere is a unique A-module homomorphism f : M1 ⊕ · · · ⊕Mk → N such that f ◦ ιi = fifor 1 ≤ i ≤ n. Explicitly,

f(m1, . . . ,mk) = f1(m1) + · · ·+ fk(mk).


The next lemma will be useful, for instance, in our understanding of the propertiesof free modules.

Lemma 7.7.22. Let a be a two-sided ideal in A and let M1, . . . ,Mk be A-modules. LetM = M1 ⊕ · · · ⊕ Mk. Then there is a natural isomorphism of A/a-modules betweenM/aM and (M1/aM1)⊕ · · · ⊕ (Mk/aMk).

Proof Recall from Proposition 7.7.15 that any A-module map between A/a-modulesis an A/a-module map. Thus, it suffices to show that aM = (aM1) ⊕ · · · ⊕ (aMk).Note that the inclusion aM ⊂ (aM1) ⊕ · · · ⊕ (aMk) follows from the fact that if a ∈ a,then a(m1, . . . ,mk) = (am1, . . . , amm) lies in (aM1) ⊕ · · · ⊕ (aMk) for each k-tuple(m1, . . . ,mk). Since (aM1) ⊕ · · · ⊕ (aMk) is closed under addition, the stated inclusionfollows.

For the opposite inclusion, note that (m1, . . . ,mk) = ι1(m1) + · · · + ιk(mk). Sinceeach ιi is an A-module homomorphism, we have ιi(mi) ∈ aM whenever mi ∈ aMi. SinceaM is closed under addition, we see that (aM1)⊕ · · · ⊕ (aMk) ⊂ aM , as claimed.

We shall also make use of infinite direct sums. Recall that if Mi is an abelian groupfor i ∈ I, then the direct sum

⊕i∈IMi is the subgroup of the product

∏i∈IMi consisting

of those I-tuples (mi | i ∈ I) such that all but finitely many of the mi are 0.The homomorphisms ιi : Mi →

⊕i∈IMi are defined by setting ιi(m) to be the I-

tuple whose i-th coordinate is m and whose other coordinates are all 0. Making use of∑i∈I mi as an alternate notation for (mi | i ∈ I), Lemma 6.5.9 gives∑

i∈Imi =

∑i∈I

ιi(mi),

where the right-hand side is the sum in⊕

i∈IMi of the finite collection of terms ιi(mi)for which mi �= 0.

Definition 7.7.23. Let Mi be an A-module for each i ∈ I. We define the direct sum⊕i∈IMi to be the A-module whose underlying abelian group is the direct sum of the

Mi and whose A-module structure is given by a∑i∈I mi =

∑i∈I ami.

The maps ιi : Mi →⊕

i∈IMi are easily seen to be A-module homomorphisms.The proof of the next result is identical to that of its analogue for Z-modules (Proposi-tion 6.5.10).

Proposition 7.7.24. Suppose given A-modules Mi for i ∈ I. Then for any collectionof A-module homomorphisms {fi : Mi → N | i ∈ I}, there is a unique A-module homo-morphism f :

⊕i∈IMi → N such that f ◦ ιi = fi for each i. Explicitly,

f

(∑i∈I

mi

)=

∑i∈I

fi(mi).

We also have an infinite sum operation for submodules of a given module.

Definition 7.7.25. Suppose given a set {Ni | i ∈ I} of submodules of an A-module M .Then the sum of the Ni is the submodule∑

i∈INi = {ni1 + · · ·+ nik | k ≥ 1, ij ∈ I and nij ∈ Nij for j = 1, . . . , k}.

Thus,∑i∈I Ni is the set of all finite sums of elements from the various submodules Ni.


Sums of submodules satisfy the following properties. The proof is left as an exercise.

Proposition 7.7.26. Let M be an A-module. If N1 and N2 are submodules of M , thenN1 +N2 is the smallest submodule of M containing both N1 and N2.

Similarly, if Ni is a submodule of M for i ∈ I, then∑i∈I Ni is the smallest submodule

of M that contains each of the Ni.If fi : Mi →M is an A-module homomorphism for each i ∈ I (with I either finite or

infinite) and if f :⊕

i∈IMi →M is the unique homomorphism such that f ◦ ιi = fi forall i, then

im f =∑i∈I

im fi.



3. Give the proof of Corollary 7.7.26.

4. Let M be an abelian group, considered as a Z-module. Show that the annihilatorAnn(M) is nonzero if and only if M has some positive exponent. If Ann(M) = (n),with n > 0, show that n is the smallest positive number which is an exponent forM .

5. Let N1 and N2 be submodules of M . Show that

Ann(N1 +N2) = Ann(N1) ∩Ann(N2).

6. Let N1 and N2 be submodules of M . Show that

Ann(N1) + Ann(N2) ⊂ Ann(N1 ∩N2).

Can you give an example where this inclusion is proper?

7. Consider Z8 as a Z-module. What is Ann((2)Z8)?

8. Let f : A → B be a ring homomorphism. In the left A-module structure on Bthat’s induced by f , what is Ann(B)?

9. Let A be a commutative ring. Let m be an A-module and let m ∈ M . Show thatAnn(m) = Ann(Am).

10. Let A be any ring and let x ∈Mn(A) be the matrix whose 11-entry is 1 and whoseother entries are all 0.

a =

⎛⎜⎜⎜⎝1 0 . . . 00 0 . . . 0

...0 0 . . . 0

⎞⎟⎟⎟⎠(a) What are the elements of AnnMn(A)(x)?


(b) What are the elements of the principal left ideal Mn(A)x?

(c) What are the elements of AnnMn(A)(Mn(A)x)?

11. Let A be a ring and let G be a finite group. Let Σ ∈ A[G] be the sum of all theelements of G: Σ =

∑g∈G g. Show that AnnA[G](Σ) = I(G), the augmentation

ideal of G.

12. Let A be a ring. Let G be a finite group and let x ∈ G. Show that an element∑g∈G agg ∈ A[G] lies in Ann(x− 1) if and only if for each left coset g〈x〉 of 〈x〉, the

coefficients of the elements in g〈x〉 are all equal (i.e., agxi = agxj for all i, j ∈ Z).Deduce that if G = 〈x〉, then Ann(x− 1) = A[G]Σ, where Σ =

∑g∈G g.

13. Let (m) be an ideal in Z. Show that for n ≥ 1, Zn/(m)Zn is isomorphic to Z(m,n),where (m,n) is the greatest common divisor of m and n. (Recall that Z1 is thetrivial group.)

14. Let M be an abelian group and let r ≥ 0. Show that (pr)M/(pr+1)M is a vectorspace over Zp.

15. Let M be a finite abelian group of exponent pr. Show that (pr−1)M is a vectorspace over Zp.

16. Let M be a finite abelian group whose order is prime to n. Show that M/(n)M = 0.

17. Let M = Zpr1 × · · · × Zprk , where 1 ≤ r1 ≤ · · · ≤ rk.(a) Show that M/(p)M is the product of k copies of Zp.

(b) Calculate, in similar terms, the Zp-vector spaces (pr)M/(pr+1)M for r ≥ 1.

(c) Calculate Ann(M).

18. Suppose given a family of submodules {Ni | i ∈ I} of M . Show that the intersection⋂i∈I Ni is a submodule of M .

19. Let N be a submodule of M . Show that N =∑x∈N Ax.

20. What is Ann(M1 ⊕ · · · ⊕Mk)?

† 21. Let A be a ring. Show that a left A-module structure on an abelian group Mdetermines and is determined by a ring homomorphism from A to EndZ(M).

22. Use the preceding problem to give a quick proof of Proposition 7.7.15.

23. Let A be a ring. Show that a right A-module structure on an abelian group Mdetermines and is determined by a ring homomorphism from the opposite ring Aop

to EndZ(M).

† 24. Let A be a ring and let M be either a right or a left A-module. Let EndA(M) ⊂EndZ(M) be the collection of A-module homomorphisms from M to itself. Showthat EndA(M) is a subring of EndZ(M).

25. Let A be a commutative ring and let M be an A-module. Show that EndA(M) isan A-algebra via the map ι : A→ EndA(M) defined by ι(a)(m) = am for all a ∈ Aand m ∈M .


26. Let A be a commutative ring. Suppose given an A-algebra B and an A-moduleM . Show that there is a one-to-one correspondence between the A-algebra ho-momorphisms from B to EndA(M) and the B-module structures on M that arecompatible with the original A-module structure.

27. For any ring A and any A-module M , we write AutA(M) for the group of unitsin EndA(M). Suppose that A is commutative and that G is a group. Show thatthere is a one-to-one correspondence between the group homomorphisms from G toAutA(M) and the A[G]-module structures on M that are compatible (with respectto the standard inclusion ι : A → A[G]) with the original A-module structure onM .

Free Modules and Generators

Finite generation is as useful a concept in module theory as it is in group theory.

Definitions 7.7.28. Let m1, . . . ,mk be elements of the A-module M . By the submod-ule generated by m1, . . . ,mk, we mean the sum Am1 + · · · + Amk. Clearly, this is thesmallest submodule of M containing the elements m1, . . . ,mk.

The elements of Am1 + · · · + Amk all have the form a1m1 + · · · + akmk, for somea1, . . . , ak ∈ A. Such a sum is called a linear combination of m1, . . . ,mk.

Similarly, the submodule of M generated by an infinite family {mi | i ∈ I} is theinfinite sum,

∑i∈I Ami, of the submodules Ami. Here, a linear combination of the mi

is a finite sum ai1mi1 + · · ·+ aikmik , with i1, . . . , ik ∈ I.If M is generated by a finite set m1, . . . ,mk, we say that M is a finitely generated

A-module. Similarly, we say that a left ideal is finitely generated if it is finitely generatedas an A-module.

Clearly, A is generated by 1, while the cyclic module Ax is generated by x. A veryimportant collection of finitely generated modules is given by the free modules An:

Definitions 7.7.29. We write An for the direct sum of n copies of A, and write ei =(0, . . . , 0, 1, 0, . . . , 0). for the element of An whose i-th coordinate is 1 and whose othercoordinates are all 0, with 1 ≤ i ≤ n. Note that (a1, . . . , an) = a1e1 + · · ·+anen, so thatAn is generated by e1, . . . , en. We call e1, . . . , en the canonical basis of An. (We adoptthe convention that A0 = 0, the 0-module, whose canonical basis is taken to be the nullset.)

We call An the standard free A-module of rank n.

The next lemma shows that An is the free A-module (in the sense of category theory)on the set {e1, . . . , en}.Lemma 7.7.30. Let M be an A-module and let m1, . . . ,mn ∈ M . Then there is aunique A-module homomorphism f : An → M with f(ei) = mi for i = 1, . . . , n. Explic-itly,

f(a1, . . . , an) = a1m1 + · · ·+ anmn.

Proof Let fi : A → M be given by fi(a) = ami. Thus, fi is the unique A-modulehomomorphism with fi(1) = mi. By Lemma 7.7.21 there is a unique A-module homomor-phism from An to N with the property that f ◦ ιi = fi for 1 ≤ i ≤ n, where ιi : A→ An


is the inclusion of the i-th summand. Since ei = ιi(1), this gives f(ei) = fi(1) = mi.The explicit formula of Lemma 7.7.21 gives

f(a1, . . . , an) = f1(a1) + · · ·+ fn(an) = a1m1 + · · ·+ anmn,

as claimed.Note that the uniqueness statement, while embedded in the above, also follows from

the fact that the e1, . . . , en generate An: Since (a1, . . . , an) = a1e1 + · · ·+ anen, we musthave f(a1, . . . , an) = a1f(e1) + · · ·+ anf(en), and hence any A-module homomorphismon An is determined by its values on e1, . . . , en.

We can generalize this to the study of arbitrary direct sums of copies of A.

Definition 7.7.31. Let I be any set, and let ei = ιi(1) ∈ ⊕i∈I A for i ∈ I. Thus, in

the I-tuples notation for⊕

i∈I A, ei is the element whose i-th coordinate is 1 and whoseother coordinates are all 0. The elements ei, i ∈ I are called the canonical basis elementsof

⊕i∈I A.

Note that an element of⊕

i∈I A may be written uniquely as a sum∑i∈I aiei, where

all but finitely many of the ai are equal to 0.The proof of the next lemma is identical to that of Lemma 7.7.30, using Proposi-

tion 7.7.24 in place of Lemma 7.7.21.

Lemma 7.7.32. Let I be any set. Let M be an A-module and let mi ∈M for all i ∈ I.Then there is a unique A-module homomorphism f :

⊕i∈I A→M such that f(ei) = mi

for all i ∈ I, where the elements ei are the canonical basis elements of⊕

i∈I A. Explicitly,

f

(∑i∈I

aiei

)=

∑i∈I

aimi.

For some purposes it is useful to get away from the use of canonical bases.

Definitions 7.7.33. An A-module M is a free module if it is isomorphic to a direct sumof copies of A.

We say that M is free of finite rank if it is isomorphic to An for some integer n ≥ 0.We say that a finite sequence x1, . . . , xn of elements of M is a basis for M if the

unique A-module homomorphism f : An → M with f(ei) = xi for i = 1, . . . , n is anisomorphism.12

If I is an infinite set, we say that a set {mi | i ∈ I} is a basis for M if the uniqueA-module homomorphism f :

⊕i∈I A→M with f(ei) = mi for i ∈ I is an isomorphism.

Note that in the case of infinitely generated free modules, a basis comes equippedwith an indexing set I. In the finitely generated case, we insist that a basis, if non-null,be indexed by one of the standard ordered sets {1, . . . , n}.Examples 7.7.34.

12By abuse of language, we shall speak of a sequence x1, . . . , xn of elements of M as an ordered subsetof M . In the case of a basis, the elements x1, . . . , xn must all be distinct, so that the sequence in questiondoes specify an ordering on an n-element subset of M .


1. Let G be a finite group, and consider A[G] as a left A-module via the canonicalinclusion ι : A→ A[G]. Then the elements of G, in any order, form a basis for A[G].The point is that every element of A[G] may be written uniquely as

∑g∈G agg, so

the A-module homomorphism from A|G| to A[G] induced by an ordering of theelements of G is a bijection.

2. For the same reason, if G is an infinite group, then {g | g ∈ G} forms a basis forthe infinitely generated free module A[G].

3. Consider Mn(A) as a left A-module via the canonical inclusion ι : A → Mn(A).Let eij be the matrix whose ij-th coordinate is 1 and whose other coordinates areall 0, where 1 ≤ i, j ≤ n. Then the matrices eij , in any order, form a basis forMn(A). Here, we make use of the obvious fact that

(aij) =∑

1≤i,j≤naijeij

for any matrix (aij) ∈Mn(A).

The very same elements form bases for A[G] and Mn(A) when we consider them asright modules via the inclusions ι. Another very important example of free modulescomes from quotients of polynomial algebras.

Proposition 7.7.35. Let A be a commutative ring and let f(X) ∈ A[X] be a polynomialof degree n > 0 whose leading coefficient is a unit. Then A[X]/(f(X)) is a free A-module with basis given by the images of 1, X, . . . ,Xn−1 under the canonical map fromA[X]. Here, (f(X)) is the principal ideal of A[X] generated by f(X), and the A-modulestructure on the quotient ring A[X]/(f(X)) comes from the A-algebra structure inducedby the standard algebra structure on A[X].

Proof Write h : An → A[X]/(f(X)) for the A-module homomorphism that carries eito Xi−1 for i = 1, . . . , n. We show that h is an isomorphism.

For g(X) ∈ A[X], the Euclidean algorithm for polynomials (Proposition 7.3.10) allowsus to write g(X) = q(X)f(X) + r(X), where deg r(X) < deg f(X). But then the imagesof g(X) and r(X) under the canonical map π : A[X] → A[X]/(f(X)) are equal, and, ifr(X) = an−1X

n−1 + · · ·+ a0, then π(r(X)) = h(a0, . . . , an−1). Thus, h is onto.Now suppose that (a0, . . . , an−1) ∈ kerh and let g(X) = an−1X

n−1 + · · ·+ a0. Thenπ(g(X)) = 0, and hence g(X) ∈ (f(X)). Thus, g(X) = q(X)f(X) for some q(X) ∈ A[X].Since the leading coefficient of f(X) is a unit, the degree of q(X)f(X) must be ≥ deg funless q(X) = 0. Thus, q(X) = g(X) = 0, and hence ai = 0 for 0 ≤ i ≤ n− 1. Thus, his injective.

The next lemma is almost immediate from the definition, but is useful nonetheless,especially in understanding matrix groups.

Lemma 7.7.36. Let f : M → N be an A-module homomorphism between the free mod-ules M and N , and let x1, . . . , xn be any basis of M . Then f is an isomorphism if andonly if f(x1), . . . , f(xn) is a basis for N .

Proof Let g : An → M be the A-module homomorphism with g(ei) = xi for i =1, . . . , n. By the definition of basis, g is an isomorphism. The result now follows fromthe fact that f ◦ g is the unique A-module homomorphism from An to N which takes eito f(xi) for i = 1, . . . , n.


Surprisingly, rank is not well defined for free modules over arbitrary rings: There arerings for which An ∼= Am for n �= m. But such rings turn out to be exotic.

We shall see shortly that rank is well defined over commutative rings. The followingcorollary, which is immediate from Lemma 7.7.22 and the definition of An, will help usdo so.

Corollary 7.7.37. Let a be a two-sided ideal in the ring A. Then An/aAn is isomorphicto (A/a)n as an A/a-module.

While free modules of finite rank may not always have a uniquely defined rank, wecan always tell the difference between the cases of finite and infinite rank.

Lemma 7.7.38. Let I be a set. Suppose that the free module M =⊕

i∈I A is finitelygenerated. Then I must be a finite set. Thus, a finitely generated free module has finiterank.

Proof Let x1, . . . , xn be a generating set for M , and write xj =∑i∈I aijei for j =

1, . . . , n. Then for each j, all but finitely many of the aij are 0. Thus, if S = {i ∈I | aij �= 0 for some j}, then S must be finite. But then, the submodule of M generatedby x1, . . . , xn must lie in the submodule generated by {ei | i ∈ S}. But this says that Mitself must be generated by {ei | i ∈ S}.

But clearly, if i is not in S, then ei is not in the submodule of M generated by{ei | i ∈ S}. Thus, S = I, and hence I is finite.

It is useful to understand what it means to be a basis.

Definitions 7.7.39. Let m1, . . . ,mn be elements of the A-module M . We say thatm1, . . . ,mn are linearly dependent if there are elements a1, . . . , an of A, not all 0, suchthat a1m1 + · · ·+ anmn = 0.

Similarly, we say that an infinite set {mi | i ∈ I} is linearly dependent if there is arelation of the form

ai1mi1 + · · ·+ aikmik = 0,

where k > 0, i1, . . . , ik are distinct elements of I and not all of the aij are 0.We say that a collection of elements is linearly independent if it is not linearly de-

pendent.

Proposition 7.7.40. Let m1, . . . ,mn be elements of the A-module M and let f : An →M be the unique A-module homomorphism such that f(ei) = mi for 1 ≤ i ≤ n. Thenthe following conditions hold.

1. The elements m1, . . . ,mn are linearly independent if and only if f is an injection.

2. The image of f is the submodule generated by m1, . . . ,mn. Thus, m1, . . . ,mn

generate M if and only if f is a surjection.

Thus, m1, . . . ,mk is a basis for M if and only if it is linearly independent and generatesM .

Similarly, if {mi | i ∈ I} is a family of elements of M , and if f :⊕

i∈I A→M is theunique homomorphism such that f(ei) = mi for all i, then f is injective if and only if{mi | i ∈ I} are linearly independent, and is surjective if and only if {mi | i ∈ I} generateM .


Proof First, consider the case of a finite subset m1, . . . ,mk. Here, an element of thekernel of f is precisely an n-tuple (a1, . . . , an) such that a1m1 + · · · + anmn = 0. Solinear independence is equivalent to f being injective.

Also, f(a1, . . . , an) = a1m1 + · · ·+ anmn. Since this gives the generic element of thesubmodule generated by m1, . . . ,mn, the second condition also holds.

In the infinitely generated case, a nonzero element of∑i∈I A may be written uniquely

in the form ai1ei1 + · · · + aikeik , where k > 0, i1, . . . , ik are distinct elements of I, andai �= 0 for all i. And f(ai1ei1 + · · · + aikeik) = ai1mi1 + · · · + aikmik . Thus, there is anonzero element of the kernel if and only if {mi | i ∈ I} is linearly dependent.

Proposition 7.7.26 shows that the image of f is∑i∈I Ami, so the result follows.

The following corollary is immediate.

Corollary 7.7.41. An A-module M is finitely generated if and only if it is a quotientmodule of An for some integer n.

Exercises 7.7.42.1. Both Z2 and Z3 are Z6-modules, as they both have exponent 6. Show that Z2⊕Z3

is a free Z6-module, but that neither Z2 nor Z3 is free over Z6.

† 2. Show that every A-module is a quotient module of a free module.

3. Show that every quotient module of a finitely generated module is finitely generated.

Exact Sequences

Definition 7.7.43. Suppose given a “sequence,” either finite or infinite, of A-modulesand module homomorphisms:

· · · →M ′ f−→Mg−→M ′′ → · · ·

We say that the sequence is exact at M if the kernel of g is equal to the image of f .We call the sequence itself exact if it is exact at every module of the sequence that sitsbetween two other modules.

One sort of exact sequence has special importance.

Definition 7.7.44. A short exact sequence of A-modules is an exact sequence of theform

0→M ′ f−→Mg−→M ′′ → 0.

Note that exactness at M ′′ shows that g has to be onto, while exactness at M ′ showsthat f is an injection. Thus, M/g(M ′) is isomorphic to M ′′, and hence the short exactsequence presents M as an extension of M ′ by M ′′.

Example 7.7.45. For any pair M,N of A-modules, the sequence

0→Mι−→M ⊕N π−→ N → 0

is exact, where ι is the standard inclusion map and π is the standard projection map.

We shall make extensive use of exact sequences, both short and otherwise. For in-stance, we can use exactness to measure the deviation of a map from being an isomor-phism.


Definition 7.7.46. Let f : M → N be an A-module homomorphism. Then the cokernelof f is the module C = N/(im f).

In particular, f is onto if and only if the cokernel of f is 0.

Lemma 7.7.47. Let f : M → N be an A-module homomorphism. Then there is anexact sequence

0→ Ki−→M

f−→ Nπ−→ C → 0,

where i is the inclusion of the kernel of f and π is the canonical map onto the cokernelof f . Conversely, given a six term exact sequence

0→ K ′ →Mf−→ N → C ′ → 0,

f is injective if and only if K ′ = 0 and is surjective if and only if C ′ = 0.

The next result can be very useful for recognizing isomorphisms. The technique usedin the proof is called a diagram chase.

Lemma 7.7.48. (Five Lemma) Suppose given a commutative diagram with exact rows

M1 M2 M3 M4 M5

N1 N2 N3 N4 N5

��

f1

��α1

��

f2

��α2

��

f3

��α3

��

f4

��α4

��

f5

��β1 ��β2 ��β3 ��β4

Then the following properties hold.

1. If f2 and f4 are injective and if f1 is surjective, then f3 is injective.

2. If f2 and f4 are surjective and if f5 is injective, then f3 is surjective.

In particular, if f2 and f4 are isomorphisms, f1 is surjective, and f5 is injective, then f3is an isomorphism.

Proof Suppose first that f2 and f4 are injective and f1 is surjective. Let m ∈ ker f3.Then (f4 ◦ α3)(m) = 0 by the commutativity of the diagram. In other words, α3(m) isin the kernel of f4. Since f4 is injective, m must be in the kernel of α3.

By exactness, m = α2(m′) for some m′ ∈ M2. Since the diagram commutes andm ∈ ker f3, this shows that f2(m′) ∈ kerβ2. By the exactness of the lower sequence,f2(m′) = β1(n) for some n ∈ N1.

Since f1 is surjective, n = f1(m′′) for some m′′ ∈M1. But then f2(α1(m′′)) = β1(n).But n was chosen so that β1(n) = f2(m′), so α1(m′′) has the same image as m′ underf2. Since f2 is injective, m′ = α1(m′′). But m′ was chosen so that α2(m′) = m. Sinceα2 ◦ α1 = 0, m must be 0, and hence f3 is injective as claimed.

Now suppose that f2 and f4 are surjective and f5 is injective, and let n ∈ N3. Sincef4 is surjective, β3(n) = f4(m) for some m ∈ M4. Since β4 ◦ β3 = 0, commutativity ofthe diagram shows that (f5 ◦ α4)(m) = 0. But f5 is injective, and hence α4(m) = 0.

By the exactness of the upper sequence, m = α3(m′) for some m′ ∈M3. Recall thatm was chosen so that f4(m) = β3(n). By commutativity of the diagram, β3 carries n andf3(m′) to the same element, so that n− f3(m′) = β2(n′) for some n′ ∈ N2 by exactness.

But f2 is surjective, so that n′ = f2(m′′) for some m′′ ∈ M2. So f3(m′ + α2(m′′)) =n− f3(m′) + f3(m′) = n, and hence f3 is surjective.


A useful illustration of the utility of the Five Lemma comes from an analysis of whena short exact sequence may be used to express the middle term as the direct sum ofits surrounding terms. Recall that a section for a homomorphism g : M → N is ahomomorphism s : N → M such that g ◦ s = 1N . Recall also that a retraction for ahomomorphism f : N → M is a homomorphism r : M → N such that r ◦ f = 1N . Weassume that all homomorphisms under discussion are A-module homomorphisms.

Proposition 7.7.49. Suppose given a short exact sequence

0→M ′ f−→Mg−→M ′′ → 0.

Then the following additional conditions are equivalent.

1. f admits a retraction.

2. g admits a section.

3. There is an isomorphism h : M ′ ⊕ M ′′ → M such that the following diagramcommutes.

0 M ′ M ′ ⊕M ′′ M ′′ 0

0 M ′ M M ′′ 0

�� ι

��h

��π ��

�� f ��g ��

Proof We first show that the existence of a retraction for f implies the existence of asection for g. Thus, let r be a retraction for f . We claim then that the restriction, g′, ofg to the kernel of r is an isomorphism of ker r onto M ′′. Note that this is sufficient toproduce a section, s, for g by setting s to be the composite

M ′′ ker r ⊂M.��(g′)−1

To prove the claim, let m ∈ ker r. If g(m) = 0, then m ∈ ker g = im f , and hencem = f(m′) for some m′ ∈ M ′. But since r ◦ f = 1M ′ , r(m) = m′. But m′ = 0, sincem ∈ ker r, and hence m = 0 as well. Thus, the restriction of g to ker r is injective, andit suffices to show that g : ker r →M ′′ is onto.

Let m′′ ∈ M ′′. Then m′′ = g(m) for some m ∈ M . Consider the element x =m− f(r(m)). Then r(x) = r(m)− r(m), since r ◦ f = 1M ′ . Thus, x ∈ ker r. But clearly,g(x) = m′′, so g : ker r

∼=−→M ′′ as claimed, and hence g admits a section.Now suppose given a section s for g. We define h : M ′ ⊕M ′′ → M by h(m′,m′′) =

f(m′) + s(m′′). This is easily seen to make the diagram that’s displayed in the thirdcondition commute. But then h is an isomorphism by the Five Lemma, and hence thethird condition holds.

It suffices to show that the third condition implies the existence of a retraction for f .But this is immediate: just take r = π1 ◦ h−1, where π1 is the projection of M ′ ⊕M ′′

onto its first factor.

Note that the specific choice of r, s, or h above determines the other two mapsprecisely.


Definition 7.7.50. If a short exact sequence

0→M ′ f−→Mg−→M ′′ → 0

satisfies the three properties listed in Proposition 7.7.49, then we say the sequence splits.A specification of a specific retraction for f , a specific section of g, or a specific isomor-phism h as in the third condition is called a splitting for the sequence.

Note that the situation with regard to splitting extensions of A-modules is cleanerthan that for nonabelian groups. Here, there is no analogue of a semidirect product, andevery split extension is a direct sum.

If the third term of a short exact sequence is free, then it always splits.


0→M ′ f−→Mg−→M ′′ → 0

with M ′′ free. Then the sequence splits.

Proof Let {m′′i | i ∈ I} be a basis for M ′′, and, for each i ∈ I, choose mi ∈ M such

that π(mi) = m′′i . We claim that there is an A-module map s : M ′′ → M such that

s(m′′i ) = mi for all i ∈ I. Once we show this we’ll be done, as then g ◦ s is the identity

on {m′′i | i ∈ I}. But any two A-module homomorphisms that agree on a generating set

for a module must be equal, so g ◦ s = 1M ′′ , and hence s is a section for g.We now show that such an s exists. Recall from Lemma 7.7.32 that if N is any

A-module and if {ni | i ∈ I} are elements of N , then there’s a unique A-module homo-morphism from

⊕i∈I A to N which carries the canonical basis vector ei to ni for each

i ∈ I.Let f :

⊕i∈I A→M ′′ be the unique A-module homomorphism that carries ei to m′′

i

for each i ∈ I. Then f is an isomorphism by the definition of a basis. Let s :⊕

i∈I A→Mbe the unique A-module homomorphism that carries ei to mi for each i ∈ I. Then thedesired homomorphism s is the composite s ◦ f−1.

Exercises 7.7.52.1. Let A be any ring and let

0→M ′ →M →M ′′ → 0

be an exact sequence of A-modules. Show that if M ′ and M ′′ are finitely generated,then so is M .

† 2. Snake Lemma Suppose given a commutative diagram with exact rows:

M ′ M M ′′ 0

0 N ′ N N ′′��

f ′

��i

��

f

��p

��f ′′

��

�� j ��q

Show that there is an exact sequence

ker f ′ → ker f → ker f ′′ δ−→ coker f ′ → coker f → coker f ′′,


where “coker” stands for cokernel, and δ is defined as follows. For m′′ ∈ ker f ′′, letm′′ = p(m). Then f(m) = j(n′) for some n′ ∈ N ′, and we let δ(m′′) be the elementof coker f ′ represented by n′. Show also that if i is injective, so is ker f ′ → ker f ,and if q is surjective, so is coker f → coker f ′′. (Hint : The key here is showingthat δ, as described, is a well defined homomorphism.)

‡ 3. Let A be an integral domain and let (x) be the principal ideal generated by x ∈ A.Show that there is a short exact sequence

0→ Afx−→ A

π−→ A/(x)→ 0

of A-modules, where fx(a) = ax.

4. Let Zn = 〈T 〉 and let A be any commutative ring. Show that there is an exactsequence of A[Zn]-modules

· · · → Ck → · · · → C0 → A→ 0,

where Ck = A[Zn] for all k ≥ 0, and the maps are given as follows. The mapC0 → A is the standard augmentation. Each map C2k+1 → C2k with k ≥ 0 ismultiplication by T − 1, and each map C2k → C2k−1 with k ≥ 1 is multiplicationby Σ = 1 + T + · · ·+ Tn−1. (In the parlance of homological algebra, the sequence· · · → Ck → · · · → C0 is called a resolution of A over A[Zn].

Rings Without Identity

Not all algebraists require rings to have multiplicative identity elements. Here, we discussthe relationship between such a theory and that of rings as we have defined them.

Definition 7.7.53. A ring without identity is an abelian group R together with an asso-ciative binary operation, to be thought of as multiplication, that satisfies the distributivelaws:

a(b+ c) = ab+ ac and (a+ b)c = ac+ bc.

When we speak of a ring, we will continue to assume that it has a multiplicativeidentity. The new notion must be named in full: rings without identity.

Example 7.7.54. Let a be a left or right ideal of a ring. Then, under the multiplicationinherited from A, a is a ring without identity.

We shall see presently that every ring without identity may be embedded as an idealin a ring. But it may be possible to embed it as an ideal in many different rings.

We define homomorphisms, modules, and ideals for rings without identity in theobvious ways. Since rings with identity may also be considered in this broader context,we shall, at the risk of redundancy, refer to the modules as “nonunital modules.”

Definitions 7.7.55. A homomorphism f : R → S of rings without identity is a homo-morphism of additive groups with the additional property that f(rs) = f(r)f(s) for allr, s ∈ R.


A nonunital module M over a ring R without identity is an abelian group togetherwith a multiplication R×M →M such that the following conditions hold:

r(m+m′) = rm+ rm′

(r + r′)m = rm+ r′mr(r′m) = (rr′)m,

for all r, r′ ∈ R and all m,m′ ∈M .A homomorphism f : M → N of nonunital modules over a ring R without identity is

a group homomorphism such that f(rm) = rf(m) for all r ∈ R and m ∈M .

The basics of module theory, such as direct sums, exact sequences, quotient modules,etc. go over to this context without change. Free modules, on the other hand, areproblematic, as R itself is not necessarily free in the categorical sense. There is no goodnotion of basis here.

Left, right, and two-sided nonunital ideals may now be defined in the obvious way,and the quotient by a two-sided nonunital ideal satisfies the analogue of Proposition 7.2.9.

There is a natural way to adjoin an identity to a ring without identity.

Definition 7.7.56. Let R be a ring without identity. The associated ring, R, of R isgiven by R = Z⊕R, as an abelian group, with the multiplication

(m, r) · (n, s) = (mn,ms+ nr + rs),

where ms and nr denote the m-th power of s and the n-th power of r, respectively, inthe additive group of R.

Write ι : R→ R for the map ι(r) = (0, r) for r ∈ R.

The reader may check the details of the following proposition.

Proposition 7.7.57. Let R be a ring without identity. Then the associated ring R isa ring with identity element (1, 0), and ι : R → R is a homomorphism of rings withoutidentity, whose image is a two-sided ideal of R.

Moreover, if A is a ring, then for each homomorphism f : R → A of rings withoutidentity, there is a unique ring homomorphism f : R→ A such that f ◦ ι = f .

In categorical language, this says that the passage from R to R is a left adjoint tothe forgetful functor that forgets that a ring has an identity element.

Given the relationship between module structures and homomorphisms into endo-morphism rings, the following proposition should not be a surprise.

Proposition 7.7.58. Let M be a nonunital module over the ring R without identity.Then there is a unique R-module structure on M whose restriction to the action byelements in the image of ι gives the original nonunital R-module structure on M :

(n, r) ·m = nm+ rm,

where nm is the n-th power of m in the additive group of M .The nonunital R-submodules of M and the R-submodules of this induced R-module

structure coincide.

This correspondence is natural in M in that it gives a functor from the category ofnonunital R-modules to the category of R-modules.


Exercises 7.7.59.1. Suppose the ring R without identity has exponent n as an additive group. Show

that there is a ring structure on Zn⊕R that extends the multiplication onR = 0⊕R.Show that the analogue of Proposition 7.7.57 holds for homomorphisms into ringsof characteristic n.

2. Let A be a commutative ring and let R be a ring without identity which is also anA-module, such that

r · as = ar · s = a(r · s)for r, s ∈ R, and a ∈ A. Construct an A-algebra associated to R and state and provean analogue of Proposition 7.7.57 for appropriate homomorphisms into A-algebras.

7.8 Chain Conditions

We return to the discussion of chain conditions that was begun in Section 7.6. Themodule theory we’ve now developed will be very useful in this regard. First, let usgeneralize the chain conditions to the study of modules and of left or right ideals.

Definitions 7.8.1. A ring is left Noetherian if it has the ascending chain condition forleft ideals. In other words, any ascending sequence

a1 ⊂ a2 ⊂ · · · ⊂ ak ⊂ · · ·of inclusions of left ideals is eventually constant in the sense that there’s an n such thatan = ak for all k ≥ n.

Similarly, left Artinian rings are those with the descending chain condition for leftideals. Here, any descending sequence

a1 ⊃ a2 ⊃ · · · ⊃ ak ⊃ · · ·of inclusions of left ideals is eventually constant in the sense that there’s an n such thatan = ak for all k ≥ n.

Right Noetherian and right Artinian rings are defined analogously via chain conditionson right ideals.

An A-module M is a Noetherian A-module if it has the ascending chain condition forits submodules. It is an Artinian module if it has the descending chain condition for itssubmodules.

Clearly, A is left Noetherian (resp. left Artinian) if and only if it is Noetherian(resp. Artinian) as a left A-module. We shall not make use of chain conditions ontwo-sided ideals. When we speak of a Noetherian or Artinian ring, without specifyingleft or right, we shall mean a commutative ring with the appropriate chain condition onits ideals.

Examples 7.8.2.1. Recall that a division ring has no left (or right) ideals other than 0 and the ring

itself. Thus, division rings are left and right Noetherian and Artinian.

2. Recall that the left ideals of a product A × B all have the form a × b, with a aleft ideal of A and b a left ideal of B. Thus, if A and B are both left Noetherian(resp. left Artinian), so is A×B.


3. Regardless of any chain conditions that hold for a nonzero ring A, the infinite directsum

⊕∞i=1A is neither Noetherian nor Artinian as an A-module.

4. The integers, Z, are not Artinian, but Zn is an Artinian Z-module for all n > 0.

5. The infinite product∏∞i=1 Zp satisfies neither chain condition on its ideals, but the

module structure on Zp obtained from the projection onto the first factor satisfiesboth chain conditions.

6. Let

A ={(

a b0 c

) ∣∣∣∣ a ∈ Q, b, c ∈ R}.

Let a ⊂ A be the set of all elements of the form ( 0 b0 0 ) with b ∈ R. Then a is a

two-sided ideal of A. Note that(a b0 c

)(0 d0 0

)=

(0 ad0 0

).

Thus, the left submodules of a may be identified with the Q-submodules of R.Since R is infinitely generated as a vector space over Q, it satisfies neither chaincondition by Example 3. (See Corollary 7.9.14 for a demonstration that R is freeas a Q-module.) Thus, A itself is neither left Artinian nor left Noetherian. But(

0 d0 0

)(a b0 c

)=

(0 dc0 0

),

so the right submodules of a may be identified with the R-submodules of R, whichconsist only of 0 and R itself. Thus, a is both right Noetherian and right Artinianas an A-module.

There is a surjective ring homomorphism f : A→ Q×R, given by

f

(a b0 c

)= (a, c).

The kernel of f is a. Since Q × R is commutative, the left and right A-modulestructures on it coincide. Note that the only proper A-submodules of Q ×R areQ× 0, 0×R, and 0. Thus, Q×R satisfies both chain conditions as an A-module.Moreover, we have a short exact sequence

0→ a→ A→ Q×R→ 0

of two-sided A-modules. As right modules, both a and Q×R are both Noetherianand Artinian. As we shall see in Lemma 7.8.7, this implies that A itself is bothright Noetherian and right Artinian. Thus, A satisfies both chain conditions onone side, and satisfies neither chain condition on the other.

The Noetherian property can be characterized in terms of finite generation.

Proposition 7.8.3. An A-module M is a Noetherian module if and only if each of itssubmodules is finitely generated.


Proof Suppose that every submodule is finitely generated and that we are given anascending chain

N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · ·of submodules of M . Let N =

⋃k≥0Nk. Then it is easy to see that N is a submodule

of M . Suppose that N is generated by n1, . . . , nk. Then we may choose m to be largeenough that n1, . . . , nk ∈ Nm. But then the submodule of M generated by n1, . . . , nkmust be contained in Nm. Since n1, . . . , nk generate N , we must have Nm = Nm+l = Nfor all l ≥ 0.

Conversely, suppose that M is a Noetherian module and let N be one of its submod-ules. Suppose by contradiction that N is not finitely generated. Then by induction wemay choose a sequence of elements ni ∈ N such that n1 �= 0 and for each i > 1, ni is notin the submodule generated by n1, . . . , ni−1.

WriteNk for the submodule generated by n1, . . . , nk. Then each inclusionNk−1 ⊂ Nkis proper, and hence the sequence

N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · ·contradicts the hypothesis that M is a Noetherian module.

Since a ring is left Noetherian if and only if it is Noetherian as a left module overitself, we obtain the following corollary.

Corollary 7.8.4. A ring A is left Noetherian if and only if every left ideal is finitelygenerated.

Note that every ideal in a principal ideal ring is a cyclic module, and hence is finitelygenerated.

Corollary 7.8.5. Principal ideal rings are Noetherian.

The next very useful lemma shows that chain conditions are inherited by submodulesand quotient modules.

Lemma 7.8.6. Let A be a ring and let M be a Noetherian (resp. Artinian) A-module.Then any submodule or quotient module of M is Noetherian (resp. Artinian) as well.

Proof An ascending (resp. descending) chain in a submodule of M is a chain in M .But then an application of the appropriate chain condition in M shows that it must holdin N as well.

For quotient modules M/N , the result follows from the usual one-to-one correspon-dence between submodules of the quotient and the submodules of M containing N .

We wish to show that the free modules An share whatever chain conditions A has.The next lemma will help us to do so.

Lemma 7.8.7. Suppose given a short exact sequence

0→M ′ f−→Mg−→M ′′ → 0

of A-modules. Then M is Noetherian (resp. Artinian) if and only if both M ′ and M ′′

are.


Proof If M satisfies the stated chain condition, then so do M ′ and M ′′ by Lemma 7.8.6.For the converse, we restrict attention to the Noetherian case. The Artinian case is

similar. Thus, suppose that M ′ and M ′′ are Noetherian modules, and suppose given anascending chain

N1 ⊂ N2 ⊂ · · · ⊂ Nk ⊂ · · ·of submodules of M . Then we obtain ascending chains

g(N1) ⊂ g(N2) ⊂ · · · ⊂ g(Nk) ⊂ · · ·in M ′′ and

f−1(N1) ⊂ f−1(N2) ⊂ · · · ⊂ f−1(Nk) ⊂ · · ·in M ′. Thus, since M ′ and M ′′ are Noetherian, for k large enough, we have g(Nk) =g(Nk+l) and f−1(Nk) = f−1(Nk+l) for all l ≥ 0. But we claim then that Nk = Nk+l.

To see this, consider the following diagram, in which the vertical maps are the in-clusion maps from the sequences above. Since f(f−1(Ni)) = Ni ∩ ker g, the rows areexact.

0 f−1(Nk) Nk g(Nk) 0

0 f−1(Nk+1) Nk+1 g(Nk+1) 0

��

��

��f

��

��g

��

��

�� f ��g ��

Since the inclusions f−1Nk ⊂ f−1Nk+l and g(Nk) ⊂ f(Nk+l) are isomorphisms, theinclusion of Nk in Nk+l is also by the Five Lemma.

Since sequences of the form

0→M →M ⊕N → N → 0

are exact, a quick induction on n gives the next corollary.

Corollary 7.8.8. If A is left Noetherian (resp. left Artinian), then An is Noetherian(resp. Artinian) as a left A-module for all n ≥ 0.

Now, a finitely generated module is a quotient module of An for some n. Since passageto quotient modules preserves chain conditions, the next proposition is immediate.

Proposition 7.8.9. Let A be a left Noetherian (resp. left Artinian) ring. Then anyfinitely generated left A-module is a Noetherian (resp. Artinian) module over A.

This now allows us to characterize the Noetherian property.

Corollary 7.8.10. A ring A is left Noetherian if and only if it has the property thatevery submodule of every finitely generated left A-module is finitely generated.

Proof If every left submodule of A itself is finitely generated, then A is left Noetherianby Corollary 7.8.4.

Conversely, if A is left Noetherian and M is a finitely generated A-module, then M isNoetherian, and hence each of its submodules is finitely generated by Proposition 7.8.3.


Proposition 7.8.9 also has immediate applications to establishing chain conditions insome of the examples we’ve seen of rings.

Corollary 7.8.11. Let f : A → B be a ring homomorphism and suppose that B isfinitely generated as a left A-module. Then if A is left Noetherian or left Artinian, so isB.

Proof Every B-module is an A-module via f . So a chain of ideals in B is a chain ofA-submodules of the finitely generated A-module B.

We obtain two immediate corollaries.

Corollary 7.8.12. Let G be a finite group. Then if A is left or right Noetherian orArtinian, then so is the group ring A[G]. In particular the group ring D[G] over adivision ring D satisfies all four chain conditions.

Corollary 7.8.13. If A is left or right Noetherian or Artinian, then so are the matrixrings Mn(A) for n ≥ 1. In particular the matrix rings Mn(D) over a division ring Dsatisfy all four chain conditions.

Of course, A[X] is infinitely generated as an A-module, so we shall have to workharder to show that it inherits the Noetherian property from A. As a result, however,we shall obtain a considerable strengthening of the Noetherian case of Corollary 7.8.11when A is commutative.

Theorem 7.8.14. (Hilbert Basis Theorem) Let A be a left Noetherian ring. Then thepolynomial ring A[X] is also left Noetherian.

Proof We show that every left ideal of A[X] is finitely generated. Let a be a left idealof A[X] and define b ⊂ A to be the set of all leading coefficients of elements of a. Notethat if a and b are the leading coefficients of f and g, respectively, and if f and g havedegrees m and n, respectively, with m ≥ n, then the leading coefficient of f +Xm−ng isa+ b. Thus, b is a left ideal of A.

Since A is left Noetherian, there is a finite generating set x1, . . . , xk for b. But thenwe can find polynomials f1, . . . , fk whose leading coefficients are x1, . . . , xk, respectively.After stabilizing by multiplying by powers of X, if necessary, we may assume that thepolynomials f1, . . . , fk all have the same degree, say m.

Let M be the A-submodule of A[X] generated by 1, X,X2, . . . , Xm−1. Thus, theelements of M are the polynomials of degree less than m. Now a ∩M is a submoduleof a finitely generated A-module. Since A is Noetherian, a ∩M is a finitely generatedA-module. We claim that every element of a may be written as the sum of an element ofA[X]f1 + · · ·+A[X]fk (i.e., the ideal generated by f1, . . . , fk) with an element of a∩M .This will be sufficient, as then a is generated as an ideal by f1, . . . , fk together with anyset of A-module generators for a ∩M .

But the claim is an easy induction on degree. Let f ∈ a have degree n. If n <m, then f ∈ a ∩ M , and there’s nothing to show. Otherwise, let a be the leadingcoefficient of f and write a = a1x1 + · · · + akxk with ai ∈ A for all i. Then a isthe leading coefficient of a1X

n−mf1 + · · · + akXn−mfk, which has degree n. But then

f − (a1Xn−mf1 + · · ·+ akX

n−mfk) has degree less than n, and hence the claim followsby induction.

Recall that an A-algebra of finite type is one which is finitely generated as an A-algebra.


Corollary 7.8.15. Let A be a commutative Noetherian ring and let B be a commutativeA-algebra of finite type. Then B is Noetherian.

Proof The hypothesis on B is that there is a surjective ring homomorphism f :A[X1, . . . , Xn] → B for some n ≥ 0. By Corollary 7.8.11, it suffices to show thatA[X1, . . . , Xn] is Noetherian. But Proposition 7.3.23 shows that A[X1, . . . , Xn] is a poly-nomial ring in one variable on A[X1, . . . , Xn−1], so the result follows from the HilbertBasis Theorem and induction.

Exercises 7.8.16.1. Let A be a nonzero ring. Show that A[X] is not left Artinian.

2. In multiplicative notation, we can write the abelian group Z as {Xn |n ∈ Z},with 1 = X0 as the identity element. For a ring A, the group ring A[Z] is some-times called the ring of Laurent polynomials in A, and is sometimes denoted byA[X,X−1]. (This is absolutely not to be confused with a polynomial ring in twovariables.) Note that A[X,X−1] contains the polynomial ring A[X] as a subring.Show that for any f ∈ A[X,X−1] there is a unit u such that uf ∈ A[X]. Use this,together with either the statement or the proof of the Hilbert Basis Theorem, toshow that if A is left Noetherian, so is A[X,X−1].

3. Since an abelian group is precisely a Z-module, any finitely generated abelian groupis a quotient group of Zn for some n. Show that if G is a finitely generated abeliangroup and if A is a left Noetherian ring, then A[G] is also left Noetherian.

4. Suppose there is a finitely generated group, say with n generators, such that Z[G]fails to be left Noetherian. Show, then, that if F is a free group on n or moregenerators, then Z[F ] must also fail to be left Noetherian.

7.9 Vector Spaces

We give some of the basic properties of vector spaces, which we shall use in studyingmodules over more general rings.

In particular, we shall show that if A is a commutative ring and if we have an A-module isomorphism Am ∼= An, then m = n. Thus, using vector spaces, we shall showthat the rank of a finitely generated free module over a commutative ring is well defined,independently of the choice of an isomorphism to one of the canonical free modules An

(i.e., independently of the choice of a basis).For the sake of generality, we will work over a division ring, rather than a field. (In

particular, we shall not treat determinants here.) Thus, “vector space” here will mean aleft module over a division ring D, and a D-linear map is a D-module homomorphism.The results, of course, will all extend to right modules as well.

Definition 7.9.1. We say that a vector space V over D is finite dimensional if it isfinitely generated as a D-module.

Recall that a finite sequence, v1, . . . , vk, of elements of a D-module V is a basisfor V if the unique D-module homomorphism f : Dk → V for which f(ei) = vi fori = 1, . . . , k is an isomorphism, where ei is the i-th canonical basis vector of Dk. Recallalso (from Proposition 7.7.40) that v1, . . . , vk is a basis if and only if v1, . . . , vk are linearlyindependent and generate V .


Proposition 7.9.2. Let V be a finite dimensional vector space over D, with generatorsv1, . . . , vn. Then there is a subset vi1 , . . . , vik of v1, . . . , vn which is a basis for V . Inaddition, if v1, . . . , vm are linearly independent for some m ≤ n, then we may choose thesubset vi1 , . . . , vik so that it contains v1, . . . , vm.

In particular, every finite dimensional D-vector space is free as a D-module.

Proof If V = 0, then the null set is a basis. Otherwise, we argue by induction onn − m. If v1, . . . , vn is linearly independent, there is nothing to prove. Otherwise, wecan find x1, . . . , xn ∈ D, not all 0, such that x1v1 + · · · + xnvn = 0. Now at leastone of xm+1, . . . , xn must be nonzero, as otherwise the elements v1, . . . , vm would belinearly dependent. For simplicity, suppose that xn �= 0. Then vn = (−x−1

n x1)v1 +· · · + (−x−1

n xn−1)vn−1. Thus, vn is in the submodule generated by v1, . . . , vn−1. Sincev1, . . . , vn generate V , V must in fact be generated by v1, . . . , vn−1. But the result nowfollows by induction.

Corollary 7.9.3. Let V be a finite dimensional vector space and let v1, . . . , vm be linearlyindependent in V . Then there is a basis of V containing v1, . . . , vm.

Proof Let v1, . . . , vn be obtained from v1, . . . , vm by adjoining any finite generating setfor V , and apply Proposition 7.9.2.

Proposition 7.9.4. Let v1, . . . , vn be a basis for the vector space V and let w1, . . . , wkbe linearly independent in V . Then k ≤ n.Proof We shall show by induction on i ≥ 0 that we may relabel the v1, . . . , vn insuch a way that w1, . . . , wi, v1, . . . , vn−i generate V . If k were greater than n, thiswould say that w1, . . . , wn is a generating set for V , and hence that wn+1 is a linearcombination of w1, . . . , wn, say wn+1 = x1w1+· · ·+xnwn, with x1, . . . , xn ∈ D. But thenx1w1 + · · ·+xnwn+(−1)wn+1 = 0, contradicting the linear independence of w1, . . . , wk.

Thus, for i ≥ 1, we suppose inductively that for some relabelling of the v’s, the setw1, . . . , wi−1, v1, . . . , vn−i+1 generates V . Thus, we may write wi as a linear combinationof these elements, say wi = x1w1 + · · ·+xi−1wi−1 +xiv1 + · · ·+xnvn−i+1. Now it cannotbe the case that xj = 0 for all j ≥ i, as then the elements w1, . . . , wi would be linearlydependent. After relabelling, if necessary, we may assume that xn �= 0, in which casevn−i+1 may be written as a linear combination of w1, . . . , wi, v1, . . . , vn−i, which thenmust generate V .

Corollary 7.9.5. Let V be a finite dimensional vector space. Then any two bases of Vhave the same number of elements.

Proof Since any basis of V is linearly independent, the number of elements in eachbasis must be less than or equal to the number in the other, by Proposition 7.9.4.

Thus, we can make the following definition.

Definition 7.9.6. Let V be a finite dimensional vector space. Then the dimension ofV , written dimV , is the number of elements in any basis of V .

By the definition of a basis, V has a basis with n elements (and hence has dimensionn) if and only if V is isomorphic to Dn.

Corollary 7.9.7. Two finite dimensional vector spaces over D have the same dimensionif and only if they are isomorphic. In particular, Dn ∼= Dm if and only if n = m.


We can now deduce an important fact about free modules over commutative rings.

Theorem 7.9.8. Let A be a commutative ring. Then An is isomorphic to Am if andonly if m = n. Thus, finitely generated free modules over a commutative ring have a welldefined rank, independent of the choice of an isomorphism to some An.

Proof Suppose that An is isomorphic to Am and let m be a maximal ideal of A. ThenAn/mAn is isomorphic as a vector space over the field A/m to Am/mAm. But for k arbi-trary, Corollary 7.7.37 provides an isomorphism of A/m-vector spaces between Ak/mAk

and (A/m)k. Thus, An/mAn and Am/mAm have dimensions n and m, respectively, asA/m-vector spaces, and hence n = m.

The following lemma is useful in detecting the difference between finite and infinitedimensional vector spaces.

Lemma 7.9.9. Let V be an infinite dimensional vector space over D. Then there arelinearly independent subsets of V of arbitrary length. By passage to the vector subspacesthat they generate, we see that V has subspaces of every finite dimension.

Proof We argue by contradiction. Suppose that the maximal length of a linearlyindependent subset of V is k, and let v1, . . . , vk be linearly independent. Let W be thesubspace of V generated by v1, . . . , vk. Then W is finite dimensional, so there must be anelement vk+1 of V that is not in W . We claim that v1, . . . , vk+1 is linearly independent,contradicting the maximality of k.

To see this, suppose that x1v1 + · · · + xk+1vk+1 = 0, with x1, . . . , xk+1 ∈ D. Ifxk+1 �= 0, then vk+1 = (−x−1

k+1x1)v1 + · · ·+ (−x−1k+1xk)vk. But this says that vk+1 ∈W ,

contradicting the choice of vk+1, so we must have xk+1 = 0. But this gives x1v1 + · · ·+xkvk = 0. Since v1, . . . , vk are linearly independent, we must have x1 = · · · = xk = 0,and hence v1, . . . , vk+1 are indeed linearly independent.

This now gives a quick proof of the following fact, which we already knew from thefact that D is left Noetherian.

Corollary 7.9.10. Let V be a finite dimensional vector space. Then any subspace of Vis finite dimensional as well.

Proof If V ′ ⊂ V , then any linearly independent subset of V ′ is a linearly independentsubset of V . By Proposition 7.9.4, the maximal length of a linearly independent subsetof V is dimV . So V ′ must be finite dimensional by Lemma 7.9.9.

We can now analyze the behavior of extensions with respect to dimension.


0→ V ′ i−→ Vπ−→ V ′′ → 0

of vector spaces over D. Then V is finite dimensional if and only if both V ′ and V ′′ arefinite dimensional, in which case dimV = dimV ′ + dimV ′′.

Proof Suppose V is finite dimensional. Since π : V → V ′′ is surjective, any set ofgenerators of V is carried onto a set of generators of V ′′ by π. Since i(V ′) is a subspaceof V that is isomorphic to V ′, finite generation of V ′ was just shown in Corollary 7.9.10.

Now suppose that V ′ and V ′′ are finite dimensional. We identify V ′ with the imageof i. Let v1, . . . , vk be a basis for V ′ and let z1, . . . , zl be a basis for V ′′. We claim that


if we’re given any choice of w1, . . . , wl ∈ V such that π(wi) = zi for 1 ≤ i ≤ k, thenv1, . . . , vk, w1, . . . , wl is a basis for V .

First, we show that v1, . . . , vk, w1, . . . , wl is linearly independent. Suppose givenx1, . . . , xk+l ∈ D with x1v1 + · · · + xkvk + xk+1w1 + · · · + xk+lwl = 0. Then apply-ing π, we see that xk+1z1 + · · · + xk+lzl = 0. Since z1, . . . , zl are linearly independent,we must have xk+1 = · · · = xk+l = 0.

But then the original equation gives x1v1+· · ·+xkvk = 0. Since v1, . . . , vk are linearlyindependent, we must have x1 = · · · = xk = 0 as well.

The proof that v1, . . . , vk, w1, . . . , wl generate V works in an exact sequence of finitelygenerated modules over any ring. Let v ∈ V . Then π(v) = xk+1z1 + · · ·+xk+lzl for somexk+1, . . . , xk+l ∈ D. But then π(v −∑l

i=1 xk+iwi) = 0, and hence (v −∑li=1 xk+iwi) ∈

kerπ. The result now follows, since kerπ = V ′ is generated by v1, . . . , vk.

We obtain a characterization of isomorphisms of finite dimensional vector spaces.

Corollary 7.9.12. Let V and W be finite dimensional vector spaces over D with thesame dimension. Let f : V → W be a D-linear map. Then the following conditions areequivalent.

1. f is an isomorphism.

2. f is injective.

3. f is surjective.

Proof Clearly, an isomorphism is both injective and surjective. We shall show that iff is either injective or surjective, then it is an isomorphism.

If f is injective, then we get an exact sequence

0→ Vf−→W

π−→ C → 0,

where C is the cokernel of f . But then dimC = dimW − dimV = 0. But then C = 0,as otherwise, C would have at least one linearly independent element in it.

If f is surjective, then a similar argument shows that the kernel of f is 0.

For infinite dimensional vector spaces, we can give an analogue of Proposition 7.9.2.

Proposition 7.9.13. Let V be an infinite dimensional vector space over D. Let {vi | i ∈I} be linearly independent in V , and suppose given a family {vi | i ∈ J} such that V isgenerated by {vi | i ∈ I ∪ J}. Then there is a subset, K, of I ∪ J containing I such that{vi | i ∈ K} is a basis for V .

Proof Let

S = {K ⊂ I ∪ J | I ⊂ K and {vi | i ∈ K} is linearly independent} .Then S is nonempty, and has a partial ordering given by inclusion of subsets. We claimthat S satisfies the hypothesis of Zorn’s Lemma.

Thus, let T = {Kα |α ∈ A} be a totally ordered subset of S and let K =⋃α∈AKα.

We claim that K is linearly independent, and hence K is an upper bound for T in S.To see this, suppose that xi1vi1 + · · · + xikvik = 0, where vi1 , . . . , vik are distinct

elements of K. Then for each j, vij ∈ Kαj for some αj ∈ A. But since T is totallyordered, there is an r ∈ {1, . . . , k} such that Kαj ⊂ Kαr for j = 1, . . . , k. But then


vi1 , . . . , vik all lie in Kαr . Since Kαr is linearly independent, this forces xij = 0 forj = 1, . . . , k. Thus, K is linearly independent as well.

Thus, by Zorn’s Lemma, there is a maximal element, K, in S. We claim that {vi | i ∈K} is a basis for V . To see this, it suffices to show that {vi | i ∈ K} generates V , whichin turn will follow if we show that each vi for i ∈ J is contained in the vector subspacegenerated by {vi | i ∈ K}.

Thus, suppose that r is contained in the complement of K in I ∪ J , and let K ′ =K∪{r}. Then {vi | i ∈ K ′} must be linearly dependent by the maximality of K. Supposegiven a relation of the form xi1vi1+· · ·+xikvik = 0, where vi1 , . . . , vik are distinct elementsof K ′, and xij is nonzero for j = 1, . . . , k.

Since {vi | i ∈ K} is linearly independent, then at least one of the ij must equal r,say r = ik. But then vr = −x−1

r xi1vi1 − · · · − x−1r xik−1vik−1 lies in the vector subspace

of V generated by {vi | i ∈ K}.Taking the set {vj | j ∈ J} to be the set of nonzero elements of V , if necessary, we

see that every vector space over D has a basis.

Corollary 7.9.14. Every module over a division ring is free.

The following corollary is now immediate from Proposition 7.7.51.

Corollary 7.9.15. Every short exact sequence

0→ V ′ → V → V ′′ → 0

of modules over a division ring splits.

7.10 Matrices and Transformations

Recall that an m × n matrix is one with m rows and n columns. We review the basicdefinitions and properties of nonsquare matrices.

Definitions 7.10.1. Addition of nonsquare matrices is defined coordinate-wise: IfM1 =(aij) and M2 = (bij) are both m × n matrices with coefficients in A, then M1 + M2 isthe m× n matrix whose ij-th coefficient is aij + bij .

Multiplication of non-square matrices is defined analogously to that of square matri-ces, except that we may form the product M1M2 whenever the number of columns ofM1 is equal to the number of rows of M2. Thus, if M1 = (aij) and M2 = (bij) are m×nand n× l matrices, respectively, then we define M1M2 to be the m× l matrix whose ij-thentry is

∑nk=1 aikbkj for 1 ≤ i ≤ m and 1 ≤ j ≤ l.

Nonsquare matrices behave analogously to square ones. The proof of the next lemmais left to the reader.

Lemma 7.10.2. Let A be a ring and let M1, M2, and M3 be m × n, n × k, and k × lmatrices over A, respectively. Then (M1M2)M3 = M1(M2M3).

In addition, the distributive laws hold: with the Mi as above, if M ′1 and M ′

2 arem × n and n × k matrices, respectively, then (M1 + M ′

1)M2 = M1M2 + M ′1M2 and

M1(M2 +M ′2) = M1M2 +M1M

′2.

Finally, for a ∈ A, let aIn ∈ Mn(A) be the matrix whose diagonal entries are all aand whose off-diagonal entries are all 0. Then if M1 = (bij) and M2 = (cij) for M1

and M2 above, then M1 · aIn is the matrix whose ij-th entry is bija, and aIn ·M2 is thematrix whose ij-th entry is acij.


The following corollary is now immediate.

Corollary 7.10.3. Right multiplication by the matrices aIn gives a right A-module struc-ture to the set of m × n matrices over A, and left multiplication by aIm gives it a leftA-module structure.

In both cases the underlying abelian group structure of the module is given by matrixaddition, but the two module structures have no direct relationship to each other if A isnot commutative. (What we have instead is an example of an A-A-bimodule, as we shallencounter in our discussions of tensor products in Section 9.4. We shall not make use ofbimodules here.)

Let us first establish notation:

Definitions 7.10.4. For m,n ≥ 1, we write Mm,n(A) for the set of m×n matrices overA. We write eij ∈Mm,n(A) for the matrix whose ij-th coordinate is 1 and whose othercoordinates are all 0, for 1 ≤ i ≤ m and 1 ≤ j ≤ n.

The next lemma generalizes a result we’ve seen for square matrices.

Lemma 7.10.5. Let A be a ring. Then as either a right or a left A-module, Mm,n(A)is a free module with basis given by any ordering of {eij | 1 ≤ i ≤ m, 1 ≤ j ≤ n}.

In the study of classical linear algebra over a field, it is shown that if K is a field, thenthe K-linear maps from Kn to Km are in one-to-one correspondence with Mm,n(K) Weshall generalize this here to show that if A is a commutative ring, then the A-modulehomomorphisms from An to Am are in one-to-one correspondence with Mm,n(A).

Thus, matrices correspond to transformations of free modules.The situation for noncommutative rings is more complicated, but only by a very little.

The (usually) standard situation, where an m× n matrix acts from the left on an n× 1column vector, gives the right A-module homomorphisms from An to Am. Left modulehomomorphisms, on the other hand, are given by the right action of the n×m matriceson the 1 × n row vectors. We now give formal treatments of these correspondences,beginning with the case of right modules.

Thus, for r ≥ 1, identify Ar with the right A-module of r × 1 column matrices (i.e.,column vectors) for all r ≥ 1. A typical element is

x =

⎛⎜⎝ a1

...ar

⎞⎟⎠ .

As shown above, the standard right module structure on Ar is given by setting xa equalto the matrix product x · aI1 for a ∈ A. We write ei for the basis element ei1 ofLemma 7.10.5, so that

ei =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

0...010...0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠.


where the 1 is in the i-th place, with 1 ≤ i ≤ n. Thus, with x as above, we havex =

∑ni=1 eiai.

Regarding An and Am are right A-modules, recall that HomA(An, Am) denotes theset of right A-module homomorphisms from An to Am. Recall also that HomA(An, Am)is an abelian group under the operation that sets (f + g)(x) = f(x) + g(x) for f, g ∈HomA(An, Am) and x ∈ An.Proposition 7.10.6. Let m,n ≥ 1. Then there is an isomorphism of abelian groupsΦm,n : Mm,n(A)

∼=−→ HomA(An, Am) defined as follows. For an m × n matrix M , wedefine Φm,n(M) : An → Am by setting (Φm,n(M))(x) to be equal to the matrix productMx for all x ∈ An.Proof Φm,n(M) is a group homomorphism from An to Am by the distributive law ofmatrix multiplication, and is an A-module homomorphism by associativity: M · (xa) =M(x · aI1) = (Mx)aI1 = (Mx)a. Thus, Φm,n : Mm,n(A) → HomA(An, Am) is a welldefined function, and by the distributive law, it is a group homomorphism.

It is easy to see, by direct calculation of matrix products, that Mei is the i-th columnof M . Thus, if Φm,n(M) and Φm,n(M ′) have the same effect on the canonical basisvectors, the matrices M and M ′ must be equal. Thus, Φm,n is injective.

Let f ∈ HomA(An, Am), and let M be the matrix whose i-th column is f(ei) fori = 1, . . . , n. Then Φm,n(M) and f have the same effect on the canonical basis elements.Since the canonical basis generates An as an A-module, Φm,n(M) and f must have thesame effect on every element of An, and hence Φm,n(M) = f . Thus, Φm,n is surjective.

As we shall see in Section 9.7, there is a natural leftA-module structure on HomA(An, Am)coming from the left action of A on Am. Similarly, there is a natural right A-modulestructure on HomA(An, Am) coming from the left action of A on An. As the interestedreader may check upon reading Section 9.7, Φm,n gives an A-module isomorphism fromthe standard left module structure on matrices to the standard left module structure onhomomorphisms, as well as from the standard right module structure on matrices to thestandard right module structure on homomorphisms.

Recall from Problem 24 of Exercises 7.7.27 that HomA(An, An) forms a ring, otherwisedenoted EndA(An), called the endomorphism ring of An over A. The addition is thesame one we’ve just considered: (f + g)(x) = f(x) + g(x). Multiplication is given bycomposition of functions: (f · g)(x) = f(g(x)).

Proposition 7.10.7. For n ≥ 1, Φn,n induces an isomorphism of rings from the matrixring Mn(A) to the endomorphism ring EndA(An). (Here, An is considered as a free rightA-module.)

More generally, if M is an m × n matrix and M ′ is an n × k matrix, then theA-module homomorphism induced by the matrix product MM ′ is the composite of thehomomorphism induced by M and the homomorphism induced by M ′:

Φm,k(MM ′) = (Φm,n(M)) ◦ (Φn,k(M ′)) .

Proof We’ve seen that Φn,n is a group homomorphism. Moreover, the multiplicativeidentity element In of Mn(A) is known to induce the identity transformation of An.Thus, Φn,n is a ring homomorphism if and only if it respects multiplication, and hencethe first statement follows from the second. But the second statement is immediate fromthe associativity of matrix multiplication:

(Φm,n(M) ◦ Φn,k(M ′)) (x) = M(M ′x) = (MM ′)x = Φm,k(MM ′)(x).


Recall that a matrix is invertible if it is a unit in Mn(A), and that the group of unitsthere is Gln(A), the n-th general linear group of A. Note that M ∈ Gln(A) if and onlyif there is a matrix M ′ ∈Mn(A) with MM ′ = M ′M = In.

There is a module theoretic analogue of the general linear group.

Definition 7.10.8. IfN is an A-module, we shall write AutA(N), the group ofA-moduleautomorphisms of N , for the unit group of EndA(N).

Since composition is the ring product in EndA(An) and the identity function is themultiplicative identity element, an A-module automorphism is precisely an isomorphismf : N → N of A-modules. Since Φn,n is an isomorphism of rings, we obtain the followingcorollary.

Corollary 7.10.9. An n × n matrix M over A is invertible if and only if the right A-module homomorphism that it induces (i.e., Φn,n(M) : An → An) is an isomorphism.In particular, Φn,n restricts to an isomorphism

Φn,n : Gln(A)∼=−→ AutA(An).

Recall from Lemma 7.7.36 that if f : N → N ′ is an A-module homomorphism betweenfree modules and if x1, . . . , xn is a basis for N , then f is an isomorphism if and only iff(x1), . . . , f(xn) is a basis for N ′. Since Mei is the i-th column of M , we obtain thefollowing corollary.

Corollary 7.10.10. An n× n matrix M over A is invertible if and only if its columnsform a basis of An.

It is useful to be able to do manipulations on the matrix level that match standardconstructions in module theory. One such construction is the block sum (sometimescalled Whitney sum13) of matrices: given an m × n matrix M and an m′ × n′ matrixM ′, the block sum M ⊕M ′ of M and M ′ is the matrix(

M 00 M ′

),

where the 0’s are 0-matrices of the appropriate size. In coordinates, the ij-th coordinateof M ⊕M ′ is given as follows. If 1 ≤ i ≤ m and 1 ≤ j ≤ n, then the ij-th coordinateof M ⊕M ′ is equal to the ij-th coordinate of M . If 1 ≤ i ≤ m and j > n, or if i > mand j ≤ n, then the ij-th coordinate of M ⊕M ′ is 0. If m + 1 ≤ i ≤ m + m′ andn+1 ≤ j ≤ n+n′, then the ij-th coordinate of M ⊕M ′ is equal to the (i−m)(j−n)-thcoordinate of M ′.

The use of the symbol “⊕” for block sum is meant to be suggestive. In module-theoretic terms, we use it as follows: If f1 : N1 → N ′

1 and g : N2 → N ′2 are A-module

homomorphisms, we write f ⊕ g : N1⊕N2 → N ′1⊕N ′

2 for the cartesian product of f andg. Thus, (f ⊕ g)(n1, n2) = (f(n1), g(n2)).

13After the topologist Hassler Whitney.


Proposition 7.10.11. Let M be an m × n matrix and let M ′ be an m′ × n′ matrixover A. Then if we identify An ⊕ An′

with An+n′in the standard way (i.e., identifying

(∑ni=1 eiai,

∑n′

i=1 eia′i) with

∑ni=1 eiai+

∑n′

i=1 ei+na′i) and identify Am⊕Am′

with Am+m′

similarly, then the A-module homomorphism represented by the block sum M ⊕M ′ isequal to the direct sum of the A-module homomorphisms represented by M and by M ′.Symbolically, we have

Φm+m′,n+n′(M ⊕M ′) = Φm,n(M)⊕ Φm′n′(M ′).

Proof Both sides agree on the canonical basis elements of An+n′.

So far we have only treated the module homomorphisms from An to Am, rather thantreating the module homomorphisms between arbitrary finitely generated free modules.The reason is that there is no way to get from module homomorphisms to matrices thatdoesn’t depend on making explicit choices of bases for the two modules: a different choiceof basis results in a different matrix to represent it.

Thus, we should think of An or Am in this discussion as a finitely generated freemodule with a specific preferred basis. Of course, the point is that if N is a finitelygenerated free module, then an ordered subset B = n1, . . . , nk of N is a basis if andonly if the unique homomorphism fB : Ak → N with fB(ei) = ni for 1 ≤ i ≤ k is anisomorphism. Thus, the bases of N with k elements are in one-to-one correspondencewith the isomorphisms from Ak to N .

Definition 7.10.12. Suppose given finitely generated free right A-modules N and N ′,and let B = n1, . . . , nk and B′ = n′

1, . . . , n′m be bases for N and N ′, respectively. Write

fB : Ak → N and fB′ : Am → N ′ for the isomorphisms induced by B and B′. Thenif g : N → N ′ is an A-module homomorphism, we define the matrix of g with respectto the bases B and B′ to be the m × k matrix defined as follows. For 1 ≤ j ≤ k, writeg(nj) =

∑mi=1 n

′iaij . Then the ij-th entry of the matrix of g is defined to be aij .

Note that the j-th column of the matrix of g with respect to the bases B and B′ isprecisely f−1

B′ g(nj). Thus, the following lemma is immediate.

Lemma 7.10.13. The matrix of g : N → N ′ with respect to the bases B and B′ is equalto the matrix of (f−1

B′ ◦ g ◦ fB) : Ak → Am, i.e., it is the matrix Φ−1m,k(f

−1B′ ◦ g ◦ fB).

Of particular interest is the use of matrices to study the A-module homomorphismsfrom a free right A-module N to itself (i.e., to study elements of EndA(N)). Since thesource and target of these homomorphisms are the same, we need only one basis to definea matrix.

Definition 7.10.14. Let B = x1, . . . , xn be a basis for the free right A-module N andlet fB : An → N be the A-module homomorphism that takes ei to xi for each i. Letg : N → N be an A-module homomorphism. Then the matrix of g with respect to thebasis B is defined to be Φ−1

n,n(f−1B ◦ g ◦ fB), the matrix whose i-th column is f−1

B (g(xi)).

The next result makes it clearer what is going on.

Lemma 7.10.15. Let B = x1, . . . , xn be a basis for the free right A-module N and letfB : An → N be the A-module homomorphism that takes ei to xi for each i. Thenthere is a ring isomorphism f∗B : EndA(N)→ EndA(An) defined by f∗B(g) = f−1

B ◦g ◦fB.


Moreover, the matrix of g with respect to the basis B is the image of g under the compositering isomorphism

EndA(N)f∗

B−→ EndA(An)Φ−1

n,n−−−→Mn(A).

In particular, the passage from a transformation to its matrix with respect to B gives anisomorphism of rings from EndA(N) to Mn(A).

Proof The function f∗B is easily seen to be a ring homomorphism. It is an isomorphismbecause the correspondence that takes g ∈ EndA(An) to fB ◦ g ◦ f−1

B gives an inverse forit. The rest follows from the definition of the matrix of g with respect to B.

It is valuable to discern which properties of a transformation may be detected by itsmatrix regardless of the basis used. Thus, it is important to understand what happensto the matrix of g when we change the basis used to define it. Of course, as we’ve statedabove, the rank of a free module is not well defined over every ring A (though it is forthe ones that we care about, as we’ll see below). We shall not consider the base-changeproblem between bases with a different number of elements, but shall restrict attentionto understanding what happens when we change between bases with the same numberof elements.

Definition 7.10.16. Suppose given bases B = x1, . . . , xn and B′ = x′1, . . . , x′n of the

free right A-module N and let fB : An → N and fB′ : An → N be the A-modulehomomorphisms that take ei to xi and x′i, respectively, for 1 ≤ i ≤ n. Then the basechange matrix MBB′ is defined to be the matrix of the identity map 1N : N → N , whereB is used as the basis for the domain of 1N , and B′ is used as the basis for the targetcopy of N . Notationally, we have MBB′ = Φ−1

n,n(f−1B′ fB).

Lemma 7.10.17. Suppose given bases B, B′, and B′′ for the free right A-module N , allof which have n elements. Then the base change matrices satisfy

MB′B′′ ·MBB′ = MBB′′ .

Additionally, MBB is the identity matrix, and hence MBB′ is invertible, with inversegiven by MB′B.

Proof We first show that the displayed formula holds. By definition, it suffices toshow that Φn,n(MB′B′′ ·MBB′) = f−1

B′′fB . But Φn,n is a ring homomorphism, so thatΦn,n(MB′B′′ ·MBB′) = Φn,n(MB′B′′) ◦ Φn,n(MBB′) = f−1

B′′ ◦ fB′ ◦ f−1B′ ◦ fB , and hence

the formula holds.Now MBB = Φ−1

n,n(1) = In, and hence MB′BMBB′ = In by setting B′′ = B in thedisplayed equation. Exchanging the roles of B and B′ now gives the desired result.

We may now express the difference between the matrices of a given homomorphismwith respect to different bases.

Corollary 7.10.18. Suppose given bases B = x1, . . . , xn and B′ = x′1, . . . , x′n of the

free right A-module N and let MBB′ be the base change matrix defined above. Then ifg ∈ EndA(N) and if M is the matrix of g with respect to the basis B, then the matrix ofg with respect to B′ is the matrix product MBB′MM−1

BB′ .


Proof Since Φn,n is a ring homomorphism, we have

Φn,n(MBB′MM−1BB′) = (f−1

B′ ◦ fB) ◦ (f−1B ◦ g ◦ fB) ◦ (f−1

B ◦ fB′).

The result now follows.

Definition 7.10.19. We say that two n × n matrices, M,M ′ ∈ Mn(A) are similar ifthere is an invertible n× n matrix M ′′ such that M ′ = M ′′M(M ′′)−1.

Corollary 7.10.20. Let N be a finitely generated free right A-module and let g ∈ EndA(N).Suppose we’re given two bases of N with the same number of elements. Then the matricesof g with respect to these two bases are similar.

Of course, we’d like to treat left modules as well. When A is commutative, the usualthing is to identify left modules with right modules, and to study homomorphisms offree left modules by the methods above. For noncommutative rings, this doesn’t work,and hence further argumentation is needed. In this case, we identify the free left moduleAn with the 1 × n row matrices (i.e., row vectors). This has the advantage that thenotation agrees with the notation we’re used to for free modules: A typical element isx = (a1, . . . , an).

Again we write ei = (0, . . . , 0, 1, 0, . . . , 0), where the 1 is in the i-th place. Here, thekey is that if M is an n ×m matrix, then the matrix product eiM is precisely the i-throw of M . Given this, and the fact that the left A-module structure on An is obtainedby setting ax equal to the matrix product (aI1)x, the proof of the next proposition isentirely analogous to that of Proposition 7.10.6, and is left to the reader.

Proposition 7.10.21. Write HomA(An, Am) for the set of left A-module homomor-phisms from An to Am. Then there is a group isomorphism

Ψn,m : Mn,m(A)→ HomA(An, Am)

defined by setting (Ψn,m(M))(x) equal to the matrix product xM .

In this case, there is a reversal between composition of functions and matrix multi-plication: if M is n×m and M ′ is k × n, then

(Ψn,m(M) ◦Ψk,n(M ′)) (x) = (xM ′)M = x(M ′M) = Ψk,m(M ′M)(x).

We obtain the following proposition.

Proposition 7.10.22. For n ≥ 1, Ψn,n induces an isomorphism from the matrix ringMn(A) to the opposite ring of the endomorphism ring of the free left A-module An.

As in the case of right modules, the block sum of matrices represents the direct sumof the associated A-module homomorphisms.

The treatment of matrices associated to different bases is analogous to that given forright modules. We leave the details to the reader.

Exercises 7.10.23.1. Let N be a free right A-module with basis B = x1, . . . , xn. Let g ∈ EndA(N) and

let M be the matrix of g with respect to the basis B. Let M ′ be an n× n matrixthat is similar to M . Show that M ′ is the matrix of g with respect to some otherbasis of N .


2. Suppose given matrices

M =(M1 M2

M3 M4

)and M ′ =

(M ′

1 M ′2

M ′3 M ′

4

),

where M1 is m× n, M2 is m× s, M3 is r × n, and M4 is r × s, while M ′1 is n× k,

M ′2 is n× t, M ′

3 is s× k, and M ′4 is s× t. Show that

MM ′ =(M1M

′1 +M2M

′3 M1M

′2 +M2M

′4

M3M′1 +M4M

′3 M3M

′2 +M4M

′4

).

3. Let M be an (n+ k)× (n+ k) matrix over A. Show that M has the form

M =(M1 M2

0 M3

)with M1 n× n, M2 n× k, and M3 k × k if and only if (Φn,n(M))(An) ⊂ An (i.e.,the standard right A-module homomorphism induced by M carries An into itself).Here An is the submodule of An+k generated by e1, . . . , en. Show also that thematrices of the above form (where n is fixed) give a subring of Mn+k(A).

4. Let B = Mn(A). Show that the matrix ring Mk(B) is isomorphic as a ring toMnk(A).

‡ 5. If A is commutative, then, since left and right A-modules may be identified witheach other, Ψ−1

n,n ◦ Φn,n provides an isomorphism from Mn(A) to its opposite ringfor each n ≥ 1. Explicitly write down the effect of Ψ−1

n,n ◦ Φn,n on a matrix (aij).

7.11 Rings of Fractions

Think for a moment about how our number system evolved. First, there were the non-negative integers. Then negative numbers were adjoined, producing the integers. Thisprocess of adjoining inverses has since been generalized to a construction that producesan abelian group from any abelian monoid. It is called the Grothendieck construction,and will be studied below.

The next stage in the development of our number system was the creation of therational numbers, by adjoining multiplicative inverses to the nonzero integers. We shallgeneralize this last process in this section. One of the consequences will be that we canthen embed any integral domain in a field.

Definition 7.11.1. Let A be a commutative ring. A multiplicative subset of A is asubset S ⊂ A such that 0 �∈ S and such that S is closed under multiplication: Fors, t ∈ S, the product st is also in S.14

Note: A ring called “A” in this section will be implicitly assumed to be commutative.Other rings, however, will not.

Let S be a multiplicative subset of the commutative ring A. We shall construct anew ring S−1A, in which the elements of S are invertible.

14Some people permit 0 to be an element of a multiplicative subset. Indeed, the entire theory of ringsof fractions generalizes to that context, but, if 0 ∈ S, then the ring of fractions S−1A is the 0 ring. Soall we’re doing is excluding this rather boring case.


The elements of S−1A will be equivalence classes of ordered pairs (a, s) with a ∈ Aand s ∈ S. We say that (a, s) is equivalent to (b, t) (written (a, s) ∼ (b, t)) if there is ans′ ∈ S such that s′(ta− sb) = 0. Notice that if A is an integral domain, then this is justthe familiar condition we use to test when two fractions a/s and b/t are equal in Q.

Lemma 7.11.2. Let S be a multiplicative subset of the commutative ring A. Then theabove relation ∼ is an equivalence relation.

Proof The relation is obviously symmetric and reflexive, so it suffices to show tran-sitivity. Suppose that (a, s) ∼ (b, t) and that (b, t) ∼ (c, u). Then there are elementss′, s′′ ∈ S with s′(ta−sb) = 0 and s′′(ub− tc) = 0. We claim now that s′s′′t(ua−sc) = 0,and hence (a, s) ∼ (c, u).

Now s′s′′tua = s′s′′usb, because s′(ta−sb) = 0. Similarly, s′s′′usb = s′s′′stc, becauses′′(ub− tc) = 0. Putting this together, we see that s′s′′t(ua− sc) = 0 as desired.

Definition 7.11.3. Let S be a multiplicative subset of the commutative ring A. Wedefine the ring of fractions S−1A as the set of equivalence classes of pairs (a, s), and,to emphasize the analogy with fractions, we write a/s, or a

s , for the element of S−1Arepresented by (a, s).

The addition and multiplication operations in S−1A are given by

a

s+b

t=ta+ sb

stand

a

s· bt

=ab

st.

We define the canonical map η : A → S−1A by setting η(a) = as/s for any elements ∈ S.

The proof of the following proposition is tedious but straightforward. We leave it tothe reader.

Proposition 7.11.4. Let S be a multiplicative subset of A. Then S−1A is a commu-tative ring under the above operations. The additive identity is 0/s for any s ∈ S, andmultiplicative identity is s/s for any s ∈ S.

The canonical map η : A→ S−1A is a ring homomorphism whose kernel is the set ofall a ∈ A that are annihilated by some s ∈ S. For s ∈ S, the element η(s) is a unit inS−1A, with inverse 1/s. The generic element of S−1A may then be characterized by

a

s= η(a) · (η(s))−1.

The reader may very well ask: Why don’t we insist that 1 ∈ S? The answer, it turnsout, is that there isn’t a good reason other than the fact that it isn’t necessary. We’ll seein Exercises 7.11.27 that if 1 �∈ S, and if we take S′ = S ∪{1}, then the natural inclusionof S−1A in S′−1

A is an isomorphism.The principal examples of rings of fractions are important enough to warrant names.

Definitions 7.11.5. 1. Let A be an integral domain and let S be the set of all nonzeroelements in A. The resulting ring S−1A is called the field of fractions, A(0), of A.

2. Let A be any commutative ring and let p be a prime ideal of A. For

S = {a ∈ A | a �∈ p} we write S−1A = Ap,

and call it the localization of A at p.


3. Let A be any commutative ring and let a ∈ A be any element that is not nilpotent.Let S = {ak | k ≥ 0}. We write

S−1A = A [1/a] ,

and call it the ring obtained by adjoining an inverse to a.

It is easy to see that Q is the field of fractions of Z, so that we have, as promised,generalized the construction of the rationals from the integers. Note that the field offractions of an integral domain A really is a field, as η(a)(η(s))−1 is a unit in A(0)

whenever a �= 0. We obtain the following corollary.

Corollary 7.11.6. The field of fractions, A(0), of an integral domain A is a field, andη : A → A(0) an embedding. Thus, a ring is an integral domain if and only if it is asubring of a field.

The following examples are important in field theory and number theory, respectively.

Examples 7.11.7.1. Let K be a field and let X1, . . . , Xn be variables. We write K(X1, . . . , Xn) for

the field of fractions of the polynomial ring K[X1, . . . , Xn]. We call it the field ofrational functions over K in the variables X1, . . . , Xn.

2. Recall from Problem 1 of Exercises 7.1.41 that the p-adic integers Zp form anintegral domain. We write Qp for its field of fractions, known as the field of p-adicrationals.

Note that a field of fractions is a special case of localization, as the ideal (0) is primein an integral domain.

If p ∈ Z is prime, then the ring Z [1/p] is sometimes called the “localization of Z awayfrom p.” As we shall see below, this is a definite abuse of the word localization.

Lemma 7.11.8. Let B be a ring and let a and b be elements of B that commute witheach other, where b ∈ B×. Then a commutes with b−1.

Proof Clearly, b annihilates ab−1 − b−1a. Since b is a unit, the result follows.

We return to the general case of a multiplicative subset of a commutative ring A.The ring of fractions S−1A satisfies a useful universal property.

Proposition 7.11.9. Let f : A → B be a ring homomorphism such that f(s) is a unitfor all s ∈ S. Then there is a unique ring homomorphism f : S−1A → B such that thefollowing diagram commutes.

A B

S−1A

��f

��

��

η

�� f

Moreover, if f gives an A-algebra structure on B, then f , if it exists, induces anS−1A-algebra structure on B. We obtain a one-to-one correspondence between the S−1A-algebra structures on B and the A-algebra structures f : A → B with the property thatf(S) ⊂ B×. And if g : B → C is an A-algebra homomorphism between S−1A-algebras,then it is also an S−1A-algebra homomorphism.


Proof Suppose there is a ring homomorphism f that makes the diagram commute.Since η(s) is a unit in S−1A for each s ∈ S, f(s) = f(η(s)) must be a unit in B.Moreover, since a/s = η(a)(η(s))−1, f must satisfy the formula f(a/s) = f(a)f(s)−1.Thus, f is uniquely defined by the commutativity of the diagram.

Conversely, if f(s) is a unit for each s ∈ S, we define f by the same formula: f(a/s) =f(a)f(s)−1. This is well defined, since if s′(ta− sb) = 0, we have f(ta− sb) = 0 by theinvertibility of f(s′). Making use of Lemma 7.11.8, if B is noncommutative, it is easy tosee that f is a ring homomorphism.

The statements about existence and uniqueness of algebra structures are straightfor-ward. To see that every A-algebra homomorphism f : B → C between S−1A-algebras isan S−1A-algebra homomorphism, write ν and μ for the S−1A-algebra structures on Band C, respectively. We want to show that f ◦ ν = μ.

Our assumption is that f is an A-algebra homomorphism, which means that f ◦ νagrees with μ if we precompose with η : A → S−1A. But two ring homomorphisms outof S−1A that agree when precomposed with η must be equal.

Now Q is the field of fractions of Z. Recall that for any ring B, there is a unique ringhomomorphism f : Z → B, where f(n) = n · 1, the n-th “power” of 1 in the additivegroup structure on B. The homomorphism f gives B a unique Z-algebra structure, andB has characteristic 0 if and only if f is injective. We obtain the following corollary.

Corollary 7.11.10. A ring B is a Q-algebra if and only if n · 1 is a unit in B foreach 0 �= n ∈ Z. The Q-algebra structure on B, if it exists, is unique, and any ringhomomorphism between Q-algebras is a Q-algebra homomorphism.

In particular, there is a unique embedding (as a ring) of Q in any field of characteristic0.

We also have the following.

Corollary 7.11.11. Let f : A → K be an embedding of an integral domain A into afield K. Then there is a unique extension of f to an embedding f : A(0) → K, where A(0)

is the field of fractions of A. The image of f is the smallest subfield of K containing theimage of f .

We now have a framework to classify the subrings of Q.

Proposition 7.11.12. Let A be a subring of Q and let S = A× ∩Z. Then A is isomor-phic as a ring to S−1Z.

Proof The elements of S become units in A. Thus, Proposition 7.11.9 provides a uniquering homomorphism f : S−1Z→ A. We claim that f is an isomorphism.

First, note that the composite S−1Zf−→ A ⊂ Q takes m/s to m/s. Since the equiva-

lence relations defining S−1Z and Q are the same (because Z is an integral domain), theabove composite is injective, and hence so is f .

Now suppose given an element of A. Since A ⊂ Q, we may write it as m/n form,n ∈ Z with n �= 0. By reducing the fraction, we may assume that m and n arerelatively prime. It suffices to show that n is a unit in A, i.e., that the element 1/n of Qlies in A.

Since m and n are relatively prime, we can find a, b ∈ Z with am+ bn = 1. Dividingboth sides by n, we see that a(m/n) + b = 1/n. Since a, m/n and b are in A, so is 1/n.


The universal property for homomorphisms out of rings of fractions gives the simplestproof of the following characterization of modules over these rings. The situation issimilar to that given by Proposition 7.7.15.

Because S−1A is an A-algebra, every S−1A-module has an underlying A-modulestructure. So an immediate question is: Which A-modules admit a compatible S−1A-module structure? And if a compatible S−1A-module structure exists, is it unique?(Compatibility here means that the A-module structure induced by the S−1A-modulestructure is the same as the one we started with.) A good answer will give us a completeunderstanding of S−1A-modules.

Proposition 7.11.13. Let M be an A-module. Then M admits a compatible S−1A mod-ule structure if and only if multiplication by s induces an isomorphism (of abelian groups)from M to M for each s ∈ S. Moreover, there is at most one S−1A-module structurecompatible with the original A-module structure on M . Finally, if the A-modules M andN admit compatible S−1A-module structures, then every A-module homomorphism fromM to N is an S−1A-module homomorphism as well.

Proof We use the correspondence given by Problem 21 of Exercises 7.7.27 betweenA-module structures on M and ring homomorphisms from A to EndZ(M). Note thatif f : A → EndZ(M) and f : S−1A → EndZ(M) are ring homomorphisms, then theS−1A-module structure on M induced by f is compatible with the A-module structurecoming from f if and only if the diagram

A EndZ(M)

S−1A

��f

��

η

�� f

commutes.For a given homomorphism f : A → EndZ(M), the homomorphisms f : S−1A →

EndZ(M) that make the diagram commute are characterized by Proposition 7.11.9: Fora given f there is at most one f with this property, and such an f exists if and only iff(s) is a unit in EndZ(M) (i.e., a group isomorphism of M) for each s ∈ S.

It remains to show that if f : M → N is an A-module homomorphism betweenthe S−1A-modules M and N , then f is in fact an S−1A-module homomorphism. Inparticular, it suffices to show that f((1/s)m) = (1/s)f(m) for s ∈ S and m ∈M .

Write s : M →M and s : N → N for the homomorphisms induced by multiplicationby s. Then the statement that f((1/s)m) = (1/s)f(m) for all m ∈ M amounts to thestatement that f ◦ −1

s = −1s ◦ f . Now, since f is an A-module homomorphism, we have

f ◦ s = s ◦ f . So the result now follows as in the proof of Lemma 7.11.8.

We’d like now to study ideals in rings of fractions. Indeed, the properties of the idealsin a localization form one of the primary motivations for studying rings of fractions. Giventhe relationship between ideals and modules, it seems reasonable to discuss modules offractions first.

Let M be an A-module and let S be a multiplicative subset of A. We shall definean S−1A-module S−1M as follows. The elements of S−1M are equivalence classes ofordered pairs (m, s) with m ∈M and s ∈ S, where (m, s) is equivalent to (n, t) (written(m, s) ∼ (n, t)) if there is an s′ ∈ S with s′(tm− sn) = 0. As expected, we write m/s forthe element of S−1M represented by the ordered pair (m, s). The proof of the followinglemma is straightforward and is left to the reader.


Lemma 7.11.14. The relation ∼ above is an equivalence relation. The resulting setS−1M of equivalence classes is an abelian group and an S−1A-module via

m

s+n

t=tm+ sn

stand

a

s· mt

=am

st,

respectively. There is a canonical A-module homomorphism η : M → S−1M given byη(m) = ms/s for any s ∈ S.

If N is an S−1A-module and f : M → N is an A-module homomorphism, thenthere is a unique S−1A-module homomorphism f : S−1M → N that makes the followingdiagram commute.

M N

S−1M

��f

��

��

�

η

�� f

We obtain an isomorphism

HomS−1A(S−1M,N)η∗−→∼= HomA(M,N)

via η∗(f) = f ◦ η.Localization forms an important special case of modules of fractions:

Definition 7.11.15. Let p be a prime ideal of A and let S be the complement of p inA. Then we write Mp for S−1M . We call it the localization of M at p.

Let S be an arbitrary multiplicative subset of A and let f : M → N be an A-modulehomomorphism. Let S−1f : S−1M → S−1N be defined by (S−1f)(m/s) = f(m)/s.Then S−1f is clearly the unique S−1A-module homomorphism that makes the followingdiagram commute.

M N

S−1M S−1N

��η

��f

��η

��S−1f

Clearly, passage from M to S−1M and from f to S−1f provides a functor from A-modules to S−1A-modules (i.e., S−1f ◦S−1g = S−1(f ◦g), and S−1(1M ) = 1S−1M , where1N denotes the identity map of the module N).

Proposition 7.11.16. Suppose given an exact sequence

· · · →M ′ f−→Mg−→M ′′ → · · ·

of A-modules. Then the induced sequence

· · · → S−1M ′ S−1f−−−→ S−1MS−1g−−−→ S−1M ′′ → · · ·

of S−1A-modules is also exact.


Proof A sequence is exact if it is exact at every module in it. We show exactness at M .First, note that S−1g ◦ S−1f = S−1(g ◦ f) = S−10, since g ◦ f is the 0-homomorphism 0(since im f ⊂ ker g). But S−10 = 0, so im(S−1f) ⊂ ker(S−1g).

Now let m/s ∈ kerS−1g. Then a glance at the equivalence relation defining S−1M ′′

shows that tg(m) = 0 for some t ∈ S. But then g(tm) = tg(m) = 0, so tm ∈ ker g = im f .Let tm = f(m′). Then m/s = (S−1f)(m

′st ). Thus, S−1f is onto.

This behavior may be abstracted as follows.

Definition 7.11.17. A functor which takes exact sequences to exact sequences is calledan exact functor.

In particular, the passage from M to S−1M is an exact functor. This has a numberof uses, since certain kinds of relationships can be described by exactness properties. Forinstance, N is a submodule of M if and only if the sequence 0→ N

⊂−→M is exact. Weimmediately obtain the following corollary.

Corollary 7.11.18. Let S be a multiplicative subset of A and let M be an A-module.Then if N is a submodule of M , the natural map from S−1N to S−1M is an inclusionof S−1A-modules.

Thus, if a is an ideal of A, then S−1a is an ideal of S−1A. We shall show that everyideal of S−1A arises this way:

Proposition 7.11.19. Let S be a multiplicative subset of A and let b be an ideal ofS−1A. Let η : A → S−1A be the canonical homomorphism and let a = η−1b. Thenb = S−1a.

Proof Let x ∈ a. Then xs/s ∈ b for any s ∈ S, and hence so is x/t = (xs)/(ts) for anyt ∈ S, since b is an ideal of S−1A. Thus, S−1a ⊂ b.

Suppose given an arbitrary element x/s ∈ b with x ∈ A and s ∈ S. It suffices to showthat x ∈ a. But this follows, since η(s) · (x/s) = η(x) (i.e., η(x) ∈ b).

Corollary 7.11.20. Let S be a multiplicative subset of A. Then the prime ideals ofS−1A are those of the form S−1p, where p is a prime ideal of A such that p ∩ S = ∅.Proof Since the preimage under a ring homomorphism of a prime ideal is prime, everyprime ideal of S−1A has the form S−1p for some prime ideal p of A. Also, since S−1pmust be a proper ideal of S−1A in order to be prime, S−1p must not contain a unit ofS−1A, and hence p ∩ S must be empty. Thus, it suffices to show that if p is any primeideal of A that doesn’t meet S, then S−1p is prime in S−1A.

Suppose given such a p, and suppose that (a/s) · (b/t) ∈ S−1p. Say (ab)/(st) = x/s′,with x ∈ p and s′ ∈ S. This says there is an s′′ ∈ S with s′′(s′ab − stx) = 0. But thens′′s′ab ∈ p. Since s′ and s′′ are not in p, one of a and b must be. Thus, one of a/s andb/t is in S−1p.

Thus, it suffices to show that S−1p is proper. Suppose that x/s = 1 in S−1A,with x ∈ p and s ∈ S. Since 1 = s′/s′ for any s′ ∈ S, we can find s′, s′′ ∈ S withs′′(s′x− s′s) = 0. But this places s′′s′s in p, which is impossible, and hence S−1p cannotbe all of S−1A.

Of course, if S−1p′ and S−1p are prime ideals of S−1A, then S−1p′ ⊂ S−1p if andonly if p′ ⊂ p, by Proposition 7.11.19.

We now are in a position to explain the term “localization.”


Definition 7.11.21. A local ring is a commutative ring that has only one maximal ideal.

Fields, of course, are local rings, as are localizations, as we shall see shortly. Anotherexample of a local ring that we’ve seen is Zp.

Example 7.11.22. Recall from Proposition 7.1.40 that the ring homomorphism

π1 : Zp → Zp

has the property that a ∈ Zp is a unit if and only if π1(a) is nonzero in Zp. Thus, anyelement of Zp which doesn’t lie in kerπ1 is a unit. Therefore, any proper ideal of Zpmust be contained in kerπ1, and hence kerπ1 is the unique maximal ideal of Zp.

Note that Problem 4 of Exercises 7.1.41 shows that kerπ1 = (p), the principal idealgenerated by p. Thus, Zp is a local ring with maximal ideal (p).

Proposition 7.11.23. Let p be a prime ideal in the commutative ring A. Then thelocalization Ap is a local ring whose maximal ideal is pp.

Proof Let S be the complement of p in A. Then p ∩ S = ∅, so that pp = S−1p is aprime ideal of Ap. Since maximal ideals are prime, it suffices to show that any otherprime ideal of Ap must be contained in pp.

But the prime ideals of Ap all have the form p′p, where p′ ∩ S = ∅. By the definitionof S, this forces p′ ⊂ p, and the result follows.

This, of course, begs the question of the relationship of the field Ap/pp to A itself.

Proposition 7.11.24. Let p be a prime ideal of A. Then Ap/pp is isomorphic to thefield of fractions of A/p.

In particular, if m is a maximal ideal of A, then Am/mm is just A/m.

Proof We have a commutative diagram

A Ap

A/p Ap/pp

��η

��

π

��

π′

��η

where π and π′ are the canonical maps. It is easy to check that p = η−1(pp), and henceη is an embedding.

By Corollary 7.11.11, η extends uniquely to an embedding, η : K → Ap/pp of thefield of fractions, K, of A/p. But η is surjective, as π′(a/s) = η(π(a)/π(s)).

If p is maximal, then A/p is a field, and hence A/p = K by Corollary 7.11.11.

Localization turns out to be a very important technique in understanding A-modules.The point is that modules over a local ring are easier to classify than those over a non-localring. This will become especially clear in our study of projective modules in Section 9.8.So it is valuable to be able to deduce information about a module M from informationabout the localizations Mp.

Proposition 7.11.25. Let M be a module over the commutative ring A. Then thefollowing conditions are equivalent.


1. M = 0.

2. Mp = 0 for each prime ideal p of A.

3. Mm = 0 for each maximal ideal m of A.

Proof Clearly, the first condition implies the second, which implies the third. We shallshow that the third implies the first. Let M be an A-module such that Mm = 0 for eachmaximal ideal m of A. Let m ∈M . It suffices to show that the annihilator of m is A.

Thus, let m be a maximal ideal of A. Then m is in the kernel of the canonical mapfrom M to Mm. Thus, m/1 = 0/1 in Mm, so there exists s in A which is not in m suchthat sm = 0. In other words, Ann(m) is not contained in m. But since Mm′ = 0 forevery maximal ideal m′ of A, we see that Ann(m) is not contained in any of the maximalideals of A. Since Ann(m) is an ideal, it must be all of A.

Corollary 7.11.26. Let f : M → N be an A-module homomorphism. Then the follow-ing conditions are equivalent.

1. f is injective.

2. fp is injective for each prime ideal p of A.

3. fm is injective for each maximal ideal m of A.

The same holds if we replace injective by surjective.

Proof Consider the exact sequence

0→ Ki−→M

f−→ Nπ−→ C → 0

of Lemma 7.7.47. Since localization is an exact functor, we see that Kp is the kernelof fp and Cp is the cokernel of fp for each prime ideal p of A. Thus, the result followsimmediately from Proposition 7.11.25.

Exercises 7.11.27.1. Show that Zn is a local ring if and only if n is a prime power.

2. Show that the localization, Z(p), of Z at (p) embeds uniquely as a subring of thep-adic integers Zp.

3. Let a be any non-nilpotent element of the commutative ring A. Show that A[1/a]is isomorphic as an A-algebra to A[X]/(aX − 1), where (aX − 1) is the principalideal generated by aX − 1 in the polynomial ring A[X].

4. Let A be an integral domain and let K be its field of fractions. Show that if0 �= a ∈ A, then the ring of fractions A[1/a] may be identified with the image ofthe evaluation map ε1/a : A[X] → K obtained by evaluating X at 1/a. Thus, thenotation A[1/a] is consistent with that for adjoining an element to A.

5. Let A be a P.I.D. and let S be a multiplicative subset of A. Show that the ring offractions S−1A is also a P.I.D.


‡ 6. Let S be a multiplicative subset of A with 1 ∈ S. Suppose that S is generatedas a monoid by a finite set s1, . . . , sk, i.e., an element is in S if and only if it hasthe form sr11 . . . srk

k where ri ≥ 0 for 1 ≤ i ≤ k. Show that S−1A is isomorphicas an A-algebra to A[X1, . . . , Xk]/(s1X1 − 1, . . . , skXk − 1). Here, X1, . . . , Xk areindeterminates over A, and (s1X1 − 1, . . . , skXk − 1) is the ideal generated bys1X1 − 1, . . . , skXk − 1. It is common to write A[1/s1, . . . , 1/sk] for S−1A in thiscontext, a usage which may be of assistance in designing the proof.

7. Let A be a commutative ring and let S ⊂ A[X] be given by S = {Xk | k ≥ 0}.Show that S−1A[X] is isomorphic to the group ring A[Z] of Z over A.

8. We say that a multiplicative subset S ⊂ A is saturated if for any s ∈ S and anya ∈ A dividing s, we must have a ∈ S. Show that if S is any multiplicative subset,then T = {a ∈ A | a divides s for some s ∈ S} is a multiplicative subset. Show alsothat T is saturated. We call T the saturation of S.

† 9. Let S be a multiplicative subset of A and let T be its saturation. Show that forany multiplicative subset S′ with S ⊂ S′ ⊂ T , the natural maps S−1A→ S′−1

A→T−1A are isomorphisms of rings. Show also that if M is an A-module, then thenatural maps S−1M → S′−1

M → T−1M are isomorphisms of S−1A-modules.

10. To complete the classification of the subrings of Q, it suffices to find all of the sat-urated multiplicative subsets of Z. Show that there is a one-to-one correspondencebetween the saturated multiplicative subsets of Z and the subsets of the set of allprimes in Z. Here, if T is a set of primes in Z, then the associated multiplicativesubset is given by

S = {±pr11 . . . prk

k | k ≥ 0, p1, . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}.

11. Let A be an integral domain and let B be any commutative ring. Show that theideal p = (0×B) of A×B is prime and that the localization (A×B)p is isomorphicto the field of fractions of A.

12. Let p be a prime number and let M be a finite abelian group. Show that the local-ization M(p) is isomorphic to the subgroup Mp of M consisting of those elementswhose order is a power of p. (Hint : Consider the restriction of the natural mapη : M →M(p) to Mp.)

13. Let S be a multiplicative subset of Z and let M be a finite abelian group. Showthat S−1M is isomorphic to the subgroup of M consisting of those elements whoseorder is relatively prime to every element in S.

14. Let S be a multiplicative subset of A. Show that if S ⊂ A×, then the natural mapη : A→ S−1A is an isomorphism.

15. Let S be a multiplicative subset of A and let M be an A-module that admits acompatible S−1A-module structure. Show that the natural map η : M → S−1Mis an isomorphism.

16. Let F be a functor from A-modules to B-modules. Show that F is an exact functorif and only if for any short exact sequence

0→M ′ f−→Mg−→M ′′ → 0


of A-modules, the induced sequence

0→ F (M ′)F (f)−−−→ F (M)

F (g)−−−→ F (M ′′)→ 0

of B-modules is also exact.

Chapter 8

P.I.D.s and Field Extensions

The theories of fields and of P.I.D.s are inextricably intertwined. If K is a field, thenthe polynomial ring K[X] is a P.I.D. We shall show in Section 8.1 that the elements ofa P.I.D. have prime decompositions similar to the ones in the integers. This is used tostudy the extension fields L of K, via the evaluation maps εα : K[X] → L obtained byevaluating X at an element α of L.

Also, we give a classification of the finitely generated modules over a P.I.D. in Sec-tion 8.9. We shall use this to classify the n× n matrices over a field K up to similarityin Section 10.6. The point is that an n× n matrix induces a K[X]-module structure onKn in which X acts on Kn by multiplication by the matrix in question.

The theory of prime factorization in P.I.D.s also gives some elementary examples ofthe behavior of primes under finite extensions of the rational numbers, Q. The point isthat if K is a finite extension field of Q, then there is a subring O(K), called the ringof integers of K, that plays a role in K similar to the role of the integers in Q. Animportant topic in number theory is the way in which the prime numbers in Z factor inO(K). In the general case, O(K) is a Dedekind domain, but not a P.I.D. In that case,we study the factorizations of the principal ideal (p) of O(K) as a product of ideals.But in some nice cases, which we shall illustrate in the exercises to Section 8.1, the ringof integers is a P.I.D., and we may study this factorization in terms of factorizations ofprime elements.

The theory of prime factorization is developed in Section 8.1 in a generalization ofP.I.D.s called unique factorization domains, or U.F.D.s.

In Section 8.2, we study the algebraic extensions of a field K. These are the extensionfields with the property that the evaluation maps εα : K[X]→ L have nontrivial kernelsfor each α ∈ L. They will be the primary objects of study in Chapter 11. We begin toanalyze them here.

Section 8.3 studies extension fields which are not algebraic. There is a notion oftranscendence degree, which measures how far a given extension is from being algebraic.There is a theory of transcendence bases, which behave toward the transcendence degreein a manner analogous to the relationship between a basis for a vector space and itsdimension.

In Section 8.4, we construct an algebraic closure for a field K. Uniqueness statementsregarding algebraic closures are deferred to Section 11.1. We shall see that every algebraicextension of a field K may be embedded in its algebraic closure. So algebraic closureswill be an important tool in the material on Galois theory.

Algebraic closures are also important in matrix theory, as the Jordan canonical form

295

CHAPTER 8. P.I.D.S AND FIELD EXTENSIONS 296

may be defined for n×n matrices over an algebraically closed field. Thus, we may studymatrices over K by passage to the matrix ring of its algebraic closure.

The various applications of prime factorization in the P.I.D. K[X] depend stronglyon our ability to actually compute these factorizations. In particular, we need to be ableto test polynomials for primality. In Section 8.5, we give criteria for primality in Q[X],or, more generally, in K[X], where K is the field of fractions of a unique factorizationdomain. In the process, we show that if A is a U.F.D., then so is A[X]. In particular,the polynomial ring K[X1, . . . , Xn] is a U.F.D. for any field K.

In Section 9.3, we make use of unique factorization in K[X1, . . . , Xn], the study oftranscendence bases, and the theory of Noetherian rings to prove Hilbert’s Nullstellen-satz, which states that if the extension field L of K is finitely generated as a K-algebra,then L is a finite extension of K. In consequence, we determine all of the maximalideals of K[X1, . . . , Xn] when K is an algebraically closed field. This is the startingpoint for the study of algebraic varieties. We proceed to define algebraic varieties andto describe the Zariski topology on affine n-space, Kn, as well as on the prime spectrumof a commutative ring. This material forms an introduction to the study of algebraicgeometry.

The next three sections study field extensions, culminating in the determination ofthe degree of the cyclotomic extension Q[ζn] over Q, as well as the determination of theminimal polynomial of ζn over Q. This establishes the most basic facts about a familyof examples that has vital connections to a number of branches of mathematics. Forinstance, we shall see that the cyclotomic rationals play a vital role in computationsof Galois theory. And their rings of integers, Z[ζn], are of crucial importance in issuesrelated to Fermat’s Last Theorem, as well as to calculations of the K-theory of integralgroup rings. The latter plays an important role in topology.

Section 8.6 introduces the Frobenius homomorphism and discusses perfect fields. Sec-tion 8.7 introduces the notion of repeated roots, which forms the basis for the study ofseparable extensions. We shall return to these ideas in Chapter 11.

8.1 Euclidean Rings, P.I.D.s, and U.F.D.s

Recall that the Euclidean algorithm was the main tool we used to analyze the structure ofthe integers. There is an analogue of it which holds in a variety of other rings, includingthe polynomial ring K[X] over a field K. The consequences, for rings that satisfy thisproperty, are very similar to the consequences that hold in Z.

Definition 8.1.1. A Euclidean domain is an integral domain A that admits a functionϕ : A→ Z with the following properties.

1. If b divides1 a and a �= 0, then ϕ(b) ≤ ϕ(a).

2. For any a, b ∈ A with b �= 0, there are elements q, r ∈ A such that a = qb + r andϕ(r) < ϕ(b).

We call the function ϕ : A→ Z a Euclidean structure map for A.

1I.e., a = bc for some c ∈ A. Once again, this just says that a ∈ (b), the principal ideal generated byb.


Examples 8.1.2.1. Z is Euclidean, with the structure map given by the absolute value function | | :

Z→ Z.

2. Let K be a field, and define ϕ : K[X]→ Z by setting ϕ(f(X)) equal to the degreeof f(X) if f(X) �= 0, and setting ϕ(0) = −1. Then Proposition 7.3.10 shows thatϕ puts a Euclidean structure on K[X].

3. In Problem 5 of Exercises 7.1.41, it is shown that every nonzero element of thep-adic integers Zp may be written as a product pku with k ≥ 0 and u ∈ Z

×p . Define

ϕ : Zp → Z by setting ϕ(pku) = k and ϕ(0) = −1. Since p is not a unit in thedomain Zp, ϕ is a well defined function. Note that if a and b are nonzero elementsof Zp, then ϕ(b) ≤ ϕ(a) if and only if b divides a. It follows that ϕ is a Euclideanstructure map for Zp.

Recall from the discussion in Example 7.11.22 that Zp is also a local ring, meaningthat it has only one maximal ideal, in this case the principal ideal (p). Thus, Zp is anexample of the following notion.

Definition 8.1.3. A discrete valuation ring, or D.V.R., is a Euclidean domain that isalso a local ring.

The Euclidean structure on Z was crucial in our analysis of the subgroups of Z. Itwill have similar value for the study of the ideals in other Euclidean domains.

There is an additional property one could ask for from the function ϕ:

Definition 8.1.4. We say that a Euclidean structure map ϕ : A→ Z is positive definiteif ϕ(a) > 0 for all 0 �= a ∈ A and ϕ(0) = 0.

Lemma 8.1.5. Let A be a Euclidean domain. Then there is a Euclidean structure mapfor A which is positive definite.

Proof We claim that if ϕ : A→ Z is any Euclidean structure map for A, then ϕ(0) <ϕ(a) for all a �= 0. Given this, it will suffice to replace ϕ by a function ϕ′ given byϕ′(a) = ϕ(a)− ϕ(0).

To prove the claim, let a �= 0 and write 0 = qa + r with ϕ(r) < ϕ(a). But thenr = (−q)a is divisible by a, which would force ϕ(a) ≤ ϕ(r) unless r = 0. But then rmust equal 0, displaying ϕ(0) < ϕ(a) as desired.

Recall that a principal ideal domain, or P.I.D., is an integral domain in which everyideal is principal.

Proposition 8.1.6. A Euclidean domain is a principal ideal domain. Indeed, let ϕ :A→ Z be a positive definite Euclidean structure map for A. Let a be a nonzero ideal ofA and let a ∈ a be such that ϕ(a) is the smallest positive value taken on by the restrictionof ϕ to a. Then a = (a).

Proof Let 0 �= b ∈ a and write b = qa+ r with ϕ(r) < ϕ(a). But r = b− qa is in a, soby the minimality of ϕ(a), we must have ϕ(r) = 0. But then r = 0 and b ∈ (a).


Lemma 8.1.7. Let A be an integral domain and let a, b ∈ A. Then (a) = (b) if and onlyif a = ub for some unit u of A. Indeed, if a �= 0 and (a) = (b), and if a = bc, then cmust be a unit.

Thus, if b �= 0 and c is not a unit, then the inclusion

(bc) ⊂ (b)

is proper.

Proof The case a = 0 is immediate, so we assume a �= 0. Here, if (a) = (b), we canfind c, d such that a = bc and b = ad. Then a = adc. By the cancellation property in adomain, dc = 1.

Since the units in K[X] (K a field) are the nonzero elements of K, we obtain animmediate corollary.

Corollary 8.1.8. Let K be a field and let a be a nonzero ideal in K[X]. Let n be thesmallest non-negative integer such that a contains an element of degree n. Then there isa unique monic polynomial, f(X), in a of degree n, and a is the principal ideal generatedby f(X). Moreover, f(X) is the unique monic polynomial generating a.

Definition 8.1.9. Let B be an algebra over the field K and let b ∈ B. Let εb : K[X]→B be the K-algebra map obtained by evaluating X at b. If εb is injective, we say thatthe minimal polynomial of b over K is 0. Otherwise, the minimal polynomial of b overK is the monic polynomial of smallest degree in ker εb. We denote it by minb(X), if thecontext is obvious, or more formally by minb/K(X).

Corollary 8.1.10. Let B be an algebra over the field K and let b ∈ B. Then K[b] isisomorphic as a K-algebra to K[X]/(minb(X)). In particular, K[b] is finite dimensionalas a K-vector space if and only if minb(X) �= 0, and in this case, the dimension of K[b]over K is equal to the degree of minb(X).

Proof If minb(X) �= 0, then the determination of the dimension of K[b] follows fromProposition 7.7.35. Otherwise, K[b] is isomorphic to K[X], which is easily seen to be aninfinite dimensional vector space over K, with basis {Xi | i ≥ 0}.

It turns out that P.I.D.s have almost all the nice properties exhibited by the integers,including those shown in Sections 2.3 and 4.4. We shall treat the material from theformer in this section and the latter in Section 8.9.

Definitions 8.1.11. Let A be a commutative ring. Then p ∈ A is irreducible if p is nota unit, but whenever p = ab, then either a or b must be a unit.

We say that p ∈ A is a prime element if p �= 0 and the principal ideal (p) is a primeideal.

Note that if A is a domain, then (0) is a prime ideal, but 0 is neither a prime elementnor an irreducible element.

There is a basic ideal theoretic consequence of irreducibility.

Lemma 8.1.12. Let p be an irreducible element in the commutative ring A and supposethat a divides p. Then either a is a unit, or p divides a.

In ideal theoretic terms, if (p) ⊂ (a), then either (a) = (p) or (a) = A. Thus, theideal generated by an irreducible element is a maximal element in the partially orderedset of proper principal ideals, ordered by inclusion.


Proof If a divides p, then p = ab for some b ∈ A. Since p is irreducible, either a or b isa unit. If b is a unit, then a = pb−1, and hence p divides a.

In terms of ideals, (p) ⊂ (a) if and only if p ∈ (a), which is equivalent to a dividingp. If a is a unit, (a) = A. Otherwise, p divides a, and hence (a) ⊂ (p).

In a domain, Lemma 8.1.12 has a converse.

Lemma 8.1.13. Let A be an integral domain. Then a nonzero nonunit p ∈ A is irre-ducible if and only if the principal ideal (p) satisfies the property that (p) ⊂ (a) impliesthat either (a) = (p) or (a) = A.

Proof Suppose that (p) satisfies the stated condition and suppose that p = ab. Then(p) ⊂ (a). If (a) = A, then a is a unit. If (p) = (a), then Lemma 8.1.7 shows b to be aunit. Thus, p is irreducible.

If every ideal is principal, then Lemma 8.1.13 may be restated as follows.

Corollary 8.1.14. Let A be a P.I.D. Then a nonzero element p ∈ A is irreducible ifand only if the principal ideal (p) is a maximal ideal of A.

Thus, since maximal ideals are prime, every irreducible element in a P.I.D. is prime.But this does not hold in a general domain. Surprisingly, the reverse implication doeshold.

Lemma 8.1.15. Let p be a prime element in the integral domain A. Then p is irre-ducible.

Proof Suppose that p = ab. Then ab ∈ (p), a prime ideal, so either a or b must lie in(p). Say a ∈ (p). But p is also in (a), since p = ab, so (a) = (p), and hence b is a unit byLemma 8.1.7.

Since maximal ideals are prime, the next corollary is immediate from Lemma 8.1.15and Corollary 8.1.14.

Corollary 8.1.16. Let A be a P.I.D. Then an element p ∈ A is irreducible if and onlyif it is a prime element. In particular, if p ∈ A is irreducible and if p divides ab, then itmust divide either a or b.

Moreover, since every nonzero prime ideal of A is generated by a prime element, everynonzero prime ideal of A is maximal, and has the form (p), where p is an irreducibleelement of A.

Thus, the longest possible chain of proper inclusions of prime ideals in a P.I.D. hasthe form

0 ⊂ (p)

where p is a prime element of A. This may only occur if A is not a field.

Corollary 8.1.17. A P.I.D. that is not a field has Krull dimension 1.

The statement in the first paragraph of Corollary 8.1.16 has a more classical proof,which is modelled on the one that we used for the integers. We shall give it to illustratethat it really uses exactly the same ideas as the proof we just gave.


Definition 8.1.18. Let A be a P.I.D. Then the greatest common divisor of a and b isthe ideal (c) = (a) + (b). We write (a, b) = (c). If (a) + (b) = A, then we say that a andb are relatively prime, and write (a, b) = 1.

Note that the notation of (a, b) for the ideal (a) + (b) is consistent with the notationthat (a, b) is the ideal generated by a and b.

We will occasionally refer to c as the greatest common divisor of a and b when(a, b) = (c). But it is really more correct for g.c.d. to refer to the ideal (c), rather thanthe generator c.

Lemma 8.1.19. Let A be a P.I.D. and let (a, b) = (c). Then x divides both a and b ifand only if x divides c.

Proof Since a and b are both in (c), c divides a and b. Thus, it suffices to show thatan element dividing both a and b must divide c. But since (c) = (a) + (b), c = ra + sbfor some r, s ∈ A. Thus, any element that divides a and b will divide c.

The entire argument above reduces to the statement that (a) + (b) is the smallestideal containing both a and b, in the context that every ideal is principal.

Second Proof of Corollary 8.1.16 Let p be an irreducible element of A. We shallshow that (p) is a prime ideal.

Suppose that p divides ab but does not divide a. Since p is irreducible, Lemma 8.1.12shows that any divisor of p is either a unit or is divisible by p. Thus, a and p cannothave any common divisors other than units. So a and p are relatively prime, and hencethere are elements r, s ∈ A with ra+ sp = 1. But then rab+ spb = b. Since p divides ab,it divides the left-hand side, so it must divide b.

Note the role that the maximality of (p) plays in the above argument.Now we would like to show that every nonzero element of a P.I.D. may be factored

in some sense uniquely as a product of primes. This may be read as a statement thatevery P.I.D. is an example of a more general type of ring called a unique factorizationdomain, or U.F.D.

Definition 8.1.20. A unique factorization domain, or U.F.D., is an integral domain inwhich every nonzero nonunit may be written as a product of prime elements.

The next lemma has already been seen to hold in P.I.D.s.

Lemma 8.1.21. Irreducible elements in a U.F.D. are prime. Thus, in a U.F.D., theprime elements and the irreducible elements coincide.

Proof Let p be an irreducible element in the U.F.D. A. Then p is a product of primeelements, and hence is divisible by a prime element, q. Say p = qr. Since q is prime, itis not a unit. But p is irreducible, so r must be a unit. Thus, (p) = (q), a prime ideal,and hence p is prime.

We now wish to show that every P.I.D. is a U.F.D. Since we’ve shown that everyirreducible element in a P.I.D. is prime, it suffices to show that every nonzero nonunit isa product of irreducible elements.

If our P.I.D. is actually a Euclidean domain, this may be shown by a quick inductionon ϕ(a), much as the proof for the integers worked. For the general P.I.D., we need towork a little bit harder. The main tool comes from the Noetherian property.


Proposition 8.1.22. Every P.I.D. is a U.F.D.

Proof Let A be a P.I.D. Since every ideal is principal, the ideals are certainly finitelygenerated, and hence A is Noetherian by Corollary 7.8.4. Let 0 �= a ∈ A be a nonunit,and suppose by contradiction that a is not a product of primes. Since a is not a unit, (a)is proper, and hence (a) is contained in a maximal ideal (p1). Thus, a = p1a1 for somea1 ∈ A.

Now p1 is irreducible by Corollary 8.1.14, and hence prime, by Corollary 8.1.16. Sincep1 is prime and a = p1a1 is not a product of primes, a1 cannot be a product of primes.Also, the inclusion (a) ⊂ (a1) must be proper by Lemma 8.1.7, as p1 and a1 are nonzerononunits.

We may continue by induction, obtaining elements ak ∈ A for k ≥ 1, such thatak−1 = pkak with pk prime, and hence ak cannot be a product of primes. Thus, weobtain an infinite sequence of proper inclusions

(a) ⊂ (a1) ⊂ · · · ⊂ (ak) ⊂ · · ·contradicting the fact that A is Noetherian.

We shall show in Section 8.5 that if A is a U.F.D., so is A[X]. Taking A to be aP.I.D. that is not a field, this will give a family of examples of U.F.D.s that are notP.I.D.s.

We shall now discuss the uniqueness properties of the factorizations of the elementsin a U.F.D. as products of primes.

Definition 8.1.23. Let p and q be prime elements in the U.F.D. A. We say that p andq are equivalent if (p) = (q).

In other words, p and q are equivalent if p = uq for some unit u. In the integers, thisdoesn’t say much, since the only units are ±1. Thus, it is easy to choose a representativefor each equivalence class of primes in Z: Just pick the positive one. That’s what we didin Section 2.3.

In the generic P.I.D. there can be infinitely many units, so choosing a representativefor each equivalence class of primes could get cumbersome. For this reason, we shall givea uniqueness statement first that uses equivalence classes, and then follow it up withanother in which actual representatives are chosen.

We say that a collection p1, . . . , pk of primes is pairwise inequivalent if pi and pj areinequivalent for i �= j.

Proposition 8.1.24. Let A be a U.F.D., and suppose given an equality

upr11 . . . prk

k = vqs11 . . . qsl

l

where u and v are units, p1, . . . , pk are pairwise inequivalent primes, q1, . . . , ql are pair-wise inequivalent primes, and the exponents ri and si are all positive. Then k = l, andafter reordering (and relabelling) the q1, . . . , ql if necessary, pi is equivalent to qi andri = si for 1 ≤ i ≤ k.Proof We argue by induction on r = r1 + · · · + rk. Here, we take the empty productto be 1, so if r = 0, then the left side is just u. But then vqs11 . . . qsl

l is a unit, and henceqs11 . . . qsl

l must be the empty product, too.If r ≥ 1, then p1 divides the left-hand side and hence divides the right as well. In

particular, it must divide qi for some i, since the ideal (p1) is prime. But since p1 is nota unit and qi is irreducible, p1 and qi must be equivalent, say qi = wp1, with w a unit.


Reorder the q’s so that i = 1. Then

p1(upr1−11 . . . prk

k − vwqs1−11 . . . qsl

l ) = 0.

Since A is an integral domain, we have

upr1−11 . . . prk

k = vwqs1−11 . . . qsl

l ,

and the result follows by induction.

By a choice of representatives for the primes in A, we mean a collection {pi | i ∈ I}of primes in A such that for each prime p of A there is a unique i ∈ I such that p isequivalent to pi.

Thus, for example, the positive primes form a choice of representatives for the primesin Z.

Corollary 8.1.25. Let A be a U.F.D. and let {pi | i ∈ I} be a choice of representativesfor the primes in A. Then any nonzero nonunit in A may be written uniquely in the formupr1i1 . . . p

rkik

, where k > 0, u is a unit, i1, . . . , ik are distinct elements of I, and ri > 0 for1 ≤ i ≤ k.Proof Every element may be written in the above form because equivalent primes differby multiplication by a unit. Thus, the u term can absorb the differences between a givenproduct of primes and a product of pi’s. The uniqueness statement now follows from thatin Proposition 8.1.24, since pi is not equivalent to pj unless i = j. Thus, if two elementsof the form upr1i1 . . . p

rkik

are equal, the primes in them, together with their exponents,must be identical. Thus, we’re reduced to the case where upr1i1 . . . p

rkik

= vpr1i1 . . . prkik

, withu and v units. But since we’re in an integral domain, we can cancel off the primes in theabove equation, leaving u = v.

Corollary 8.1.26. Let A be a U.F.D. and let a ∈ A. Suppose that a = upr11 . . . prk

k ,where u is a unit, p1, . . . , pk are pairwise inequivalent primes, and ri ≥ 0 for i = 1, . . . , k.Then an element b divides a if and only if b = vps11 . . . psk

k , where v is a unit and 0 ≤si ≤ ri for 1 ≤ i ≤ k.Proof If b divides a, then any prime that divides b will also divide a. If a = bc,then c also divides a, and the same reasoning applies. Thus, b = vps11 . . . psk

k , andc = wpt11 . . . ptkk , where v and w are units and the exponents are all non-negative.

But then a = bc = vwps1+t11 . . . psk+tkk . By the uniqueness of factorization, si+ti = ri

for i = 1, . . . , k, and hence si ≤ ri as claimed.Conversely, if si ≤ ri for i = 1, . . . , k and if v is a unit, then vps11 . . . psk

k is easily seento divide a.

We’d like to be able to talk about greatest common divisors in a U.F.D., but thedefinition that we used in P.I.D.s fails for U.F.D.s that are not P.I.D.s. The next corollarywill help us.

Corollary 8.1.27. Let A be a U.F.D. and let 0 �= a, b ∈ A. Let p1, . . . , pk be a set ofrepresentatives for the set of primes dividing either a or b. Thus, p1, . . . , pk are pairwiseinequivalent, and we may write a = upr11 . . . prk

k and b = vps11 . . . psk

k , where u and v areunits, and the exponents are non-negative.

Let ti = min(ri, si) for 1 ≤ i ≤ k and let c = pt11 . . . ptkk . (Note that c = 1 if no primeelement divides both a and b.) Then c divides both a and b. In addition, if d is anyelement that divides both a and b, then d divides c.


In the language of ideals, this says that (a) + (b) ⊂ (c), and if d is any other elementsuch that (a) + (b) ⊂ (d), then (c) ⊂ (d). In particular, (c) is the smallest principal idealthat contains a and b.

Proof This is almost immediate from Corollary 8.1.26: If d divides both a and b, thend = wpl11 . . . p

lkk , where, if 1 ≤ i ≤ k, then 0 ≤ li ≤ ri, because d divides a, and 0 ≤ li ≤ si,

because d divides b. But then li ≤ ti for all i, so d divides c. Clearly, c divides a and b,so the result follows.

Definitions 8.1.28. Let A be a U.F.D. and let a, b ∈ A. Then the greatest commondivisor of a and b is the smallest principal ideal that contains (a) + (b). (Such an idealexists by Corollary 8.1.27.) If the greatest common divisor of a and b is c, we write

gcd(a, b) = (c),

or sometimes, by abuse of notation, gcd(a, b) = c.We say that a and b are relatively prime if their greatest common divisor is (1).

Notice that if A is a P.I.D., then these definitions agree with the old ones. Of course,Corollary 8.1.27 tells us exactly how to calculate greatest common divisors in a U.F.D.We state the result formally here:

Corollary 8.1.29. Let A be a U.F.D. and let 0 �= a, b ∈ A. Let p1, . . . , pk be a set ofrepresentatives for the set of primes dividing either a or b. Thus, p1, . . . , pk are pairwiseinequivalent, and we may write a = upr11 . . . prk

k and b = vps11 . . . psk

k , where u and v areunits, and the exponents are non-negative.

Let ti = min(ri, si) for 1 ≤ i ≤ k. Then the greatest common divisor of a and b is(c), where c = pt11 . . . ptkk . In particular, a and b are relatively prime if and only if noprime element of A divides both a and b.

More generally, we can talk about the greatest common divisor of a set of elementsin a U.F.D.

Definition 8.1.30. Let a1, . . . , an be elements of the U.F.D. A. Then the greatest com-mon divisor, gcd(a1, . . . , an) of these elements is the smallest principal ideal containing(a1) + · · ·+ (an). By abuse of notation, we also write gcd(a1, . . . , an) for any generatorof this ideal.

We say that a1, . . . , an are relatively prime if gcd(a1, . . . , an) = 1.

Notice the difference between saying that a1, . . . , an are relatively prime and sayingthat they are pairwise relatively prime. The reader should supply the proof of the nextcorollary.

Corollary 8.1.31. Let a1, . . . , an be nonzero elements of the U.F.D. A. Let p1, . . . , pkbe a set of representatives for the set of all primes that divide the elements a1, . . . , an,and write

ai = uipri11 . . . prik

k

for 1 ≤ i ≤ n, with ui a unit in A.Let tj = min(r1j , . . . , rnj) for 1 ≤ j ≤ k. Then

gcd(a1, . . . , an) =(pt11 . . . ptkk

).


Another construction which makes sense in a U.F.D. is the least common multiple.Once again we shall need a computation to set it up.

Proposition 8.1.32. Let A be a U.F.D. and let a1, . . . , an ∈ A. Define an elementa ∈ A as follows. If any of the ai = 0, then set a = 0. Otherwise, let p1, . . . , pk be a setof representatives for the primes that divide at least one of the ai, and let

ai = uipri11 . . . prik

k

for 1 ≤ i ≤ n, where ui is a unit and the exponents rij are all non-negative. Then definetj = max(r1j , . . . , rnj), and set

a = pt11 . . . ptkk .

Then the intersection of the principal ideals generated by the ai is the principal idealgenerated by a:

n⋂i=1

(ai) = (a).

Proof Here,⋂ni=1(ai) is the collection of elements divisible by each of the ai, so the

result follows immediately from Corollary 8.1.26.

Definition 8.1.33. Let A be a U.F.D. and let a1, . . . , an ∈ A. By the least commonmultiple of a1, . . . , an we mean the intersection,

⋂ni=1(ai), of the principal ideals they

generate, or, by abuse of notation, any generator of⋂ni=1(ai).

We also obtain a sharpening of the Chinese Remainder Theorem.

Corollary 8.1.34. (Chinese Remainder Theorem, second form) Let A be a P.I.D. andlet a1, . . . , ak ∈ A, such that ai and aj are relatively prime for i �= j. Let a = a1 . . . ak.Then there is an A-algebra isomorphism

f : A/(a)∼=−→ A/(a1)× · · · ×A/(ak)

given by f(x) = (x, . . . , x) for all x ∈ A.

Proof Since ai and aj are relatively prime for i �= j, (ai)+(aj) = A for i �= j. Thus, thefirst form of the Chinese Remainder Theorem (Proposition 7.2.23) applies, showing thatthe A-algebra homomorphism f : A→ A/(a1)× · · · ×A/(ak) given by f(a) = (a, . . . , a)is a surjection whose kernel is

⋂ki=1(ai). But

⋂ki=1(ai) = (a) by Proposition 8.1.32.

If a U.F.D. A is not a P.I.D., then its prime elements generate prime ideals which willnot, in general, be maximal ideals. In fact, they turn out to be minimal nonzero primeideals of A.

Definition 8.1.35. Let A be a domain. A minimal nonzero prime ideal of A is a nonzeroprime ideal with the property that the only prime ideal which is properly contained init is 0.

Proposition 8.1.36. Let A be a U.F.D. Then a prime ideal p of A is a minimal nonzeroprime ideal if and only if p = (p) for a prime element p of A.


Proof To show that every minimal nonzero prime ideal of A is principal, it suffices toshow that every nonzero prime ideal q contains a prime element. But if 0 �= a ∈ q, leta = p1 . . . pk, a product of (not necessarily distinct) prime elements of A. Because a ∈ qand q is prime, at least one of the pi must lie in q.

Conversely, let p be a prime element and suppose that q is a prime ideal that’s properlycontained in (p). Let K be the fraction field of A and let a = (1/p)q = {(1/p)a | a ∈ q} ⊂K. Since every element of q is divisible by p, a ⊂ A, and is easily seen to be an ideal ofA.

But now q = pa = (p)a. Since q is prime, either (p) ⊂ q or a ⊂ q by Lemma 7.6.9.Since q is properly contained in (p), we must have a ⊂ q.

But the fact that q = (p)a shows that q ⊂ a, so q = a. In other words, q = (p)q = pq.But by induction, we see that q = pnq for all n, and hence each a ∈ q is divisible by pn

for all n. By the uniqueness of factorization in A, no element but 0 can be divisible byall powers of a prime element, so q must be the 0 ideal. Thus, the only prime ideal thatcan be properly contained in a principal prime ideal is 0.

Exercises 8.1.37.1. Let A be a U.F.D. Suppose that a divides bc in A, where a and b are relatively

prime. Show that a divides c.

2. Let p1, . . . , pk be pairwise inequivalent primes in the U.F.D.A and let a = upr11 . . . prk

k

for some unit u. Show that a divides b if and only if prii divides b for 1 ≤ i ≤ k.

3. Let K be a field. Show that every degree 1 polynomial of K is irreducible in K[X].If f(X) has degree 1, what is K[X]/(f(X))?

4. Let K be a field and let f(X) ∈ K[X] be a polynomial whose degree is either 2 or3. Show that f(X) is irreducible if and only if it has no roots.

5. Let K be the field of fractions of the U.F.D. A. Let a/b ∈ K be a root of f(X) =amX

m + · · · + a0 ∈ A[X], where am �= 0 and where a, b ∈ A are relatively prime.Show that b must divide am and a must divide a0.

6. The simplest procedure for finding primes in Z is to proceed inductively, testingeach number by dividing it by all primes smaller than its square root. If it has noprime divisor less than or equal to its square root, it is prime. One could adopt asimilar strategy in F [X] for a finite field F . Here, a polynomial f(X) is irreducibleif it has no irreducible factors of degree ≤ 1

2 deg f . Carry out this procedure forF = Z2 to find all irreducible polynomials of degree ≤ 4.

7. Let A be a P.I.D. and let B be a subring of the field of fractions of A with A ⊂ B.Show that B = S−1A for a suitable multiplicative subset S of A.

8. Let A be a U.F.D. and let S be a saturated multiplicative subset of A. Let T bethe set of prime elements of A that lie in S. Show that

S = {upr11 . . . prk

k | k ≥ 0, u ∈ A×, p1, . . . , pk ∈ T and ri ≥ 0 for i = 1, . . . , k}.

9. Let A be a U.F.D. with only one prime element, p, up to equivalence of primes.Show that there is a Euclidean structure map ϕ : A → Z obtained by settingϕ(0) = −1 and setting ϕ(a) = k if pk is the highest power of p that divides thenonzero element a. Deduce that a ring is a discrete valuation ring if and only if itis a U.F.D. with at most one prime element up to equivalence of primes.


10. Let A be a U.F.D. and let S be a multiplicative subset of A. Show that S−1A is aU.F.D.

11. Let A be a U.F.D. and let p be a prime element of A. Show that the localizationA(p) of A at the prime ideal (p) is a discrete valuation ring.

12. Let A be a U.F.D. that has infinitely many equivalence classes of prime elements.Show that the field of fractions of A is infinitely generated as an A-algebra.

13. Show that if A is not a field, then A[X] is not a P.I.D.

14. Recall the ring of Gaussian integers, Z[i], introduced in Example 7.3.6: Z[i] is thesubring of C consisting of the elements a+ bi with a, b ∈ Z.

(a) Show that Z[i] is Euclidean, via the function ϕ(a+ bi) = a2 + b2.

(b) Show that complex conjugation restricts to an isomorphism of rings from Z[i]to itself. Show also that ϕ(z) = zz for z ∈ Z[i], where z is the complexconjugate of z. Deduce that ϕ(zw) = ϕ(z)ϕ(w) for z, w ∈ Z[i].

(c) Show that the only units in Z[i] are ±i and ±1.

(d) Show that an element z ∈ Z[i] is prime if and only if it has no divisors w with1 < ϕ(w) ≤√

ϕ(z).

(e) Show that 1 + i is a prime and that 1 + i = −i · (1 + i). Show also thatϕ(1 + i) = 2, so that

2 = (1 + i) · (1 + i) = −i · (1 + i)2.

(In the language of algebraic number theory, this implies that 2 ramifies inZ[i].) Deduce that 2 is not prime in Z[i].

(f) Show that if z = a+bi ∈ Z[i] with a, b ∈ Z is a prime element such that neithera nor b is zero and (z) �= (1 + i), then z is a prime that is not equivalent to z.Deduce that every nonzero element n ∈ Z has a prime decomposition in Z[i]of the form

n = u · (1 + i)2r · pr11 . . . prk

k · zs11 zs11 . . . zsl

l zsl

l ,

where u is a unit, the pi are primes in Z that remain prime in Z[i], and the ziare primes in Z[i] of the form a+ bi, where a and b are positive integers withb < a.

(g) Let z = a + bi where a and b are nonzero integers. Show that z is prime inZ[i] if and only if ϕ(z) is prime in Z.

(h) Let p be a prime in Z that is not prime in Z[i]. Show that p = ϕ(z) for somez ∈ Z[i]. Deduce that there are integers a and b with p = a2 + b2.

(i) Show that a prime in Z that is congruent to 3 mod 4 remains prime in Z[i].

(j) Suppose that a prime p in Z remains prime in Z[i]. Deduce that Z[i]/(p) isa field of order p2, which is additively a vector space over Zp of dimension2, with a basis given by the images of 1 and i under the canonical map fromZ[i]. Deduce that the only elements of order 4 in the unit group (Z[i]/(p))×

must lie outside the image of the unique ring homomorphism from Zp toZ[i]/(p). Deduce that Zp× has no elements of order 4, and hence that p mustbe congruent to 3 mod 4.


(k) Let p be a prime in Z that is congruent to 1 mod 4. Deduce from the resultsabove that p = zz for some prime z of Z[i]. Show that the field Z[i]/(z)is of characteristic p, and is generated additively by the images of 1 and iunder the canonical map from Z[i]. Deduce from the fact that Zp has unitsof order 4 that the image of i under the canonical map must lie in the imageof the unique ring homomorphism from Zp to Z[i]/(z). Deduce that Z[i]/(z)is isomorphic to Zp.

15. Characterize the collection of integers that may be written as the sum of twosquares.

16. We study the subring Z[ζ3] of C.

(a) Show that the kernel of the evaluation map εζ3 : Z[X]→ Z[ζ3] is the principalideal of Z[X] generated by X2 + X + 1. Deduce from Proposition 7.7.35that the elements of Z[ζ3] may be written uniquely in the form m+ nζ3 withm,n ∈ Z. Write down the multiplication formula for the product of two suchelements.

(b) Using the fact that ζ3 = ζ−13 = ζ2

3 , where ζ3 is the complex conjugate of ζ3,show that setting ϕ(α) = αα defines a Euclidean structure map ϕ : Z[ζ3]→ Z.

(c) Show that Z[ζ3]× = {±1,±ζ3,±ζ23}, where ζ2

3 = −1−ζ3 and−ζ23 = ζ6 = 1+ζ3.

(d) Show that 3 ramifies in Z[ζ3] in the sense that 3 = up2, where u ∈ Z[ζ3]× andp is a prime of Z[ζ3].

(e) Show that an element of the form m+ nζ3, where m and n are both nonzeroelements of Z, is prime in Z[ζ3] if and only if its image under ϕ is a primein Z.

(f) Show that a prime in Z remains prime in Z[ζ3] if and only if it is not in theimage of ϕ : Z[ζ3]→ Z.

(g) Show that a prime p �= 3 in Z remains prime in Z[ζ3] if and only if p is notcongruent to 1 mod 3.


17. Show that Z[i√

2] is a Euclidean domain via the square of the complex norm, asabove. What are the units of Z[i

√2]? Find congruences that will guarantee that a

prime in Z remains prime in Z[i√

2].

18. Show that Z[√

2] admits a Euclidean structure map.

8.2 Algebraic Extensions

Definitions 8.2.1. Let K be a field. By an extension of K, or, more formally, anextension field of K, we mean a field L that contains K as a subfield.

We say that L is a finite extension of K if it is finite dimensional as a vector spaceover K.

If L is a finite extension of K, then the degree of L over K, written [L : K], isthe dimension of L as a K-vector space. If L is an infinite extension of K, we write[L : K] =∞.

Throughout this section, K is a field.

Lemma 8.2.2. Let L be an extension of K and let L′ be an extension of L. Then L′ isfinite over K if and only if both L′ is finite over L and L is finite over K, in which case

[L′ : K] = [L′ : L] · [L : K].

Proof If L′ is finite over K, then the other two extensions are clearly finite as well.Conversely, if y1, . . . , yn is a basis for L′ as a vector space over L, and x1, . . . , xm is abasis for L as a vector space over K, then any ordering of {xiyj | 1 ≤ i ≤ m, 1 ≤ j ≤ n}will give a basis for L′ over K.

We now consider the elements of an extension field L of K. For each α ∈ L, there is aunique K-algebra homomorphism εα : K[X]→ L that takes X to α. Here, εα evaluatespolynomials at α:

εα

(n∑i=0

aiXi

)=

n∑i=0

aiαi.

Thus, f(X) ∈ ker εα if and only if α is a root of f . In particular, ker εα �= 0 if and onlyif α is a root of some nonzero polynomial over K.

Recall from Corollary 8.1.10 that the kernel of εα is the principal ideal generated byminα(X), the minimal polynomial of α. Here, minα(X) = 0 if εα is injective, and is themonic polynomial of lowest degree in ker εα otherwise. In particular, minα(X) dividesany polynomial that has α as a root.

The image of εα, of course, is K[α], the K-subalgebra of L generated by α.

Definition 8.2.3. Let L be an extension field of K and let α ∈ L. We say that α isalgebraic over K if α is a root of a nonzero polynomial over K, and hence minα(X) �= 0.

If α is not a root for any nonzero polynomial over K, we say that α is transcendentalover K.

Examples 8.2.4.1. Let ζn = ei·

2πn ∈ C be the standard primitive n-th root of unity in C. Since ζn has

order n in C×, it is a root of the polynomial Xn−1 ∈ Z[X], and hence is algebraicover Q.


2. For positive integers n and a, the real number n√a is a root of the integer polynomial

Xn − a, and hence is algebraic over Q.

3. The real numbers π and e are known to be transcendental over Q. The standardproofs of this use analytic number theory.

Let α be an element of the extension field L of K. Then K[X]/(minα(X)) ∼= K[α],which, as a subring of L, is an integral domain. Thus, (minα(X)) is a prime ideal inK[X]. But by Corollary 8.1.16, the nonzero prime ideals of the P.I.D. K[X] are allmaximal, generated by irreducible elements of K[X]. Thus, if α is algebraic over K,then K[α] ∼= K[X]/(minα(X)) is a field. The following proposition is now immediatefrom Corollary 8.1.10.

Proposition 8.2.5. Let L be an extension field of K and let α ∈ L be algebraic over K.Then K[α] is a finite extension field whose degree over K is equal to the degree of theminimal polynomial of α over K:

[K[α] : K] = deg minα(X).

Recall that K[α] is the smallest K-subalgebra of L that contains α.

Definition 8.2.6. Let L be an extension field of K and let α ∈ L. We write K(α) forthe smallest subfield of L that contains both K and α.

More generally, if {αi | i ∈ I} is any family of elements in L, we write K(αi | i ∈ I) forthe smallest subfield of L containing both K and {αi | i ∈ I}. For a finite set α1, . . . , αnof elements of L, the notation for this subfield, of course, becomes K(α1, . . . , αn).

In particular, Proposition 8.2.5, when α is algebraic over K, or Corollary 7.11.11,when α is transcendental, now gives the following corollary.

Corollary 8.2.7. Let L be an extension field of K and let α ∈ L.If α is algebraic over K, then K(α) = K[α].If α is transcendental over K, then K(α) is isomorphic to the field of fractions of

K[α], and consists of the elements of the form f(α)/g(α), where f(X), g(X) ∈ K[X],with g(α) �= 0. Since α is transcendental over K, this just says that g(X) �= 0, and henceK(α) is isomorphic to K(X), the field of rational functions over K.

The extensions K(α) with α algebraic over K are of sufficient interest to warrant aname.

Definitions 8.2.8. An extension field L of K is simple if L = K(α) for some α ∈ Lthat is algebraic over K.

An element α that is algebraic over K and satisfies L = K(α) is called a primitiveelement for L over K.

An alternative name for a simple extension is a primitive extension.We shall see in Chapter 11, using the Fundamental Theorem of Galois Theory, that

if K has characteristic 0, then every finite extension of K is simple. For fields of nonzerocharacteristic, we shall see that every finite separable extension is simple.

The degree of an extension is a useful number to calculate. Thus, to study a simpleextension K(α) of K, it is useful to be able to determine the minimal polynomial of α


over K. In practice, what we’re often given is some polynomial f(X) ∈ K[X] of whichα is a root (e.g., ζn is a root of Xn − 1, but we don’t yet know the minimal polynomialof ζn over Q, nor, indeed, the degree [Q(ζn) : Q]). All we know is that minα(X) is oneof the prime divisors of f(X).

Thus, it is quite valuable to be able to compute prime decompositions in K[X],and, in particular, to test the elements of K[X] to see if they are irreducible. Tests ofirreducibility will be a topic in Section 8.5.

Even in the case of a simple extension, it is sometimes difficult to find a primitiveelement. Thus, it is valuable to be able to handle extensions that are generated asalgebras by more than one element. Recall that if B is a commutative K-algebra and ifb1, . . . , bn ∈ B, then we writeK[b1, . . . , bn] for the image of theK-algebra homomorphism

εb1,...,bn : K[X1, . . . , Xn]→ B

obtained by evaluating Xi at bi for i = 1, . . . , n.It is easy to see that K[b1, . . . , bn] is the smallest K-subalgebra of B containing

b1, . . . , bn. The next lemma gives a partial generalization of Proposition 8.2.5.

Lemma 8.2.9. Let L be an extension field of K and let α1, . . . , αn ∈ L be algebraic overK. Then K[α1, . . . , αn] is a field. Indeed, K[α1, . . . , αn] is a finite extension of K.

Proof We argue by induction on n, with the case n = 1 being given by Proposition 8.2.5.Suppose then by induction that K[α1, . . . , αn−1] is a finite extension field of K.

Since K[α1, . . . , αn] is the smallest K-subalgebra of L that contains the elementsα1, . . . , αn, it must be the case that

K[α1, . . . , αn] = K[α1, . . . , αn−1][αn].

Now αn is algebraic over K, and hence is a root of a nonzero polynomial f(X) ∈ K[X].But f may be regarded as a polynomial over K[α1, . . . , αn−1], so αn is algebraic overK[α1, . . . , αn−1] as well. But Proposition 8.2.5 now shows that K[α1, . . . , αn] is a finiteextension of K[α1, . . . , αn−1], and hence also of K by Lemma 8.2.2.

Corollary 8.2.10. Let L be an extension of K and let α1, . . . , αn ∈ L be algebraic overK. Then

K(α1, . . . , αn) = K[α1, . . . , αn]

is a finite extension of K.

We would like to study the situation of adjoining infinitely many algebraic elementsto a field K.

Definition 8.2.11. Let B be a commutative A-algebra and let {bi | i ∈ I} be a familyof elements of B. We write A[bi | i ∈ I] for the set of all elements in B lying in one ofthe subalgebras A[bi1 , . . . , bik ] for a finite subset {i1, . . . , ik} ⊂ I.

Clearly, A[bi | i ∈ I] is the smallest A-subalgebra of B that contains {bi | i ∈ I}.Proposition 8.2.12. Let L be an extension field of K and let {αi | i ∈ I} be elementsof L that are algebraic over K. Then K[αi | i ∈ I] is a field, and hence

K(αi | i ∈ I) = K[αi | i ∈ I]is the smallest K-subalgebra of L containing {αi | i ∈ I}.


Proof Every nonzero element β ∈ K[αi | i ∈ I] lies in an extension K[αi1 , . . . , αik ],which is a field, since the α’s are algebraic over K. So β is a unit in K[αi | i ∈ I].

We shall be especially interested in extension fields that are algebraic over K:

Definitions 8.2.13. An extension L of K is algebraic if every element of L is algebraicover K.

An extension that is not algebraic is said to be transcendental.

The next proposition characterizes the algebraic elements in an extension field L.

Proposition 8.2.14. Let L be an extension field of K and let α ∈ L. Then the followingconditions are equivalent.

1. α is algebraic over K.

2. K[α] is a finite extension of K.

3. There is a field L′ ⊂ L, containing both K and α, such that L′ is a finite extensionof K.

Proof The first condition implies the second by Proposition 8.2.5, while the secondimplies the first by Corollary 8.1.10. The second implies the third by taking L′ = K[α],so it suffices to show that the third implies the second.

But if L′ is a finite extension of K containing α, then K[α] ⊂ L′, and hence K[α] isalso finite over K.

Corollary 8.2.15. An extension L of K is algebraic if and only if every element of Lis contained in a finite extension of K. In particular, every finite extension of K isalgebraic.

Corollary 8.2.16. Let L be an extension field of K. Then L is a finite extension of Kif and only if L = K(α1, . . . , αn), where α1, . . . , αn are algebraic over K.

Proof If L = K(α1, . . . , αn), where α1, . . . , αn are algebraic over K, then L is finiteover K by Corollary 8.2.10.

Conversely, if L is finite over K, then every element of L is algebraic over K byProposition 8.2.14. If α1, . . . , αn is a basis for L over K, then clearly, L = K[α1, . . . , αn],so the result follows from Corollary 8.2.10.

Thus, if α1, . . . , αn are algebraic over K, then K(α1, . . . , αn) is an algebraic extensionof K. Indeed, we may extend this to infinitely generated extensions.

Corollary 8.2.17. An extension L of K is algebraic if and only if L = K(αi | i ∈ I) forsome collection of elements {αi | i ∈ I} that are algebraic over K.

Proof Let L = K(αi | i ∈ I), where each αi is algebraic over K. Then L = K[αi | i ∈ I]by Proposition 8.2.12. In particular, every element of L lies in a finitely generatedK-subalgebra of the form K[αi1 , . . . , αik ]. Since the αi are algebraic over K, thesesubalgebras are finite extension fields of K, and hence L is algebraic over K by Propo-sition 8.2.14.

Conversely, if L is algebraic over K and if {αi | i ∈ I} is any basis for L as a vectorspace over K, then L = K(αi | i ∈ I).


We can also use Proposition 8.2.14 to obtain an important closure property for alge-braic extensions.

Corollary 8.2.18. Let L be an algebraic extension of K and let L1 be any extension ofL. Suppose that α ∈ L1 is algebraic over L. Then α is algebraic over K as well.

In particular, if L is an algebraic extension of K, then any algebraic extension of Lis an algebraic extension of K.

Proof Let α ∈ L1 be algebraic over L. Then α is a root of a nonzero polynomial f(X)over L. Say f(X) = αnX

n + · · ·+ α0, with αi ∈ L for 0 ≤ i ≤ n.Since L is algebraic over K, so is αi for 0 ≤ i ≤ n, and hence K(α0, . . . , αn) is a finite

extension of K by Corollary 8.2.10.But α is algebraic over K(α0, . . . , αn). Thus, K(α0, . . . , αn)(α) is a finite extension

of K(α0, . . . , αn), and hence of K as well. So α is contained in a finite extension of K,and hence is algebraic over K by Proposition 8.2.14.

We shall be primarily interested in finite extensions, but some results regarding alge-braic closures will require information about infinite algebraic extensions. The next resultis trivial for finite extensions and false for transcendental extensions (see Corollary 8.6.7).Thus, it shows the strength of the condition that an extension is algebraic.

Proposition 8.2.19. Let L be an algebraic extension of K and let ν : L→ L be a homo-morphism of fields that restricts to the identity map on K. Then ν is an automorphismof L.

Proof Every homomorphism of fields is injective. So it suffices to show ν is onto. If Lis a finite extension, this is immediate from a dimension argument. Otherwise, let α ∈ L.

Let f(X) ∈ K[X] be the minimal polynomial of α over K, and let α1, . . . , αn be theset of all roots of f in L, with α = α1. Then K(α1, . . . , αn) is a finite extension of K,and it suffices to show that ν carries K(α1, . . . , αn) into itself.

Since K(α1, . . . , αn) is the smallest K-subalgebra of L containing α1, . . . , αn, it suf-fices to show that ν carries each αi into {α1, . . . , αn}.

Write f(X) = amXm + · · ·+ a0, with a0, . . . , am ∈ K. Then f(αi) = 0, so

0 = ν(f(αi)) = am(ν(αi))m + · · ·+ a0 = f(ν(αi)),

since ν acts as the identity on K. In particular, ν(αi) is a root of f in L. And the set ofall such roots is precisely {α1, . . . , αn}.

We shall next show that in any extension L of K, there is a largest subfield K1 of Lsuch that K1 is an algebraic extension of K.

Definition 8.2.20. Let L be an extension of K. Then the algebraic closure of K in Lis the set of all elements of L that are algebraic over K.

Proposition 8.2.21. Let L be an extension of K. Then the algebraic closure of K inL is a subfield of L containing K. In particular, it is an algebraic extension of K.

Proof If α is algebraic over K, then K(α) is a finite extension of K, and hence everyelement of K(α) must be algebraic over K by Proposition 8.2.14. In particular, −α andα−1 are algebraic over K.

If α and β are both algebraic over K, then β is a root of a nonzero polynomial overK, and hence also over K(α). Thus, β is contained in a finite extension, L′, of K(α),


which is then finite over K by Lemma 8.2.2. But α+β and αβ lie in L′, and hence α+βand αβ are algebraic over K.

Thus, the elements of L that are algebraic over K form a subfield of L. This subfieldincludes K, as each a ∈ K is a root of X − a.

Definition 8.2.22. Let L be an extension of K. We say that K is algebraically closedin L if the only elements of L that are algebraic over K are those of K itself.

Proposition 8.2.23. Let L be an extension of K and let K1 be the algebraic closure ofK in L. Then K1 is algebraically closed in L.

Proof Let α ∈ L be algebraic over K1. Then α is algebraic over K by Corollary 8.2.18,and hence an element of K1, as K1 is the algebraic closure of K in L.

Exercises 8.2.24.1. Let α ∈ C with α �∈ R. Show that both α+ α and α · α are real. Deduce that

minα/R(X) = (X − α)(X − α) = X2 − (α+ α)X + αα.

2. Find the minimal polynomial of i over Q.

3. Find the minimal polynomial of ζ3 over Q.

4. Show that minζ8/Q(X) = X4 + 1. (Hint : Show that Q(ζ8) is a degree 2 extensionof Q(i).)

5. Let n be any integer that is not a perfect cube (i.e., not the cube of an integer).Show that Q( 3

√n) is a degree 3 extension of Q.

6. Let L be an extension of K of degree p, with p prime. Show that every element αof L that is not in K is a primitive element for L over K (i.e., that L = K(α)).

7. Let K be a field of characteristic �= 2 and let L be a quadratic extension of K (i.e.,[L : K] = 2). Show that there is a primitive element, α, for L over K such that theminimal polynomial of α over K has the form X2 − d with d ∈ K. Thus, we maywrite L = K(

√d). (Hint : Suppose that L has a primitive element whose minimal

polynomial is aX2 + bX+ c. Show that d = b2−4ac has a square root, α, in L andthat L = K(α).)

8. Let K be the field of fractions of a domain A of characteristic �= 2, and let L bea quadratic extension of K. Show that L = K(

√d) for some d ∈ A. (Hint : If

L = K(√a/b), with a, b ∈ A, take d = ab.)

If A is a U.F.D., show that we may additionally assume that the element d ∈ A issquare-free, in the sense that it is not divisible by the square of any prime elementof A.

9. Let K be any field and let K(X) be the field of fractions of K[X]. Show that K isalgebraically closed in K(X).

10. Deduce from the preceding problem that if K is finite, then the only finite subfieldsof K(X) are those contained in K itself.


8.3 Transcendence Degree

Some extensions are more transcendental than others. We shall discuss this measurementhere. There’s a notion of transcendence basis analogous to that of a basis for a vectorspace. We obtain a notion of finite transcendence degree, measured by the number ofelements in a transcendence basis.

Definitions 8.3.1. Let A be a subring of the commutative ring B. We say that theelements b1, . . . , bn of B are algebraically independent over A if the evaluation mapεb1,...,bn : A[X1, . . . , Xn]→ B that evaluates each Xi at bi is injective.

Let L be an extension field of K We say that α1, . . . , αn is a transcendence basis forL if they are algebraically independent and L is algebraic over K(α1, . . . , αn). (Infinitetranscendence bases may be defined analogously.)

We say that an extension L of K has finite transcendence degree over K if there existsa finite transcendence basis for L over K.

Examples 8.3.2.1. A single element α ∈ L is algebraically independent if and only if it is transcendental

over K.

2. The null set forms a transcendence basis for an algebraic extension of K. In par-ticular, algebraic extensions have finite transcendence degree.

3. Let α = f(X)/g(X) be any element of K(X) that’s not in K. Then α is a tran-scendence basis for K(X) over K. The point is that X is algebraic over K(α), asit is a root of the polynomial αg(T )− f(T ) in K(α)[T ]. Thus, α is transcendentalover K, as otherwise, X would be algebraic over K.

4. Of course, X1, . . . , Xn is a transcendence basis for the ring of rational functionsK(X1, . . . , Xn).

Recall the isomorphism A[X1, . . . , Xn] ∼= A[X1, . . . , Xn−1][Xn]. Under this isomor-phism, we may write any polynomial f(X1, . . . , Xn) uniquely as a polynomial in Xn withcoefficients in A[X1, . . . , Xn−1]:

f(X1, . . . , Xn) = fk(X1, . . . , Xn−1)Xkn + · · ·+ f0(X1, . . . , Xn−1).

Definitions 8.3.3. Suppose given a polynomial


in A[X1, . . . , Xn]. We say that f has degree k in Xn if fk(X1, . . . , Xn−1) �= 0 in theequation above.

Let A be a subring of the commutative ring B and b1, . . . , bn ∈ B. Suppose given apolynomial


in A[X1, . . . , Xn]. We say that f has degree m in bn when evaluated at b1, . . . , bn if thepolynomial

fk(b1, . . . , bn−1)Xkn + · · ·+ f0(b1, . . . , bn−1)

has degree m as a polynomial in Xn over A[b1, . . . , bn−1]. Here, by abuse of notation, weshall say that f(b1, . . . , bn) has degree m in bn.


Lemma 8.3.4. Let L be an extension of K and let α1, . . . , αn ∈ L be algebraicallyindependent over K. Then an element β ∈ L is algebraic over K(α1, . . . , αn) if and onlyif α1, . . . , αn, β are algebraically dependent over K.

Proof Suppose that β is algebraic over K(α1, . . . , αn), so it satisfies an equation of theform

fm(α1, . . . , αn)gm(α1, . . . , αn)

βm + · · ·+ f0(α1, . . . , αn)g0(α1, . . . , αn)

= 0,

with positive degree in β. Now, multiplying both sides by the product

g0(α1, . . . , αn) . . . gm(α1, . . . , αn),

we obtain an algebraic dependence relation between α1, . . . , αn, β.Conversely, if α1, . . . , αn, β are algebraically dependent, let f(X1, . . . , Xn+1) be a

nonzero polynomial in A[X1, . . . , Xn+1] that vanishes when evaluated at α1, . . . , αn, β.Write

f(X1, . . . , Xn+1) = fk(X1, . . . , Xn)Xkn+1 + · · ·+ f0(X1, . . . , Xn).

Since α1, . . . , αn are algebraically independent, fi(α1, . . . , αn) is nonzero whenever fi isa nonzero polynomial. Thus, f has positive degree in β, and hence β is algebraic overK(α1, . . . , αn).

A new characterization of transcendence bases is now immediate.

Corollary 8.3.5. Let L be an extension of K and let α1, . . . , αn ∈ L be algebraicallyindependent over K. Then α1, . . . , αn is a transcendence basis for L over K if and onlyif there is no element β in L such that α1, . . . , αn, β are algebraically independent overK.

The theory of transcendence bases now proceeds much like the theory of bases inordinary linear algebra.

Proposition 8.3.6. Let L be an extension of K and let α1, . . . , αn in L such that Lis algebraic over K(α1, . . . , αn). Suppose in addition that α1, . . . , αm are algebraicallyindependent for some m < n. Then we can find i1, . . . , ik ∈ {m + 1, . . . , n} such thatα1, . . . , αm, αi1 , . . . , αik is a transcendence basis for L over K.

Proof By induction on n−m, we may choose i1, . . . , ik so that α1, . . . , αm, αi1 , . . . , αikis algebraically independent, but is not properly contained in any other algebraicallyindependent subset of α1, . . . , αn. For simplicity, we reorder the α’s if necessary so thatα1, . . . , αm+k is our maximal algebraically independent subset containing α1, . . . , αm.

Now, for i > m+k, α1, . . . , αm+k, αi is algebraically dependent, so Lemma 8.3.4 showsαi to be algebraic over K(α1, . . . , αm+k). Corollary 8.2.10 now shows that K(α1, . . . , αn)is algebraic over K(α1, . . . , αm+k), and hence L is also.

Recall that the total degree of a monomial aXi11 . . . Xin

n is i1 + · · · + in and thatthe total degree of a polynomial f(X1, . . . , Xn) is the largest of the total degrees of thenonzero monomials in it.

Definition 8.3.7. Let A be a subring of a commutative ring B and let b1, . . . , bn ∈B. We say that a nonzero polynomial f(X1, . . . , Xn) ∈ A[X1, . . . , Xn] gives a minimalalgebraic dependence relation between b1, . . . , bn if f(b1, . . . , bn) = 0 and if no nonzeropolynomial whose total degree is less than that of f vanishes when evaluated on b1, . . . , bn.


Lemma 8.3.8. Let A be a subring of a commutative ring B and let b1, . . . , bn ∈ B. Sup-pose that f(X1, . . . , Xn) ∈ A[X1, . . . , Xn] gives a minimal algebraic dependence relationbetween b1, . . . , bn. Suppose also that f has degree k > 0 in Xn. Then f(b1, . . . , bn) hasdegree k in bn.

Proof Suppose that

f(X1, . . . , Xn) = fk(X1, . . . , Xn−1)Xkn + · · ·+ f0(X1, . . . , Xn−1).

Then fk(X1, . . . , Xn−1) is a nonzero polynomial whose total degree is k less than that off . If fk(b1, . . . , bn−1) = 0, then fk may be regarded as an algebraic dependence relationfor b1, . . . , bn, contradicting the minimality of f . Thus, fk(b1, . . . , bn−1) �= 0.

Proposition 8.3.9. Let α1, . . . , αn be a transcendence basis for L over K and let β1, . . . , βk ∈L be algebraically independent over K. Then k ≤ n, and if k = n, then β1, . . . , βn is atranscendence basis for L over K.

Proof Suppose that k ≥ n. We shall show by induction on i that we may reorder theα’s in such a way that L is algebraic over K(β1, . . . , βi, α1, . . . , αn−i) for each i ≤ n.When i = n, this says that L is algebraic over K(β1, . . . , βn), so that β1, . . . , βn is atranscendence basis for L over K. But if k > n, Lemma 8.3.5 then shows that β1, . . . , βkcannot be algebraically independent over K, so the result will follow from our inductionon i.

Suppose, inductively, that L is algebraic over K(β1, . . . , βi−1, α1, . . . , αn−i+1). ThenProposition 8.3.6 shows that we may reorder the α’s if necessary so that β1, . . . , βi−1, α1, . . . , αris a transcendence basis for L over K for some r ≤ n− i+ 1.

Thus, Lemma 8.3.4 shows that β1, . . . , βi, α1, . . . , αr are algebraically dependent. Letf(X1, . . . , Xr+i) be a minimal algebraic dependence relation between β1, . . . , βi, α1, . . . , αr.By Lemma 8.3.8, f must have positive degree in at least one of the α’s, as otherwise, fonly involves the variables X1, . . . , Xi, and hence gives an algebraic dependence relationfor the β’s.

Renumbering the α’s, if necessary, we may assume that f has positive degree inαr. But then αr is algebraic over K(β1, . . . , βi, α1, . . . , αr−1). Since L is algebraic overK(β1, . . . , βi, α1, . . . , αr), we have established the inductive step.

Corollary 8.3.10. Let L be an extension of finite transcendence degree over K. Thenany two transcendence bases of L over K have the same number of elements.

Thus, we may define transcendence degree.

Definition 8.3.11. Let L be an extension of finite transcendence degree over K. Thenthe transcendence degree of L over K is the number of elements in any transcendencebasis of L over K.

Exercises 8.3.12.1. Suppose that L has transcendence degree n over K and that L is algebraic overK(α1, . . . , αn). Show that α1, . . . , αn is a transcendence basis for L over K.

2. Suppose given extensions K ⊂ L ⊂ L1 such that L has finite transcendence degreeover K and L1 has finite transcendence degree over L. Show that L1 has finitetranscendence degree over K, and that the transcendence degree of L1 over K isthe sum of the transcendence degree of L over K and the transcendence degree ofL1 over L.


8.4 Algebraic Closures

Definition 8.4.1. A field K is algebraically closed if every polynomial in K[X] of degree≥ 1 has a root in K.

We shall show in Section 11.6, in what’s known as the Fundamental Theorem of Al-gebra, that the complex numbers, C, are an algebraically closed field. Every argumentfor this relies at some point on topology. Perhaps the simplest proof makes use of fun-damental groups, but requires no ring or field theory beyond the definitions of complexmultiplication and of polynomials. However, the topology would require too much de-velopment to present here. So we shall give a proof that is more algebraic, and relies onthe Fundamental Theorem of Galois Theory, in Chapter 11.

Lemma 8.4.2. Let K be an algebraically closed field. Then every polynomial f(X) ofdegree ≥ 1 in K[X] breaks up as a product

f(X) = a(X − a1) . . . (X − an)in K[X], where a, a1, . . . , an ∈ K.

Proof Since every polynomial of degree ≥ 1 has a root in K, it has a degree 1 factor.So no polynomial of degree > 1 is irreducible. So the irreducibles are precisely the degree1 polynomials, and the result follows from unique factorization in K[X].

A simpler argument, by induction on degree, would have sufficed, of course. We shallmake use of the next lemma in Section 9.3.

Lemma 8.4.3. Every algebraically closed field is infinite.

Proof First note that Z2 is not algebraically closed, as X2 +X + 1 has no roots in Z2.Let K be a finite field with more than two elements and let a be a non-identity

element of K×. Let n be the order of K×. Then every element of K× has exponent n,and hence Xn − a has no roots in K.

Often, problems over R are most easily solved by passing to C and first solving theproblem there. In general, it can be very useful to study a given field by making use ofan algebraically closed field in which it embeds.

Definition 8.4.4. Let K be a field. An algebraic closure of K is an algebraic extension,L, of K, such that L is algebraically closed.

Notice that this is a quite different notion from the algebraic closure of K in someextension, L, of K (Definition 8.2.20), though that concept can be useful in constructingsome algebraic closures.

Proposition 8.4.5. Suppose L is an extension of K that is algebraically closed. LetK1 ⊂ L be the algebraic closure of K in L (i.e., the set of elements of L that arealgebraic over K.) Then K1 is an algebraic closure of K.

Proof We’ve seen in Proposition 8.2.21 that K1 is an algebraic extension of K. So itsuffices to show that K1 is algebraically closed.

Thus, suppose that f(X) ∈ K1[X] has positive degree. Since L is algebraically closed,f has a root, α, in L. It suffices to show that α is in K1.

But α is algebraic over K1, and by Proposition 8.2.23, the only elements of L thatare algebraic over K1 are those of K1.


Thus, once we’ve established the Fundamental Theorem of Algebra, we’ll know thatthe algebraic closure of Q in C is an algebraic closure of Q. In particular, it will beour preferred model for the algebraic closure of Q, as we’ve become accustomed tovisualizing the elements of C. It also allows the use of complex analysis to study thealgebraic extensions of Q, and forms the basis for the field of analytic number theory.

Here, we shall give an abstract proof that algebraic closures exist. We shall treat theuniqueness of algebraic closures in Chapter 11.

Lemma 8.4.6. Let f1(X), . . . , fk(X) be polynomials over K. Then there exists an ex-tension of K in which each of the polynomials fi has a root.

Proof By induction, it suffices to treat the case of a single polynomial f(X). Andby factoring f , if necessary, we may assume that f(X) is irreducible in K[X]. Thus,L = K[X]/(f(X)) is a field, containing K as a subfield. Moreover, if α ∈ L is the imageof X under the canonical map from K[X], then f(α) = 0, and hence α is a root of f inL.

The construction we shall use relies on polynomial rings in infinitely many variables.Here, if A is a ring and S is a set of variables, the elements of the polynomial ring A[S]are simply polynomials in finitely many of the variables of S with coefficients in A. Weadd and multiply these as we would in a polynomial ring in finitely many variables, andmay do so because for any finite collection of elements of A[S], there is a finite subset{X1, . . . , Xn} such that each of the polynomials in this collection lies in A[X1, . . . , Xn] ⊂A[S]. (In categorical language, A[S] is the direct limit of the polynomial rings on thefinite subsets of S.)

Polynomials in infinitely many variables behave analogously to polynomials in finitelymany variables. In particular, there is an inclusion map i : S → A[S] which, as the readermay verify, is universal in the following sense.

Proposition 8.4.7. Let A be a commutative ring and let B be a commutative A-algebra.Let S be a set of variables and let f : S → B be any function. Then there is a uniqueA-algebra homomorphism εf : A[S] → B such that εf ◦ i = f . The image of εf isA[f(Y ) |Y ∈ S], the smallest A-subalgebra of B containing the image of f .

We shall refer to εf as the A-algebra homomorphism obtained by evaluating eachY ∈ S at f(Y ). The next lemma is the key step in constructing algebraic closures.

Lemma 8.4.8. Let K be a field. Then there is an algebraic extension K1 of K such thatevery polynomial in K[X] has a root in K1.

Proof It suffices to construct K1 so that each monic irreducible polynomial of degree> 1 in K[X] has a root in K1. Thus, for each monic irreducible polynomial f(X) ofdegree > 1 in K[X], we define a variable Xf , and we let S be the set of all such Xf .Then substituting Xf for the variable X, we obtain the polynomial f(Xf ) ∈ K[S].

Let a be the ideal of K[S] generated by the set {f(Xf ) |Xf ∈ S}. We claim that a isa proper ideal of K[S].

To see this, we argue by contradiction. Suppose that 1 ∈ a. Then there aremonic irreducible polynomials f1(X), . . . , fk(X) of degree > 1 in K[X] and polynomialsg1, . . . , gk ∈ K[S] such that

1 = g1f1(Xf1) + · · ·+ gkfk(Xfk).


By Lemma 8.4.6, there is an extension L of K in which each of f1(X), . . . , fk(X) has aroot. Say αi ∈ L is a root of fi(X) for 1 ≤ i ≤ k. Let h : S → L be any function suchthat h(Xfi

) = αi for 1 ≤ i ≤ k and let εh : K[S] → L be the K-algebra map obtainedby evaluating each Xf in S at h(Xf ).

Applying εh to the displayed equation above, we get

1 = εh(g1)f1(α1) + · · ·+ εh(gk)fk(αk)= 0,

as αi is a root of fi for 1 ≤ i ≤ k.Thus, a is a proper ideal of K[S], and hence there is a maximal ideal m of K[S] that

contains it. So let K1 = K[S]/m. Now write βf for the image in K1 of Xf under thecanonical map π : K[S] → K1 for each Xf ∈ S. Then f(βf ) = π(f(Xf )) = 0, andhence each monic irreducible polynomial f(X) ∈ K[X] of degree > 1 has a root in K1,as claimed. Also, each βf is algebraic over K. Since K1 = K[βf |Xf ∈ S], it is algebraicover K by Proposition 8.2.12 and Corollary 8.2.17.

Now, we can construct algebraic closures.

Proposition 8.4.9. Let K be a field. Then K has an algebraic closure.

Proof By induction, using Lemma 8.4.8, construct a sequence

K = K0 ⊂ K1 ⊂ · · · ⊂ Kn ⊂ · · ·of field extensions such that for i ≥ 0, Ki+1 is an algebraic extension of Ki with theproperty that each polynomial in Ki[X] has a root in Ki+1. Note that an inductiveapplication of Corollary 8.2.18 now shows that Ki is an algebraic extension of K for alli.

Using set theory, we may construct a set in which the Ki all embed, compatiblywith their embeddings in each other. For instance, the direct limit lim−→Ki constructed inSection 6.8 has that property. We set L equal to the union of the Ki in such a set.

Thus, each element of L lies in Ki for some i, and for any finite set of elements in L,we can find a Ki that contains them all. Thus, the addition and multiplication operationson the Ki induce addition and multiplication operations in L, and the fact that each Ki

is a field implies that L is a field under these operations.Since every element of L lies in some Ki, each element of L is algebraic over K.

Also, if f(X) ∈ L[X], then there is some Ki that contains all of the coefficients of f . Inparticular, f(X) ∈ Ki[X] ⊂ L[X], and hence has a root in Ki+1, and hence in L. Thus,L is algebraically closed, and hence an algebraic closure for K.

Exercises 8.4.10.1. Show that the algebraic closure of Q in C (and hence any algebraic closure of Q,

once we have the uniqueness statement) is countable.

2. Show that if K is an algebraically closed field and if L is any extension of K, thenK is algebraically closed in L.

3. Let K be an algebraically closed field and let D be a division ring that is also a K-algebra. Suppose that D is finite dimensional as a vector space over K, under thevector space structure induced by its K-algebra structure. Show that the inclusionK ⊂ D given by the K-algebra structure map is a bijection, so that K = D.


8.5 Criteria for Irreducibility

We wish to study the irreducibility of polynomials over Q. For this purpose, it is usefulto note that if f(X) = αnX

n+ · · ·+α0 with α0, . . . , αn ∈ Q, and if m is divisible by theleast common multiple of the denominators (when placed in lowest terms) of α0, . . . , αn,then mf(X) has coefficients in Z.

We can use this technique to reduce questions about Q[X] to questions about Z[X].Along the way, we shall show that Z[X] is a U.F.D.

Let p be a prime number in Z. Then the divisors of p in Z[X] all lie in Z, and hencep is irreducible in Z[X]. Thus, the following result is necessary for Z[X] to be a U.F.D.

Lemma 8.5.1. Let p be a prime number and let f(X), g(X) ∈ Z[X] be such that pdivides f(X) · g(X). Then p must divide at least one of f and g.

In other words, the principal ideal generated by p in Z[X] is a prime ideal.

Proof Write f(X) =∑ni=0 aiX

i and g(X) =∑mi=0 biX

i. Suppose that p does notdivide f . Then there is a smallest index i such that ai is not divisible by p. Similarly, ifg is not divisible by p, there is a smallest index j such that bj is not divisible by p.

Now the i + j-th coefficient of f(X)g(X) is∑i+jk=0 akbi+j−k. If k < i, ak is divisible

by p. If k > i, bi+j−k is divisible by p. Thus, the i + j-th coefficient of f(X)g(X) isdivisible by p if and only if aibj is divisible by p. Since neither ai nor bj is divisible byp, this cannot be. Thus, either f or g must have been divisible by p.

The greatest common divisor of a collection of integers is the largest integer thatdivides all of them. It can be computed in terms of primes in the usual way (Corol-lary 8.1.31).

Definitions 8.5.2. The content, c(f), of a polynomial f(X) ∈ Z[X] is the greatestcommon divisor of the coefficients of f .

We say that f(X) is primitive if it has content 1.

The next result is a corollary of Lemma 8.5.1.

Corollary 8.5.3. Let f(X), g(X) ∈ Z[X]. Then c(f(X)g(X)) = c(f) · c(g).Proof Dividing, we can write f(X) = c(f)f1(X) and g(X) = c(g)g1(X), where f1(X)and g1(X) are primitive. Multiplying these, we see that it suffices to show that theproduct of two primitive polynomials is primitive.

But if the product of two polynomials is not primitive, then it must be divisible bysome prime number p, in which case p must divide one of the two polynomials.

This now sets up the very useful Gauss Lemma.

Lemma 8.5.4. (Gauss Lemma) Let f(X) ∈ Z[X] be a polynomial of degree > 1. Thenf(X) is reducible in Q[X] if and only if f factors in Z[X] as the product of two polyno-mials of degree ≥ 1.

Proof Suppose that f(X) = g(X)h(X) in Q(X), with both g and h of positive degree.By first clearing the denominators and then factoring out contents, we can write

g(X) =m

ng1(X) and h(X) =

k

lh1(X),

where g1 and h1 are primitive polynomials in Z[X] and k, l,m, n ∈ Z.Thus, nlf(X) = mkg1(X)h1(X). Equating the contents of both sides, we see nlc(f) =

±mk, and hence nl divides mk. In particular, mk = nlr for some r ∈ Z, and hencef(X) = rg1(X)h1(X) does indeed factor in Z[X] as claimed.


A variant of the above argument is used in the proof of the next lemma.

Lemma 8.5.5. Let f(X) ∈ Z[X] be a primitive polynomial that is irreducible as anelement of Q[X]. Then f(X) is a prime element in Z[X].

Proof Suppose given g(X), h(X) ∈ Z[X] such that f divides g(X)h(X) in Z[X]. Thenf also divides g(X)h(X) in Q[X]. Since f(X) is a prime element in the P.I.D. Q[X], itmust divide one of g and h there. Let’s say that g(X) = f(X)g1(X) in Q[X].

Write g1(X) = mn g2(X), where g2(X) is a primitive polynomial in Z[X] and m,n ∈ Z

with (m,n) = 1. Then

ng(X) = mf(X)g2(X).

Comparing contents, we see that nc(g) = ±m. Since (m,n) = 1, this gives n = ±1, andhence g(X) = ±mf(X)g2(X) is divisible by f(X) in Z[X].

The next lemma adds a little clarity.

Lemma 8.5.6. Let f(X) ∈ Q[X]. Then up to sign, there is a unique primitive polyno-mial in Z[X] that generates the same principal ideal of Q[X] as f does. In particular, iff is irreducible, then up to sign there is a unique primitive polynomial in Z[X] equivalentto f as a prime in Q[X].

Proof It suffices to show that if g(X) and h(X) are primitives in Z[X] that differ bya unit in Q[X], then g = ±h. But if g and h differ by a unit in Q[X], then there areintegers m,n such that mg(X) = nh(X). Comparing contents, we see that m = ±n, andhence g = ±h.

To get a unique choice of primitive polynomial in Z[X] equivalent to a given primein Q[X], we may choose the one whose leading coefficient is positive.

We now obtain the prime decompositions in Z[X].

Proposition 8.5.7. Let f(X) be a polynomial of degree > 0 over Z and suppose thatthe prime decomposition of f(X) in Q[X] is

f(X) =m

n(p1(X))r1 . . . (pk(X))rk ,

where p1(X), . . . , pk(X) are pairwise inequivalent prime elements in Q[X]. Then if qi(X)is the unique primitive polynomial in Z[X] with positive leading coefficient that is equiv-alent to pi in Q[X], we have

f(X) = ±c(f) (q1(X))r1 . . . (qk(X))rk

in Z[X].

Proof By uniqueness of prime decomposition in Q[X], we have

f(X) =k

l(q1(X))r1 . . . (qk(X))rk

for some k, l ∈ Z, and hence

lf(X) = k (q1(X))r1 . . . (qk(X))rk .

Comparing contents, we see that lc(f) = ±k, so the result follows by dividing l into k.


This leads directly to the verification that Z[X] is a U.F.D.

Corollary 8.5.8. The polynomial ring Z[X] is a U.F.D. whose prime elements are, upto sign, the primes in Z, together with the primitive polynomials that are irreducible inQ[X].

Proof We’ve already seen that the above elements are prime elements of Z[X]. Byfactoring c(f) in Z, Proposition 8.5.7 shows that every nonzero nonunit in Z[X] may bewritten as a product of these primes. Thus, Z[X] is a U.F.D.

The uniqueness statement for prime decompositions in a U.F.D. now shows that everyprime element in Z[X] must be equivalent to one of these primes. As the units in Z[X]are ±1, the primes here are determined up to sign.

Actually, the proofs given apply to a much more general setting. The reader mayverify the following proposition.

Proposition 8.5.9. Let A be a U.F.D. with fraction field K. Then A[X] is a U.F.D.whose prime elements are the primes in A, together with the primitive polynomials ofA[X] that are irreducible in K[X].

The following criterion for irreducibility shows the value of the Gauss Lemma.

Proposition 8.5.10. (Eisenstein’s Criterion) Let f(X) =∑ni=0 aiX

i ∈ Z[X] and let pbe a prime. Suppose that the leading coefficient an is relatively prime to p and that pdivides ai for 0 ≤ i < n. Suppose also that p2 does not divide the constant term a0. Thenf(X) is irreducible in Q[X].

Proof If f(X) is reducible in Q[X], then the Gauss Lemma provides a factorizationf(X) = g(X)h(X) in Z[X], where g and h both have positive degree. Write g(X) =∑ki=0 biX

i and h(X) =∑n−ki=0 ciX

i.Since p divides the constant term of f but p2 does not, p divides the constant term

of exactly one of the polynomials g and h. Let’s say that it divides the constant term ofg.

We claim now by induction that p must divide every coefficient of g. Thus, supposethat p divides b0, . . . , bi−1 for i ≤ k. Then we have ai =

∑ij=0 bjci−j . Now p divides ai,

since i ≤ k and k < n. Also, for j < i, p divides bj , and hence divides the summandbjci−j . Thus, p must divide the remaining summand bic0. Since p does not divide c0, itmust divide bi.

In particular, we obtain that p divides the leading coefficient bk. But an = bkcn−k,contradicting the fact that p does not divide an. Thus, f must not be reducible.

We shall see that there are lots of irreducible polynomials that do not satisfy Eisen-stein’s criterion, but many do.

Corollary 8.5.11. Let m be a positive integer that’s divisible by a prime p but is notdivisible by p2. Then Xn −m is irreducible in Q[X] for all n ≥ 1. In particular,

[Q( n√m) : Q] = n.

Proof The polynomial Xn −m clearly satisfies Eisenstein’s criterion for the prime p,and hence is irreducible over Q. Since n

√m is a root of Xn−m, its minimal polynomial is

a prime divisor of Xn−m. But Xn−m is prime, so it must be the minimal polynomialof n√m.


Corollary 8.5.12. For each n ≥ 2, there is a simple extension of Q of degree n.

In some cases, we may apply Eisenstein’s criterion via a change of variables.

Corollary 8.5.13. For r ≥ 1, X2r−1+ 1 is the minimal polynomial of ζ2r over Q. In

particular,

[Q(ζ2r ) : Q] = 2r−1.

Proof Write f(X) = X2r−1+ 1. Since ζ2r has order 2r in C×, and since −1 is the

unique element of C× of order 2, ζ2r is a root of f . Thus, it suffices to show that f(X)is irreducible over Q.

Let f(X + 1) be the polynomial obtained by evaluating f at X + 1. Note that forany polynomial g(X), g(X +1) has the same degree as g(X). Thus, if f(X) is reducible,f(X + 1) will be reducible also. So it suffices to show that f(X + 1) is irreducible.

Now f(X + 1) = (X + 1)2r−1

+ 1, and (X + 1)2r−1

can be expanded by the BinomialTheorem (Theorem 4.5.16). We have

(X + 1)2r−1

=2r−1∑i=0

(2r−1

i

)Xi.

In particular, (X + 1)2r−1

is monic, and hence so is f(X + 1), while the constant term of(X + 1)2

r−1is 1.

Thus, the constant term of f(X+1) is 2. Thus, the result will follow from Eisenstein’scriterion for p = 2, provided that the coefficients

(2r−1

i

)are even for 0 < i < 2r−1. But

this follows immediately from Lemma 4.5.17.

Exercises 8.5.14.1. Show that Xn − 1 = (X − 1)(Xn−1 +Xn−2 + · · ·+ 1) for all n > 1. Deduce that

the minimal polynomial of ζn divides (Xn−1 +Xn−2 + · · ·+ 1).

† 2. Let p be an odd prime and let f(X) = Xp−1 + Xp−2 + · · · + 1. The precedingproblem gives (X+1)p−1 = X ·f(X+1). Deduce from Eisenstein’s criterion thatf(X + 1) is irreducible, and hence f(X) is the minimal polynomial of ζp. Deducethat

[Q(ζp) : Q] = p− 1.

3. Suppose that f(X) ∈ Z[X] has a leading coefficient prime to p, and that f(X) isirreducible in Zp[X]. Show that f(X) must be irreducible in Q[X].

4. Show that X4 + 1 is not irreducible in Z2[X]. Deduce that the converse of thepreceding problem is false.

5. Let f(X) ∈ Z[X] be primitive, and write Z[X] · f(X) and Q[X] · f(X) for theprincipal ideals of Z[X] and Q[X], respectively, that are generated by f . Showthat

(Q[X] · f(X)) ∩ Z[X] = Z[X] · f(X).

6. Show that Eisenstein’s criterion holds in the field of fractions of any U.F.D.


8.6 The Frobenius

The material here is very basic, but it is important enough to deserve emphasis.Let p be a prime and let A be any commutative ring of characteristic p. By Propo-

sition 7.2.14, this means that the unique ring homomorphism Z → A factors throughan embedding of Zp in A. But since Zp is a field, any ring homomorphism out of Zpis injective, and hence the rings of characteristic p are precisely the Zp-algebras. Noticethat the Zp-algebra structure on a ring A of characteristic p is unique, and that A has aunique subring isomorphic to Zp. We identify this subring with Zp.

Definition 8.6.1. Let p be a prime and let A be a commutative ring of characteristicp. Then the Frobenius map ϕ : A→ A is defined by ϕ(a) = ap for all a ∈ A.

Proposition 8.6.2. Let A be a commutative ring of characteristic p. Then the Frobeniusmap ϕ : A→ A is a ring homomorphism.

Proof This is an immediate application of the Binomial Theorem. We have

(a+ b)p =p∑i=0

(p

i

)aibp−i.

But for 0 < i < p, Lemma 4.5.17 shows(pi

)to be divisible by p. But every element

in a ring of characteristic p has additive exponent p (Corollary 7.7.16), and hence thesummands with 0 < i < p all vanish. We are left with (a+b)p = ap+bp, so the Frobeniusis a homomorphism of additive groups. But the rest is immediate.

Corollary 8.6.3. Let K be a field of characteristic p. Then the Frobenius map ϕ : K →K is an embedding. In particular, if K is finite, then ϕ is an automorphism of the fieldK.

For K either finite or infinite, the elements fixed by ϕ (i.e., the elements x ∈ K withϕ(x) = x are precisely the elements of Zp.

Proof Of course, every homomorphism of fields is an embedding. As far as the fixedpoints are concerned, note that x is a fixed point of ϕ if and only if x is a root of Xp−X.As a polynomial of degree p over the field K, Xp −X has at most p roots in K. So itsuffices to show that every element of Zp is fixed by ϕ.

Now Z×p is a (cyclic) group of order p− 1, so xp−1 = 1 for all 0 �= x ∈ Zp. So xp = x

for all x ∈ Zp.

Definition 8.6.4. Let K be a field of characteristic p. We say that K is perfect if theFrobenius map ϕ : K → K is onto, and hence is an automorphism of the field K.

We also say that every field of characteristic 0 is perfect.

The next corollary is now immediate from Corollary 8.6.3.

Corollary 8.6.5. All finite fields are perfect.

Since Zp[X] has characteristic p, it has a Frobenius homomorphism, ϕ.

Lemma 8.6.6. The Frobenius homomorphism ϕ : Zp[X]→ Zp[X] is given by ϕ(f(X)) =f(Xp), the effect of evaluating f at Xp.


Proof Since ϕ is a homomorphism, we have

ϕ

(n∑i=0

aiXi

)=

n∑i=0

ϕ(ai)ϕ(X)i

=n∑i=0

ai(Xp)i

since ϕ is the identity on Zp. But this last expression is f(Xp).

We now show that the field of rational functions, Zp(X), of Zp is imperfect.

Corollary 8.6.7. The Frobenius map on Zp(X) gives an isomorphism from Zp(X) ontoZp(Xp), the field of fractions of the subring Zp[Xp]. In particular, Zp(X) is not perfect.

Proof The elements of Zp(X) have the form f(X)/g(X), where f(X), g(X) ∈ Zp[X]with g(X) �= 0. Clearly,

ϕ (f(X)/g(X)) = ϕ (f(X)) /ϕ (g(X))= f(Xp)/g(Xp),

so the image of ϕ is Zp(Xp), as claimed. To see that this is not all of Zp(X), note thatif f(Xp)/g(Xp) = X/1 in Zp(X), then f(Xp) = Xg(Xp), which is impossible due to thecongruences mod p of the exponents that appear.

Exercises 8.6.8.1. Show that every algebraically closed field is perfect.

2. Let k be a perfect field and let f(X) ∈ k[X]. Show that there is a polynomialg(X) ∈ k[X] such that f(Xp) = (g(X))p.

3. Let K be any field of characteristic p �= 0. Show that K(X) is not perfect andcalculate the image of ϕ.

4. Let A be a ring of characteristic 2. Show that the Frobenius function ϕ(a) = a2

gives a ring homomorphism from A to A if and only if A is commutative.

8.7 Repeated Roots

The material here is essential in understanding Galois theory for fields of nonzero char-acteristic. We shall make use of it in studying separable extensions. For now, we look atthe elementary properties involved in the issues of repeated roots.

Definitions 8.7.1. Let L be an extension of K and let α ∈ L be a root of the polynomialf(X) ∈ K[X]. We say that α is a repeated root, or multiple root, of f if (X−α)2 dividesf in L[X].

We say that α is a root of multiplicity k for f(X) if

f(X) = (X − α)kg(X)

in L[X], where g(α) �= 0.


On the surface, this would seem to depend on the field L, but in fact, it only dependson f and α.

Lemma 8.7.2. Let L be an extension field of K and let f(X) ∈ K[X]. Suppose thatα ∈ L is a root of multiplicity k ≥ 1 of f(X). Then there is a factorization

f(X) = (X − α)kg(X)

in K(α)[X], with g(α) �= 0. In particular, α is a root of multiplicity k of f whenconsidered as an element of any extension field of K(α).

Proof Since f(α) = 0, the Euclidean algorithm inK(α)[X] gives a factorization f(X) =(X − α)f1(X) in K(α)[X]. And unique factorization in L[X] says that α is a root ofmultiplicity k − 1 of f1(X). The result follows by induction.

Example 8.7.3. Over Q(i), X4 + 2X2 + 1 factors as (X − i)2(X + i)2. So i and −i areboth multiple roots of the rational polynomial X4 + 2X2 + 1.

It is a little more difficult to produce an example of an irreducible polynomial withrepeated roots.

Example 8.7.4. Let L = Zp(T ) be the field of fractions of the polynomial ring Zp[T ].Let K = Zp(T p) be the field of fractions of the subring Zp[T p] of Zp[T ].

Then T ∈ L is a root of Xp−T p ∈ K[X]. Moreover, by consideration of the Frobeniushomomorphism in L[X], we obtain a factorization

Xp − T p = (X − T )p

in L[X]. So T is a repeated root of Xp − T p.Since X−T is prime in L[X], an examination of constant terms shows that no proper

factor of (X − T )p can have coefficients in K. So Xp − T p must be irreducible in K[X].

We shall see later that this shows L to be an inseparable extension of K.There is a formal derivative for polynomials, used to detect multiple roots.

Definition 8.7.5. Let A be a commutative ring and let f(X) =∑ni=0 aiX

i be a poly-nomial in A. The formal derivative of f is defined as follows:

d

dX(f(X)) = f ′(X) =

n∑i=1

iaiXi−1.

Lemma 8.7.6. Let A be a commutative ring. Then the formal derivative ddX : A[X]→

A[X] is an A-module homomorphism satisfying the product rule:

d

dX(f(X)g(X)) = f ′(X) · g(X) + f(X) · g′(X).

A morphism with these properties is often called an A-linear derivation.The formal derivative detects multiple roots in the following manner.

Lemma 8.7.7. Let L be an extension field of K and let f(X) ∈ K[X]. Then an elementα ∈ L is a repeated root of f if and only if α is a root of both f(X) and f ′(X).


Proof If α is a repeated root of f , then we have f(X) = (X − α)2g(X) in L[X]. Theproduct rule then gives

f ′(X) = (X − α) [2g(X) + (X − α)g′(X)]

and hence α is a root of f ′(X) also.Conversely, if α is a root of both f(X) and f ′(X), then f(X) = (X−α)g(X) in L[X],

and it suffices to show that α is a root of g(X). Here, the product rule gives

f ′(X) = g(X) + (X − α)g′(X),

and hence f ′(α) = g(α). Since α is a root of f ′, the result follows.

This allows us to characterize the irreducible polynomials with multiple roots.

Proposition 8.7.8. Let L be an extension field of K and let f(X) be an irreducibleelement of K[X] with a root in L. Then f has a repeated root in L if and only if theformal derivative f ′(X) is the zero polynomial, in which case every root of f(X) is arepeated root.

If K has characteristic 0, this cannot happen at all. If K has characteristic p, then thederivative f ′(X) vanishes if and only if f(X) = g(Xp) for some irreducible polynomialg(X) in K[X]. Inductively, if f(X) has a repeated root, we may write f(X) = h(Xpk

)for some k > 0, where h(X) is irreducible in K[X] and h′(X) �= 0.

Proof If α ∈ L is a repeated root of f , then α must also be a root of f ′(X). But f isirreducible, and hence must be, up to multiplication by a scalar, the minimal polynomialof α. Since α is a root of f ′, f must divide f ′. But the degree of f ′ is less than that off , so this can only happen if f ′ is the zero polynomial. Note that if f ′ = 0, then everyroot of f is also a root of f ′ and hence is a multiple root of f .

Since f is irreducible, it must have degree at least one. Say f(X) =∑ni=0 aiX

i withn > 0 and an �= 0. So if K has characteristic 0, the term nanX

n−1 in the expansion off ′(X) is nonzero, and hence f ′(X) is nonzero. Thus, α could not have been a repeatedroot of f .

If K has characteristic p, then f ′(X) vanishes if and only if the only nonzero coeffi-cients of f(X) are those in degrees divisible by p. This then gives f(X) =

∑ki=0 biX

pi =g(Xp), where bi = api, k = n

p and g(X) =∑ki=0 biX

i.Since any factorization of g would induce a factorization of f(X) = g(Xp), we see

that g is irreducible.For the final statement, note that if g does not have repeated roots, we may take

h(X) = g(X) and k = 1. Otherwise, the result follows by induction on degree, as thedegree of g is less than that of f .

8.8 Cyclotomic Polynomials

We characterize the minimal polynomial of ζn over Q, for each n.First, consider the subgroup of Q(ζn)× generated by ζn:

〈ζn〉 = {1, ζn, . . . , ζn−1n },

a cyclic group of order n. Thus, the order of ζkn is n/(n, k), and hence ζkn generates 〈ζn〉if and only if (n, k) = 1.


Definitions 8.8.1. The primitive n-th roots of unity in Q(ζn) (or in C) are the gener-ators of 〈ζn〉.

We define the Euler φ function by setting φ(n) equal to the number of primitive n-throots of unity for n ≥ 1.

Thus, φ(n) is the number of generators of a cyclic group of order n. As noted above,m generates Zn if and only if (m,n) = 1. We see (e.g., from Lemma 4.5.3) that φ(n) isthe order of the group of units Z×

n of the ring Zn. As such, a calculation of φ(n) is givenin Section 4.5, where it is shown that φ(n) is the order of the automorphism group ofthe cyclic group Zn, which is then calculated in Corollary 4.5.9:

Corollary 8.8.2. Let n = pr11 . . . prk

k , where p1, . . . , pk are distinct primes and the expo-nents are all positive. Then

φ(n) = pr1−11 (p1 − 1) . . . prk−1

k (pk − 1).

Notice that since 〈ζn〉 ⊂ Q(ζn)× has order n, every element in 〈ζn〉 is a root of Xn−1.In particular, this means that there are n distinct roots of Xn − 1 in Q(ζn). Factoringout the associated degree one polynomials and equating leading coefficients, we obtain afactorization of Xn − 1.

Lemma 8.8.3. In Q(ζn)[X], we have

Xn − 1 =n−1∏k=0

(X − ζkn).

Corollary 8.8.4. In Z[X], there is a decomposition

Xn − 1 = p1(X) . . . pk(X)

where p1(X), . . . , pk(X) are distinct monic polynomials in Z[X] that are irreducible inQ[X].

Proof By Proposition 8.5.7, we may write

Xn − 1 = ±p1(X) . . . pk(X)

in Z[X], where p1(X), . . . , pk(X) are (not necessarily distinct) primitive polynomials inZ[X] that are irreducible in Q[X]. We may assume that the pi(X) all have positiveleading coefficients. But in this case, since Xn − 1 is monic, each pi(X) must be monicas well, and the sign in the displayed equation must be positive. So it suffices to showthat the pi are all distinct.

Suppose that pi(X) = pj(X) for some i �= j. As a factor of Xn−1 of positive degree,pi(X) must have at least one root, say α, in Q(ζn) by Lemma 8.8.3. But then α is alsoa root of pj , and hence is a repeated root of Xn − 1. Since Xn − 1 has n distinct rootsin Q(ζn), this is impossible.

Since each pi(X) is monic and is irreducible in Q[X], it must be the minimal polyno-mial over Q of any of its roots in Q(ζn). Since each ζkn is a root of Xn − 1, its minimalpolynomial must be one of the pi.


Definition 8.8.5. We write Φn(X) for the minimal polynomial over Q of ζn. We callΦn(X) the n-th cyclotomic polynomial over Q.

By the above, we see that Φn(X) is a monic polynomial in Z[X].

Examples 8.8.6.1. Clearly, Φ1(X) = X − 1, and Φ2(X) = X + 1.

2. We’ve seen in Corollary 8.5.13 that

Φ2r (X) = X2r−1+ 1.

3. In Problem 2 of Exercises 8.5.14, it is shown that if p is an odd prime, then

Φp(X) = 1 +X + · · ·+Xp−1.

We now obtain our main result.

Proposition 8.8.7. The roots of Φn(X) in Q(ζn) are precisely the primitive n-th rootsof unity. In particular, [Q(ζn) : Q] = φ(n), and

Φn(X) =∏

0<k<n(n,k)=1

(X − ζkn)

in Q(ζn)[X].

Proof Suppose that ζkn is not a primitive n-th root of unity. Then it is a root of Xd−1for some d < n that divides n, and hence the minimal polynomial of ζkn over Q mustdivide Xd−1. Thus, none of the roots of the minimal polynomial of ζkn may be primitiven-th roots of unity, and, in particular, ζkn cannot be a root of Φn(X).

Thus, it suffices to show that every primitive n-th root of unity is a root of Φn(X).Since the primitive roots of unity all have the form ζkn with (n, k) = 1, induction on thesum of the exponents in the prime decomposition of k shows that it suffices to show thatif η is a root of Φn(X) and if p is a prime that does not divide n, then ηp is also a rootof Φn(X).

Thus, given a root η of Φn(X) and a prime p that does not divide n, suppose that ηp

is not a root of Φn(X). Then let f(X) be the minimal polynomial of ηp over Q. By thediscussion above, f(X) is a monic polynomial in Z[X], and we may write

Xn − 1 = Φn(X) · f(X) · g(X)

for some g(X) ∈ Z[X].Since ηp is a root of f(X), η is a root of f(Xp). Thus, Φn(X) divides f(Xp) in Q[X],

and hence also in Z[X], by Proposition 8.5.7.Now let us reduce the coefficients of all these polynomials mod p. In Zp[X], Lemma 8.6.6

shows f(Xp) to be equal to the p-th power (f(X))p. Thus, Φn(X) divides (f(X))p inZp[X], and hence Φn(X) and f(X) must have a common prime factor, h(X), in theirprime decompositions in Zp[X].

Since Φn(X)f(X) divides Xn − 1, (h(X))2 must divide Xn − 1 in Zp[X]. But thenXn− 1 will have a repeated root in the field Zp[X]/(h(X)). But this is impossible, since0 is the only root of d

dX (Xn−1) = nXn−1. Since 0 is not a root of Xn−1, Lemma 8.7.7shows that Xn − 1 cannot have repeated roots in any field of characteristic prime ton.


And in particular, the next corollary will now give an inductive, though tedious, wayto calculate the polynomials Φn(X) explicitly.

Corollary 8.8.8. The prime decomposition of Xn − 1 in either Q[X] or Z[X] is givenby

Xn − 1 =∏d|n

Φd(X)

where the product is taken over all positive integers d that divide n.

Proof Let d divide n, say, n = dr. Then

Xn − 1 = (Xd − 1)(1 +Xd + · · ·+Xd(r−1)).

Thus, Xd−1 divides Xn−1, and hence so does Φd(X). Assembling this over the variousd, we see that

∏d|n Φd(X) divides Xn − 1.

Thus, it suffices to show that each ζkn is a root of Φd(X) for some d that divides n.Thus, let r = (n, k) and let d = n

r . Then ζrn = ζd, and hence ζkn = ζld, where l = kr is

relatively prime to d. In particular, ζkn is a primitive d-th root of unity, and hence a rootof Φd(X).

As a dividend, we obtain a determination of the ring structure on the group ringQ[Zn].

Corollary 8.8.9. There is an isomorphism of rings

f : Q[Zn]∼=−→

∏d|n

Q(ζd)

where the product ranges over all positive integers d that divide n. Here, if Zn = 〈T 〉,then the coordinate of f(T ) corresponding to a given d dividing n is ζd.

Proof We shall construct f to make the following diagram commute.

Q[X]/(Xn − 1)∏d|n

Q[X]/(Φd(X))

Q[Zn]∏d|n

Q(ζd)

��γ∼=

��

εT ∼=��

∏d|n εζd

∼=

��f

Here, γ is the isomorphism coming out of the second form of the Chinese Remainder The-orem (Corollary 8.1.34), while the vertical maps are induced by the indicated evaluationmaps.

The right-hand vertical map is an isomorphism by the definition of minimal polyno-mial. The left-hand vertical map is an isomorphism by Problem 16 of Exercises 7.5.10.We define f by traversing the diagram. It is easily seen to have the indicated effect onthe generator T .


The analogous result for Z[Zn] fails. The maps in the square are all well defined, andthe vertical maps are isomorphisms, but the map analogous to γ fails to be surjective.The point is that Z[X] is a U.F.D., but not a P.I.D. In a U.F.D., elements a and b maybe relatively prime without (a) + (b) being equal to the whole ring. So the key step inthe proof of the first form of the Chinese Remainder Theorem (Proposition 7.2.23) doesnot apply here.

Exercises 8.8.10.1. Calculate Φ6(X), Φ9(X), Φ12(X), and Φ15(X).

2. Give a general formula for Φpr (X) for odd primes p.

3. Give a general formula for Φ2p(X) for odd primes p.

4. Let (m,n) = 1. Show that [Q(ζmn) : Q(ζm)] = φ(n). Deduce that Q(ζn)∩Q(ζm) =Q.

5. Show that Q(ζn) ∩R = Q(ζn + ζn) = Q(ζn + ζ−1n ). Show that if n > 2, then

[Q(ζn) : Q(ζn + ζ−1n )] = 2.

6. Find the minimal polynomial of ζ8 + ζ−18 over Q.

7. Show that Φp(1) = p for all primes p.

(a) Deduce that

p = (1− ζp)(1− ζ2p) . . . (1− ζp−1

p )

in Z[ζp].

(b) Recall from Problem 20 of Exercises 7.3.16 that if 1 < k < p, then (ζkp −1)/(ζp−1) = 1+ζp+· · ·+ζk−1

p is a unit in Z[ζp]. Deduce that p = u(1−ζp)p−1

for some unit u ∈ Z[ζp]×.

8. Let p be any prime and let r > 0. Show that Φpr (1) = p. Deduce that

p =∏

0<k<pr

(k,p)=1

(1− ζkpr ) = u(1− ζpr )(p−1)pr−1

for some unit u ∈ Z[ζpr ]×.

9. Show that Z[X]/(Φn(X))∼=−→ Z[ζn]. Deduce that Z[ζn] is a free abelian group with

basis 1, ζn, . . . , ζφ(n)−1n .

10. Let Z2 = 〈T 〉 and let

f : Z[Z2]→ Z[ζ1]× Z[ζ2] = Z× Z

be the ring homomorphism induced by

f(T ) = (ζ1, ζ2) = (1,−1).

Show that f is an injection whose image has index 2 in Z× Z.


8.9 Modules over P.I.D.s

We give a generalization of the Fundamental Theorem of Finite Abelian Groups. Here,we shall generalize it to include the case of finitely generated abelian groups as well asfinite ones. We also generalize from the case of abelian groups (Z-modules) to finitelygenerated modules over an arbitrary P.I.D.

Let A be an integral domain. The analogue for A-modules of finiteness in Z-modulesis given by finitely generated torsion modules.

Definitions 8.9.1. Let A be any ring and let M be an A-module. An element m ∈Mis a torsion element if there exists an a ∈ A, a �= 0, such that am = 0.

An A-module M is a torsion module if every element in M is a torsion element.On the other hand, we say that an A-module M is torsion-free if 0 is the only torsion

element in M .

Thus, a nonzero torsion element in A itself is a right zero divisor, so if A is an integraldomain, then A is torsion-free as an A-module.

In general, m ∈ M is a torsion element if and only if the annihilator Ann(m) �= 0,but a module can be a torsion module even if no nonzero element of A annihilates everyelement of M . Also, there are examples of finitely generated modules over a commutativering that are generated by torsion elements, but are not torsion modules (e.g., considerZ6 as a Z6-module). However, over an integral domain, finitely generated modules arewell behaved with respect to torsion phenomena.

Lemma 8.9.2. Let A be an integral domain and let M be a finitely generated moduleover A. Suppose that M has a generating set m1, . . . ,mk consisting of torsion elements.Then M is a torsion module, and

Ann(M) =k⋂i=1

Ann(mi)

is nonzero.

Proof Clearly, any element in⋂ki=1 Ann(mi) annihilates M , so it suffices to show that⋂k

i=1 Ann(mi) is nonzero. But if 0 �= ai ∈ Ann(mi) for 1 ≤ i ≤ k, then a1 . . . ak is anonzero element of

⋂ki=1 Ann(mi).

Recall from Proposition 8.1.32 that in a U.F.D., the intersection of a collection ofprincipal ideals is generated by the least common multiple of their generators.

Corollary 8.9.3. Let M be a finitely generated torsion module over the P.I.D. A, withgenerators m1, . . . ,mk. Let Ann(mi) = (ai) for 1 ≤ i ≤ k. Then Ann(M) is the idealgenerated by the least common multiple of a1, . . . , ak.

In studying the finitely generated modules over a P.I.D., it turns out that we canstudy the torsion modules and the torsion-free modules separately and then assemblethe information. The first step in showing this is the next lemma.

Lemma 8.9.4. Let M be a module over an integral domain A, and let Tors(M) be theset of torsion elements in M . Then Tors(M) is a submodule of M , and the quotientmodule M/Tors(M) is torsion-free.


Proof Let m,n ∈ Tors(M). Then the submodule generated by m and n is a torsionmodule by Lemma 8.9.2, and therefore is contained in Tors(M). In particular, Tors(M)is a submodule of M .

Let m ∈ M/Tors(M) be a torsion element, where m represents the image of m ∈M under the canonical map. Say am = 0 for a �= 0. Then am ∈ Tors(M), sinceam = am = 0, while Tors(M) is the kernel of the canonical map. Since Tors(M) is thetorsion subgroup of M , we have bam = 0 for some b �= 0. But then ba �= 0, and hencem ∈ Tors(M). So m = 0 in M/Tors(M). Thus, M/Tors(M) is torsion-free.

Let M be an A-module. We say that m ∈M is divisible by a nonunit if we may writem = am′ where a is not a unit in A.

Lemma 8.9.5. Let M be a finitely generated torsion-free module over the P.I.D. A andlet 0 �= m ∈M . Then we may write m = am′, where m′ is not divisible by a nonunit.

Proof A is Noetherian by Corollary 7.8.5. Since M is finitely generated, it is a Noethe-rian A-module by Proposition 7.8.9. We shall argue by contradiction.

Suppose that m cannot be written as a product am′ where m′ is not divisible by anonunit. Then m itself must be divisible by a nonunit. Say m = am1, where a is not aunit in A. We claim that the inclusion Am ⊂ Am1 is proper.

If not, then m1 = bm for some b ∈ A. But then (ab−1)m = 0. Since M is torsion-freeand m �= 0, this forces ab = 1, contradicting the assumption that a is not a unit.

But now m1 may not be written as a product a′m′ such that m′ is not divisible by anonunit, as then m would have such a decomposition. By induction, we may constructan infinite sequence

Am ⊂ Am1 ⊂ · · · ⊂ Amk ⊂ · · ·

of proper inclusions of submodules of M , contradicting the fact that M is a NoetherianA-module.

Our next result is the key in being able to treat all finitely generated modules, insteadof just the torsion modules.

Proposition 8.9.6. Let A be a P.I.D. and let M be a finitely generated torsion-freeA-module. Then M is isomorphic to An for some n.

Proof We argue by induction on the number of elements needed to generate M . IfM is generated by one element, m, then the unique A-module map fm : A → M withfm(1) = m is an isomorphism, since Ann(m) = 0 and Am = M .

Thus, suppose that M may be generated by n elements and that any torsion-freemodule with fewer than n generators is free. Let m1, . . . ,mn be a generating set for M .Note that if m1 = am′, then m′,m2, . . . ,mn generate M . Thus, using Lemma 8.9.5 toreplace m1 if necessary, we may assume that m1 is not divisible by a nonunit.

We claim now that M/Am1 is torsion-free. Thus, we claim that if 0 �= a ∈ A andif am = 0, with m ∈ A/Am1, then m = 0. But this says that if 0 �= a ∈ A, thenmultiplication by a gives an injective homomorphism from M/Am1 to itself.

If a is a unit, this is immediate, as multiplication by a gives an isomorphism, whoseinverse is given by multiplication by a−1. But if a is not a unit, we may write it as aproduct of primes, in which case multiplication by a is the composite of the multiplica-tions by these primes. By induction, it suffices to show that multiplication by a primeelement p ∈ A gives an injective homomorphism from A/Am1 to itself.


Thus, suppose that pm = 0 for some m ∈ M/Am1. But then pm ∈ kerπ = Am1 inM , and hence pm = qm1 for some q ∈ A. Since M is torsion-free, q �= 0. Suppose that pand q are relatively prime, and let rp+ sq = 1 for r, s ∈ A. Then m1 = rpm1 + sqm1 =rpm1 + spm = p(rm1 + sm). But that would imply that m1 is divisible by the nonunitp, contradicting our assumption about m1. Thus, p and q cannot be relatively prime.Since p is prime, this means that q = pr for some r ∈ A. Thus, pm = prm1, and hencem = rm1 ∈ Am1. But then m = 0 as desired, and hence M/Am1 is torsion-free.

Since M/Am1 is generated by m2, . . . ,mn, it is free by induction. Let f : Ak →M/Am1 be an isomorphism and let ni = f(ei) for i = 1, . . . , k. Let f : Ak+1 → M bethe A-module homomorphism obtained by setting f(ei) = ni for i = 1, . . . , k and settingf(ek+1) = m1. We claim that f is an isomorphism.

Clearly, the elements n1, . . . , nk,m1 generate m, so f is onto. But if π : M →M/Am1

is the canonical map, then

(π ◦ f)(a1, . . . , ak+1) = f(a1, . . . , ak).

Thus, if (a1, . . . , ak+1) is in the kernel of f , then (a1, . . . , ak) is in the kernel of f . Sincef is injective, any element of ker f is a multiple of ek+1. But f(aek+1) = am1, so f is anisomorphism.

Corollary 8.9.7. Let M be a finitely generated module over the P.I.D. A. Then M isisomorphic to Tors(M)⊕An for some n.

Proof Since M/Tors(M) is torsion-free, there is an isomorphism

f : An∼=−→M/Tors(M).

Let ei be the canonical basis element of An for 1 ≤ i ≤ n and write f(ei) = mi formi ∈M .

Define g : Tors(M) ⊕ An → M as follows. The restriction of g to Tors(M) is justthe inclusion i : Tors(M) ⊂ M , while g(ei) = mi for i = 1, . . . , n. Thus, there is acommutative diagram

0 Tors(M) Tors(M)⊕An An 0

0 Tors(M) M M/Tors(M) 0

�� ι

��

g

��π ��

��

f ∼=

�� i ��π ��

where the maps ι and π in the upper sequence are the natural inclusion and projectionmaps of the direct sum. But now g is an isomorphism by the Five Lemma.

To study the torsion modules over A, we need to develop a little information aboutcyclic modules. Recall that a cyclic module is a module generated by one element, say,M = Am. Recall that there is an isomorphism Am ∼= A/Ann(m).

If A is commutative, Ann(m) = Ann(Am), or, starting out with an ideal rather thana module, Ann(A/a) = a. Notice, then, that over a commutative ring, the annihilator ofm plays the same role that the order of m plays for a torsion element in an abelian group.(The annihilator of a module is the analogue of the exponent of an abelian group.)


This means that we have to find substitutes for some of the counting arguments weused for abelian groups: We can’t count in order to calculate annihilators. The nextlemma gives the analogue of the calculation of the orders of the elements in a cyclicgroup.

Lemma 8.9.8. Let A be a P.I.D. and let x ∈ A/(a) be the image of x ∈ A underthe canonical map. Then Ann(x) may be calculated as follows. Let (c) be the greatestcommon divisor of a and x and let a = bc. Then Ann(x) = (b). Additionally, if Ax is thesubmodule of A/(a) generated by x, then the quotient module (A/(a))/Ax is isomorphicto A/(c).

Proof Let fx : A→ A be the A-module homomorphism with fx(1) = x. Then Ann(x)

is the kernel of the composite Afx−→ A

π−→ A/(a), where π is the canonical map. Notethat the second Noether Theorem gives an isomorphism from (x)/[(x)∩(a)] to Ax. Thus,Ann(x) = {y ∈ A | yx ∈ (x) ∩ (a)}.

By Proposition 8.1.32, (x)∩ (a) is the principal ideal generated by the least commonmultiple, d, of x and a. But then it is easy to see that (d) = (bx). And unique factorizationnow shows that yx ∈ (bx) if and only if y ∈ (b), so Ann(x) = (b).

For the last statement, note that Ax = [(x)+(a)]/(a). But A is a P.I.D., so (x)+(a) =(c), the greatest common divisor of x and a.

Corollary 8.9.9. Let A be a principal ideal domain and let a, x ∈ A. Let M be thecyclic module M = A/(a). Then the quotient M/(x)M is isomorphic to A/(c), where(c) is the greatest common divisor of a and x.

Proof Let x ∈M = A/(a) be the image of x under the canonical map π : A→ A/(a).Then clearly, Ax = (x)M . The rest now follows from Lemma 8.9.8.

We shall need an analogue of the order of a group, as that was the key for inductionarguments for abelian groups. We shall use something a little cruder, as it depends on agenerating set, rather than a module.

Definitions 8.9.10. Let M be a module over a P.I.D. A and let m be a torsion elementof M . Let (a) = Ann(m) and let a = upr11 . . . prk

k with u a unit and p1, . . . , pk primes inA. Then we say that the weight of m is r1 + · · ·+ rk.

If m1, . . . ,mk is a collection of torsion elements in M (the most interesting examplebeing a set of generators of M if M is a torsion module), we say that the weight of theset m1, . . . ,mk is the sum of the weights of the elements in it.

Notice that the second form of the Chinese Remainder Theorem, while phrased interms of A-algebras, also expresses the interrelationship among certain cyclic modules:If a1, . . . , ak are pairwise relatively prime elements of the P.I.D. A, and if a = a1 . . . ak,then the cyclic module A/(a) breaks up as a direct sum:

A/(a) ∼= A/(a1)⊕ · · · ⊕A/(ak).The next lemma shows that use of the Chinese Remainder Theorem will not affect

calculations of weight.

Lemma 8.9.11. Let a1, . . . , ak ∈ A be pairwise relatively prime and let a = a1 . . . ak.Let mi ∈ M = A/(a1) ⊕ · · · ⊕ A/(ak) be the image in M of the standard generator ofA/(ai) for i = 1, . . . , k. Then the generating set m1, . . . ,mk of M has the same weightas the standard generator of the cyclic module A/(a).


We could now classify finitely generated torsion modules in exactly the same way thatwe classified finite abelian groups. We shall use a different method, relying on assemblingsmaller cyclic groups into larger ones.

Lemma 8.9.12. Let M be a module over the P.I.D. A and let m and n be torsionelements of M . Let (a) and (b) be the annihilators of m and n, respectively, and supposethat a and b are relatively prime. Then the submodule Am + An generated by m and nis cyclic, with annihilator equal to (ab).

Proof By the Chinese Remainder Theorem, it suffices to show that Am + An is iso-morphic to Am ⊕ An. We claim that Am + An is the internal direct sum of Am andAn, i.e., that the map μ : Am ⊕ An → Am + An given by μ(cm, dn) = cm + dn is anisomorphism.

Clearly, μ is surjective, and (cm, dn) is in the kernel if and only if cm = −dn. But thegenerator of Ann(cm) divides a and the generator of Ann(−dn) divides b by Lemma 8.9.8.Since a and b are relatively prime, this forces Ann(cm) to be A, and hence cm = dn = 0,so μ is injective.

Corollary 8.9.13. Let m1, . . . ,mk be a set of generators for a torsion module M overthe P.I.D. A. Then there is another generating set n1, . . . , nl of M , whose weight is thesame as that of m1, . . . ,mk, such that the annihilator of nl is the same as that of M .

Proof First, apply the Chinese Remainder Theorem to each of the generators mi,replacing Ami with a direct sum of cyclic modules whose annihilators are generated byelements of the form pj , where p is a prime in A. As shown in Lemma 8.9.11, replacingmi with the set of generators of these cyclic modules does not change the weight of thegenerating set. Thus, we may as well assume that we started out with a generatingset m1, . . . ,mk in which the annihilator of each mi is generated by a power of a primeelement in A.

Also, by Corollary 8.9.3, the annihilator of M is the least common multiple of theannihilators of the mi. Thus, if Ann(M) = (a), where a = pr11 . . . prk

k , with p1, . . . , pka pairwise inequivalent set of primes and with the exponents ri all positive, then theremust be a subcollection mi1 , . . . ,mik of the generating set such that Ann(mij ) = (prj

j )for 1 ≤ j ≤ k.

Since the annihilators of the mij are pairwise relatively prime, an inductive applica-tion of Lemma 8.9.12 shows that the submodule generated by mi1 , . . . ,min is a cyclicmodule, generated by an element m, whose weight is the same as that of mi1 , . . . ,min .Moreover, the annihilator of m is precisely (a). Now take n1, . . . , nl−1 to be the set ofelements of m1, . . . ,mk that are not in mi1 , . . . ,min , and take nl = m.

We shall use the following analogue of Lemma 4.4.17.

Lemma 8.9.14. Let M be a finitely generated torsion module over the P.I.D. A. Sup-pose given m ∈ M with Ann(m) = Ann(M). Let x ∈ M/Am. Then there existsy ∈ M with y = x (i.e., x is the image of y ∈ M under the canonical map), suchthat Ann(y) = Ann(x).

Proof Let (a) = Ann(M). Since a annihilates every element of M , it is divisible bythe annihilator of every element of M .

If x ∈M maps to x under the canonical map π, let Ann(x) = (b). Then as just noted,b divides a. Also, since π(x) = x, b annihilates x, so that if Ann(x) = (c), then c dividesb.


If (b) = (c), we just take y = x. Otherwise, it suffices to find an element y = x+a′m,for some a′ ∈ A, such that Ann(y) = (c). Note that for any element of the formy = x+ a′m, Ann(y) ⊂ (c) because Ann(π(y)) = (c). Thus, it suffices to find an a′ suchthat c annihilates y = x+ a′m.

Since π(cx) = cx = 0, cx ∈ kerπ = Am, say cx = dm with d ∈ A. But then it sufficesto show that d is divisible by c, as, if d = cd′, we can take a′ = −d′ to form y as above.

Write b = cb′. Then Ann(cx) = (b′) by Lemma 8.9.8. But another applicationof the same lemma shows that if a′′ is the greatest common divisor of a and d, thenAnn(cx) = Ann(dm) = (a/a′′). Since Ann(cx) = (b′), we may choose the greatestcommon divisor a′′ so that b′a′′ = a. But since b = b′c divides a, c must divide a′′, whichdivides d.

We are now ready to show how finitely generated torsion modules over P.I.D.s breakup as direct sums of cyclic modules.

Theorem 8.9.15. Let M be a finitely generated torsion module over the P.I.D. A. IfM is nonzero, then there is an isomorphism

M ∼= A/(a1)⊕ · · · ⊕A/(ak)

for a collection of ideals (ai) which satisfies

(a1) ⊂ (a2) ⊂ · · · ⊂ (ak),

where ak is not a unit in A. In particular, the annihilator of M is (a1).

Proof We argue by induction on the weight of a generating set for M . If M has agenerating set of weight 1, then M must be cyclic, and the result holds.

Suppose, then, that m1, . . . ,mn is a generating set for M and that the theorem istrue for all torsion modules that admit generating sets of lesser weight.

By Corollary 8.9.13, we may assume that Ann(m1) = Ann(M). We write Ann(m1) =Ann(M) = (a1). If Am1 = M , then M is cyclic, and the theorem holds. Otherwise,M/Am1 is generated by the images, m2, . . . ,mn, of m2, . . . ,mn under the canonicalmap. Since the generator of Ann(mi) divides the generator of Ann(mi), the generatingset m2, . . . ,mn of M/Am1 has weight strictly less than the weight of m1, . . . ,mn, andhence the induction hypothesis holds in M/Am1.

Thus, by induction, there is an isomorphism M/Am1∼= A/(a2) ⊕ · · · ⊕ A/(ak) for

ideals (ai) such that (a2) ⊂ (a3) ⊂ · · · ⊂ (ak). Then clearly, Ann(M/Am1) = (a2). Notethat since M/Am1 is a quotient of M , (a1) = Ann(M) must annihilate M/Am1. Butthat says (a1) ⊂ (a2). We claim that M is isomorphic to A/(a1)⊕ · · · ⊕A/(ak).

To see this, let f : A/(a2) ⊕ · · · ⊕ A/(ak) → M/Am1 be an isomorphism and letm′i ∈M/Am1 be the image under f of the standard generator of A/(ai) for i = 2, . . . , k.

But Lemma 8.9.14 says that we may choose an m′′i ∈M whose image under the canonical

map is m′i, such that the annihilator of m′′

i is the same as that of m′i (which is (ai)).

Define g : A/(a1)⊕ · · · ⊕A/(ak)→M by

g(x1, . . . , xk) = x1m1 + x2m′′2 + · · ·+ xkm

′′k .

Since the annihilator of m′′i is (ai), g is a well defined homomorphism. Moreover, abbre-


viating with N = A/(a2)⊕ · · · ⊕A/(ak), we have a commutative diagram

0 A/(a1) A/(a1)⊕N N 0

0 Am1 M M/Am1 0

��

��

∼=

��⊂

��

g

��π

��

f ∼=

��

�� ⊂ ��π ��

where the maps π are the appropriate canonical maps. The result now follows from theFive Lemma.

Of course, a free A-module is a direct sum of copies of A/(0), and we may now applyCorollary 8.9.7.

Corollary 8.9.16. Let M be any finitely generated module (not necessarily a torsionmodule) over the P.I.D. A, with M �= 0. Then there is an isomorphism

M ∼= A/(a1)⊕ · · · ⊕A/(ak)for a collection of ideals (ai) which satisfies

(a1) ⊂ (a2) ⊂ · · · ⊂ (ak),

where ak is not a unit in A.

We can now state and prove the Fundamental Theorem.

Theorem 8.9.17. (Fundamental Theorem of Finitely Generated Modules over a P.I.D.)Let M be a finitely generated module over the P.I.D. A, with M �= 0. Then M may bewritten uniquely as a direct sum

M ∼= A/(a1)⊕ · · · ⊕A/(ak)for a collection of ideals (ai) which satisfies

(a1) ⊂ (a2) ⊂ · · · ⊂ (ak),

where ak is not a unit in A.Here, uniqueness means that if we’re given a collection of ideals

(b1) ⊂ (b2) ⊂ · · · ⊂ (bl)

such that bl is not a unit and

A/(a1)⊕ · · · ⊕A/(ak) ∼= A/(b1)⊕ · · · ⊕A/(bl),then k = l and (ai) = (bi) for i = 1, . . . , k.

Proof The existence of such an isomorphism M ∼= A/(a1) ⊕ · · · ⊕ A/(ak) is shown inCorollary 8.9.16, so it suffices to show uniqueness.

Thus, let M = A/(a1)⊕ · · · ⊕A/(ak) and let N = A/(b1)⊕ · · · ⊕A/(bl), with the a’sand b’s as above, and suppose given an isomorphism f : M → N .

Now A-module homomorphisms always carry torsion elements to torsion elements.Applying this to both f and f−1, we see that f induces an isomorphism f : Tors(M)→


Tors(N). Moreover, the induced homomorphism f : M/Tors(M) −→ N/Tors(N) will bean isomorphism, with inverse given by the similarly induced map coming from f−1.

Notice that Tors(M) is the direct sum of those terms A/(ai) for which ai �= 0, andhence M/Tors(M) is isomorphic to the direct sum of those terms A/(ai) for which ai = 0;similarly for N . By Theorem 7.9.8, isomorphic free modules have the same rank. Sincef is an isomorphism, we may conclude that the same number of a’s as b’s are equal to0. Thus, since f restricts to an isomorphism of torsion submodules, it suffices to assumethat both M and N are torsion modules, and hence that a1 and b1 are nonzero.

Let p be a prime of A which divides ak. Then p divides ai for all i ≤ k, and hence[A/(ai)]/(p)[A/(ai)] is isomorphic to A/(p) by Corollary 8.9.9. In this case, M/(p)M isisomorphic to (A/(p))k, and hence has dimension k as a vector space over A/(p).

On the other hand, if p does not divide ak, [A/(ak)]/(p)[A/(ak)] is isomorphic toA/(1) = 0, since (p, ak) = (1). In this case, M/(p)M has dimension strictly less than kas a vector space over A/(p). Note that one way or the other, the dimension of M/(p)Mas a vector space over A/(p) is less than or equal to k. Of course, the same sort ofreasoning applies to calculating the dimension of N/(p)N as a vector space over A/(p).

Thus, suppose that p divides ak. Then M/(p)M is a k-dimensional vector space overA/(p). But the dimension of N/(p)N (which is isomorphic to M/(p)M), must be lessthan or equal to l, and hence k ≤ l. But applying the same argument to a prime q thatdivides bl shows l ≤ k, and hence l = k.

Note that this argument also shows that the primes dividing ak are precisely theprimes dividing bk. Let p be one such.

For an A-module M ′, write M ′[p] for the set of elements of M ′ annihilated by p.

Thus, M ′[p] = {m ∈ M ′ | pm = 0}. Note that since (p) is a maximal ideal of A, a

nonzero element m ∈M ′ is annihilated by p if and only if Ann(m) = (p).Consider the cyclic module A/(d), where d = pc with c �= 0. By Lemma 8.9.8, we

have Ann(x) = (p) if and only if (x, d) = (c), in which case Ax = Ac. In particular,(A/(d))[p] = Ac, which is isomorphic to A/(p), as Ann(c) = (p). Moreover, we have

(A/(d))/(A/(d))[p] ∼= A/(c).

Now p divides ak and bk, and hence it divides all the other a’s and b’s. Say ai =pa′i and bi = pb′i for i = 1, . . . , k. Clearly (M ′ ⊕M ′′)[p] = M ′

[p] ⊕M ′′[p]. Thus, the

isomorphism induced by f from M/M[p] to N/N[p] gives

A/(a′1)⊕ · · · ⊕A/(a′k) ∼= A/(b′1)⊕ · · · ⊕A/(b′k).Note that a′1 has lower weight than a1, so an argument by induction on the weight

of a1 is possible. Here, if a1 has weight 1, then M = (A/(p))k, and hence M and N areboth annihilated by p. Thus, N = N/(p)N = (A/(p))k and the result follows.

Thus, we may assume inductively that a′i = b′i for 1 ≤ i ≤ k. But then ai = bi asdesired.

This form of the Fundamental Theorem has an appealing formulation, but there’sanother form that is perhaps better from a practical standpoint. The point is that theChinese Remainder Theorem allows us to break up any cyclic module into a direct sumof cyclic modules whose annihilators are prime powers.

Definitions 8.9.18. Let p be a prime element of the P.I.D. A and letM be an A-module.We say that m ∈M is a p-torsion element if m is annihilated by some power of p.

We write Mp ⊂M for the collection of p-torsion elements in M .We say that M is a p-torsion module if M = Mp. By a primary torsion module, we

mean a p-torsion module for some prime p of A.


Lemma 8.9.19. Let p be a prime element of the P.I.D. A and let M be an A-module.Then a nonzero element m ∈ M is in Mp if and only if Ann(m) = (pr) for somer ≥ 1. Moreover, Mp is a submodule of M . Finally, if f : M → N is an A-modulehomomorphism, then f(Mp) ⊂ Np.

In particular, a cyclic module A/(a) is a p-torsion module if and only if (a) = (pr)for some r.

Theorem 8.9.20. (Fundamental Theorem of Finitely Generated Modules over a P.I.D.,second form) Let M be a finitely generated torsion module over the P.I.D. A. Then Mhas a unique decomposition as a direct sum of primary torsion cyclic modules.

Here, uniqueness means that if M and N are two different direct sums of primarytorsion cyclic modules, then M and N are isomorphic if and only if for each prime p ofA and each r > 0, M and N have the same number of summands isomorphic to A/(pr).

Proof The existence of such a decomposition may be obtained by applying the Chi-nese Remainder Theorem to break up the cyclic modules of the decomposition fromTheorem 8.9.15 into direct sums of primary torsion modules. Thus, it suffices to showuniqueness.

Note that Lemma 8.9.19 shows that if f : M → N is an isomorphism of A-modules,then it restricts to an isomorphism f : Mp → Np for each prime p of A. Thus, wemay assume that M = A/(pr1) ⊕ · · · ⊕ A/(prk), where r1 ≥ · · · ≥ rk ≥ 1, and N =A/(ps1)⊕ · · · ⊕A/(psl), where s1 ≥ · · · ≥ sl ≥ 1. We are given an isomorphism from Mto N , and must deduce that k = l and ri = si for 1 ≤ i ≤ k.

But now (pr1) ⊂ · · · ⊂ (prk) and (ps1) ⊂ · · · ⊂ (psl), so we may apply the uniquenessstatement from the first form of the Fundamental Theorem.

Exercises 8.9.21.1. Show that Q/Z is a torsion module over Z, but that no element of Z annihilates

all of Q/Z. (Of course, Q/Z is not finitely generated as a Z-module.)

2. Show that a finitely generated torsion module M over the P.I.D. A is cyclic if andonly if there are no primes p of A for which M contains a submodule isomorphicto A/(p)⊕A/(p).

Chapter 9

Radicals, Tensor Products, andExactness

In this chapter, we develop some important tools for studying rings and modules. Weshall use a number of these tools in Chapter 12, where we shall study some particulartypes of rings in detail. More generally, these tools are essential for anyone who worksin fields involving rings and modules.

In Section 9.1, we develop the Jacobson and nil radicals and the radical of an ideal ina commutative ring. We also give Nakayama’s Lemma, an essential ingredient in certainexactness arguments and in the study of local rings.

Section 9.2 develops the theory of primary decomposition, a weak analogue of theunique factorization of ideals occurring in a P.I.D.

In Section 9.3, Hilbert’s Nullstellensatz is proven, using many of the tools so fardeveloped, and is used to define affine algebraic varieties.

Section 9.4 gives the basic theory of tensor products. Extension of rings is emphasizedas an application. Then, in Section 9.5, we use extension of rings and the theory of rad-icals to extend some of the basic results regarding homomorphisms of finite dimensionalvector spaces to the study of free modules over more general rings. In particular, if Ais a ring that admits a ring homomorphism to a division ring, we show that Am ∼= An

if and only if m = n. We also show that if f : Am → An, then the following conditionshold.

1. If f is surjective, then m ≥ n.

2. If A is commutative, f is surjective, and m = n, then f is an isomorphism.

3. If A is commutative and f is injective, then m ≤ n.

We also discuss flat modules and characterize the flat modules over a P.I.D.Section 9.6 gives the tensor product of algebras and verifies its universal property.

Then, in Section 9.7, we develop the module structures and exactness properties of theHom functors and describe their relationship to tensor products.

In Sections 9.8 and 9.9, we develop the theory of projective modules. These arethe modules P for which the functor HomA(P,−) preserves exact sequences, and are ageneralization of free modules. They play a crucial role in our analysis of semisimplerings and Dedekind domains in Chapter 12, and hence are important in the study of

341

CHAPTER 9. RADICALS, TENSOR PRODUCTS, AND EXACTNESS 342

group representations and in number theory. They are also of fundamental importancein homological algebra.

Section 9.9 introduces algebraic K-theory with the functor K0, which classifies thefinitely generated projective A-modules modulo a sort of stabilization process. It playsan important role in numerous mathematical problems.

Section 9.10 introduces the tensor algebra, TA(M), on an A-module M . This is thefree A-algebra on M . We use it to construct the symmetric algebra and exterior algebraon M , and to construct the Clifford algebra on a finite set of generators. The exterioralgebras are related to determinant theory, while the Clifford algebras have applicationsin both topology and analysis. In the process of analyzing these algebras, we developthe theory of skew commutative graded algebras, which has applications in homologicalalgebra and topology.

We shall assume a familiarity here with the notions of categories and functors, asdeveloped in the first two sections of Chapter 6. We shall also make use of some otherideas from that chapter, but will not assume a familiarity with them.

9.1 Radicals

We return to the study of maximality considerations for ideals, but this time in thecontext of noncommutative rings.

Definitions 9.1.1. A maximal left ideal is a maximal element of the partially orderedset of proper left ideals. Maximal right ideals are defined analogously, and a maximaltwo-sided ideal is a maximal element in the partially ordered set of proper two-sidedideals.

The next lemma is proven by the argument that was given for Corollary 7.6.14.

Lemma 9.1.2. Let A be a nonzero ring. Then any proper left ideal of A is containedin a maximal left ideal, any proper right ideal is contained in a maximal right ideal, andany proper two-sided ideal is contained in a maximal two-sided ideal.

Maximal left ideals have important consequences to module theory. They relate tothe analogue for modules of simple groups.

Definition 9.1.3. A nonzero left A-module is called simple, or irreducible, if it has nosubmodules other than 0 and itself.

The maximal left ideals are the key to understanding simple modules.

Lemma 9.1.4. Let M be a simple left A-module and let 0 �= m ∈ M . Then Ann(m) isa maximal left ideal m of A, and there is a module isomorphism between A/m and M .

Conversely, if m is a maximal left ideal of A, then A/m is a simple left A-module.

Proof Given 0 �= m ∈ M , the submodule Am generated by m is nonzero, and henceequal to M , since M is simple. But if m is the annihilator of m, we have an isomorphismfm : A/m → Am = M , inducing a one-to-one correspondence between the submodulesof M and the left ideals of A containing m. Thus, the only left ideals containing m arem and A, so that m is a maximal left ideal.

The converse comes precisely from the fact that there are only two left ideals con-taining a maximal left ideal m: m itself and A.

We shall also make use of the following fact about maximal left ideals.


Lemma 9.1.5. Let A be a ring and let x ∈ A. Then x lies outside all the maximal leftideals if and only if x has a left inverse.

Proof Consider the principal left ideal Ax generated by x. Then x lies in no maximalleft ideal if and only if Ax lies in no maximal left ideal. But this is the case if and onlyif 1 ∈ Ax, which happens if and only if 1 = yx for some y ∈ A.

Definition 9.1.6. Let A be a ring. The Jacobson radical, R(A), is the intersection ofthe maximal left ideals of A.

Many of the rings we’re familiar with have R(A) = 0. Thus, it is valuable to keep inmind that if A is a local ring, with maximal ideal m, then R(A) = m.

The next proposition gives a useful characterization of elements in R(A).

Proposition 9.1.7. Let A be a ring and let x ∈ A. Then the Jacobson radical of A is atwo-sided ideal, and the following statements are equivalent.

1. x is in the Jacobson radical, R(A).

2. For every simple left A-module M , x ∈ Ann(M).

3. For any a, b ∈ A, 1− axb is a unit in A.

Proof Since the annihilator of a module is a two-sided ideal, the fact that R(A) istwo-sided will follow from the equivalence of Conditions 1 and 2. We shall establish thisfirst.

We first show that the first condition implies the second. Let x ∈ R(A) and let Mbe a simple left A-module. It suffices to show that for 0 �= m ∈ M , xm = 0. But theannihilator of m is a maximal left ideal of A because M is simple. Since R(A) is theintersection of the maximal left ideals, x must annihilate m as required.

But now we can see that the second condition implies the first: if x annihilates A/m,with m a maximal left ideal, then it must annihilate the element 1 ∈ A/m. Since theannihilator of 1 is m, x ∈ m.

We now show that the first two conditions imply the third. Let x ∈ R(A) and leta, b ∈ A. Since R(A) is a two-sided ideal, axb ∈ R(A) as well. But then 1− axb cannotlie in any maximal left ideal, as that would force 1 to be in the ideal as well.

By Lemma 9.1.5, this says that 1− axb has a left inverse, y. But then y − yaxb = 1,and hence −yaxb = 1 − y. But −yaxb is in R(A), and hence lies in every maximal leftideal of A. But since 1−y is in every maximal left ideal, y can be in no maximal left ideal,and hence y has a left inverse z. But a standard argument now shows that z = 1− axb,and hence y is a two-sided inverse for 1− axb.

Finally, we show that the third condition implies the first. Suppose that 1− axb is aunit for each a, b ∈ A and let m be a maximal left ideal of A. Suppose that x is not inm. Then we must have Ax+ m = A. But then 1 = ax+m for some a ∈ A and m ∈ m,so that 1 − ax ∈ m. But this contradicts the fact that 1 − ax is a unit, so x must havebeen in m.

The following use of radicals is valuable in the study of local rings.

Proposition 9.1.8. (Nakayama’s Lemma) Let A be a ring and let a be a left ideal con-tained in R(A). Suppose that aM = M for some finitely generated left A-module M .Then M = 0.


Proof Since a is contained in R(A), we may as well use R(A) in place of a. Weargue by contradiction. Assume that M �= 0, and let m1, . . . ,mk be a set of A-modulegenerators for M , which is minimal in the sense that no proper subset will generate M .Since R(A)M = M , we have m1 ∈ R(A)M , and hence m1 = a1n1 + · · · + arnr, withai ∈ R(A) and ni ∈ M for all i. But since since each ni is a linear combination of thegenerators m1, . . . ,mk, and since R(A) is a right ideal, we may collect terms, writingm1 = b1m1 + · · ·+ bkmk, with bi ∈ R(A) for all i.

Rearranging terms, we get (1 − b1)m1 = b2m2 + · · · + bkmk. But since b1 ∈ R(A),1− b1 is a unit, so that m1 is in the submodule of M generated by m2, . . . ,mk. But thenm2, . . . ,mk generate M , contradicting the minimality of the generating set m1, . . . ,mk.

In the commutative case, there is another radical which is of interest.

Lemma 9.1.9. Let A be a commutative ring. Then the set of all nilpotent elements inA forms an ideal, N(A), called the nil radical of A.

Proof Let x ∈ N(A) and let a ∈ A. Then xn = 0 for some n ≥ 1, and hence(ax)n = anxn = 0, also. It suffices to show that N(A) is closed under addition. Thus,let x, y ∈ N(A), with xn = ym = 0 for n,m ≥ 0. We claim that (x+ y)n+m−1 = 0.

To see this, recall that the Binomial Theorem (Theorem 4.5.16) holds in any commu-tative ring. Thus, (x + y)n+m−1 =

∑n+m−1i=0

(n+m−1

i

)xiyn+m−i−1. But if xi �= 0, then

i < n, and hence n+m− i− 1 ≥ m, so that yn+m−i−1 = 0.

The nil radical also has a characterization related to that of the Jacobson radical.

Proposition 9.1.10. Let A be a commutative ring. Then the nil radical of A is theintersection of all the prime ideals of A.

Proof Let x ∈ N(A) and let p be a prime ideal of A. We claim that x ∈ p. To see this,let n > 0 be the smallest exponent such that xn = 0. Then xn ∈ p. If n = 1, there’snothing to show. Otherwise, either x ∈ p or xn−1 ∈ p, so in either case, xn−1 ∈ p. Butthen a downward induction shows that x ∈ p.

It suffices to show that if x ∈ A is not nilpotent, then there is a prime ideal thatdoesn’t contain x. Let X = {xk | k ≥ 1} and let S be the set of all ideals a such thata ∩X = ∅. Since x is not nilpotent, 0 ∈ S, and hence S is nonempty. By an argumentsimilar to the one that shows the existence of maximal ideals, Zorn’s Lemma implies thatthere are maximal elements in S. We claim that if p is a maximal element of S, then pis prime.

It suffices to show that if a and b are in the complement of p, then so is ab. Since ais not in p, p + (a) strictly contains p, and hence xm = p + ya for some p ∈ p, y ∈ A,and m > 0. Similarly, there are p′ ∈ p, y′ ∈ A, and n > 0, with xn = p′ + y′b. Butmultiplying these two elements together, we get xn+m = p′′ +yy′ab for some p′′ ∈ p. Butsince p does not meet X, ab cannot lie in p.

We can generalize the nil radical to define a radical for any ideal of a commutativering.

Definition 9.1.11. Let a be an ideal in the commutative ring A. The radical rad(a) isgiven by

rad(a) = {x ∈ A |xn ∈ a for some n > 0}.


Notice that if π : A→ A/a is the canonical map, then rad(a) = π−1(N(A/a)). Thus,rad(a) is an ideal of A. The following proposition is now immediate from the one-to-onecorrespondence between the prime ideals of A/a and the primes of A that contain a.

Proposition 9.1.12. Let a be a proper ideal in the commutative ring A. Then rad(a) isthe intersection of the prime ideals of A containing a.

It is useful to be able to characterize the ideals a in a commutative ring with rad(a) =a. As the reader may easily check, rad(rad(a)) = rad(a) for any ideal a. We obtain thefollowing lemma.

Lemma 9.1.13. An ideal a of a commutative ring A is its own radical if and only ifa = rad(b) for some ideal b of A.

Of course, if p is prime, then the intersection of the prime ideals containing it isprecisely p. Proposition 9.1.12 gives the following corollary.

Corollary 9.1.14. A prime ideal is its own radical.

The next lemma is easy and its proof is left as an exercise. It use is sufficientlycommon to warrant a formal statement.

Lemma 9.1.15. Let a1, . . . , an be ideals in a commutative ring A. Then rad(⋂ni=1 ai) =⋂n

i=1 rad(ai).

Corollary 9.1.14 now gives a corollary.

Corollary 9.1.16. A finite intersection of prime ideals is its own radical.

We shall show in Corllary 9.2.11 that a proper ideal in a Noetherian commutativering is its own radical if and only if it is a finite intersection of prime ideals.

Exercises 9.1.17.1. What is R(A×B)? What is N(A×B)?

2. Let A be a P.I.D. Show that R(A) = 0 if and only if A has infinitely many primeideals.

3. Let A be the ring of Example 6 of Examples 7.8.2. What is R(A)?

4. Let A be a ring. Show that R(A) is equal to the intersection of the maximal rightideals in A.

5. Let A be a ring. Show that R(A) is contained in the intersection of the maximaltwo-sided ideals in A.


7. Let n > 1 in Z. What is rad((n))?

8. Let A be a commutative ring with only one prime ideal. Show that any injectiveA-module homomorphism from A to itself is an isomorphism.


9.2 Primary Decomposition

Primary decomposition of ideals is a weak analogue of the unique factorization of idealsoccurring in a P.I.D. The theory was first developed for polynomial rings by the chessmaster Emanuel Lasker. Emmy Noether then showed that primary decompositions existfor all proper ideals of a Noetherian commutative ring. We shall give her argument here.

The theory works as follows: There is a generalization of prime ideals called primaryideals. In a Noetherian commutative ring, every proper ideal is a finite intersection ofprimary ideals. But the decomposition is not unique, even if we insist that none of theprimary ideals contain the intersection of the others. However, under that assumption,the set of ideals which are radicals of the primary ideals in the decomposition is unique.

As an application of this theory, we shall show that an ideal a in a Noetheriancommutative ring satisfies a = rad(a) if and only if a is a finite intersection of primeideals. We shall make use of this in Section 9.3 to give decompositions of affine varietiesover an algebraically closed field.

Definition 9.2.1. Let A be a commutative ring. A proper ideal q of A is primary ifxy ∈ q implies that either x ∈ q or y ∈ rad(q) (i.e., yn ∈ q for some n > 0).

If a is an ideal of A, then a primary decomposition of a is a presentation

a =m⋂i=1

qi

of a as a finite intersection of primary ideals qi for i = 1, . . . ,m.

Clearly, any prime ideal is primary. (Recall from Corollary 9.1.14 that a prime idealis its own radical.) We shall give further examples below.

Throughout this section, A denotes a commutative ring.

Since rad(q) is the inverse image under the canonical map of the nil radical of A/q,the next lemma is immediate.

Lemma 9.2.2. A proper ideal q of A is primary if and only if every zero-divisor in A/qis nilpotent.

The reader should verify the following lemma.

Lemma 9.2.3. Let q be a primary ideal of A. Then rad(q) is prime.

Definition 9.2.4. If q is primary with radical p, we say that q is p-primary, or that p isthe prime associated to q.

The converse to Lemma 9.2.3 is false: there are nonprimary ideals whose radical isprime. Having a maximal ideal as radical is stronger.

Proposition 9.2.5. Let q be an ideal of A whose radical is a maximal ideal. Then q isprimary.

Proof Let m = rad(q). Since the radical is the intersection of the prime ideals containingq (Proposition 9.1.12), any prime ideal containing q must contain m. Thus, m is the onlyprime ideal of A containing q, and hence m/q is the only prime ideal in A/q.

Thus, any element of A/q outside m/q is a unit. So every zero-divisor of A/q lies inm/q. But since m = rad(q), m/q is the nil radical of A/q, so all its elements are nilpotent.Thus, q is primary by Lemma 9.2.2.


Examples 9.2.6.1. Let m be a maximal ideal of A and let mk = m . . .m be the ideal product of m

with itself k times. (See Definitions 7.2.21.) Let q be any proper ideal containingmk. Then m ⊂ rad(q). Since q is proper, so is its radical. So rad(q) = m, and q isprimary by Proposition 9.2.5.

2. If K is a field and r1, . . . , rn > 0, let q = (Xr11 , . . . , Xrn

n ) ⊂ K[X1, . . . , Xn], theideal generated by Xr1

1 , . . . , Xrnn . Then rad(q) contains the maximal ideal m =

(X1, . . . , Xn). Since q is proper, it is m-primary.

3. Let A be a P.I.D. and let a = pr11 . . . prk

k , where p1, . . . , pk are pairwise inequivalentprimes of A. Then an element lies in rad((a)) if and only if it is divisible byp1, . . . , pk. Thus, rad((a)) = (p1 . . . pk). Thus, a nonzero ideal (a) is primary if andonly if (a) = (pr) = (p)r for some prime element p of A and r > 0.

The ideal 0 is also primary, being prime.

We shall now show that every proper ideal in a Noetherian commutative ring has aprimary decomposition. We shall make use of the following operation on ideals.

Definition 9.2.7. Let a and b be ideals of A, Then the ideal quotient (a : b) is the ideal

(a : b) = {x ∈ A |xb ⊂ a}.Thus, x ∈ (a : b) if and only if xb ∈ a for all b ∈ b.

For a principal ideal (x) ⊂ A, we write (a : (x)) = (a : x). Note that y ∈ (a : x) ifand only if xy ∈ a.

We shall actually show that every proper ideal in a Noetherian commutative ring isa finite intersection of irreducible ideals.

Definition 9.2.8. A proper ideal a of A is irreducible if a = b ∩ c implies that eithera = b or a = c.

Lemma 9.2.9. If A is Noetherian, then every irreducible ideal of A is primary.

Proof Let q be an irreducible ideal of A and let xy ∈ q with y �∈ q. Then we have anascending chain of ideals

(q : x) ⊂ (q : x2) ⊂ · · · ⊂ (q : xk) ⊂ · · · .Since A is Noetherian, the sequence is eventually constant, and hence (q : xn) = (q :xn+1) for some n > 0. We claim that (q + (y)) ∩ (q + (xn)) = q.

To see this, note that xy ∈ q implies that q + (y) ⊂ (q : x). Thus, if z+ axn ∈ q + (y)with z ∈ q, then zx+axn+1 = w ∈ q. So a ∈ (q : xn+1) = (q : xn). But then z+axn ∈ q,as desired.

Thus, (q+(y))∩(q+(xn)) = q. Since q is irreducible and y �∈ q, this forces q+(xn) = q,and hence x ∈ rad(q). Thus, q is primary.

We can now show the existence of primary decompositions in a Noetherian commu-tative ring. The technique we shall use is called Noetherian induction. We shall makefurther use of it in Chapter 12.

Theorem 9.2.10. Every proper ideal in a Noetherian commutative ring A is a finiteintersection of irreducible ideals. In particular, Lemma 9.2.9 now shows that any properideal of A has a primary decomposition.


Proof Recall from Proposition 7.6.16 that if S is any nonempty set of ideals in aNoetherian ring, ordered by inclusion, then S has maximal elements. Here, a is maximalin S if a ∈ S and any ideal in S containing a must equal a.

Thus, we may argue by contradiction: Assuming the theorem false, let S be the setof proper ideals which are not finite intersections of irreducible ideals and let a be amaximal element in S.

Since a is in S, it cannot be irreducible, so we may write a = b ∩ c, where b and cboth properly contain a (and hence b and c are both proper ideals). Since a is a maximalelement in S, b and c cannot lie in S, and hence each of them is a finite intersection ofirreducible ideals. But a = b∩ c, so this contradicts the assumption that a is not a finiteintersection of irreducible ideals.

Corollary 9.2.11. A proper ideal a in a Noetherian commutative ring A is its ownradical if and only if a is a finite intersection of prime ideals.

Proof Corollary 9.1.16 shows that a finite intersection of prime ideals is its own radical,so it suffices to show the converse. As noted in Lemma 9.1.13, this amounts to showingthat the radical of any proper ideal of A is a finite intersection of prime ideals.

Let a be a proper ideal of A. Theorem 9.2.10 gives a decomposition a =⋂ni=1 qi

where qi is primary for all i. But then rad(a) =⋂ni=1 rad(qi) (Lemma 9.1.15). The result

follows, since the radical of a primary ideal is prime (Lemma 9.2.3).

We shall now discuss the uniqueness properties of primary decompositions. As weshall assume the existence of such a decomposition, A need not any longer be Noetherian.

Theorem 9.2.10 itself provides an example of nonuniqueness, in that not every primaryideal is irreducible.

Example 9.2.12. Let K be a field and let m = (X,Y ), the ideal of K[X,Y ] generatedby X and Y . Then m2 = (X2, XY, Y 2) is primary by Proposition 9.2.5. As the readermay check,

m2 = (m2 + (X)) ∩ (m2 + (Y )),

the intersection of two ideals properly containing it. Also, both m2 + (X) and m2 + (Y )have radical m, and hence are primary.

We can eliminate this particular sort of nonuniqueness to our decompositions if weinsist that each of the primary ideals being intersected have a distinct radical. Thefollowing lemma allows us to do this.

Lemma 9.2.13. Let q1, . . . , qm be p-primary ideals in A for some prime ideal p (i.e.,each qi is primary with rad(qi) = p). Then q =

⋂mi=1 qi is p-primary as well.

Proof By Lemma 9.1.15, rad(q) =⋂mi=1 rad(qi) = p, so it suffices to show q is primary.

Suppose xy ∈ q and y �∈ rad(q). It suffices to show that x ∈ q. Since q is the intersectionof the qi, xy ∈ qi for all i. But y �∈ rad(qi) = rad(q). Since each qi is primary, x must liein qi for all i, and hence in the intersection, q.

Definition 9.2.14. A primary decomposition a =⋂mi=1 qi of an ideal a is reduced if

1. The prime ideals rad(q1), . . . , rad(qm) are all distinct.

2. No qi contains the intersection of the other q’s.


Using Lemma 9.2.13 to obtain the first condition and then induction to obtain thesecond, Theorem 9.2.10 gives the following corollary.

Corollary 9.2.15. Every proper ideal in a Noetherian commutative ring has a reducedprimary decomposition.

Even reduced primary decompositions are not unique.

Example 9.2.16. Let a = (X2, XY ) ⊂ K[X,Y ] with K a field. Let m = (X,Y ).Then (X2, Y ) and m2 = (X2, XY, Y 2) both have radical m, and hence are primary byProposition 9.2.5. Thus, the following equalities, which the reader should verify, give twodistinct primary decompositions for a.

a = (X) ∩m2 anda = (X) ∩ (X2, Y ).

Note that the radicals of the primary ideals in these two decompositions are given by(X) and (X,Y ). Note that (X) ⊂ (X,Y ), even though there is no analogous inclusionbetween the corresponding primary ideals in these decompositions.

We shall now develop some positive results regarding uniqueness of primary decompo-sitions. First note that prime ideals are better behaved than primary ideals with respectto irreducibility.

Lemma 9.2.17. Prime ideals are irreducible.

Proof Let p = a∩b with p prime. Since p contains a∩b, either a or b must be containedin p by Corollary 7.6.10. But p is contained in both a and b, and hence must equal oneof them.

Lemma 9.2.18. Let p be a prime ideal of A and let q be p-primary. Then for x ∈ A,the ideal quotient (q : x) is given by

(q : x) =

⎧⎨⎩ A if x ∈ qq if x �∈ pa p-primary ideal otherwise.

Proof The case x ∈ q is true for any ideal, while that of x �∈ p is immediate from thedefinition of a primary ideal: If xy ∈ q and x �∈ p = rad(q), then y must lie in q.

Thus, it suffices to show that if x �∈ q then (q : x) is p-primary. Note, then that ifynx ∈ q, then yn ∈ rad(q) = p, since x �∈ q and q is primary. But then y ∈ p, since p isprime. Thus, rad((q : x)) ⊂ p. But q ⊂ (q : x), and hence p = rad(q) ⊂ rad((q : x)), sop = rad((q : x)), and it suffices to show that (q : x) is primary.

Thus, suppose that ab ∈ (q : x) with a �∈ rad((q : x)) = p. We wish to show b ∈ (q : x).We have abx ∈ q, so bx ∈ (q : a) = q by the second case above. But then b ∈ (q : x).

We can use the ideal quotients to show that the primes associated to the primaryideals in a primary decomposition are unique.

Proposition 9.2.19. Let a =⋂mi=1 qi be a reduced primary decomposition of the ideal a

and let pi = rad(qi) for i = 1, . . . ,m. Then {p1, . . . , pm} is the set of prime ideals whichoccur among the ideals {rad((a : x)) |x ∈ A}.

Thus, if a =⋂ki=1 q′i is another reduced primary decomposition of a, then k = m, and

we may reindex the q′i so that rad(q′i) = pi for i = 1, . . . ,m.


Proof As the reader may easily check, (⋂mi=1 qi : x) =

⋂mi=1(qi : x). Since radicals also

commute with intersections (Lemma 9.1.15), Lemma 9.2.18 shows that

rad((a : x)) ={ ⋂

{i | x∈qi} pi if x �∈ a =⋂mi=1 qi

A if x ∈ a.

If rad((a : x)) is prime, then x �∈ a, and rad((a : x)) is the finite intersection⋂x∈qi

pi.But prime ideals are irreducible (Lemma 9.2.17), so rad((a : x)) must equal one of thepi.

Conversely, since the decomposition is reduced, for each i we can find an elementxi which lies in

⋂j =i qj but not in qi. But then the displayed equation shows that

rad((a : xi)) = pi.

Definitions 9.2.20. Let a be an ideal admitting a primary decomposition. Then theprimes associated to a (or, more simply, the primes of a) are the radicals of the primaryideals in any reduced primary decomposition of a.

Let p1, . . . , pm be the primes of a. If pi contains pj for some j �= i, we say that pi isan embedded prime of a. Otherwise, we say that pi is an isolated prime of a.

If a =⋂mi=1 qi is a reduced primary decomposition of a, with pi the associated prime

of qi, we call the qi the primary components of the decomposition, and say that qi isisolated (resp. embedded) if pi is isolated (resp. embedded).

Thus, Example 9.2.16 shows that (X) is an isolated prime of (X2, XY ) ⊂ K[X,Y ]and (X,Y ) is embedded.

The word “embedded” expresses the fact that if K is an algebraically closed field andif p ⊂ p′ is a proper inclusion of prime ideals of K[X1, . . . , Xn], then the affine varietyV(p′) induced by p′ embeds as a subvariety of V(p). We shall investigate this situationin Section 9.3.

We can obtain a uniqueness result for the isolated primary components of a decom-position.

Proposition 9.2.21. Let a =⋂mi=1 qi be a reduced primary decomposition for a with

associated primes p1, . . . , pm. Suppose that pi is an isolated prime of a. Then

qi = {x ∈ A | (a : x) �⊂ pi}.Since the right hand side is independent of the choice of reduced primary decomposition,so is qi.

Proof Write S = {x ∈ A | (a : x) �⊂ pi}. We first show that S ⊂ qi. Since a ⊂ qi,(a : x) ⊂ (qi : x) for all x. If x �∈ qi, then (qi : x) ⊂ pi by Lemma 9.2.18. But then(a : x) ⊂ pi, and hence x �∈ S. So S must be contained in qi.

Since pi is isolated, it cannot contain⋂j =i pj : If it did, then it would have to contain

some pj for j �= i by Corollary 7.6.10. Let y be an element of⋂j =i pj not in pi. Since

pj = rad(qj) for all j, we can choose an k > 0 such that yk ∈ ⋂j =i qj . Since pi is a prime

not containing y, yk �∈ p.Let x ∈ qi. Then xyk lies in all of the q’s, and hence in a. Thus, yk ∈ (a : x). Since

yk �∈ pi, (a : x) �⊂ pi, and hence x ∈ S. Thus, qi ⊂ S, and hence qi = S.

The situation for embedded primary components is much more complicated. Indeed,if A is Noetherian, then one may make infinitely many different choices for each embeddedcomponent.1

1See p. 231 of Zariski and Samuel, Commutative Algebra, Vol. I , Van Nostrand, 1958.


But reduced primary decompositions are unique if all primes are isolated. The mostobvious case is for a p-primary ideal, q. Here, there is only one associated prime, p, andhence q = q is the only reduced primary decomposition of q.

A more interesting case is that of a finite intersection of prime ideals. (Recall fromCorollary 9.2.11 that if A is Noetherian, then an ideal has this form if and only if it is itsown radical.) Here, if a =

⋂mi=1 pi, then we can obtain a reduced primary decomposition

for a by inductively eliminating primes containing the intersection of all the other primes.The result is a primary decomposition in which every primary ideal is prime and everyprime is isolated.

Corollary 9.2.22. Let a be a finite intersection of prime ideals. Then a may be writtenuniquely as a finite intersection of prime ideals none of which contain the intersection ofthe others. Here, the uniqueness is up to permutation of the primes.

9.3 The Nullstellensatz and the Prime Spectrum

Recall that an A-algebra of finite type is one which is finitely generated as an A-algebra.

Lemma 9.3.1. Suppose given inclusions A ⊂ B ⊂ C of commutative rings, where ANoetherian and C is an A-algebra of finite type. Suppose that C is finitely generated asa B-module. Then B is finitely generated as an A-algebra.

Proof Suppose that C = A[c1, . . . , cn] and that d1, . . . , dm generate C as a B-module.Then we can find elements bij and bijk in B such that

cj =m∑i=1

bijdi for j = 1, . . . , n and

djdk =m∑i=1

bijkdi for 1 ≤ j, k ≤ m.

Write A = A[bij , birs | 1 ≤ i, r, s ≤ m, 1 ≤ j ≤ n] ⊂ B. Since A is finitely generatedas an A-algebra, it is Noetherian by Corollary 7.8.15.

Using the displayed equations and induction, it is easy to see that any monomial inthe cj with coefficients in A lies in the A-submodule of C generated by 1, d1, . . . , dm.Thus, C is finitely generated as an A-module.

But then C is a Noetherian A-module (Proposition 7.8.9), so its submodule B is alsofinitely generated as an A-module (Proposition 7.8.3). Since A is finitely generated asan A-algebra, B must be, also.

We obtain a recognition principle for finite extensions.

Theorem 9.3.2. (Hilbert’s Nullstellensatz) Suppose the extension field L of K is finitelygenerated as a K-algebra. Then L is a finite extension of K.

Proof Let L = K[α1, . . . , αn]. If each αi is algebraic over K, then we are done byLemma 8.2.9.

Otherwise, we wish to derive a contradiction. By Proposition 8.3.6, we may reorderthe α’s so that α1, . . . , αr is a transcendence basis for L over K for some r ≤ n. Thus,L is algebraic over K1 = K(α1, . . . , αr). Since L is obtained by adjoining finitely manyalgebraic elements to K1, Lemma 8.2.9 shows that L is a finite extension of K1.


By Lemma 9.3.1,K1 is finitely generated as aK-algebra. We writeK1 = K[β1, . . . , βs].Since α1, . . . , αr are algebraically independent, K[α1, . . . , αr] is a unique factorizationdomain, by Proposition 8.5.9. Since K1 is its field of fractions, we may write βi =fi(α1, . . . , αr)/gi(α1, . . . , αr), where fi and gi are relatively prime for i = 1, . . . , s.

But then any polynomial in β1, . . . , βs has a denominator whose prime factors alldivide the product of the gi. Thus, it suffices to show that K[α1, . . . , αr] has infinitelymany primes, as, if h has a prime factor that does not divide any of the gi, then theelement 1/h of K(α1, . . . , αr) cannot lie in K[β1, . . . , βs].

But if p1, . . . , pk are primes in K[α1, . . . , αr], then p1 . . . pk + 1 is relatively prime toeach pi, and hence has a prime factor distinct from the pi.

The Nullstellensatz has a number of important consequences for commutative algebrasof finite type over a field.

Corollary 9.3.3. Let K be a field and let f : A → B be a K-algebra homomorphismbetween commutative K-algebras of finite type. Then for each maximal ideal m of B, theinverse image f−1(m) is maximal in A.

Proof Let m be a maximal ideal of B and let p = f−1(m). Then there is an inclusionof K-algebras A/p ⊂ B/m. But the Nullstellensatz shows B/m to be a finite extensionof K. Since A/p is a subalgebra of B/m, every element α ∈ A/p is algebraic over K(Proposition 8.2.14). Thus, (Proposition 8.2.5) K[α] ⊂ A/p is a field, and hence α isinvertible in A/p. Since α was arbitrary, A/p is a field, and hence p = f−1(m) is maximalin A.

This allows us to deduce an important fact about radicals in commutative algebrasof finite type over a field.

Proposition 9.3.4. Let K be a field and let a be a proper ideal in K[X1, . . . , Xn]. Thenthe radical, rad(a), of a is the intersection of the maximal ideals of K[X1, . . . , Xn] con-taining a.

In consequence, if A is a commutative algebra of finite type over a field K, then theJacobson radical and nil radical of A coincide:

N(A) = R(A).

Proof The second assertion is immediate from the first: if A is a commutative algebra offinite type overK and if f : K[X1, . . . , Xn]→ A is a surjectiveK-algebra homomorphism,then f−1(N(A)) = rad(ker f), while f−1(R(A)) is the intersection of the maximal idealsof K[X1, . . . , Xn] containing ker f .

Thus, let a be a proper ideal of K[X1, . . . , Xn]. By Proposition 9.1.12, rad(a) is theintersection of the prime ideals of K[X1, . . . , Xn] containing a. It suffices to show that iff(X1, . . . , Xn) is not in rad(a), then there is a maximal ideal containing a which doesn’tcontain f .

Since f is not in rad(a), there is a prime ideal p containing a but not f . Thus, frepresents a nonzero element f of the domain A = K[X1, . . . , Xn]/p. Let B = A[1/f ],the subring of the field of fractions of A obtained by adjoining 1/f to A. Write g :K[X1, . . . , Xn]→ B for the composite

K[X1, . . . , Xn]π−→ K[X1, . . . , Xn]/p = A ⊂ B,

where π is the canonical map. Then g is a K-algebra homomorphism between commu-tative K-algebras of finite type. Thus, if m is a maximal ideal of B, then g−1(m) is a


maximal ideal of K[X1, . . . , Xn] by Corollary 9.3.3. Since g(f) is a unit in B, f �∈ g−1(m).Since a ⊂ ker g ⊂ g−1(m), the result follows.

In a domain A, there are no nonzero nilpotent elements, and hence N(A) = 0.

Corollary 9.3.5. Let A be an integral domain finitely generated as an algebra over afield K. Then R(A) = 0.

We can use the Nullstellensatz to give a complete determination of the maximal idealsof a polynomial ring over an algebraically closed field. This consitutes the starting pointfor algebraic geometry.

Corollary 9.3.6. (Nullstellensatz, weak version) Let K be an algebraically closed fieldand let m be maximal ideal of K[X1, . . . , Xn]. Then m is the kernel of the evaluationmap

εa1,...,an: K[X1, . . . , Xn]→ K

for some a1, . . . , an ∈ K.Thus, by Proposition 7.3.22, m = (X1 − a1, . . . , Xn − an), the ideal generated by

X1 − a1, . . . , Xn − an.Proof Let L = K[X1, . . . , Xn]/m. Since the polynomial ring maps onto L, L is aK-algebra of finite type, and hence is a finite extension of K, by the Nullstellensatz.

Since K is algebraically closed, it has no irreducible polynomials of degree > 1. Thus,any element algebraic over K has a minimal polynomial of degree 1, and hence lies inK. Thus, the inclusion of K in L is an isomorphism.

We obtain a K-algebra isomorphism f : K[X1, . . . , Xn]/m∼=−→ K. Note that m is the

kernel of the composite

K[X1, . . . , Xn]π−→ K[X1, . . . , Xn]/m

f−→ K,

where π is the canonical map. But if f ◦ π(Xi) = ai for i = 1, . . . , n, then f ◦ π mustbe the evaluation map εa1,...,an : K[X1, . . . , Xn] → K by the universal property of apolynomial algebra (Corollary 7.3.20).

Definitions 9.3.7. Let A be a commutative ring. We write Spec(A) for the set of primeideals of A and write Max(A) for the set of maximal ideals of A. We call them the primespectrum and the maximal ideal spectrum of A, respectively.

If K is a field and (a1, . . . , an) ∈ Kn, we write ma1,...,an∈ Max(A) for the kernel of

the evaluation map εa1,...,an : K[X1, . . . , Xn]→ K:

ma1,...,an = {f ∈ K[X1, . . . , Xn] | f(a1, . . . , an) = 0}.By Proposition 7.3.22, ma1,...,an = (X1 − a1, . . . , Xn − an), the ideal generated by X1 −a1, . . . , Xn − an.

We write ν : Kn → Max(K[X1, . . . , Xn]) for the map given by ν(a1, . . . , an) =ma1,...,an .

Proposition 9.3.8. The maps ν : Kn → Max(K[X1, . . . , Xn]) are injective for any fieldK and are surjective if K is algebraically closed.

Proof If ma1,...,an = mb1,...,bn , then Xi − ai and Xi − bi lie in this ideal for each i. But(Xi − ai)− (Xi − bi) is a unit unless ai = bi, so ν is injective.

If K is algebraically closed, then ν is surjective by the weak Nullstellensatz.


If K is algebraically closed, we can think of ν as a way of providing a geometry for themaximal ideals of K[X1, . . . , Xn]. In particular, when K = C,2 this provides a standardsort of geometry to associate to algebraically significant subsets of the set of maximalideals. Some subsets of Kn which have algebraic significance are the affine varieties.

Definitions 9.3.9. Let K be an algebraically closed field. We shall refer to Kn as affinen-space over K. If S ⊂ K[X1, . . . , Xn] is any subset, then the affine variety, V(S),determined by S is the subset of affine n-space consisting of the common zeroes of theelements of S:

V(S) = {(a1, . . . , an) ∈ Kn | f(a1, . . . , an) = 0 for all f ∈ S}.

As we shall verify in Lemma 9.3.12, the affine variety V(S) is mapped by ν : Kn ∼=Max(K[X1, . . . , Xn]) to the set of maximal ideals of K[X1, . . . , Xn] containing S.

There is a geometry to affine varieties intrinsic to the algebra. It does not coincidewith ordinary complex geometry when K = C. It is called the Zariski topology, and maybe defined not only on affine varieties, but also on Spec(A) for any commutative ring A.

Definitions 9.3.10. Let A be a commutative ring and let S ⊂ A. The Zariski open setU(S) of Spec(A) is given by

U(S) = {p ∈ Spec(A) |S �⊂ p}.

The Zariski topology on Spec(A) is the one in which the open sets are precisely theZariski open sets. The Zariski topology on Max(A) is the subspace topology of that onSpec(A).

Let K be an algebraically closed field. Then Proposition 9.3.8 shows that

ν : Kn → Max(K[X1, . . . , Xn])

is a bijection. We define the Zariski topology on Kn by declaring ν to be a homeomor-phism.

That this gives a topology, of course, requires justification.

Proposition 9.3.11. Let A be a commutative ring. Then the Zariski open sets form atopology on Spec(A). Moreover, they have the following properties.

1. U({1}) = Spec(A).

2. U({0}) = ∅.3. If S ⊂ T ⊂ A, then U(S) ⊂ U(T ).

4. If a is the ideal generated by S, then U(S) = U(a).

5. If Si ⊂ A for i ∈ I, then⋃i∈I U(Si) = U(

⋃i∈I Si).

6. U({s}) ∩ U({t}) = U({st}) for all s, t ∈ A.

7. For S, T ⊂ A, U(S) ∩ U(T ) = U({st | s ∈ S and t ∈ T}).2We shall show in Theorem 11.6.7 that C is algebraically closed.


Proof The first condition holds because the elements of Spec(A) are proper ideals. Thenext four conditions are clear, and the sixth is a restatement of the definition of a primeideal: st ∈ p if and only if either s ∈ p or t ∈ p. The last condition follows from theothers.

If K is algebraically closed, then the Zariski topology on affine n-space over K isdetermined by the affine varieties.

Lemma 9.3.12. Let K be an algebraically closed field. Then the affine varieties are pre-cisely the closed subspaces in the Zariski topology on the affine n-space Kn. Specifically,if S ⊂ K[X1, . . . , Xn], then the homeomorphism ν : Kn → Max(K[X1, . . . , Xn]) carriesthe affine variety V(S) onto the complement of the Zariski open set U(S).

Proof An n-tuple (a1, . . . , an) lies in V(S) if and only if f(a1, . . . , an) = 0 for allf ∈ S. But f(a1, . . . , an) = 0 if and only if f lies in the maximal ideal ν(a1, . . . , an) =ker εa1,...,an , so ν(V(S)) is the set of maximal ideals of K[X1, . . . , Xn] containing S. Butthis is precisely the complement of U(S).

By analogy, for an arbitrary commutative ring A and a subset S ⊂ A, we shall writeV (S) for the complement of U(S) in either Spec(A) or Max(A). In this more generalcontext, the spaces are called schemes, rather than varieties.

From a topologist’s point of view, the topologies on Spec(A) and Max(A) are bizarre.Note that a point p in Spec(A) is closed if and only if there is a set S such that pcontains S, but no other prime ideal contains S. Clearly, such an S exists if and only ifp is maximal.

Thus, Max(A) is a T1 space, but Spec(A) is not. So why bother with Spec(A)?For one thing, prime ideals have some interest in their own right. For another, as weshall see presently, Spec is a functor of commutative rings, but Max is not. However,Corollary 9.3.3 shows that Max is a functor on algebras of finite type over a field K:

Definition 9.3.13. Let f : A → B be a homomorphism between commutative rings.Then the induced map f∗ : Spec(B)→ Spec(A) is given by f∗(p) = f−1(p).

If K is a field and f : A → B is a K-algebra homomorphism between commutativeK-algebras of finite type, then the induced map f∗ : Max(B) → Max(A) is given byf∗(m) = f−1(m).

The next lemma is left as an exercise.

Lemma 9.3.14. Let f : A → B be a homomorphism of commutative rings. Then f∗ :Spec(B)→ Spec(A) is continuous. Moreover, if g : B → C is another homomorphism ofcommutative rings, then f∗ ◦ g∗ = (g ◦ f)∗. Thus, Spec is a contravariant functor fromthe category of commutative rings to the category of topological spaces.

The same conclusions hold for the maps on Max induced by K-algebra homomor-phisms between commutative algebras of finite type over a field K.

Let a be an ideal of the commutative ring A and let π : A → A/a be the canonicalmap. Then π∗ : Spec(A/a) → Spec(A) is a closed map, inducing a homeomorphism ofSpec(A/a) onto the closed subset V (a) of Spec(A). Moreover, since π−1(p) is maximalin A if and only if p is maximal in A/a, π∗ restricts to a homeomorphism of Max(A/a)onto the closed subset V (a) of Max(A).

The next lemma is obvious, but important.


Lemma 9.3.15. Let a be the ideal generated by S in A, then V (S) = V (a) in Spec(A).Thus, the closed subspaces of Spec(A) are the homeomorphic images of the prime spectraof the quotient rings of A.

Of course, the ideals in a Noetherian ring are all finitely generated.

Corollary 9.3.16. Let A be a Noetherian ring. Then every closed subset of Spec(A) orMax(A) has the form V ({a1, . . . , ak}) for some finite subset {a1, . . . , ak} of A.

Similarly, if K is an algebraically closed field, then every affine variety has the formV({f1, . . . , fk}) for some finite subset {f1, . . . , fk} of K[X1, . . . , Xn].

Examples 9.3.17.1. Let A be a P.I.D. and let 0 �= a ∈ A. Then V ({a}) is the set of equivalence classes

of prime elements that divide a, and hence is finite. In fact, we can see that theproper closed subsets of Spec(A) are precisely the finite subsets of Max(A).

2. If A is a P.I.D. with only one nonzero prime ideal (e.g., A = Z(p) or A = Zp), thenSpec(A) is a space consisting of two points, one of which is open and the other not.This space is sometimes called Sierpinski space.

3. Let K be algebraically closed. Then in affine n-space Kn, we have V({Xn}) ={(a1, . . . , an) ∈ Kn | an = 0}, the image of the standard inclusion of Kn−1 in Kn.In particular, not all Zariski closed sets are finite.

Remarks 9.3.18. There is an important interplay between the study of affine varietiesover the complex numbers and differential topology. Given a finite collection f1, . . . , fkof C∞ functions from Rm to R, let f : Rm → Rk be given by f(x) = (f1(x), . . . , fk(x))for x ∈ Rn. Then the set of common zeros of f1, . . . , fk is precisely f−1(0). Using theImplicit Function Theorem of advanced calculus, one can show that f−1(0) is a smooth(i.e., C∞) submanifold of Rm, provided that the Jacobian matrix of f has rank k atevery point of f−1(0).

This construction is generic for smooth manifolds: Every smooth manifold is diffeo-morphic (i.e., equivalent as a smooth manifold) to a manifold obtained by the aboveprocedure.

Now, a complex affine variety is the set of common zeros of a finite collection ofcomplex polynomials f1, . . . , fk ∈ C[X1, . . . , Xn], and hence is the zero set of the functionf : Cn → Ck whose i-th component function is fi for i = 1, . . . , n.

Regarding f as a function from R2n to R2k, we can compute the Jacobian matrixof f . If it has rank 2k at each point of V(f1, . . . , fk), then V(f1, . . . , fk) is a smoothmanifold. (Here, we use the subspace topology from R2n, of course, and not the Zariskitopology.)

In this manner, complex varieties have been an important source of examples ofmanifolds. Such examples have been useful in a number of contexts, especially in realdimension 4, where some of the standard techniques of smooth handle theory fail.

Complex varieties that are not manifolds are also interesting topologically as a sortof generalized manifolds.

From the other side, topological techniques provide invariants for complex varieties.These invariants are generally weaker than their structure as algebraic varieties, but canbe useful nevertheless.

Returning to the study of algebraic varieties, let K be an algebraically closed fieldand let a be an ideal in K[X1, . . . , Xn]. By Lemma 9.3.14, we may identify the affine


variety V(a) with Max(K[X1, . . . , Xn]/a), the maximal ideal spectrum of a commutativeK-algebra of finite type. Since Max is a functor on such algebras, we see that thestructure of the affine variety V(a) depends only on the K-algebra K[X1, . . . , Xn]/a.Indeed, if A is a commutative K-algebra of finite type, we can think of Max(A) as anabstract affine variety: Any choice of K-algebra generators, a1, . . . , an, of A induces asurjective K-algebra homomorphism εa1,...,an : K[X1, . . . , Xn] → A, and hence inducesa homeomorphism ε∗a1,...,an

: Max(A)→ V (ker εa1,...,an) ∼= V(ker εa1,...,an

).Abstract varieties may be compared by means of algebraic maps.

Definitions 9.3.19. Let K be an algebraically closed field and let A and B be commu-tative K-algebras of finite type. An algebraic map Max(B) → Max(A) is a function ofthe form f∗ for some K-algebra homomorphism f : A→ B. Here, the map depends onlyon its effect as a function Max(B) → Max(A), and not on the homomorphism inducingit.

We need to take a slightly different approach in order to define mappings betweenaffine varieties. The point is that while there’s a standard identification of V(a) ⊂ Kn

with Max(K[X1, . . . , Xn]/a), the variety V(a) does not determine the ideal a: Thereexist proper inclusions a ⊂ b of ideals such that V(b) = V(a). To resolve this, we shalldevelop the notions of polynomial mappings and coordinate rings.

Definition 9.3.20. Let K be an algebraically closed field and let a be an ideal ofK[X1, . . . , Xn]. Write X ⊂ Kn for the affine variety determined by a:

X = V(a) = {(a1, . . . , an) | f(a1, . . . , an) = 0 for all f ∈ a}.

We define the ideal of X, written IX, to be the set of all polynomials which vanish onevery element of X:

IX = {f ∈ K[X1, . . . , Xn] | f(a1, . . . , an) = 0 for all (a1, . . . , an) ∈ X}.

Then the coordinate ring, O(X), of X is defined by

O(X) = K[X1, . . . , Xn]/IX.

The coordinate ring catalogues the polynomial functions from X to K. Recall fromLemma 7.3.25 that the passage from polynomials f(X1, . . . , Xn) ∈ K[X1, . . . , Xn] tothe induced functions f : Kn → K gives a ring homomorphism φ : K[X1, . . . , Xn] →Map(Kn,K), where Map(Kn,K) is the ring of all functions from Kn to K. Indeed,restricting the effect of these polynomials to the affine variety X ⊂ Kn gives a ringhomomorphism

φX : K[X1, . . . , Xn]→ Map(X,K).

The ideal, IX, of X is easily seen to be the kernel of φX. Thus, we may identify O(X)with the image of φX, the ring of functions from X to K induced by polynomials. Thus,two polynomials, f and g, are equal in O(X) if and only if they agree as functions whenrestricted to X.

By definition of X = V(a), we see that a ⊂ IX.

Proposition 9.3.21. Let X = V(a) be an affine variety over the algebraically closedfield K and let IX be the ideal of X. Then IX is the intersection of all the maximalideals of K[X1, . . . , Xn] containing a. Thus, by Proposition 9.3.4, IX = rad(a).


Moreover, V(IX) = V(a) = X, and the canonical map

π : K[X1, . . . , Xn]/a→ K[X1, . . . , Xn]/IX = O(X)

induces a homeomorphism from Max(O(X)) to Max(K[X1, . . . , Xn]/a).

Proof Recall that the element (a1, . . . , an) ∈ Kn corresponds to the maximal idealma1,...,an ⊂ K[X1, . . . , Xn], which is the kernel of the evaluation map εa1,...,an

: K[X1, . . . , Xn]→K. An element (a1, . . . , an) lies in X if and only if every element of a vanishes on(a1, . . . , an), meaning that a ⊂ ma1,...,an

. In particular, since K is algebraically closed,the points of X correspond to the maximal ideals of K[X1, . . . , Xn] containing a.

But an element of K[X1, . . . , Xn] lies in IX if and only if it vanishes on each point ofX, and hence is contained in every maximal ideal of K[X1, . . . , Xn] containing a.

Since a ⊂ IX, V(IX) ⊂ V(a). But since every element of IX kills every element ofX = V(a), the two varieties coincide. The result now follows from Lemma 9.3.14.

In particular, an ideal b has the form IX for an affine variety X if and only if b = rad(a)for some ideal a. Since rad(rad(a)) = rad(a), this is equivalent to saying that b is its ownradical (Lemma 9.1.13).

Theorem 9.3.22. Let K be an algebraically closed field. There is a one-to-one corre-spondence between the affine varieties in Kn and the ideals in K[X1, . . . , Xn] which aretheir own radicals. The correspondence takes a variety X to IX and takes an ideal a thatis its own radical to V(a).

Since each direction of this correspondence is order reversing, we see that X ⊂ Yif and only if IY ⊂ IX, and if either inclusion is proper, so is the other. Similarly, ifeach of a and b is its own radical, then a ⊂ b if and only if V(b) ⊂ V(a), and if eitherinclusion is proper, so is the other.

Proof If a is its own radical, then IV(a) = rad(a) = a. Conversely, if X is a variety inKn, then X = V(IX) by Proposition 9.3.21.

It is easy to see (Corollary 9.1.16) that a finite intersection of prime ideals is its ownradical.

Corollary 9.3.23. Let K be an algebraically closed field and let a be a finite intersectionof prime ideals of K[X1, . . . , Xn]. Then O(V(a)) = K[X1, . . . , Xn]/a.

In particular, affine n-space Kn is V(0). Since 0 is prime, we obtain a calculation ofO(Kn).

Corollary 9.3.24. O(Kn) = K[X1, . . . , Xn] for any algebraically closed field K. Thus,polynomials in n variables over K are determined by their effect as functions on Kn.

We could have given simpler argument using Proposition 7.3.26 and Lemma 8.4.3.SinceK[X1, . . . , Xn] is Noetherian, the theory of primary decomposition shows (Corol-

lary 9.2.11) that an ideal a of K[X1, . . . , Xn] is its own radical if and only if it is a finiteintersection of prime ideals. Using this, we can show that every affine variety over analgebraically closed field has a unique decomposition as a union of irreducible varieties.

Definition 9.3.25. An affine variety X over an algebraically closed field is irreducibleif whenever we have affine varieties Y and Z with X = Y ∪ Z, either X = Y or X = Z.

Recall from Corollary 7.6.10 that a prime ideal contains an intersection a∩b of idealsif and only if it contains at least one of a and b. We obtain the following lemma.


Lemma 9.3.26. Let a and b be ideals in the commutative ring A. Then V (a ∩ b) =V (a) ∪ V (b) in either Spec(A) or Max(A).

If a and b are ideals in K[X1, . . . , Xn] for K algebraically closed, then the affinevarieties V(a ∩ b) and V(a) ∪V(b) are equal.

Proposition 9.3.27. An affine variety X over an algebraically closed field K is irre-ducible if and only if IX is prime.

Proof Suppose that IX is prime and that X = Y ∪ Z. Clearly, IY∪Z = IY ∩ IZ. Butprime ideals are irreducible ideals (Lemma 9.2.17), so either IX = IY or IX = IZ. ByTheorem 9.3.22, either X = Y or X = Z, so X is irreducible.

Conversely, suppose that X is irreducible. By Corollary 9.2.11 and Theorem 9.3.22,X = V(a), where a =

⋃mi=1 pi, where pi is prime for i = 1, . . . ,m. Now Lemma 9.3.26

gives X =⋃mi=1 V(pi). Since X is irreducible, X = V(pi) for some i, and hence IX = pi

by Theorem 9.3.22.

Corollary 9.2.22 shows that every self-radical ideal of K[X1, . . . , Xn] may be writtenuniquely as an intersection

⋂mi=1 pi of prime ideals such that no pi contains

⋂j =i pj .

Theorem 9.3.22 and Lemma 9.3.26 give the following proposition.

Proposition 9.3.28. Every affine variety X over the algebraically closed field has aunique decomposition

X =m⋃i=1

Xi

where Xi is irreducible for i = 1, . . . ,m, and no Xi is contained in⋃j =iXj. Here, the

uniqueness is up to a reordering of the Xi.

We now define the mappings appropriate for comparing two affine varieties.

Definition 9.3.29. Let K be an algebraically closed field and let X ⊂ Kn be an affinevariety. A polynomial function f : X→ Km is a function of the form

f(x) = (f1(x), . . . , fm(x))

for all x ∈ X, where f1, . . . , fm ∈ O(X). If the image of f is contained in an affine varietyY ⊂ Km, we may consider f to be a polynomial function from X to Y.

The next lemma is quite elementary, but occurs frequently in the subsequent material.

Lemma 9.3.30. Let A be a commutative ring and let g(X1, . . . , Xm) ∈ A[X1, . . . , Xm].Suppose given polynomials f1, . . . , fm ∈ A[X1, . . . , Xn]. Then evaluating each Xi at figives a polynomial g(f1, . . . , fm) ∈ A[X1, . . . , Xn]. The evaluation of this polynomial atan element x ∈ An is given by

g(f1, . . . , fm)(x) = g(f1(x), . . . , fm(x)),

where the right hand side is the evaluation of g at the element (f1(x), . . . , fm(x)) ∈ Am.In other words, if εf1,...,fm : A[X1, . . . , Xm]→ A[X1, . . . , Xn] is the A-algebra homo-

morphism obtained by evaluating Xi at fi for i = 1, . . . ,m, then for g ∈ A[X1, . . . , Xm],the function from An to A induced by εf1,...,fm

(g) is given by

(εf1,...,fm(g))(x) = (g ◦ f)(x),

where f : An → Am is the function given by f(x) = (f1(x), . . . , fm(x)) for all x ∈ An.


Proof If f : B → C is a homomorphism of commutative A-algebras and b1, . . . , bm ∈ B,then the composite

A[X1, . . . , Xm]εb1,...,bm−−−−−−→ B

f−→ C

is just the evaluation map εf(b1),...,f(bm). Taking f to be the map from A[X1, . . . , Xn]obtained from evaluation at x ∈ An, the result follows.

A first application is to composites of polynomial functions.

Corollary 9.3.31. Let K be an algebraically closed field and let X ⊂ Kn and Y ⊂ Km

be affine varieties. Then for any polynomial mapping f : X → Y, there is a polynomialmapping f : Kn → Km making the following diagram commute.

X Y

Kn Km��

∩

��f

��

∩

��f

If Z ⊂ Kk is another affine variety and if g : Y→ Z is a polynomial mapping, thenthe composite g ◦ f is a polynomial mapping as well.

Finally, let ξi be the image of Xi under the canonical map π : K[X1, . . . , Xn]→ O(X)for i = 1, . . . , n. Then ξi : X → K is the restriction to X of the projection of Kn ontoits i-th factor. Thus, the polynomial mapping ι : X→ Kn given by

ι(x) = (ξ1(x), . . . , ξn(x))

is just the inclusion map of X in Kn. The image of ι is, of course X, so the identitymap of X is a polynomial mapping.

Proof Since f is a polynomial mapping, there are elements f1, . . . , fm ∈ O(X) suchthat f(x) = (f1(x), . . . , fm(x)) for all x ∈ X. But O(X) = K[X1, . . . , Xn]/IX, sothere are polynomials fi ∈ K[X1, . . . , Xn] representing fi for i = 1, . . . , n. Settingf(x) = (f1(x), . . . , fm(x)) for all x ∈ Kn, we obtain a polynomial mapping f : Kn → Km

making the diagram commute.Similarly, there are polynomials gi ∈ K[X1, . . . , Xm] for i = 1, . . . , k such that the

polynomial mapping g : Km → Kk given by g(y) = (g1(y), . . . , gk(y)) for y ∈ Km

restricts on Y to g : Y→ Z.The component functions of g ◦ f are the maps gi ◦ f , which satisfy

(gi ◦ f)(x) = gi(f1(x), . . . , fm(x))

for all x ∈ Kn. By Lemma 9.3.30, this is the map induced by the polynomial gi(f1, . . . , fm) ∈K[X1, . . . , Xn]. Thus, the composite g ◦ f : Kn → Kk is a polynomial mapping whichclearly restricts on X to the composite function (g ◦ f) : X→ Z. Since the restriction toX of any element of K[X1, . . . , Xn] determines an element of O(X), the the polynomialmappings are closed under composites.

Let ξi be the image in O(X) of the polynomial Xi. Then Xi : Kn → K restricts toξi on X. But Xi(a1, . . . , an) = ai, and the result follows.


In particular, Corollary 9.3.31 shows that the affine varieties and polynomial mappingsform a category.

Definition 9.3.32. Let K be an algebraically closed field. We write VarK for thecategory whose objects and are the affine varieties over K and whose morphisms aregiven as follows. If X and Y are affine varieties over K, then the set of morphisms,MorVarK

(X,Y), from X to Y in VarK is the set of polynomial mappings from X to Y.

We now show that polynomial mappings induce homomorphisms of coordinate rings.

Proposition 9.3.33. Let X ⊂ Kn and Y ⊂ Km be affine varieties over the algebraicallyclosed field K and let f : X→ Y be a polynomial mapping, given by

f(x) = (f1(x), . . . , fm(x))

for x ∈ X, where fi ∈ O(X) for i = 1, . . . ,m. Then there is a unique K-algebrahomomorphism O(f) : O(Y) → O(X) with the property that O(f)(ξi) = fi for i =1, . . . ,m, where ξi is the image of Xi under the canonical map π : K[X1, . . . , Xm] →O(Y).

If g : Y → Z is another polynomial mapping, then O(f) ◦ O(g) = O(g ◦ f). Also,O(1X) = 1O(X). Thus, O is a contravariant functor from the category, VarK , of affinevarieties and polynomial mappings over K to the category of commutative K-algebras offinite type and K-algebra homomorphisms.

Proof Since ξ1, . . . , ξm generate O(Y) as a K-algebra, there is at most one K-algebrahomomorphism carrying ξi to fi for all i. We must show that such a homomorphismexists.

For each i, let fi be a lift of fi under the canonical map π : K[X1, . . . , Xn]→ O(X).Then there is a K-algebra homomorphism εf1,...,fm

: K[X1, . . . , Xm] → K[X1, . . . , Xn]

obtained by evaluating Xi at fi for i = 1, . . . ,m. By the universal property of quotientrings, it suffices to show that εf1,...,fm

(IY) ⊂ IX: We may then define O(f) to be theunique homomorphism making the following diagram commute.

K[X1, . . . , Xm] K[X1, . . . , Xn]

O(Y) O(X)��

π

��εf1,...,fm

��π

��O(f)

Thus, let h(X1, . . . , Xm) ∈ IY. We wish to show that εf1,...,fm(h) ∈ IX, meaning

that it vanishes on each element of X. By Lemma 9.3.30, εf1,...,fm(h) = h ◦ f as a

function on Kn, where f(x) = (f1(x), . . . , fm(x)) for x ∈ Kn. Since fi is a lift of fi toK[X1, . . . , Xn], fi(x) = fi(x) for all x ∈ X. Thus, as a function on X, f = f . Sincef(X) ⊂ Y and since h vanishes on Y, εf1,...,fm

(h) ∈ IX, as desired.If g : Y → Z ⊂ Kk is a polynomial map, write g(y) = (g1(y), . . . , gk(y)) for y ∈ Y,

where gi ∈ O(Y) for all i. Let gi be a lift of gi to K[X1, . . . , Xm] for i = 1, . . . , k. Thenthe following diagram commutes.

K[X1, . . . , Xk] K[X1, . . . , Xm] K[X1, . . . , Xn]

O(Z) O(Y) O(X)��

π

��εg1,...,gk

��π

��εf1,...,fm

��π

��O(g)

��O(f)


Thus, O(f) ◦ O(g)(ξi) is the image in O(X) of εf1,...,fm(gi) = gi(f1, . . . , fm). Now, an

element of O(X) is determined by its value as a function on X. But for x ∈ X, we have

gi(f1(x), . . . , fm(x)) = gi(f(x))= gi(f(x)),

since f(x) ∈ Y. Since they have the same effect on elements of X, O(f) ◦O(g)(ξi) mustbe equal to the i-th component function of g ◦ f . Thus, the uniqueness statement usedto define O(g ◦ f) gives O(f) ◦O(g) = O(g ◦ f).

Finally, Corollary 9.3.31 shows that the identity map of X is given by 1X(x) =(ξ1(x), . . . , ξn(x)) for all x ∈ X, so O(1X) is characterized by O(1X)(ξi) = ξi for i =1, . . . , n. This forces O(1X) = 1O(X).

Proposition 9.3.34. Let X ⊂ Kn and Y ⊂ Km be affine varieties over the algebraicallyclosed field K. Then the passage from f : X→ Y to O(f) : O(Y)→ O(X) gives a one-to-one correspondence between the polynomial mappings from X to Y and the K-algebrahomomorphisms from O(Y) to O(X).

Proof By Proposition 9.3.33, it suffices to show that every K-algebra homomorphismγ : O(Y)→ O(X) is equal to O(f) for some polynomial mapping f : X→ Y.

Thus, let γ : O(Y) → O(X) be a K-algebra homomorphism, and let fi = γ(ξi) ∈O(X) for i = 1, . . . ,m. (As above, ξi is the image of Xi under the canonical mapπ : K[X1, . . . , Xm]→ O(Y).) Let f : X→ Km be the polynomial mapping given by

f(x) = (f1(x), . . . , fm(x))

for x ∈ X. By the characterization of the maps O(f) given in Proposition 9.3.33, itsuffices to show that f(X) ⊂ Y.

Now, Y = V(IY) is the set of elements of Km which vanish under every element ofthe ideal of Y, so it suffices to show that if g ∈ IY and x ∈ X, then g vanishes on f(x).

Let fi be a lift of fi to K[X1, . . . , Xn] for i = 1, . . . ,m and let f : Kn → Km bethe polynomial mapping whose i-th component function is fi for i = 1, . . . ,m. Thenf(x) = f(x) for all x ∈ X, so it suffices to show that the polynomial inducing g ◦ f liesin IX.

Lemma 9.3.30 shows that g◦ f is induced by εf1,...,fm(g). Note that the fi were chosen

to make the following diagram commute.

K[X1, . . . , Xm] K[X1, . . . , Xn]

O(Y) O(X)��

π

��εf1,...,fm

��π

��γ

The kernels of the vertical maps π are precisely IY and IX. Since g ∈ IY, its imageunder εf1,...,fm

must lie in IX.

Thus, the relationships between affine varieties given by polynomial mappings andthe relationships between their corrdinate rings given by K-algebra homomorphisms areprecise mirrors of one another.


Corollary 9.3.35. Let X and Y be affine varieties over the algebraically closed field K.Then there is a polynomial isomorphism between X and Y (i.e., a bijective polynomialmapping f : X → Y whose inverse function is polynomial) if and only if the coordinaterings O(X) and O(Y) are isomorphic as K-algebras.

Proof If f : X→ Y is a polynomial isomorphism, then O(f) is an isomorphism by thefunctoriality of the coordinate ring: If f−1 : Y → X is the inverse function of f , thenO(f) ◦O(f−1) = O(1X) = 1O(X) and O(f−1) ◦O(f) = O(1Y) = 1O(Y).

If O(f) is an isomorphism, then Proposition 9.3.34 shows that O(f)−1 = O(g) forsome polynomial mapping g : Y→ X. But then O(f ◦ g) = 1O(Y), so Proposition 9.3.34shows that f ◦ g = 1Y. Similarly, g ◦ f = 1X.

We next show that the polynomial functions between affine varieties may be identifiedwith the algebraic maps between the maximal ideal spectra of their coordinate rings.

Proposition 9.3.36. Let X ⊂ Kn and Y ⊂ Km be affine varieties over the algebraicallyclosed field K. Let f : X → Y be a polynomial mapping. Then there is a commutativediagram

Max(O(X)) Max(O(Y))

X Y��

∼=

��O(f)∗

��∼=��f

where the vertical maps are the standard isomorphisms. Thus, the algebraic maps fromMax(O(X)) to Max(O(Y)) correspond precisely under the standard isomorphisms to thepolynomial mappings from X to Y.

Proof The standard isomorphism from Max(O(X)) to X is the composite

Max(O(X)) π∗−→∼= V (IX) ν−1

−−→∼= X,

where π : K[X1, . . . , Xn] → O(X) is the canonical map. The analogous result holds forY.

Let f : Kn → Km be a polynomial mapping which restricts to f on X. Then the topand bottom squares of the following diagram commute.

Max(O(X)) Max(O(Y))

Max(K[X1, . . . , Xn]) Max(K[X1, . . . , Xm])

Kn Km

X Y

��π∗

��O(f)∗

��π∗

��O(f)∗

ν ∼=��f

ν ∼=

∪

��f

∪


It suffices to show that the middle square commutes.For x = (a1, . . . , an) ∈ Kn, ν(x) is the kernel of the evaluation map εa1,...,an , and

hence O(f)∗(ν(x)) is the kernel of εa1,...,an ◦ O(f). Let fi be the i-th component func-tion of f . Then O(f) is precisely the evaluation map εf1,...,fm

: K[X1, . . . , Xm] →K[X1, . . . , Xn]. Thus, εa1,...,an

◦ O(f) is the evaluation map from K[X1, . . . , Xm] → K

which evaluates Xi at fi(a1, . . . , an). The kernel of this evaluation map is preciselyν(f(x)), so the middle square commutes.

Corollary 9.3.37. Polynomial mappings are continuous in the Zariski topology.

The polynomial mappings give a much more rigid notion of equivalence of varietiesthan is provided by the Zariski topology alone. Indeed, it is quite possible for varieties tobe homeomorphic as spaces but inequivalent as varieties. Consider the following example.

Example 9.3.38. Let K be an algebraically closed field. Let κ : K[X,Y ] → K[X] bethe K-algebra homomorphism taking X to X2 and Y to X3, and let p = kerκ. Sinceimκ is a domain, p is prime, so O(V(p)) = K[X,Y ]/p by Corollary 9.3.23. (We leave itas an exercise to show that p = (Y 2 −X3).)

Write κ : O(V(p)) = K[X,Y ]/p → K[X] = O(K) for the homomorphism inducedby κ. Then Proposition 9.3.34 shows that κ = O(f), where f : K → V(p) is given byf(x) = (x2, x3) for all x ∈ K. By inspection, f : K → V(p) is injective. We claim it isa homeomorphism. But the coordinate ring of V(p) is isomorphic to imκ = K[X2, X3],which cannot be generated as a K-algebra by a single element. Thus, Corollary 9.3.35shows that V(p) is not isomorphic to K as a variety (i.e., through polynomial mappings).

The coordinate ring O(K) = K[X] is a P.I.D. Thus, as shown in Example 1 ofExamples 9.3.17, the proper closed subsets of K are precisely the finite subsets of K.Since affine varieties are T1 spaces, f(S) is closed in V(p) for every proper closed subspaceS of K. Thus, to show that f is a homeomorphism, it suffices to show it is onto. Thus,by Proposition 9.3.36, it suffices to show that i∗ : Max(K[X]) → Max(K[X2, X3]) isonto, where i : K[X2, X3] ⊂ K[X] is the inclusion.

For this, we shall make use of the theory of integral dependence developed in Sec-tion 12.7. Note first that since K[X] is generated by 1, X as a K[X2, X3]-module,it is integral over K[X2, X3] by Proposition 12.7.2. Now Proposition 12.7.22 showsthat for each maximal ideal m of K[X2, X3] there is a prime ideal p of K[X] withm = p ∩K[X2, X3]. But every nonzero prime ideal of K[X] is maximal, so m lies in theimage of i∗ : Max(K[X])→ Max(K[X2, X3]).


2. Let A be a commutative ring. Show that A[X2, X3] is free as an A[X2]-module,with basis 1, X3. Deduce that the kernel of the evaluation map εX2,X3 : A[X,Y ]→A[X2, X3] is the principal ideal generated by Y 2 −X3.

3. Let K be a field and fix n > 0. Suppose the map ν : Kn → Max(K[X1, . . . , Xn])is onto. Show that K is algebraically closed.

4. Let K be any field. Show that a maximal ideal m lies in the image of ν : Kn →Max(K[X1, . . . , Xn]) if and only if there is an element (a1, . . . , an) ∈ Kn such thatf(a1, . . . , an) = 0 for all f ∈ m.


5. Show that we may identify Max(R[X]) with the upper half plane in C. What isthe map Max(C[X])→ Max(R[X]) induced by the natural inclusion?

6. What is Spec(A×B)?

7. Let A be a commutative ring. Show that the topology on Spec(A) is T0, i.e., thatfor any pair of points p, q, at least one of the two points is open in the subspacetopology on {p, q}.

8. Let A be a commutative ring. Show that the sets U({a}) for a ∈ A form a basisfor the Zariski topology on Spec(A). It is customary to write X = Spec(A) andXa = U({a}).

9. Let A be a commutative ring. Show that Spec(A) is compact in the sense that anyopen cover of Spec(A) has a finite subcover. (Many writers call a non-Hausdorffspace with this property quasi-compact.)

9.4 Tensor Products

The theory of tensor products is one of the most important tools in module theory. Wehave already seen it used implicitly in more than one argument. One especially importantapplication, which we shall emphasize here, is extension of rings, whereby, if M is anA-module and f : A → B is a ring homomorphism, we create an associated B-modulefrom the tensor product B ⊗A M . This construction is functorial, and has importantapplications to exactness arguments.

Tensor products are a good example of an important concept that really cannot beadequately understood except via universal mapping properties.

We shall use some somewhat nonstandard terminology in the next definition, as thereare two definitions of bilinearity common in the literature, and we wish to distinguishbetween them.

Definitions 9.4.1. Let A be a ring and suppose given a right A-module M and a leftA-module N . Let G be an abelian group. A function f : M × N → G is said to beweakly A-bilinear if

f(m+m′, n) = f(m,n) + f(m′, n)f(m,n+ n′) = f(m,n) + f(m,n′) and

f(ma, n) = f(m, an)

for all choices of m,m′ ∈M , n, n′ ∈ N , and a ∈ A.If A is commutative, then we need not distinguish between left and right modules.

Let M1, M2, and N be A-modules. We say that f : M1 ×M2 → N is A-bilinear if foreach m1 ∈M1 and each m2 ∈M2, the maps

f(m1,−) : M2 → N and f(−,m2) : M1 → N

are A-module homomorphisms.

If A is commutative, then an A-bilinear map is easily seen to be weakly A-bilinear.Note also that for A-arbitrary, any weakly A-bilinear map is Z-bilinear.

We shall give a formal construction of the tensor product, but its purpose is thatit is the universal group for bilinear functions. Thus, we shall have an abelian group


M ⊗AN and a weakly A-bilinear map ι : M ×N →M ⊗AN with the following universalproperty: For each abelian group G and each weakly A-bilinear map f : M × N → G,there is a unique group homomorphism f : M⊗AN → G such that the following diagramcommutes.

M ×N G

M ⊗A N

��f

��

ι

�� f

Surprisingly, it turns out that if A is commutative, then M ⊗AN is an A-module that isuniversal in exactly the same manner for A-bilinear mappings. Thus, the two differentnotions of bilinearity are “solved” by exactly the same construction.

Indeed, almost everything we know about tensor products will be derived from thisuniversal property, rather than the actual construction of the tensor product. Probablythe most important fact coming out of the construction itself is that M⊗AN is generatedas an abelian group by the image of ι.

We now construct the tensor product. First, let F (M,N) be the free abelian group(i.e., free Z-module) on the set M ×N . Thus, using the notation of Section 7.7,

F (M,N) =⊕

(m,n)∈M×NZ.

Continuing this notation, we write e(m,n) for the canonical basis element correspondingto the (m,n)-th summand. We then have a function i : M × N → F (M,N) withi(m,n) = e(m,n).

Now let H ⊂ F (M,N) be the subgroup generated by

{e(m+m′,n) − e(m,n) − e(m′,n) |m,m′ ∈M,n ∈ N}∪{e(m,n+n′) − e(m,n) − e(m,n′) |m ∈M,n, n′ ∈ N}∪{e(ma,n) − e(m,an) |m ∈M,n ∈ N, a ∈ A}

Definition 9.4.2. With the conventions above, we set the tensor product M ⊗A N

equal to F (M,N)/H, and set ι : M ×N →M ⊗A N equal to the composite M ×N i−→F (M,N) π−→M ⊗A N , where π is the canonical map onto F (M,N)/H.

We write m⊗ n for the element ι(m,n) ∈M ⊗A N .

The map ι : M × N → M ⊗A N is easily seen to be weakly A-bilinear. That isimportant in showing that the following proposition gives universality for weakly bilinearmaps.

Proposition 9.4.3. Suppose given a right A-module M , a left A-module N , and a weaklyA-bilinear map f : M×N → G. Then there is a unique group homomorphism f : M⊗ANsuch that the following diagram commutes.

M ×N G

M ⊗A N

��f

��

ι

�� f


Proof The function f : M × N → G is defined on the generators of the free abeliangroup F (M,N). By the universal property of free abelian groups (Lemma 7.7.32), thereis a unique group homomorphism f : F (M,N) → G such that f(e(m,n)) = f(m,n) forall (m,n) ∈M ×N .

Since f is weakly A-bilinear, f is easily seen to vanish on the generators of H ⊂F (M,N). So the Noether Isomorphism Theorem, which gives the universal property offactor groups, shows that there is a unique group homomorphism f : M ⊗AN → G withf ◦ π = f . Here, π : F (M,N) → M ⊗A N is the canonical map. The composite of theuniqueness of f with respect to f and the uniqueness of f with respect to f gives thedesired result.

Warning: Not every element of M ⊗A N has the form m⊗ n. The generic element hasthe form m1 ⊗ n1 + · · ·+mk ⊗ nk for k ≥ 0 (the empty sum being 0), with mi ∈M andni ∈ N for 1 ≤ i ≤ k.Remarks 9.4.4. In practice, in defining homomorphisms out of tensor products, oneoften neglects to explicitly write down a weakly A-bilinear function from M ×N to G,but does so implicitly, by a statement such as “setting f(m ⊗ n) = 〈expression〉 definesa homomorphism f : M ⊗A N → G.” Of course, f ◦ ι is the desired weakly A-bilinearfunction, and is specified in a statement such as the above, since m⊗ n = ι(m,n). It isleft to the reader to verify that the expression that is supplied in such a statement doesin fact define a weakly A-bilinear function.

We shall follow this convention of implicitly defined bilinear functions from now on,other than in explicit statements about universal properties of the tensor product. Wereiterate, as pointed out in the above warning, that specifying f on terms of the formm⊗n only gives the value of f on a generating set for M⊗AN , and not on every element.

The following fundamental observation is immediate from the fact that 0 + 0 = 0 ineither M or N . We shall use it without explicit mention.

Lemma 9.4.5. Let M and N be right and left A-modules, respectively. Then for m ∈Mand n ∈ N , the elements m⊗ 0 and 0⊗ n are 0 in M ⊗A N .

The next lemma says that the tensor product is a functor of two variables.

Lemma 9.4.6. Let f : M → M ′ and g : N → N ′ be left and right A-module ho-momorphisms, respectively. Then there is a homomorphism, which we shall denote byf ⊗ g : M ⊗A N →M ′ ⊗A N ′ whose effect on generators is given by

(f ⊗ g)(m⊗ n) = f(m)⊗ g(n).

If f ′ : M ′ → M ′′ and g′ : N ′ → N ′′ are left and right A-module homomorphisms,respectively, then

(f ′ ⊗ g′) ◦ (f ⊗ g) = (f ′ ◦ f)⊗ (g′ ◦ g).Finally, 1M ⊗ 1N is the identity map of M ⊗N .

Proof That setting (f ⊗ g)(m ⊗ n) = f(m) ⊗ g(n) gives rise to a well defined homo-morphism follows precisely as in Remarks 9.4.4. The rest follows since any two homo-morphisms that agree on the generators of a group must be equal.

The functoriality of tensor products is all that we need to establish their universalityfor A-bilinear mappings when A is commutative.


Proposition 9.4.7. Let A be a commutative ring and let M and N be A-modules. Thenthere is a natural A-module structure on M ⊗A N which is defined on generators by

a · (m⊗ n) = (am)⊗ n.Under this convention, ι : M × N is A-bilinear, and is universal with respect to thatproperty: For any A-bilinear map f : M ×N → N ′, there is a unique A-module homo-morphism f : M ⊗A N → N ′ such that the following diagram commutes.

M ×N N ′

M ⊗A N

��f

��

ι

�� f

Proof For a ∈ A, write μa : M → M for the map induced by multiplication by a.Because A is commutative, μa is an A-module homomorphism for each a ∈ A. Thus,μa ⊗ 1N : M ⊗A N → M ⊗A N is a group homomorphism. We use this to definemultiplication by a on M ⊗AN . That this gives an A-module structure on M ⊗AN maybe deduced from the basic properties of tensor products and of A-module structures,together with the fact that

(μa ⊗ 1N ) ◦ (μb ⊗ 1N ) = (μa ◦ μb)⊗ 1N = μab ⊗ 1N .

We leave the rest as an exercise for the reader.

There is a very useful generalization of Proposition 9.4.7 in which M has modulestructures coming from two different (and possibly non-commutative) rings. It will giveus a way to turn A-modules into B-modules, induced by any ring homomorphism fromA to B.

Definition 9.4.8. LetA andB be rings. We say thatM isB-A-bimodule if the followinghold.

1. M is a left B-module.

2. M is a right A-module.

3. These two module structures interact as follows. For m ∈ M , b ∈ B, and a ∈ A,we have b(ma) = (bm)a.

Examples 9.4.9.1. A itself is an A-A-bimodule via ordinary multiplication. Its A-A-submodules are

its two-sided ideals.

2. If f : A→ B is a ring homomorphism, then any B-B-bimodule may be pulled backto an A-B, a B-A, or an A-A-bimodule, via f .

3. If A is commutative, then any A-moduleM is an A-A-bimodule, where both actionsare just the usual action on M .

4. For any ring A, identify An with the space of 1 × n row matrices. Then An is aright module over Mn(A) by setting x ·M equal to the matrix product xM forx ∈ An and M ∈Mn(A). This gives An the structure of an A-Mn(A)-bimodule.


If M is a B-A-bimodule and N is a left A-module, we shall put a left B-modulestructure onM⊗AN which will be universal for weaklyA-bilinear maps with the followingadditional property.

Definition 9.4.10. Let M be a left B-module and let N be any set. We say that afunction f : M ×N → N ′ is B-linear in M if N ′ is a left B-module and for each n ∈ Nthe function

f(−, n) : M → N ′

is a B-module homomorphism.

Let M be a B-A-bimodule and let b ∈ B. Then multiplication by b, μb : M → Mis easily seen to be a homomorphism of right A-modules. Thus, the proof of the nextproposition proceeds along the lines of that of Proposition 9.4.7.

Proposition 9.4.11. Let M be a B-A-bimodule and let N be a left A-module. Thenthere is a natural B-module structure on M ⊗A N which is defined on generators by

b · (m⊗ n) = (bm)⊗ n.Under this convention, ι : M ×N is B-bilinear in M and satisfies the following universalproperty: if N ′ is a left B-module and if f : M ×N → N ′ is weakly A-bilinear as well asbeing B-linear in M , then there is a unique B-module homomorphism f : M ⊗AN → N ′

such that the following diagram commutes.

M ×N N ′

M ⊗A N

��f

��

ι

�� f

Similarly, if M is a right A-module and N is an A-B-bimodule, then the tensorproduct M ⊗A N inherits a right B-module structure and satisfies a similar universalproperty.

An important application of the role of bimodule structures in tensor products comesfrom the B-A-bimodule structure on B that arises from a ring homomorphism f : A→ B.Given a left A-module M , we obtain a left B-module B ⊗A M . The passage from Mto B ⊗AM is functorial, and is known as “extension of rings” or “base change,” amongother names. It satisfies an important universal property:

Proposition 9.4.12. Let f : A → B be a ring homomorphism. Let M be a left A-module and let N be a left B-module. Regarding N as an A-module via f , let g : M → Nbe an A-module homomorphism. Then there is a unique B-module homomorphism g :B ⊗AM → N such that the following diagram commutes.

M N

B ⊗AM��

αM

��g�� g

Here, αM : M → B ⊗AM is the A-module homomorphism defined by αM (m) = 1 ⊗mfor m ∈M . Explicitly, g is defined on generators by setting

g(b⊗m) = bg(m).


Proof Setting g(b⊗m) = bg(m) is easily seen to induce a left B-module homomorphismfrom B ⊗AM to N . Uniqueness follows from the fact that 1⊗m = αM (m), and hencethe commutativity of the diagram requires that g(1⊗m) = g(m).

Recall that if M and N are A-modules, then HomA(M,N) is the group (under addi-tion of homomorphisms) of A-module homomorphisms. The following may be expressedin category theoretic language by saying that passage from M to B⊗AM is a left adjointto the forgetful functor from B-modules to A-modules induced by f : A→ B.

Corollary 9.4.13. Let f : A→ B be a ring homomorphism. Let M be a left A-moduleand N a left B-module. Let αM : M → B⊗AM be the A-module homomorphism definedby αM (m) = 1⊗m for all m ∈M . Then there is an isomorphism of abelian groups

HomB(B ⊗AM,N)α∗

M−−→∼= HomA(M,N)

defined by α∗M (h) = h ◦ αM .

Additionally, the isomorphism is natural in the sense that if g : M ′ → M is anA-module homomorphism and h : N → N ′ is a B-module homomorphism, then thefollowing diagram commutes.

HomB(B ⊗AM,N) HomA(M,N)

HomB(B ⊗AM ′, N ′) HomA(M ′, N ′)

��α∗M

��β

��γ

��α∗M ′

Here, β and γ are defined by

β(k) = h ◦ k ◦ (1B ⊗ g) and γ(l) = h ◦ l ◦ gfor k ∈ HomB(B ⊗AM,N) and l ∈ HomA(M,N).

Proof Since αM is an A-module homomorphism, α∗M is well defined, and is clearly

a homomorphism between the Hom groups. Let h ∈ HomB(B ⊗AM,N). Then thefollowing diagram commutes.

M N

B ⊗AM��

αM

��α∗M (h)

�� h

So α∗M is an isomorphism by Proposition 9.4.12. The reader may verify the naturality

statement.

The simplest type of base change is the trivial one: B = A and f is the identity map.Here, Corollary 9.4.13 gives us an isomorphism

HomA(A⊗AM,N)α∗

M−−→∼= HomA(M,N)

for any pair of A-modules M and N . For formal reasons, this implies that αM : M∼=−→

A⊗AM for all A-modules M . We can also prove it directly, with a direct constructionof the inverse homomorphism:


Lemma 9.4.14. Let M be a left A-module and let αM : M → A⊗AM be the A-modulehomomorphism given by αM (m) = 1 ⊗m for all m ∈ M . Then αM is an isomorphismwhose inverse, εM : A ⊗A M → M , is defined on generators by εM (a ⊗ m) = am.Additionally, the isomorphisms εM and their inverses are natural in the sense that iff : M →M ′ is an A-module homomorphism, then the following diagram commutes.

A⊗AM M

A⊗AM ′ M ′

��εM

��

1A ⊗ f��

f

��εM ′

Proof It is easy to see that εM is well defined and that εM ◦ αM = 1M . So εM andαM are inverse isomorphisms, providing that αM is onto. But each generator a ⊗m isequal to αM (am), so the image of αM contains a generating set for A⊗AM . The rest isleft to the reader.

We’ve already seen a couple of examples of extension of rings without knowing it.

Corollary 9.4.15. Let S be a multiplicative subset of the commutative ring A. Thenfor any A-module M there is an isomorphism of S−1A-modules

ϕM : S−1A⊗AM → S−1M

whose effect on generators is given by ϕM ((a/s) ⊗ m) = am/s for a ∈ A, s ∈ S, andm ∈M .

This isomorphism is natural in the sense that if f : M → N is an A-module homo-morphism, then the following diagram commutes,

S−1A⊗AM S−1M

S−1A⊗A N S−1N

��ϕM

��

1⊗ f��

S−1(f)

��ϕN

where 1 is the identity of S−1A.

Proof The point is that Lemma 7.11.14 shows that the canonical map η : M → S−1Msatisfies exactly the same universal property in terms of A-module maps into S−1A-modules that Proposition 9.4.12 shows for the mapping αM : M → S−1A ⊗A M . Inparticular, the universal property for η shows that there exists a unique S−1A-modulehomomorphism ψ : S−1M → S−1A⊗AM such that the following diagram commutes.

M S−1A⊗AM

S−1M

��αM

��

η�� ψ

And ϕM : S−1A⊗AM → S−1M , if placed in the diagram in the opposite direction fromthat of ψ, would make it commute as well.


Thus, the composite S−1A-module homomorphisms ϕM ◦ ψ and ψ ◦ ϕM restrict tothe identity maps on the images of αM and η, respectively. Since these images generateS−1A⊗AM and S−1M , respectively, as modules over S−1A, ϕM and ψ must be inverseisomorphisms.

The naturality statement holds by an easy diagram chase.

Similarly, Lemma 7.7.19 shows that if a is a two-sided ideal in the ring A, then thepassage from a left A-module M to the A/a-module M/aM satisfies the same universalproperty as the extension of rings. We leave the proof of the following to the reader.

Corollary 9.4.16. Let a be a two-sided ideal in the ring A. Then for any left A-moduleM , there is an isomorphism of A/a-modules

βM : A/a⊗AM →M/aM

whose effect on generators is given by βM (a⊗m) = am for a ∈ A and m ∈M .This isomorphism is natural in the sense that if f : M → N is an A-module homo-

morphism, then the following diagram commutes,

A/a⊗AM M/aM

A/a⊗A N N/aN

��βM

��1⊗ f

��f

��βN

where 1 is the identity map of A/a, and f is the map induced by f .

We may now reinterpret Nakayama’s Lemma in terms of extension of rings.

Corollary 9.4.17. (Nakayama’s Lemma, second form) Let a be a two-sided ideal whichis contained in the Jacobson radical of the ring A and let M be a finitely generatedA-module. Then A/a⊗AM = 0 if and only if M = 0.

Proof M/aM = 0 if and only if M = aM . Now apply Nakayama’s Lemma.

We’d like to know that extension of rings takes free modules to free modules. Forthis purpose, it suffices to understand what tensoring does to a direct sum.

Proposition 9.4.18. Let N1, . . . , Nk be left A-modules and let ιi : Ni →⊕k

i=1Ni bethe canonical inclusion for i = 1, . . . , k. Then for any right A-module M , there is anisomorphism

ω :k⊕i=1

(M ⊗A Ni) −→M ⊗A(

k⊕i=1

Ni

)whose restriction to M ⊗A Ni is 1M ⊗ ιi for i = 1, . . . , k. If M is a B-A-bimodule, thenω is an isomorphism of left B-modules; if each Ni is an A-B-bimodule, then ω is anisomorphism of right B-modules.

Similarly, given right A-modules M1, . . . ,Mk and a left A-module N , there is anisomorphism ω′ :

⊕ki=1(Mi ⊗A N) → (

⊕ki=1Mi) ⊗A N whose restriction to Mi ⊗A N

is ιi ⊗ 1N for i = 1, . . . , k. Once again, appropriate bimodule structures give moduleisomorphisms.

These isomorphisms are natural in all variables.


Proof Let β : M ⊗A (⊕k

i=1Ni)→⊕k

i=1(M ⊗ANi) be the homomorphism whose effecton generators is given by

β(m⊗ (n1, . . . , nk)) = (m⊗ n1, . . . ,m⊗ nk).

Then the composites ω ◦ β and β ◦ ω are easily seen to give the identity maps on gen-erating sets for M ⊗A (

⊕ki=1Ni) and

⊕ki=1(M ⊗A Ni), and hence ω and β are inverse

isomorphisms.We leave the rest to the reader.

Corollary 9.4.19. Let f : A → B be a ring homomorphism. Then B ⊗A An is a freeB-module of rank n with basis 1⊗ e1, . . . , 1⊗ en, where e1, . . . , en is the canonical basisof An.

Similarly, if we regard An as a right A-module, then An⊗AB is a free right B-modulewith basis e1 ⊗ 1, . . . , en ⊗ 1.

Proof We have B⊗AAn ∼= (B⊗AA)n. Now notice that the isomorphism B⊗AA ∼= Bfrom Lemma 9.4.14 is one of left B-modules, and takes 1⊗ 1 to 1.

The proof for right modules is identical.

Recall that Mm,n(A) is the set of m×n matrices over A. Given a ring homomorphismf : A→ B, we write f∗ : Mm,n(A)→Mm,n(B) for the function that takes a matrix (aij)to the matrix whose ij-th coordinate is f(aij).

Recall from Section 7.10 that if g : An → Am is a homomorphism of right A-modules(or left ones, of course, if A is commutative), then g is induced by left-multiplying acolumn vector by the m× n matrix whose i-th column is g(ei).

Similarly, if g : An → Am is a left A-module map, then g is induced by right-multiplying a row vector by the n × m matrix whose i-th row is g(ei). In particular,there are two different conventions possible for the matrix of g if A is commutative.

It is now an easy verification to show the following.

Corollary 9.4.20. Let f : A → B be a ring homomorphism. Let g : An → Am be ahomomorphism of right A-modules, represented by the matrix M . Then the matrix ofg ⊗ 1B with respect to the bases coming from Corollary 9.4.19 is precisely f∗(M).

Similarly, if the left A-module homomorphism g : An → Am is represented (underwhichever convention one chooses to use in the case where both A and B are commutative)by the matrix M , then the matrix of 1B ⊗ g with respect to the bases of Corollary 9.4.19is f∗(M).

Suppose given ring homomorphisms Af−→ B

g−→ C. We’d like to know that theextension of rings from A to B and then from B to C coincides with the extension inone step from A to C. The verification of this depends on an understanding of theassociativity properties of iterated tensor product.

In the study of cartesian products, the existence of a single n-fold product greatlysimplifies the discussion of iterated products and their associativity properties. We shallmake use of the same device in the study of iterated tensor products.

Definitions 9.4.21. Let A1, . . . , An−1 be rings. Suppose given a right A1-module M1,a left An−1-module Mn, together with Ai−1-Ai-bimodules, Mi for 2 ≤ i ≤ n− 1. Let Gbe an abelian group. Then a function

f : M1 × · · · ×Mn → G


is weakly A1, . . . , An−1-multilinear if the maps

f(m1, . . . ,mi−1,−,−,mi+2, . . . ,mn) : Mi ×Mi+1 → G

are weakly Ai-bilinear for each i = 1, . . . , n − 1 and for all possible choices of elementsmj ∈Mj for j �= i, i+ 1.

If A is commutative and if M1, . . . ,Mn are A-modules, then a function

f : M1 × · · · ×Mn → N

is said to be A-multilinear if N is an A-module and the maps

f(m1, . . . ,mi−1,−,mi+1, . . . ,mn) : Mi → N

are A-module homomorphisms for each i = 1, . . . , n and all possible choices of elementsmj ∈Mj for j �= i.

A-multilinear maps are easily seen to be weakly A, . . . , A-multilinear. Note that theword “weakly” may be safely omitted in the A1, . . . , An-multilinear case if the ringsA1, . . . , An are not all equal, as there’s no stronger concept with which to confuse it.

Note that ifA is commutative, then a weaklyA, . . . , A-multilinear map isA-multilinearif and only if it is A-linear on M1 in the sense of Definition 9.4.10.

The construction of the n-fold tensor product has no surprises. Suppose given a rightA1-module M1, a left An−1-module Mn, and Ai−1-Ai-bimodules Mi for i = 2, . . . , n −1. Let F (M1, . . . ,Mn) be the free abelian group on M1 × · · · ×Mn, where e(m1,...,mn)

is the basis element corresponding to (m1, . . . ,mn) ∈ M1 × · · · ×Mn. Let H be thesubgroup of F (M1, . . . ,Mn) generated by all elements of the form e(m1,...,mi+m′

i,...,mn)−e(m1,...,mi,...,mn)−e(m1,...,m′

i,...,mn), for i = 1, . . . , n and all possible choices of mj ∈Mj forj = 1, . . . , n and m′

i ∈Mi, together with all elements of the form e(m1,...,mia,mi+1,...,mn)−e(m1,...,mi,ami+1,...,mn), for i = 1, . . . , n−1, and all possible choices of a ∈ Ai, andmj ∈Mj

for j = 1, . . . , n.

Definition 9.4.22. Under the conventions above, set

M1 ⊗A1 · · · ⊗An−1 Mn = F (M1, . . . ,Mn)/H

and write ι : M1×· · ·×Mn →M1⊗A1 · · ·⊗An−1Mn for the map that takes (m1, . . . ,mn)to the image in M1 ⊗A1 · · · ⊗An−1 Mn of the basis element e(m1,...,mn).

The proof of the next proposition is analogous to the proofs of Propositions 9.4.3,9.4.7, and 9.4.11.

Proposition 9.4.23. Suppose given a right A1-module M1, a left An−1-module Mn,together with Ai−1-Ai-bimodules, Mi for i = 2, . . . , n− 1. Let f : M1× · · · ×Mn → G bea weakly A1, . . . , An−1-multilinear function. Then there is a unique group homomorphismf : M1 ⊗A1 · · · ⊗An−1 Mn → G such that the following diagram commutes.

M1 × · · · ×Mn G

M1 ⊗A1 · · · ⊗An−1 Mn

��f

��

ι

�� f

Suppose in addition M1 is an A0-A1-bimodule. Then M1 ⊗A1 · · · ⊗An−1 Mn is a leftA0-module by an action in which

a ·m1 ⊗ · · · ⊗mn = (am1)⊗ · · · ⊗mn.


Moreover, if G is a left A0-module and f is A0-linear in M1, then the induced mapf : M1 ⊗A1 · · · ⊗An−1 Mn → G is a homomorphism of left A0-modules.

The analogous statement holds for the case where Mn is an An−1-An-bimodule andf is An-linear in Mn.

In particular, if A is commutative and M1, . . . ,Mn are A-modules, then there is anA-module structure on M1 ⊗A · · · ⊗AMn induced by

a ·m1 ⊗ · · · ⊗mn = (am1)⊗ · · · ⊗mn.

If G is an A-module and f : M1 × · · · ×Mn is an A-multilinear map, then the inducedmap f : M1 ⊗A · · · ⊗AMn → G is an A-module homomorphism.

As in the bilinear case, we shall generally define our multilinear maps implicitly,rather than explicitly.

Corollary 9.4.24. Suppose given a right A-module M1, a left B-module M3, and anA-B-bimodule M2. Then there are isomorphisms

(M1 ⊗AM2)⊗B M3γ1←−−∼= M1 ⊗AM2 ⊗B M3

γ2−→∼= M1 ⊗A (M2 ⊗B M3)

whose effects on generators are given by

γ1(m1 ⊗m2 ⊗m3) = (m1 ⊗m2)⊗m3 andγ2(m1 ⊗m2 ⊗m3) = m1 ⊗ (m2 ⊗m3).

Proof The formulæ above induce well defined homomorphisms γ1 and γ2 as stated, soit suffices to construct their inverse homomorphisms. We shall confine ourselves to thecase of γ1.

Here, we note that for m3 ∈ M3, there is a homomorphism from M1 ⊗A M2 toM1⊗AM2⊗BM3 that carries each generator m1⊗m2 to m1⊗m2⊗m3. This is enoughto construct a homomorphism δ : (M1⊗AM2)⊗BM3 →M1⊗AM2⊗BM3 whose effecton generators is given by δ((

∑ki=1mi ⊗m′

i) ⊗m′′) =∑ki=1(mi ⊗m′

i ⊗m′′). And δ iseasily seen to give an inverse to γ1.

Note that ifM1 is a C-A-bimodule, then γ1 and γ2 are isomorphisms of left C-modules.Similarly, we get right module isomorphisms if M3 is a B-C-bimodule.

Corollary 9.4.25. Suppose given ring homomorphisms Af−→ B

g−→ C. Then there is anatural isomorphism of left C-modules

C ⊗B (B ⊗AM) ∼= C ⊗AMfor A-modules M .

Proof The isomorphisms γi are clearly natural in all three variables. It suffices to notethat

(C ⊗B B)⊗AM ∼= C ⊗AMby Lemma 9.4.14.

The theory of localization, together with Nakayama’s Lemma, now gives a valuableapplication of extension of rings.


Corollary 9.4.26. Let M be a finitely generated module over the commutative ring A.Then M = 0 if and only if A/m⊗AM = 0 for every maximal ideal m of A.

Proof Let m be a maximal ideal of A. Recall from Proposition 7.11.24 that A/m isisomorphic to the quotient ring Am/mm of the localization Am of A. Thus,

A/m⊗AM ∼= Am/mm ⊗AM ∼= Am/mm ⊗Am (Am ⊗AM) = Am/mm ⊗Am Mm.

Since M is finitely generated over A, there is a surjection Ak →M for some k ≥ 0. Sincelocalization is an exact functor, we see that Mm is finitely generated over Am. Thus, ifA/m⊗AM = 0, then Mm = 0 by the second form of Nakayama’s Lemma.

Proposition 7.11.25 shows that M = 0 if and only if Mm = 0 for every maximal idealof M , so the result follows.

Exercises 9.4.27.1. Let a and b be two-sided ideals of A. Show that

A/a⊗A A/b ∼= A/(a + b).

2. Let f : A→ B be a ring homomorphism and let M and N be a right and a left B-module, respectively. Then we may consider them to be A-modules as well. Showthat there is a natural surjection from M ⊗AN onto M ⊗BN . In particular, everytensor product is a quotient group of the appropriate tensor product over Z.

† 3. Let a be a two-sided ideal of A, and let M and N be a right and a left A/a module,respectively. Show that the natural map M ⊗A N → M ⊗A/a N of the precedingproblem is an isomorphism.

† 4. Show that if A and B are commutative rings and f : A → B is a ring homomor-phism, then there is a natural B-module isomorphism from (B⊗AM)⊗B (B⊗AN)to B ⊗A (M ⊗A N) for A-modules M,N .

5. Let A be a commutative ring and let m,n > 0. Show that Am ⊗A An is a freeA-module of rank mn, with basis ei ⊗ ej for 1 ≤ i ≤ m and 1 ≤ j ≤ n. Here,e1, . . . , em and e1, . . . , en are the canonical bases of Am and An, respectively.

† 6. Give the generalization of Proposition 9.4.18 to infinite direct sums, and deducethat extension of rings carries infinitely generated free modules to free modules.

7. Let f : A→ B be a ring homomorphism. Show that passage from a left A-moduleM to the extended B-module B⊗AM provides a left adjoint to the forgetful functorfrom left B-modules to left A-modules (i.e., the functor that considers a B-moduleto be an A-module via f).

8. Let A be a ring. Show that if M is a left A-module, then the action of A on Minduces a homomorphism α : A⊗ZM →M . Conversely, if M is an abelian group,then any homomorphism α : A⊗ZM →M defines the function (α◦ι) : A×M →M ,which we may then investigate to see if it gives a module action.Show that α : A ⊗Z M → M induces an A-module structure on M if and only ifthe following diagrams commute:

A⊗Z A⊗Z M A⊗Z M

A⊗Z M M

��1A ⊗ α

��μ⊗ 1M

��α

��α

Z⊗Z M A⊗Z M

M

��ν ⊗ 1M

��

εM��α


Here, μ is induced by the multiplication of A, ν is the unique ring homomorphismfrom Z to A, and εM is the isomorphism induced by the action of Z on M as inLemma 9.4.14.

9.5 Tensor Products and Exactness

Here, we shall explore the relationships between tensor products and exact sequences,with particular attention paid to extension of rings. As an application, we shall studythe A-module homomorphisms f : An → Am between finitely generated free modules,and extend, as much as we can, the sort of results that are known to hold for finitelygenerated vector spaces over a division ring.

We’ve seen one example, tensoring with S−1A, where S is a multiplicative subset ofthe commutative ring A, where extension of rings is an exact functor. However, extensionof rings is often not an exact functor. For instance, as we shall see in Exercises 9.5.21,extension of rings from Z to Z2 fails to be exact. Thus, the next definition has significance.

Definition 9.5.1. A right A-module M is flat if the passage from left A-modules N toM ⊗A N and from A-module homomorphisms f to 1M ⊗ f gives an exact functor.

Similarly, a left A-module N is flat if −⊗A N gives an exact functor.

As we noted above, Corollary 9.4.15, together with the proof that passage to modulesof fractions is exact, gives the following corollary.

Corollary 9.5.2. Let S be a multiplicative subset of the commutative ring A. ThenS−1A is a flat A-module.

Similarly, the next corollary follows from Lemma 9.4.14.

Corollary 9.5.3. A itself is flat as either a left or a right A-module.

While extension of rings is not in general an exact functor, it does have some exactnessproperties.

Definitions 9.5.4. Let F be a covariant functor between two categories of modules.

1. We say that F is right exact if for each short exact sequence

M ′ f−→Mg−→M ′′ → 0

in the domain category of F , the resulting sequence

F (M ′)F (f)−−−→ F (M)

F (g)−−−→ F (M ′′)→ 0

is exact.

2. We say that F is left exact if for each short exact sequence

0→M ′ f−→Mg−→M ′′

in the domain category of F , the resulting sequence

0→ F (M ′)F (f)−−−→ F (M)

F (g)−−−→ F (M ′′)

is exact.


One can also study partial exactness properties of contravariant functors. We shalldo so (without giving formal definitions of what constitutes “left” and “right”) in ourstudy of the Hom functors in Section 9.7.

Proposition 9.5.5. Let M be a right A-module. Then the functor which takes N toM ⊗A N and takes f to 1M ⊗ f is a right exact functor from the category of left A-modules to abelian groups.

Similarly, if N is a left A-module, then − ⊗A N is a right exact functor on rightA-modules.

Proof We treat the case of M ⊗A−. The other is similar. Suppose given a short exactsequence

N ′ f−→ Ng−→ N ′′ → 0.

We wish to show that

M ⊗A N ′ 1M⊗f−−−−→M ⊗A N 1M⊗g−−−−→M ⊗A N ′′ → 0

is exact.Consider first the exactness at M ⊗A N ′′. This amounts to showing that 1M ⊗ g is

onto. But if n′′ = g(n), then m ⊗ n′′ = (1M ⊗ g)(m ⊗ n) for all m ∈ M , and hence theimage of 1M ⊗ g contains a set of generators for M ⊗A N ′′, so 1M ⊗ g is onto.

Clearly, (1M ⊗ g) ◦ (1M ⊗ f) = 0. Thus, if B ⊂M ⊗AN is the image of 1M ⊗ f , thenB is contained in the kernel of 1M ⊗ g. We obtain an induced map

h : (M ⊗A N)/B →M ⊗A N ′′.

It suffices to show that h is an isomorphism.To show this, we construct a map the other way. As usual, we use the universal

mapping property of the tensor product. This time, we shall explicitly define a weaklyA-bilinear function k : M×N ′′ → (M⊗AN)/B, as follows. For a pair (m,n′′) ∈M×N ′′,choose n ∈ N with g(n) = n′′, and set k(m,n′′) = m⊗ n, the element of (M ⊗A N)/Brepresented by m ⊗ n. To see that k(m,n′′) does not depend on which n ∈ g−1(n′′)we choose, suppose that g(n) = g(n1) = n′′. By the exactness of the original sequence,n− n1 = f(n′) for some n′ ∈ N ′. But then

(m⊗ n)− (m⊗ n1) = m⊗ (n− n1) = (1M ⊗ f)(m⊗ n′)

lies in B. Thus, k is a well defined function, and is easily seen to be weakly A-bilinear.Thus, there is a homomorphism k : M ⊗A N ′′ → (M ⊗A N)/B extending k. Now if

g(n) = n′′ and m ∈M , then

(h ◦ k)(m⊗ n′′) = h(m⊗ n) = (1M ⊗ g)(m⊗ n) = m⊗ n′′.

Since h ◦ k is the identity on a set of generators for M ⊗A N ′′, it must be the identitymap everywhere.

For m⊗ n ∈ (M ⊗AN)/B, we have (k ◦h)(m⊗ n) = k(m⊗g(n)) = m⊗ n, since anylift to N of g(n) may be used in defining k. So k ◦h is the identity on a set of generators,and hence k and h are inverse to one another.

The right exactness of extension of rings has some immediate applications, but let usfirst develop some tools for recognizing flat modules.


Proposition 9.5.6. Let F be a covariant functor from A-modules to abelian groups.Then F is exact if and only if it takes short exact sequences to short exact sequences.

Proof To see that an exact functor, F , preserves short exact sequences, it suffices toshow that F (0) = 0. Note that 0 0−→ 0 0−→ 0 is an exact sequence in which the morphismsare both isomorphisms. Since any functor preserves isomorphisms, the same is true of

F (0)F (0)−−−→ F (0)

F (0)−−−→ F (0). Thus, F (0) = 0, as claimed.

For the converse, suppose given an exact sequence M ′ f−→ Mg−→ M ′′ of A-modules.

Then we have a commutative diagram

0 0

M/ ker g

M ′ M M ′′

M ′/ ker f

0 0

��

�� g

��

π′

��f

��π

��g�� f

��

in which all the straight lines are exact.Suppose then that F takes short exact sequences to short exact sequences. Then it

is easy to see that it must take injections to injections and take surjections to surjec-tions. Thus, if we apply F to the above diagram, the diagonal straight lines will remainexact. Thus, imF (f) = imF (f) and kerF (g) = kerF (π), and hence the exactness

of F (M ′)F (f)−−−→ F (M)

F (g)−−−→ F (M ′′) follows from the exactness of five-term diagonalsequence.

Corollary 9.5.7. Let F be a right exact functor from A-modules to abelian groups. ThenF is exact if and only if for any injective A-module homomorphism f : M ′ → M , theinduced map F (f) is also injective.

The following consequence of Corollary 9.5.7 has an easy direct proof, as well.

Proposition 9.5.8. Let N1 and N2 be left A-modules. Then N1 ⊕N2 is a flat moduleif and only if both N1 and N2 are flat. The analogous statement holds for right modulesas well.

Proof Suppose given an injective right A-module homomorphism f : M ′ → M . Thenaturality of the isomorphism of Proposition 9.4.18 shows that we may identify f ⊗A1(N1⊕N2) with (f ⊗A 1N1)⊕ (f ⊗A 1N2). But clearly, a direct sum of homomorphisms isinjective if and only if each one of them is injective, and hence the result follows.

The next corollary now follows from Lemma 9.4.14 and induction on n.

Corollary 9.5.9. For n ≥ 1, the free module An is flat as either a right or a left A-module.

The next lemma can be useful in recognizing flatness.


Lemma 9.5.10. Let M and N be left and right A-modules, respectively, and suppose that∑ki=1mi⊗ni = 0 in M⊗AN . Then there are finitely generated submodules M0 ⊂M and

N0 ⊂ N such that mi ∈M0 and ni ∈ N0 for all i, and∑ki=1mi ⊗ ni = 0 in M0 ⊗A N0.

Proof As in Section 9.4, we write F (M,N) for the free abelian group on M ×N andwrite e(m,n) for the basis element corresponding to the ordered pair (m,n) ∈ M × N .Then our construction of tensor products shows that

∑ki=1 e(mi,ni) must be equal to a

sum of integral multiples of terms of the form e(xi+x′i,yi) − e(xi,yi) − e(x′

i,yi), e(xi,yi+y′i) −e(xi,yi)−e(xi,y′i), and e(xiai,yi)−e(xi,aiyi). So takeM0 to be the submodule ofM generatedby m1, . . . ,mk together with the x’s and x′’s and take N0 to be the submodule of Ngenerated by n1, . . . , nk together with the y’s and y′’s.

Corollary 9.5.11. Let N be a left A-module with the property that every finitely gener-ated submodule of N is flat. Then N is flat.

Proof Let f : M ′ → M be an injective homomorphism of right A-modules and let∑ki=1mi ⊗ ni be an element of the kernel of f ⊗ 1N . Then Lemma 9.5.10 provides a

finitely generated submodule N0 containing n1, . . . , nk such that∑ki=1 f(mi)⊗ni = 0 in

M ⊗A N0.By our hypothesis, N0 is flat, and hence

∑ki=1mi ⊗ ni = 0 in M ′ ⊗A N0, and hence

also in M ′ ⊗A N .

We can now characterize the flat modules over a P.I.D.

Corollary 9.5.12. Let A be a P.I.D. Then an A-module is flat if and only if it is torsion-free.

Proof Let M be a torsion-free A-module. Then every finitely generated submoduleof M is also torsion-free, and hence free by Proposition 8.9.6. By Corollary 9.5.9, thefinitely generated submodules of M are flat, so M is flat by Corollary 9.5.11.

Conversely, suppose that M is a flat A-module. Then M ⊗A A 1⊗η−−→ M ⊗A K isinjective, where η : A → K is the canonical inclusion of A in its field of fractions. ByCorollary 9.4.15, we may identify 1 ⊗ η with the canonical map η : M → M(0) from Mto its localization at (0). But the kernel of η is easily seen to be the torsion submoduleof M , and hence M is torsion-free.

There are times when tensoring with even a non-flat module will preserve short ex-actness. Recall that a short exact sequence

0→ N ′ f−→ Ng−→ N ′′ → 0

splits if f admits a retraction, meaning an A-module homomorphism r : N → N ′ suchthat r ◦ f = 1N ′ .

Corollary 9.5.13. Suppose given a split short exact sequence of left A-modules:

0→ N ′ f−→ Ng−→ N ′′ → 0.

Then for any right A-module M , the sequence

0→M ⊗A N ′ 1⊗f−−→M ⊗A N 1⊗g−−→M ⊗A N ′′ → 0

is exact and splits.


Proof Note that any map that admits a retraction is injective. Thus, since tensoringwith M is right exact, it suffices to show that 1 ⊗ f admits a retraction. But if r is aretraction for f , then 1⊗ r is a retraction for 1⊗ f .

We now consider the applications of this material to homomorphisms of free modules.We’ve already seen, in Theorem 7.9.8, that the rank of a free module is well defined overa commutative ring. The argument there implicitly used extension of rings from A tothe field A/m, where m is a maximal ideal of A. Now, we’re given a general extension ofrings functor, so we can give a more general result:

Proposition 9.5.14. Let A be any ring that admits a ring homomorphism f : A→ D,with D a division ring. Then the free left A-modules Am and An are isomorphic if andonly if m = n.

Proof Let g : Am → An be an A-module isomorphism and let g−1 be the inverseisomorphism. Since extension of rings is a functor, 1D⊗g is aD-module isomorphism fromDm to Dn, with inverse 1D ⊗ g−1. Since dimension is well defined for finite dimensionalvector spaces over a division ring (Corollary 7.9.5), m = n.

This does indeed cover the case of commutative rings by precisely the same argumentas that given earlier, as the commutative ring A maps to the field A/m if m is a maximalideal of A.

Here is another result about vector spaces that extends to a more general context.

Proposition 9.5.15. Let A be a commutative ring, or more generally, any ring thatadmits a ring homomorphism f : A→ D, with D a division ring. Let g : Am → An be asurjective homomorphism. Then m ≥ n.

Proof By the right exactness of the tensor product, 1D ⊗ g : Dm → Dn is surjective.Now apply Proposition 7.9.2.

Of course, ifD is a division ring and if f : Dm → Dn is surjective, then this determinesthe isomorphism type of ker f . In particular, any surjection from Dn to itself must bean isomorphism. We shall extend this latter result to commutative rings:

Proposition 9.5.16. Let A be a commutative ring. Then any surjective homomorphismf : An → An is an isomorphism.

Proof Let K = ker f . Then the exact sequence

0→ Ki−→ An

f−→ An → 0

splits by Proposition 7.7.51. Thus, for each maximal ideal m of A, the sequence

0→ A/m⊗A K 1⊗i−−→ A/m⊗A An 1⊗f−−→ A/m⊗A An → 0

is exact by Corollary 9.5.13. In particular, the dimension of the kernel of 1⊗ f must be0, and hence A/m⊗A K = 0 for each maximal ideal m of A.

Note that if r : An → K is a retraction for i, then r is surjective, and hence K isfinitely generated. Thus, K = 0 by Corollary 9.4.26.


We shall now show that if A is a commutative ring, and if f : Am → An is injective,then m ≤ n.

If A is an integral domain, then this is almost immediate, because tensoring withits field of fractions gives an exact functor to a category of vector spaces. If A is nota domain, we shall need some sort of substitute for the field of fractions. The thing tonotice is that in an integral domain, 0 is minimal in the partially ordered set of all primeideals under inclusion.

So the technique we shall use in general is to localize at a minimal prime ideal.Before we can do that we need to know that minimal primes exist. If A has finite Krulldimension, this is obvious. For the general case, we need further argument.

Lemma 9.5.17. A commutative ring A has minimal prime ideals.

Proof A minimal prime is a maximal element in the partially ordered set of prime idealsunder reverse inclusion. We apply Zorn’s Lemma. Given a totally ordered collection ofprime ideals, an “upper bound” in this ordering would be given by their intersection,provided it is prime.

Normally, intersections of prime ideals are not prime. (Look at Z, for instance.) Butif we’re intersecting a totally ordered collection, say p =

⋂i∈I pi, suppose that ab is in p.

Then if a is not in p, it must not be in pi for some i. But then b ∈ pj whenever pj ⊂ pi.But since the collection {pj | j ∈ I} is totally ordered, for j ∈ I we either have pj ⊂ pior pi ⊂ pj , so in either case b ∈ pj . So b must be in p, the intersection of the pj . So p isprime, and the hypotheses of Zorn’s Lemma are satisfied.

If f : Am → An is injective, so is its localization at a minimal prime ideal. Thus,we may assume that A has exactly one prime ideal, p, which therefore is its nil radical(Proposition 9.1.10). In other words, every element of p is nilpotent. The following willbe useful.

Lemma 9.5.18. Let A be a commutative ring with only one prime ideal, p. Let p1, . . . , pkbe nonzero elements of p. Then there is a nonzero element p ∈ p which annihilates eachof the pi.

Proof We claim that p may be taken to have the form pr11 . . . prk

k , with ri ≥ 0 for1 ≤ i ≤ k. We show this by induction on k. If k = 1, just take p = ps−1

1 , where s is thesmallest positive integer for which ps1 = 0. For k > 1, assume by induction that we’regiven an element p′ = pr11 . . . p

rk−1k−1 , such that p′ �= 0, and p′ annihilates p1, . . . , pk−1.

Let t be the smallest integer such that ptk = 0. If p′pt−1k �= 0, then that’s our value

of p. Otherwise, let rk be the largest non-negative integer such that p′prk

k �= 0, and setp = p′prk

k .

Proposition 9.5.19. Let A be a commutative ring with only one prime ideal. Let f :Am →M be an injective A-module homomorphism for some A-module M . Then 1⊗ f :A/p⊗A Am → A/p⊗M is also injective.

Proof We may identify 1⊗ f with the unique A/p-module homomorphism that makesthe following diagram commutative.

Am M

(A/p)m M/pM,

��f

��πm

��π

��f


Where π : A→ A/p and π : M →M/pM are the canonical maps.Suppose that (a1, . . . , am) ∈ ker f , where ai = π(ai) for 1 ≤ i ≤ m. Then f(a1, . . . , am) ∈

pM . Thus, we can write f(a1, . . . , am) = p1m1+ · · ·+pkmk, where p1, . . . , pk are nonzeroelements of p, and m1, . . . ,mk ∈M . By Lemma 9.5.18, there is a nonzero element p ∈ pwhich annihilates p1, . . . , pk. But then pf(a1, . . . , am) = 0.

Since f is injective, this says p annihilates (a1, . . . , am). But since A is local, withmaximal ideal p, every zero divisor of A lies in p. In particular, we must have ai ∈ p forall i.

But then π(ai) = 0 for all i, and hence (a1, . . . , am) = πn(a1, . . . , am) = 0. Thus, thekernel of f is trivial.

We can now put it all together.

Theorem 9.5.20. Let A be a commutative ring and let f : Am → An be an injectiveA-module homomorphism. Then m ≤ n.

Proof Since localization is exact, we may localize at a minimal prime ideal. Thus,we may assume that A is a local ring with a unique prime ideal p. But now, Proposi-tion 9.5.19 shows that the homomorphism obtained by extension of rings to A/p is alsoinjective. Since A/p is a field, the result follows.

Exercises 9.5.21.1. Deduce that Z2 is not a flat Z-module by tensoring it with the exact sequence

0→ Zf2−→ Z π−→ Z2 → 0

from Problem 3 of Exercises 7.7.52. Is Zn a flat Z-module for n > 2?

2. Show that Z2 is not a flat Z2r -module for r > 1. Note that Z2r is a local ring withonly one prime ideal.

3. Let A be a domain. For which ideals a is A/a flat as an A-module?

† 4. Show that an infinite direct sum of flat modules is flat. Deduce that all free modulesare flat.

5. Suppose given a direct system M0f0−→ M1

f1−→ · · · of right A-modules, togetherwith a left A-module M .

(a) Show that lim−→(Mi ⊗A N) is naturally isomorphic to (lim−→Mi)⊗A N .

(b) Deduce that if each Mi is flat, so is lim−→Mi.

9.6 Tensor Products of Algebras

Let A be a commutative ring and let B and C be A-algebras, via homomorphisms f : A→B and g : A→ C. We shall show that B ⊗A C is then an A-algebra, and in fact satisfiesan important universal mapping property with respect to A-algebra homomorphisms outof B and C.

A ring called A, in this section, will be assumed to be commutative.


Proposition 9.6.1. Let B and C be A-algebras, via f : A → B and g : A → C. Thenthere is an A-algebra structure on B ⊗A C whose product is given on generators by

(b⊗ c) · (b′ ⊗ c′) = bb′ ⊗ cc′.

In particular, B ⊗A C is commutative if and only if both B and C are commutative.In addition, if we define ι1 : B → B ⊗A C and ι2 : C → B ⊗A C by ι1(b) = b ⊗ 1

and ι2(c) = 1 ⊗ c, then the ιi are A-algebra homomorphisms, and satisfy the followinguniversal property.

1. Every element of the image of ι1 commutes with every element of the image of ι2.

2. Suppose given an A-algebra D, together with A-algebra homomorphisms h1 : B →D and h2 : C → D such that every element of the image of h1 commutes withevery element of the image of h2. Then there is a unique A-algebra homomorphismh : B ⊗A C → D such that the following diagram commutes.

B B ⊗A C C

D

��ι1

��

��

��

�

h1

��

h

�� ι2

��

��

��

h2

If B, C, and D are commutative, this says simply that B ⊗A C, together with the mapsιi, is the coproduct of B and C in the category of commutative A-algebras.

Proof First note that the A-algebra structures on B and C imply that there is anA-module homomorphism

μ : B ⊗A C ⊗A B ⊗A C → B ⊗A C

whose effect on generators is given by μ(b⊗ c⊗ b′⊗ c′) = bb′⊗ cc′. By the associativity ofthe tensor product (e.g., Corollary 9.4.24), the iterated tensor product B⊗AC⊗AB⊗ACis naturally isomorphic to (B⊗AC)⊗A (B⊗AC), so that μ induces a product operationon B ⊗A C via x · y = μ(x ⊗ y) for x, y ∈ B ⊗A C. By the laws of the tensor product(B ⊗A C)⊗A (B ⊗A C), this product satisfies both distributive laws.

We claim that this product makes B ⊗A C a ring, with multiplicative identity 1⊗ 1.On generators, we have (1⊗ 1) · (b⊗ c) = b⊗ c = (b⊗ c) · (1⊗ 1). Since 1⊗ 1 acts as theidentity on a set of generators, it must act as the identity everywhere. Similarly, it sufficesto show associativity of products of generators, where it follows from the associativity ofthe products in B and C.

The homomorphisms ιi defined above are easily seen to be ring homomorphisms. Notethat if ν and ν′ give the A-algebra structures of B and C, respectively, then ιi ◦ ν(a) =a ⊗ 1 = 1 ⊗ a = ι2 ◦ ν′(a) for all a ∈ A, and that this element lies in the center ofB ⊗A C. Since the ιi, ν, and ν′ are ring homomorphisms, this composite is also, andgives B⊗AC an A-algebra structure such that the ιi are A-algebra homomorphisms. Wehave (b⊗ 1) · (1⊗ c) = b⊗ c = (1⊗ c) · (b⊗ 1), and hence the elements of the image of ι1commute with the elements of the image of ι2.

Let h1 : B → D and h2 : C → D be A-algebra homomorphisms with the propertythat every element of the image of h1 commutes with every element of the image of h2.


Then, as the reader may check, there is an A-module homomorphism h : B ⊗A C → Dwhose effect on generators is given by h(b⊗ c) = h1(b)h2(c).

The fact that each h1(b) commutes with each h2(c) shows that h(x · y) = h(x)h(y)as x and y range over a set of generators of B ⊗A C. Since h is a homomorphism ofabelian groups, this is enough to show that h(xy) = h(x)h(y) for all x, y ∈ B ⊗A C.Since h(1 ⊗ 1) = 1, h is a ring homomorphism, which is easily seen to be an A-algebrahomomorphism as well. It is easy to see that h is the unique A-algebra homomorphismthat makes the diagram commute.

We’ve already seen some examples of tensor products of algebras.

Proposition 9.6.2. Let A be a commutative ring and let B be an A-algebra via f : A→B. Then there is an isomorphism of A-algebras

η : B ⊗AMn(A)→Mn(B)

which makes the following diagram commute.

B B ⊗AMn(A) Mn(A)

Mn(B)

��ι1

��

ι��

η

�� ι2

��

f∗

Here ι : B →Mn(B) is given by ι(b) = bIn, while f∗ takes (aij) to the matrix whose ij-thentry is f(aij). The maps ιi are the structure maps of the tensor product of algebras.

Proof Because the square

A B

Mn(A) Mn(B)��

ι

��f

��ι

��f∗

commutes, Mn(B) has an A-algebra structure such that both f∗ and ι : B →Mn(B) areA-algebra homomorphisms.

It is easy to see that every element in ι(B) commutes with every element in the imageof f∗. So the universal property of the tensor product of algebras provides an A-algebrahomomorphism η : B ⊗AMn(A)→ Mn(B) that makes the above diagram commute. Itsuffices to show that η is an isomorphism.

Write eij for the matrix whose ij-th coordinate is 1 and whose other coordinates areall 0. Recall from Lemma 7.10.5 that if C is any ring, then {eij | 1 ≤ i, j ≤ n}, in anyorder, gives a basis for Mn(C) as either a left or a right C-module.

By extension of rings, B⊗AMn(A) is free as a left B-module, with basis {1⊗eij | 1 ≤i, j ≤ n}. But η(1⊗ eij) = eij , so, viewing η as a left B-module homomorphism, we seethat it carries a basis of B ⊗AMn(A) onto a basis of Mn(B). By Lemma 7.7.36, this issufficient to conclude that η is an isomorphism.

There is an analogous result for monoid rings, which we shall treat as an exercise.We shall see other examples later.


Exercises 9.6.3.

‡ 1. Let B be an A-algebra, via f : A → B, and let M be a monoid. Show that thereis a natural A-algebra isomorphism from B ⊗A A[M ] to B[M ].

2. Combining Proposition 9.6.2 with the preceding problem, we see that if A is com-mutative and M is a monoid, then there is an isomorphism between Mn(A[M ]) andMn(A)[M ]. Write down the isomorphism explicitly and give a proof of isomorphismthat doesn’t use tensor products.

3. Let M and M ′ be monoids. Show that there is a natural A-algebra isomorphismfrom A[M ]⊗A A[M ′] to A[M ×M ′].

4. Let K be a subfield of the fields L and L′. Is L⊗K L′ a field?

5. Let B be an A-algebra with structure map ν : A → B. Show that there is ahomomorphism of A-modules (but not necessarily of A-algebras), μ : B⊗AB → Binduced by μ(b⊗ b′) = bb′. Show also that the following diagrams commute.

B ⊗A B ⊗A B B ⊗A B

B ⊗A B B

��μ⊗ 1

��

1⊗ μ��

μ

��μ

A⊗A B B ⊗A B B ⊗A A

B

��ν ⊗ 1

��

εB��

μ

��1⊗ ν

��

εB

Here, the maps εB are the natural isomorphisms of Lemma 9.4.14.

Conversely, suppose given an A-module B, together with A-module homomor-phisms ν : A → B and μ : B ⊗A B → B such that the above diagrams commute.Show that B is an A-algebra with structure map ν and with ring product given bybb′ = μ(b⊗ b′).

6. Let B be an A-algebra and let μ : B⊗AB → B be induced by the ring product of B.Show that B is a commutative ring if and only if μ is an A-algebra homomorphism.

9.7 The Hom Functors

Let M and N be A-modules. Recall that HomA(M,N) is the abelian group of A-modulehomomorphisms from M to N . Here, the group structure on HomA(M,N) is given by(f+g)(m) = f(m)+g(m). Without extra hypotheses (e.g., A being commutative), thereis no module structure on HomA(M,N).

We shall use the same notation HomA(M,N) whether we are discussing the homomor-phisms of left modules or of right modules, and the theory proceeds identically. Thus,in general, we shall not make reference to the sidedness of the module structures. Inapplications, it must either be made clear by the context or be explicitly stated.


We should immediately discuss the functoriality of Hom, as it is good to keep itin mind as we go. Here, for an A-module M , HomA(M,−) gives a covariant functorfrom A-modules to abelian groups. For f : N → N ′, the induced map HomA(M,f) :HomA(M,N)→ HomA(M,N ′) is generally written as f∗, and is given by f∗(g) = f ◦ g.

Similarly, for an A-module N , HomA(−, N) is a contravariant functor. Here, if f :M ′ → M , the induced map HomA(f,N) : HomA(M,N) → HomA(M ′, N) is generallywritten as f∗, and is given by f∗(g) = g ◦ f .

We shall first establish the exactness properties of the Hom functors.

Lemma 9.7.1. For any A-module M , the functor HomA(M,−) is left exact; i.e., ifwe’re given an exact sequence

0→ N ′ f−→ Ng−→ N ′′,

then the induced sequence

0→ HomA(M,N ′)f∗−→ HomA(M,N)

g∗−→ HomA(M,N ′′)

is exact.

Proof Clearly, f∗ is injective because f is, and g∗ ◦ f∗ = (g ◦ f)∗ = 0. Let h : M → Nbe in the kernel of g∗. Then imh ⊂ ker g. But the kernel of g is equal to the image of f .Thus, since f is injective, the induced map f : N ′ → ker g is an isomorphism. Thus, h is

the image under f∗ of the composite M h−→ ker gf−1

−−→ N ′.

The contravariant direction of Hom also has exactness properties.

Lemma 9.7.2. Let N be an A-module and let

M ′ f−→Mg−→M ′′ → 0

be an exact sequence of A-modules. Then the following sequence is exact.

0→ HomA(M ′′, N)g∗−→ HomA(M,N)

f∗−→ HomA(M ′, N)

Proof The injectivity of g∗ follows directly from the surjectivity of g. It is also imme-diate that f∗ ◦ g∗ = 0. Let h : M → N be in the kernel of f∗. Thus, h ◦ f = 0. Thus,h vanishes on the image of f . By the exactness of the original sequence, the image of fis the kernel of g. But then h factors through the canonical map from M to M/(ker g).But the Noether Theorem says we may identify this canonical map with g itself. In otherwords, there is a homomorphism h : M ′′ → N such that h ◦ g = h. But this is justanother way of saying that h = g∗h for some h ∈ HomA(M ′′, N).

We have so far avoided a discussion of module structures on the Hom groups. Theproof of the next lemma is easy, and is left to the reader.

Lemma 9.7.3. Bimodule structures may be used to put module structures on Hom groupsin any of the following ways.

1. If M is an A-B-bimodule and N is a left A-module, then there is a left B-modulestructure on HomA(M,N) via (b · f)(m) = f(mb).

2. If M is a B-A-bimodule and N is a right A-module, then there is a right B-modulestructure on HomA(M,N) via (f · b)(m) = f(bm).


3. If N is an A-B-bimodule and M is a left A-module, then there is a right B-modulestructure on HomA(M,N) via (f · b)(m) = f(m)b.

4. If N is a B-A-bimodule and M is a right A-module, then there is a left B-modulestructure on HomA(M,N) via (b · f)(m) = bf(m).

If A is commutative, then any A-module is an A-A-bimodule. So each of the pro-cedures above puts an A-module structure on HomA(M,N). In fact, each of the aboveprocedures puts the same A-module structure on HomA(M,N).

In particular, for any ring A, if M is a left A-module, then so is HomA(A,M), by thefirst item above. Similarly, if M is a right A-module, then the second item puts a rightA-module structure on HomA(A,M). Recall from Corollary 7.7.9 that the evaluationmap ε : HomA(A,M)→ M is an isomorphism of abelian groups, where ε(f) = f(1). Infact, it is more:

Proposition 9.7.4. The evaluation map ε : HomA(A,M)→M gives a natural isomor-phism of A-modules.

The various module structures on Hom allow us to demonstrate the relationshipbetween Hom and the tensor products.

Proposition 9.7.5. Suppose given a right B-module M1, a B-A-bimodule M2, and aright A-module M3. Then there is an isomorphism of abelian groups, natural in all threevariables:

ϕ : HomB(M1,HomA(M2,M3))→ HomA(M1 ⊗B M2,M3).

Explicitly, if f ∈ HomB(M1,HomA(M2,M3)), then the effect of ϕ(f) on generators isgiven by

ϕ(f)(m1 ⊗m2) = f(m1)(m2).

If M1 is a C-B-bimodule, then ϕ is an isomorphism of right C-modules.

Proof For f ∈ HomB(M1,HomA(M2,M3)), it is easy to see that the displayed formulafor ϕ(f) gives rise to a well defined homomorphism. And ϕ itself is then easily seen tobe a homomorphism.

To see that ϕ is an isomorphism, we construct its inverse. Define

ψ : HomA(M1 ⊗B M2,M3)→ HomA(M1,HomA(M2,M3))

as follows. For g ∈ HomA(M1 ⊗B M2,M3), we set

((ψ(g))(m1))(m2) = g(m1 ⊗m2).

We leave it to the reader to check that ϕ and ψ are inverse isomorphisms, and toverify the claims regarding naturality and C-module structures.

Corollary 9.7.6. Let A be a commutative ring and let Mi be an A-module for i = 1, 2, 3.Then there is a natural isomorphism

ϕ : HomA(M1,HomA(M2,M3))→ HomA(M1 ⊗AM2,M3)

of A-modules.


As in the case of tensor products, it is useful to understand the behavior of Homgroups with respect to direct sums.

Proposition 9.7.7. Let M,M1, . . . ,Mk and N,N1, . . . , Nk be left A-modules. Thenthere are natural isomorphisms

α : HomA(k⊕i=1

Mi, N)→k⊕i=1

HomA(Mi, N),

whose i-th coordinate map is ι∗i , where ιi is the natural inclusion of Mi in⊕k

i=1Mi, and

β : HomA(M,

k⊕i=1

Ni)→k⊕i=1

HomA(M,Ni),

whose i-th coordinate map is (πi)∗, where πi is the natural projection map from⊕k

i=1Nionto Ni.

These isomorphisms are natural in all variables. If each Mi is an A-B-bimodule orif N is an A-B-bimodule, then α is an isomorphism with respect to the appropriate B-module structure. Similarly, if each Ni is an A-B-bimodule or if M is an A-B-bimodule,then β is an isomorphism with respect to the appropriate B-module structure.

Proof In fact, these follow from the universal properties of the coproduct and productin the category of A-modules. Thus, α is an isomorphism because a homomorphism outof a direct sum is determined uniquely by its restrictions to the individual summands.Similarly, β is an isomorphism because finite direct sums coincide with finite directproducts. Thus, a homomorphism into a finite direct sum is determined uniquely by itscomponent functions. We leave the statements about naturality and module structuresto the reader.

Definitions 9.7.8. Let M be an A-module. Define the dual module, M∗, by M∗ =HomA(M,A). Since A is an A-A-bimodule, M∗ is a right A-module when M is a leftA-module, and M∗ is a left A-module when M is a right A-module. In either case, thedual of M∗, M∗∗ = HomA(HomA(M,A), A) is defined, and is an A-module from thesame side that M is.

The next proposition is fundamental in our understanding of the duality of free mod-ules.

Proposition 9.7.9. Let M be a finitely generated free left A-module, with basis m1, . . . ,mn.Then the dual module M∗ is a free right A-module, with basis m∗

1, . . . ,m∗n, where m∗

i (a1m1+· · ·+ anmn) = ai. We call m∗

1, . . . ,m∗n the dual basis of m1, . . . ,mn.

Proof Let g : An →M be the isomorphism defined by the basism1, . . . ,mn: g(a1, . . . , an) =a1m1 + · · · + anmn. Then g∗ : M∗ → (An)∗ is an isomorphism (because (g−1)∗

is its inverse). But Proposition 9.7.7 provides an isomorphism α : (An)∗ → (A∗)n.And A∗ = HomA(A,A) is isomorphic to A (Proposition 9.7.4) by the correspondence,ε : HomA(A,A)→ A, which takes a left A-module homomorphism f : A→ A to f(1).

In Proposition 9.7.4 it was shown that ε is a left A-module map with respect to theleft module structure on A∗ induced by the A-A-bimodule structure on the A serving asthe domain. Here, we are using the right module structure on A∗ which comes from thebimodule structure of the A that receives the maps f ∈ HomA(A,A). But we see that


ε is a module map with respect to this structure also: ε(fa) = (fa)(1) = f(1)a by thedefinition of fa. But this last is (ε(f))a.

Thus, we have a composite isomorphism of right A-modules:

M∗ g∗−→ (An)∗ α−→ (A∗)n εn

−→ An.

But this composite is easily seen to carry the dual basis m∗1, . . . ,m

∗n to the canonical

basis of An.

Note that the homomorphism m∗i depends on the entire basis m1, . . . ,mn, and not

just on the element mi. A different basis whose i-th term is also mi could give rise to adifferent homomorphism m∗

i . Explicitly, m∗i is the composite

Mg−1

−−→∼= Anπi−→ A

where g is the isomorphism induced by the basis m1, . . . ,mn, and πi is the projectiononto the i-th factor.

We see from Proposition 9.7.9 that if A is commutative and ifM is a finitely generatedfree A-module, then M and M∗ are isomorphic. However, even in the case where A is afield, there is no natural isomorphism between M and M∗: if we’re given two differentbases for M and map each one of them to its dual basis, we will get two differentisomorphisms from M to M∗. But if we dualize twice, some naturality returns. Thereader may verify the following lemma.

Lemma 9.7.10. Let M be an A-module for any ring A. Then there is an A-modulehomomorphism κM : M → M∗∗ defined as follows. For m ∈ M , κM (m) is the homo-morphism from M∗ to A defined by evaluating at m: κM (m)(f) = f(m).

In addition, κM is natural in M in the sense that if g : M → N is an A-modulehomomorphism, then the following diagram commutes.

M N

M∗∗ N∗∗

��g

��

κM

��

κN

��g∗∗

Here, for h ∈M∗∗ and f ∈ N∗, g∗∗(h)(f) = h(f ◦ g).We shall borrow a definition from functional analysis.

Definition 9.7.11. An A-module M is reflexive if κM : M →M∗∗ is an isomorphism.

Proposition 9.7.12. Finitely generated free modules are reflexive.

Proof Let M be a finitely generated free module with basis m1, . . . ,mn. Then the dualmodule M∗ is free, with basis m∗

1, . . . ,m∗n, by Proposition 9.7.9.

But then M∗∗ is free on the dual basis, m∗∗1 , . . . ,m

∗∗n , to m∗

1, . . . ,m∗n. Here, m∗∗

i :M∗ → A is given by m∗∗

i (m∗1a1 + · · ·+m∗

nan) = ai.Thus, M and M∗∗ are both free modules of the same rank, and hence are abstractly

isomorphic. But as the reader may check, κM (mi) = m∗∗i . Thus, κM carries a basis of

M to a basis of M∗∗, and hence is an isomorphism.


The fact that M is finitely generated here is essential. Infinitely generated free mod-ules are not reflexive, because their duals are infinite direct products of copies of A,rather than infinite direct sums.

The proof of Proposition 9.7.12 brings up a useful notation that originated in exactlythis context, called the Kronecker delta.

Definition 9.7.13. The Kronecker delta, δij , is defined to be 0 or 1, depending on thevalues of i and j. Explicitly,

δij ={

1 if i = j0 otherwise.

Thus, if m1, . . . ,mn is the basis of a free module and if m∗1, . . . ,m

∗n is its dual basis,

then m∗j (mi) = δij . Indeed, this fact uniquely determines the dual basis m∗

1, . . . ,m∗n.

Exercises 9.7.14.1. Let a be a two-sided ideal of A and let M be a left A-module. Show that there is

a natural isomorphism of left A-modules

ε : HomA(A/a,M)→ {m ∈M | a ⊂ Ann(m)}

given by ε(f) = f(1).

Note the use of the modules HomA(A/(a),M) in the proof of the FundamentalTheorem of Finitely Generated Modules Over a P.I.D.

2. LetM,M ′ andN beA-modules, withM ′ a submodule ofM . Show that HomA(M/M ′, N)is isomorphic to the subgroup of HomA(M,N) consisting of those homomorphismsthat factor through the canonical map π : M →M/M ′.

3. Give an example of a ring A and an A-module M for which HomA(M,−) is not anexact functor.

4. Give an example of a ring A and an A-module N for which HomA(−, N) is not anexact functor.

5. Let A be an integral domain and let M be an A-module. Let π : M →M/Tors(M)be the canonical map. Show that the induced map of dual modules, π∗ : (M/Tors(M))∗ →M∗ is an isomorphism.

6. Let M be a finitely generated module over the principal ideal domain A. Showthat the double dual module M∗∗ is naturally isomorphic to M/Tors(M).

7. Suppose given A-modules M , {Mi | i ∈ I}, N and {Ni | i ∈ I}.(a) Show that there is a natural isomorphism

HomA(⊕i∈I

Mi, N) α−→∏i∈I

HomA(Mi, N).

Here, the i-th factor of α(f) is f ◦ ιi, where ιi is the canonical inclusion of Mi

in⊕

i∈IMi.


(b) Show that there is a natural isomorphism

HomA(M,∏i∈I

Ni)β−→

∏i∈I

HomA(M,Ni).

Here, the i-th factor of β(f) is πi ◦ f , where πi is the canonical projection of∏i∈I Ni onto its i-th factor.

(c) Suppose that M is finitely generated. Show that the preceding map β restrictsto an isomorphism

HomA(M,⊕i∈I

Ni)β−→

⊕i∈I

HomA(M,Ni).

8. Interpret the relationship between Hom and tensor in terms of adjoint functors.Which of the results on these functors would then have automatic proofs?

9.8 Projective Modules

Unless it’s explicitly stated to the contrary, all modules in this section are left A-modules,and all homomorphisms are A-module homomorphisms.

Recall that a short exact sequence

0→M ′ f−→Mg−→M ′′ → 0

splits if g admits a section, meaning an A-module homomorphism s : M ′′ → M suchthat g ◦ s = 1M ′′ . Recall from Proposition 7.7.49 that if the sequence above splits, weobtain an A-module isomorphism from M ′ ⊕M ′′ to M .

Definition 9.8.1. An A-module P is projective if every short exact sequence

0→M ′ f−→Mg−→ P → 0

of A-modules splits.

Note that this is equivalent to the statement that every surjective A-module homo-morphism g : M → P admits a section. There is another characterization of projectivesthat is often used.

Proposition 9.8.2. Let P be an A-module. Then the following are equivalent:

1. P is projective.

2. Given a diagram

P

M M ′′ 0��

h

��g ��


where the horizontal row is exact, there is a lift of h to M , i.e., a map h′ : P →Mmaking the following diagram commute.

P

M M ′′ 0��

h

��

��

��h′

��g ��

3. HomA(P,−) is an exact functor.

Proof The second property says that if g : M →M ′′ is onto, then g∗ : HomA(P,M)→HomA(P,M ′′) is onto. But since HomA(N,−) is left exact for any A-module N , thesecond property is equivalent to the third by Proposition 9.5.6.

The second property also easily implies the first. Given a surjection g : M → P , asection of g is a lift of the identity map of P to M .

Thus, it suffices to show that the first property implies the second. Suppose given asurjective homomorphism g : M → M ′′ and a homomorphism h : P → M ′′. Recall thatthe pullback of g and h is the subgroup M ×M ′′ P ⊂ M × P specified by M ×M ′′ P ={(m, p) | g(m) = h(p)}. Since h and g are A-module homomorphisms, M ×M ′′ P is anA-submodule of M × P . The projections of M × P onto its factors provide A-modulehomomorphisms h : M ×M ′′ P → M and g : M ×M ′′ P → P making the followingdiagram commute.

M ×M ′′ P P

M M ′′ 0

��g

��

h

��

g

��h ��

We claim that g is surjective.To see this, let p ∈ P . Since g is surjective, h(p) = g(m) for some m ∈ M . But this

says that (m, p) ∈M ×M ′′ P , and hence p = g(m, p).Since P is projective and g is surjective, there is a section s for g. But then h ◦ s

gives the desired lift of h.

We’ve already seen some projective modules in action. Indeed, Proposition 7.7.51translates precisely to the following statement.

Corollary 9.8.3. Free modules are projective.

The reader may now ask whether all projective modules are free. A quick answeris “no,” but there is then a rich field of study in determining what all the projectivemodules are. In a certain stabilized sense, this question is answered for the finitelygenerated projectives by the study of the group K0(A) introduced in Section 9.9.

Definition 9.8.4. We say that a submodule N ⊂M is a direct summand of M if M isthe internal direct sum of N and some other submodule N ′ ⊂ M . As was the case forgroups, this means that the homomorphism

μ : N ⊕N ′ →M given by μ(n, n′) = n+ n′

is an isomorphism.


It is important to note the distinction between saying that N is isomorphic to a directsummand of M and that it in fact is a direct summand. For instance, the ideal (2) ⊂ Zis isomorphic to Z itself, but is not a direct summand of Z, as may be seen by the nextlemma: the sequence

0→ (2) ⊂−→ Z→ Z2 → 0

fails to split.

Lemma 9.8.5. Suppose given a submodule N ⊂M . Then the following are equivalent.

1. N is a direct summand of M .

2. The inclusion i : N ⊂M admits a retraction.

3. The sequence

0→ N⊂−→M →M/N → 0

splits.

Proof Suppose that N is a direct summand of M with complementary summand N ′,and let μ : N ⊕N ′ →M be the isomorphism given by μ(n, n′) = n+ n′. Then π1 ◦ μ−1

gives a retraction for i, where π1 : N ⊕ N ′ → N is the projection map. Thus, the firstcondition implies the second.

The second and third conditions are equivalent by Proposition 7.7.49, which alsoshows that these two conditions imply the existence of an isomorphism h : N ⊕M/N

∼=−→M such that h(n, 0) = i(n) for all n ∈ N . But then M is easily seen to be the internaldirect sum of N and h(0⊕M/N).

We obtain an important characterization of projective modules.

Proposition 9.8.6. A module is projective if and only if it is a direct summand of afree module. In particular, a direct summand of a projective module is projective.

An A-module is a finitely generated projective module if and only if it is a directsummand of a finitely generated free module.

Proof Suppose P is projective, and construct a short exact sequence

0→ Q→ F → P → 0,

where F is free, and is finitely generated if P is finitely generated. (We may do this inthe finitely generated case by Corollary 7.7.41, and in the infinitely generated case byProblem 2 of Exercises 7.7.42.)

In either case, since P is projective, the exact sequence splits, giving an isomorphismP ⊕ Q ∼= F by Proposition 7.7.49. Since any module isomorphic to a free module isfree, P ⊕Q is free, and is finitely generated if P is finitely generated. And P is a directsummand of P ⊕Q.

Notice that a direct summand N of a finitely generated module M is always finitelygenerated: If r : M → N is a retraction for N ⊂M , then r is surjective.


Thus, it suffices to show that a direct summand of a projective module is projective.Suppose that N ⊕N ′ is projective and that we’re given a commutative diagram

N

M M ′′ 0��

h

��g ��

where the bottom row is exact. Since N⊕N ′ is projective, there is a map k : N⊕N ′ →Mthat makes the following diagram commute.

N ⊕N ′ N

M M ′′ 0

��π

��

k

��

h

��g ��

Here, π is the projection onto the first factor. But then k ◦ ι1 : N →M gives a lift of h,where ι1 is the inclusion of the first factor in N ⊕N ′.

We can now give an example of a non-free projective.

Example 9.8.7. Let A and B be nonzero rings. Then the ideals A × 0 and 0 × B aredirect summands of the ring A×B, and hence are projective A×B-modules. Note thateven if A is isomorphic to A × B as a ring, A × 0 is not free as an A × B-module, asAnnA×B(A× 0) = 0×B, while AnnA×B(A×B) = 0.

And more applications of Proposition 9.8.6 follow:

Corollary 9.8.8. Every finitely generated projective module over a P.I.D. is free.

Proof If A is a P.I.D., then any submodule of a free A-module is torsion-free. Butthe Fundamental Theorem of Finitely Generated Modules over a P.I.D. shows that anyfinitely generated torsion-free A-module is free.

We shall show in Section 12.4 that every infinitely generated projective module overa P.I.D. is free. We return to giving corollaries of Proposition 9.8.6.

Corollary 9.8.9. Projective modules are flat.

Proof By Proposition 9.5.8, a direct summand of a flat module is flat. By Corol-lary 9.5.9 in the finitely generated case, or Problem 4 of Exercises 9.5.21 in the generalcase, free modules are flat.

Corollary 9.8.10. Let f : A → B be a ring homomorphism and let P be a projectiveleft A-module. Then B⊗A P is a projective B-module. Also, if P is finitely generated asan A-module, then B ⊗A P is finitely generated as a B-module.

Proof Because P is projective, there is an A-module Q such that P ⊕Q is free over A.But then B⊗A (P ⊕Q), which is isomorphic to (B⊗AP )⊕ (B⊗AQ), is a free B-module,via Corollary 9.4.19 in the finitely generated case, or Problem 6 of Exercises 9.4.27 in


the general case. Thus, B ⊗A P is a direct summand of a free B-module, and hence isprojective.

As for finite generation, if P is finitely generated, there is a surjective A-modulehomomorphism f : An → P . But then 1B ⊗ f gives a surjection from Bn ∼= B ⊗A Anonto B ⊗A P , since tensoring is right exact.

The next result should, by now, be expected.

Proposition 9.8.11. Any direct sum of projective modules is projective.

Proof The simplest argument is just to apply Proposition 9.8.2. Let P =⊕

i∈I Pi,where each Pi is projective. Suppose given a surjective A-module homomorphism g :M → M ′′ and an A-module map h : P → M ′′. Let ιi : Pi → P be the canonicalinclusion and let hi = (h◦ιi) : Pi →M ′′. Since Pi is projective, there is a homomorphismh′i : Pi →M with g ◦ h′i = hi.

By the universal property of the direct sum, there is a unique homomorphism h′ :P → M such that h′ ◦ ιi = h′i for each i ∈ I. But then g ◦ h′ agrees with h on eachsummand Pi, and hence g ◦ h′ = h.

We are primarily interested in finitely generated projective modules. There are anumber of reasons for this, including the fact that many of the interesting applicationsof projectives are restricted to finitely generated phenomena. It is also the case thatif we allow the consideration of infinitely generated phenomena, then the behavior ofprojectives becomes in a sense too simple. For instance, consider the following.

Proposition 9.8.12. (Eilenberg Swindle) Let P be a projective module. Then there isa free module F such that P ⊕ F is free.

Proof Since P is projective, it is isomorphic to a direct summand of a free module.Thus, there is a module Q (which must also be projective) such that P ⊕Q is free. Let Fbe a direct sum of infinitely many copies of P ⊕Q: F = (P ⊕Q)⊕ (P ⊕Q)⊕ · · · . SinceP ⊕Q ∼= Q⊕ P , we have P ⊕ F ∼= P ⊕ (Q⊕ P )⊕ (Q⊕ P )⊕ · · · . But if we reassociatethis sum, we get (P ⊕Q)⊕ (P ⊕Q)⊕ · · · , which is just F . Explicitly, P ⊕ F ∼= F .

We wish to be able to describe the projective modules over certain types of rings.Here, we consider local rings.

Lemma 9.8.13. Let A be a local ring with maximal ideal m. Let f : M → N bean A-module homomorphism, with N finitely generated. Then f is onto if and only if1⊗ f : A/m⊗AM → A/m⊗A N is onto.

Proof We have an exact sequence

Mf−→ N

π−→ C → 0,

where π is the canonical map onto the cokernel of f . Since the sequence remains exactwhen tensored with A/m, it suffices that C = 0 if and only if A/m ⊗A C = 0. But thisfollows from Nakayama’s Lemma, since the Jacobson radical of a local ring is its maximalideal.

Projective modules over local rings are well behaved:

Proposition 9.8.14. Let A be a local ring. Then every finitely generated projectiveA-module is free.


Proof Let P be a finitely generated projective module over A and let x1, . . . , xn be aminimal generating set for P in the sense that no proper subset of it generates P . Letf : An → P be the homomorphism that takes the canonical basis element ei to xi fori = 1, . . . , n. Then f is surjective. Let K be its kernel.

We obtain an exact sequence

0→ Ki−→ An

π−→ P → 0,

which splits, because P is projective. Thus, by Corollary 9.5.13, the sequence

0→ A/m⊗A K 1⊗i−−→ A/m⊗A An 1⊗π−−→ A/m⊗A P → 0

is also split exact, where m is the maximal ideal of A.If a proper subset of 1 ⊗ x1, . . . , 1 ⊗ xn were to generate A/m ⊗A P , then a proper

subset of x1, . . . , xn would generate P , by Lemma 9.8.13. Thus, 1⊗ x1, . . . , 1⊗ xn mustbe a basis for the vector space A/m ⊗A P . But this says that 1 ⊗ f is an isomorphism,and hence A/m⊗A K = 0. But then K = 0 by Nakayama’s Lemma, and hence P is thefree module with basis x1, . . . , xn.

In fact, it is also true that every infinitely generated projective module over a localring is free, but we shall not prove it here. Note that in the infinitely generated case, wewould not be able to use Nakayama’s Lemma in our argument.

Recall that if p is a prime ideal of a commutative ring A and if P is a finitely generatedprojective module over A, then Pp = Ap⊗AP is a finitely generated projective Ap-moduleby Corollary 9.8.10. Thus, we may make the following.

Definitions 9.8.15. Let P be a finitely generated projective module over the commu-tative ring A, and let p be a prime ideal of A. Then the p-rank of P is the rank of thefinitely generated free module Pp over Ap.

We say that P has constant rank if the p-rank of P is the same for each prime idealp of A.

The relationship between the ranks at various primes depends in part on the waythat the prime ideals are nested.

Proposition 9.8.16. Let A be an integral domain. Then every finitely generated pro-jective A-module has constant rank.

Proof Let P be a finitely generated A-module and let p be a prime ideal of A. Let Kbe the field of fractions of A. Suppose that Pp is free of rank r over Ap. Then K ⊗Ap Pp

is an r-dimensional vector space over K. But

K ⊗Ap Pp = K ⊗Ap Ap ⊗A P = K ⊗A P = P(0),

so the rank of P at (0) is r, also.

Exercises 9.8.17.1. Let A be a local ring with only one prime ideal, and let x1, . . . , xm be a generating

set for An. Suppose also that x1, . . . , xk are linearly independent. Show that thereis a subset of x1, . . . , xm that contains x1, . . . , xk and is a basis for An.

† 2. Let A and B be rings. Show that every submodule of (A×B)n has the form M×N ,where M and N are submodules of An and Bn, respectively. Show also that everyfinitely generated projective module of A × B has the form P × P ′, where P andP ′ are finitely generated projective over A and B, respectively.


3. What are all the finitely generated projective modules over Zn?

4. Give an example of a commutative ring A and a finitely generated projective moduleP over A whose rank is not constant.

5. Let P be a finitely generated projective left A-module. Show that the dual moduleP ∗ is a finitely generated projective right A-module.

6. Let M = M1 ⊕M2 be a direct sum of two A-modules. Show that M is reflexiveif and only if M1 and M2 are reflexive. Deduce that finitely generated projectivemodules are reflexive.

9.9 The Grothendieck Construction: K0

We show here how the finitely generated projective modules can be used to construct afunctor, K0, from rings to abelian groups, which detects, in a stable sense, the deviationof a finitely generated projective module from being free.

Consider first the set FP (A) of isomorphism classes of finitely generated left projectivemodules. One may construct this set by first taking the set consisting of the directsummands of An for all n ≥ 0 (which is a set because the collection of all subsets of An

is a set), and identifying any two such direct summands that are isomorphic. If P is afinitely generated projective over A, we denote its isomorphism class by [P ] ∈ FP (A).

Notice that FP (A) is an abelian monoid under the operation [P ] + [Q] = [P ⊕ Q].The identity element is the 0 module. The group we shall define here, called K0(A),is an abelian group which satisfies a universal mapping property with respect to theabelian monoid FP (A): There is a monoid homomorphism η : FP (A) → K0(A) withthe property that if f : FP (A) → G is a monoid homomorphism, with G an abeliangroup, then there is a unique group homomorphism f : K0(A) → G such that thefollowing diagram commutes.

FP (A) G

K0(A)

��f

��

η

�� f

The universal mapping property is solved by a general construction called the Grothendieckconstruction. There are two different ways to build it. The one we give first is similar tothe construction of rings of fractions.

Suppose given an abelian monoid M , written in additive notation. The elements ofthe Grothendieck construction G(M) are formal differences m−n of elements m,n ∈M ,in the same way that a ring of fractions consists of formal quotients a/s. Here, the elementm − n represents the equivalence class of the ordered pair (m,n) under the equivalencerelation that sets (m,n) ∼ (m′, n′) if there is an r ∈M such that m+n′+r = m′+n+r.We leave it to the reader to show that this is, in fact, an equivalence relation. Given thatit is, we see that m− n = m′ − n′ in G(M) if and only if m+ n′ + r = m′ + n+ r in Mfor some r ∈M .

As the reader may verify, setting (m−n)+(m′−n′) = (m+m′)−(n+n′) gives a welldefined binary operation on G(M). Under this operation, G(M) is an abelian monoidwith identity element 0 = 0− 0. But then n−m is an inverse for m− n, so that G(M)is an abelian group. We also have a monoid homomorphism η : M → G(M) defined byη(m) = m− 0.


It is common to refer to G(M) as the Grothendieck group of M .Note that m − 0 = m′ − 0 in G(M) if and only if there is an m′′ ∈ M such that

m+m′′ = m′ +m′′ in M . This sets up the following definition and lemma.

Definition 9.9.1. Let M be an abelian monoid. We say that elements m,m′ are stablyequivalent in M if there is an element m′′ ∈M such that m+m′′ = m′ +m′′ in M .

Lemma 9.9.2. Let M be an abelian monoid and let η : M → G(M) be the canonicalmap to its Grothendieck group. Then elements m,m′ ∈M become equal in G(M) if andonly if they are stably equivalent in M .

The Grothendieck construction satisfies the following universal property.

Proposition 9.9.3. Let M be an abelian monoid. Then for any monoid homomorphismf : M → G with G an abelian group, there is a unique group homomorphism f : G(M)→G such that the following diagram commutes.

M G

G(M)

��f

��

��

η

�� f

Proof Note that m− n = η(m)− η(n), so if f : G(M)→ G is a group homomorphismsuch that f ◦ η = f , then we must have f(m− n) = f(m)− f(n).

Thus, it suffices to show that if f : M → G is a monoid homomorphism, thenf(m−n) = f(m)−f(n) gives a well defined group homomorphism, as then automatically,(f ◦ η)(m) = f(m) − 0 = f(m) for all m ∈ M . But this is an easy check and is left tothe reader.

In fact, the group G in the last proposition need not be abelian.There is an alternative construction of the Grothendieck group which has a certain

appeal, so we give it. Thus, for an abelian monoid M , let F (M) be the free abeliangroup generated by M , with canonical basis {em |m ∈ M}. Let H ⊂ F (M) be thesubgroup generated by the elements {em1+m2 − em1 − em2 |m1,m2 ∈ M}. We defineG′(M) = F (M)/H, and define η′ : M → G′(M) by η′(m) = em. Note the fact that thegenerators of H have been set equal to 0 implies that η′ is a homomorphism.

Proposition 9.9.4. The homomorphism η′ : M → G′(M) satisfies the same universalproperty as η: For any monoid homomorphism f : M → G with G an abelian group,there is a unique group homomorphism f : G′(M)→ G such that the following diagramcommutes.

M G

G′(M)

��f

��

��

�

η′

�� f

In consequence, there is an isomorphism η : G′(M)→ G(M) with η ◦ η′ = η.

Proof Since f : M → G is a function and G is abelian, there is a unique grouphomomorphism, f ′, from the free abelian group F (M) to G such that f ′(em) = f(m)


for all m ∈ M . The fact that f is a monoid homomorphism shows that f ′ vanishes onthe generators of H, and hence H ⊂ ker f ′. Thus, there is a unique homomorphismf : G′(M) → G such that f ◦ π = f ′, where π : F (M) → G′(M) is the canonical map.But now f ◦ η′ = f as desired. The uniqueness of such an f follows from the universalproperties of free abelian groups and of factor groups.

Since η : M → G(M) and η′ : M → G′(M) are monoid homomorphisms, the universalproperties of η′ and η give us unique homomorphisms η : G′(M) → G(M) and η′ :G(M) → G′(M) such that η ◦ η′ = η and η′ ◦ η = η′, respectively. By the uniquenessstatements for the universal properties, η ◦ η′ = 1G(M) and η′ ◦ η = 1G′(M).

Definitions 9.9.5. Let A be a ring. Then K0(A) is the Grothendieck group of themonoid FP (A) of isomorphism classes of finitely generated projective left A-modules.

We write K0(A) for K0(A)/H, where H is the subgroup generated by [A] = [A]− 0.K0(A) is sometimes known as the projective class group of A, and sometimes as the

reduced projective class group, as some authors use the former term for K0(A). Withoutambiguity, we can refer to K0(A) as the reduced K0 group of A.

We should immediately establish the functoriality of these groups.

Proposition 9.9.6. Extension of rings makes K0 a functor from rings to abelian groups.Here, if f : A→ B is a ring homomorphism, K0(f)([P ]) = [B ⊗A P ].

Since K0(f)([A]) = [B], there is an induced map K0(f) : K0(A)→ K0(B).

Proof Let P be a finitely generated projective A-module. Then Corollary 9.8.10 showsus that B ⊗A P is finitely generated and projective over B. By the commutativity ofdirect sums with tensor products (Proposition 9.4.18), the passage from [P ] to [B⊗A P ]induces a monoid homomorphism from FP (A) to FP (B). The universal property ofthe Grothendieck construction then gives a unique homomorphism K0(f), which has thestated effect on the generators, [P ].

Lemma 9.4.14 shows that K0(1A) is the identity map of K0(A), so it suffices to showthat K0(g) ◦ K0(f) = K0(g ◦ f) for any ring homomorphism g : B → C. But this isimmediate from Corollary 9.4.25.

K0(A) fits into a general theory called algebraic K-theory, in which functors Ki(A)are defined for each i ∈ Z. We shall study K1 in Section 10.9.

K0 and K0 occur in many situations of importance in algebra, topology, and analysis.For instance, if G is a finite group, and K is a field whose characteristic does not

divide the order of G, then, using the theory of semisimple rings, one may show thatK0(K[G]) is isomorphic (as an abelian group) to the representation ring RK(G) of Gover K.

And we shall show in Chapter 12 that if A is a Dedekind domain, then K0(A) is theclass group, Cl(A), which plays an important role in number theory.

The groups K0(Z[G]) have numerous applications in topology. For instance, Wallhas shown that for appropriate spaces X, there is a finiteness obstruction, σ(X) ∈K0(Z[π1(X)]), which vanishes if and only ifX is homotopy equivalent to a finite simplicialcomplex. As a result, the groups K0(Z[G]) play an important role in the calculation ofthe surgery obstruction groups, which are used to classify manifolds.

These last two examples have a useful connection. We shall see in Chapter 12 thatZ[ζp] is a Dedekind domain if p is prime. We shall not prove the following theorem here.


Theorem (Rim’s Theorem) Let p be a prime and write Zp = 〈T 〉. Let

f : Z[Zp]→ Z[ζp]

be the ring homomorphism specified by f(T ) = ζp. Then K0(f) is an isomorphism.

Note that every finitely generated projective module P determines an element [P ] =[P ]−0 in K0(A). Since K0(A) is an abelian group, it may actually be possible to computewith it, and detect whether [P ] = [Q] in K0(A) for the finitely generated projectivemodules P and Q. In the best of all possible worlds, this would imply that P and Q wereisomorphic (as would be the case if [P ] were equal to [Q] in FP (A), rather than K0(A)).For instance, we shall see in Chapter 12 that if A is a semisimple ring, then [P ] = [Q] inK0(A) if and only if P ∼= Q.

However, for general rings, we must fall back on Lemma 9.9.2, which tells us that[P ] = [Q] in K0(A) if and only if P and Q are stably equivalent in FP (A), meaning thatP ⊕ P ′ ∼= Q⊕ P ′ for some finitely generated projective module P ′ of A. We shall makea formal definition for this relationship.

Definitions 9.9.7. We say that the finitely generated projective modules P and Q arestably isomorphic if there is a finitely generated projective module P ′ such that P ⊕P ′ ∼=Q⊕ P ′.

We say that a finitely generated projective module P is stably free if P ⊕ An ∼= Am

for some m,n ≥ 0.

Notice that P can be stably free without being stably equivalent to a free module,if it should happen that P ⊕ An ∼= Am with n > m. But extension of rings tells usthat this cannot happen if A admits a ring homomorphism to a division ring D, as thendim(D ⊗A P ) + n = m.

The following lemma gives a characterization of stable isomorphism and discusses itsrole in K0(A) and K0(A). We have already verified its initial statement.

Lemma 9.9.8. Let P and Q be finitely generated projective A-modules. Then [P ] = [Q]in K0(A) if and only if P and Q are stably isomorphic.

Moreover, P and Q are stably isomorphic if and only if there is an isomorphismP ⊕An ∼= Q⊕An for some n ≥ 0.

In consequence, the classes of P and Q become equivalent in K0(A) if and only ifP ⊕ Am ∼= Q ⊕ An for some nonnegative integers m and n. Thus, [P ] = 0 in K0(A) ifand only if P is stably free.

Proof Suppose that P and Q are stably isomorphic, say P ⊕P ′ ∼= Q⊕P ′ for a finitelygenerated projective module P ′ over A. But then P ⊕ P ′ ⊕ Q′ ∼= Q ⊕ P ′ ⊕ Q′ for anyQ′. Since P ′ is a direct summand of a finitely generated free module, we must haveP ⊕An ∼= Q⊕An for some n.

Now P and Q become equivalent in K0(A) if and only if [P ] − [Q] = [Ar] − [As] inK0(A) for some r and s, meaning that P ⊕As and Q⊕Ar are stably isomorphic.

It is hard to conceive of the fact that P and Q may be stably isomorphic but notisomorphic. But there do exist examples where this is so. And it is often very difficultto verify that stable isomorphism implies isomorphism for a given ring. For instance,the following theorem, proven independently by Quillen and Suslin, was a longstandingconjecture, special cases of which were solved by some eminent mathematicians. It isstill known as the Serre Conjecture, despite the fact that it has been proven.


Theorem (Serre Conjecture) LetK be a field. Then every finitely generated projectivemodule over a polynomial ring K[X1, . . . , Xn] is free.

In this case, using the theory of K0-groups, it is not hard to show that every finitelygenerated module over K[X1, . . . , Xn] is stably free. The hard part is showing that thestably free modules are all free. We shall not prove it here.

Note, in light of the Eilenberg Swindle of Section 9.8, how important it is that P andQ are being summed with An, rather than an infinitely generated free module in thedefinition of stable isomorphism.

So far we haven’t made any computations of K0. First, consider the following.

Lemma 9.9.9. Let N be the monoid of non-negative integers under addition. Thenthere is a commutative diagram

N Z

G(N)

��i

��

��

η

��

ı∼=

where η : N→ G(N) is the canonical map to the Grothendieck construction, and i is theusual inclusion of N in Z.

Proof Since i : N ⊂ Z is a monoid homomorphism, the universal property of theGrothendieck construction produces a group homomorphism ı : G(N)→ Z which makesthe diagram commute. Since Z is the free abelian group on the generator 1, there is agroup homomorphism j : Z → G(N) such that j(1) = η(1) = 1 − 0. Then ı ◦ j(1) = 1,so that ı ◦ j = 1Z. But since the elements of N generate G(N), and since 1, in turn,generates N, j must be onto. Thus, ı and j are inverse isomorphisms.

A slicker proof follows from the fact that N is the free monoid on 1, while Z is thefree group on 1. In consequence i : N ⊂ Z satisfies the same universal property asη : N→ G(N).

Recall that the rank of a free module is not well defined for all rings (i.e., there arerings for which Am ∼= An for some m �= n), but that it is well defined, for instance, if Aadmits a ring homomorphism to a nonzero division ring.

Corollary 9.9.10. Suppose that every finitely generated projective left A module is free,or, more generally, that every finitely generated projective left A module is stably free.Suppose in addition that the rank of a free A-module is well defined. Then K0(A) isisomorphic to Z, with generator [A], and hence K0(A) = 0.

Proof We assume throughout the argument that the rank of a free A-module is welldefined.

If every finitely generated projective module is free, then FP (A) is isomorphic to themonoid N of non-negative integers, and the result follows from Lemma 9.9.9.

If the finitely generated projectives are only stably free, the generic element of K0(A)has the form [P ] − [Q] = η([P ]) − η([Q]). But since P and Q are stably isomorphicto free modules, this element may be written in the form [An] − [Am] for some m,n ≥0. (Here, A0 is the 0 module, whose class gives the identity element of FP (A), andhence also of K0(A).) Thus, [A] generates K0(A) as a group. It suffices to show thatthe homomorphism Z → K0(A) which takes 1 to [A] is injective. Since every nonzero


subgroup of Z is cyclic, generated by a positive element, it suffices to show that [An] �= 0in K0(A) for n > 0.

Lemma 9.9.8 shows that [An] = 0 in K0(A) if and only if An+m ∼= Am for somem ≥ 0. But this would contradict the statement that rank is well defined.

Corollary 9.8.8 and Proposition 9.8.14 show that every finitely generated projectivemodule over a P.I.D. or a local ring, respectively, is free.

Corollary 9.9.11. If A is a P.I.D., a local ring, or a division ring, K0(A) is the infinitecyclic group generated by [A] and K0(A) = 0.

We shall make some additional computations of K0 in Chapter 12.We get some additional structure on K0(A) if A is commutative.

Proposition 9.9.12. Let A be a commutative ring. Then K0(A) is also a commutativering, with the multiplication given by [P ] · [Q] = [P ⊗A Q]. In addition, if f : A → Bis a homomorphism of commutative rings, then K0(f) is a ring homomorphism as well.Thus, K0 gives a functor from commutative rings to commutative rings.

Proof The key is to show that if P and Q are finitely generated projectives, then so isP ⊗A Q. To see this, suppose that P ⊕ P ′ ∼= An and Q⊕Q′ ∼= Am. Then (P ⊕ P ′)⊗A(Q⊕Q′) is isomorphic to Amn by a couple of applications of the commutativity of tensorproducts and direct sum (Proposition 9.4.18). But the commutativity of tensor productswith direct sums also shows that P ⊗A Q is a direct summand of (P ⊕ P ′)⊗A (Q⊕Q′).

Proposition 9.4.18 also shows that for fixed Q, the map which sends [P ] to [P ⊗A Q]gives a monoid homomorphism from FP (A) to itself. By the universal property ofthe Grothendieck construction, this extends to a group homomorphism r[P ] : K0(A) →K0(A). Since P ⊗A Q is isomorphic to Q ⊗A P , we see that if [P ] − [Q] = [P ′] − [Q′]in K0(A), then r[P ] − r[Q] agrees with r[P ′] − r[Q′] on the image of η : FP (A)→ K0(A).Since K0(A) is generated by the image of η, we see that the tensor product induces acommutative binary operation on K0(A) that satisfies the distributive law.

Associativity of this multiplication follows from Corollary 9.4.24, and [A] is a mul-tiplicative identity by Lemma 9.4.14. Finally, if f : A → B is a homomorphism ofcommutative rings, then K0(f) is a ring homomorphism by Problem 4 of Exercises 9.4.27.

Corollary 9.9.13. Let A be a P.I.D. or a local ring. Then K0(A) is isomorphic to Zas a ring.

We mentioned earlier that if G is a finite group and K is a field whose characteristicdoes not divide the order of G, then K0(K[G]) is naturally isomorphic as an abeliangroup to the representation ring RK(G) of G over K. Thus, if G is abelian, we have ringstructures on K0(K[G]) coming from either its structure as a representation ring or asK0 of a commutative ring. These ring structures are quite different. For instance, themultiplicative identity of the representation ring is K (with the trivial G-action), whilethe multiplicative identity of K0(K[G]) is K[G].

Proposition 9.9.14. Let A be a ring that admits a homomorphism to a division ring.Then the sequence

0→ K0(Z) i∗−→ K0(A)→ K0(A)→ 0

is split short exact, and hence K0(A) ∼= Z⊕ K0(A).If A is commutative and f : A → K is a homomorphism to a field, then we may

identify K0(A) with the kernel of f∗, which is an ideal of K0(A).


Proof Let f : A → D be a homomorphism to a division ring. Then f ◦ i : Z → Dinduces an isomorphism of K0, so f∗ may be used to construct a retraction for i∗. Nowapply Proposition 7.7.49.

Note that the section K0(A)→ K0(A) induced by this retraction has image equal tothe kernel of f∗. The result follows.

We shall see below that if A is a product of fields, say A = K1 × K2, then theprojection maps A → Ki have different kernels. Thus, there is no preferred splitting ofthe exact sequence above, nor is there an unambiguous identification of K0(A) with anideal of K0(A) for general commutative rings. If A is an integral domain, however, thisproblem does not occur.

The next result can be useful.

Proposition 9.9.15. Let A and B be rings. Then K0(A × B) is naturally isomorphicto K0(A) ×K0(B). If A and B are commutative, then this isomorphism is a ring iso-morphism.

Proof By Problem 2 of Exercises 9.8.17, every finitely generated projective moduleover A × B has the form P × P ′, where P is a finitely generated projective over A andP ′ is a finitely generated projective over B. But this says that FP (A×B) is isomorphicto FP (A) × FP (B) as a monoid. Since finite products and coproducts agree in thecategories of abelian groups and abelian monoids, it is easy to see that the Grothendieckconstruction on a product M×M ′ of abelian monoids is naturally isomorphic to G(M)×G(M ′).

If A and B are commutative, then the natural map

K0(A×B)(K0(π1),K0(π2))−−−−−−−−−−→ K0(A)×K0(B)

is a ring homomorphism, since K0(f) is a ring homomorphism for any homomorphism fof commutative rings. Thus, it suffices to show that (K0(π1),K0(π2)) coincides with theisomorphism constructed above, i.e., that K0(π1)([P×P ′]) = [P ] and K0(π2)([P×P ′]) =[P ′] if P and P ′ are finitely generated projectives over A and B, respectively.

We have K0(π1)([P × P ′]) = [A⊗A×B (P × P ′)].

A⊗A×B (P × P ′) ∼= (A×B/0×B)⊗A×B (P × P ′)∼= (P × P ′)/(0×B)(P × P ′)∼= P

Here, the second isomorphism is from Corollary 9.4.16. The case of π2 is analogous.

Notice that K0(A×B) does not split as a product the way that K0(A × B) does.Thus, even if one is primarily interested in phenomena regarding the reduced K0 groups,it is sometimes useful to consider the unreduced groups as well.

There is an alternative description of K0(A) which can be useful. We define anequivalence relation ≈ on FP (A) by setting [P ] ≈ [Q] if P ⊕An is isomorphic to Q⊕Amfor some m and n. Write FP (A) for the set of equivalence classes on FP (A) given bythis relation, and write [[P ]] ∈ FP (A) for the equivalence class of [P ].

Lemma 9.9.16. FP (A) is an abelian group under the operation [[P ]]+[[Q]] = [[P ⊕Q]].


Proof The direct sum operation in FP (A) is easily seen to respect equivalence classeswith respect to the relation ≈, and therefore induces a binary operation on FP (A). Itinherits commutativity, associativity, and an identity element from FP (A), so it sufficesto show that every element has an inverse. But if [P ] ∈ FP (A), then there is a finitelygenerated projective module Q such that P ⊕ Q is isomorphic to An for some n. Since[[An]] = [[0]] = 0 in FP (A), [[Q]] is the inverse of [[P ]].

A nice application of this is the next corollary, which only requires the uniqueness ofinverses in a group.

Corollary 9.9.17. Let P be a finitely generated projective module over A, and let Qand Q′ be finitely generated projective modules such that P ⊕Q and P ⊕Q′ are both free.Then Q⊕An ∼= Q′ ⊕Am for some m,n ≥ 0.

Note that FP is a functor via extension of rings.

Proposition 9.9.18. There is a natural isomorphism between the functors K0 and FP .

Proof Since [An] = 0 in K0(A), there is a homomorphism ν : FP (A)→ K0(A) definedby ν([[P ]]) = [P ]. Clearly, this is natural in A. Since K0(A) is generated by the elementsof FP (A), ν is onto. Let [[P ]] be in the kernel of ν. Then Lemma 9.9.8 shows that P isstably free. But that says that [P ] ≈ [0], and hence that [[P ]] = 0 in FP (A).

Exercises 9.9.19.1. Calculate K0(Zn).

2. Let f : A → B be a ring homomorphism, where A is commutative. Show thatK0(B) is a module over K0(A).

3. Let A be a commutative ring and let p be a prime ideal of A. Show that we mayidentify K0(A) with the kernel of K0(η), where η : A→ Ap is the canonical map.

4. Let A be a commutative ring and let P be a finitely generated projective A-module.Show that [P ] = 0 in K0(A) if and only if P is the 0-module. (Hint : Show thatPp = 0 for each prime ideal p.) Note that this gives us no information about theproperties of stable isomorphism between nonzero modules.

5. Verify, using the first construction of the Grothendieck group, that if f : M → G isa monoid homomorphism, where M is abelian, but G is not, then there is a uniquegroup homomorphism f : G(M)→ G such that f ◦ η = f .

6. Show that the second construction of the Grothendieck group may be (drasti-cally) modified to work in the nonabelian world. Thus, if M is a possibly non-abelian monoid, we can construct a group G′′(M) and a monoid homomorphismη′′ : M → G′′(M) which is universal in the sense that if f : M → G is amonoid homomorphism into a group G, then there is a unique group homomor-phism f : G′′(M) → G such that f ◦ η′′ = f . Now using the preceding exercise,show that if M is abelian, there is a group isomorphism η′′ : G(M)→ G′′(M) suchthat η′′ ◦ η = η′′.


9.10 Tensor Algebras and Their Relatives

Throughout this section, A is a commutative ring.Here, we describe the tensor algebra of an A-module M , together with some algebras

derived from it. The tensor algebra is the free A-algebra on an A-module M .

Definition 9.10.1. Let M be an A-module. Then the tensor algebra, TA(M), of Mover A is given by

TA(M) =⊕k≥0

M⊗k,

where M⊗k =

⎧⎨⎩A if k = 0

M ⊗A · · · ⊗AM︸︷︷︸k times

if k > 0.

The product in TA(M) is induced by setting

(m1 ⊗ · · · ⊗mk) · (m′1 ⊗ · · · ⊗m′

l) = m1 ⊗ · · · ⊗mk ⊗m′1 ⊗ · · · ⊗m′

l

and by letting A act via its usual module structure on M⊗k.

One way of interpreting this product is by noting that it may be obtained from thenatural isomorphisms

M⊗k ⊗AM⊗l ∼= M⊗k+l.

In particular, it is well defined, and satisfies the distributive laws. It is easy to see thatTA(M) is an A-algebra via this product.

The easiest example to analyze is the tensor algebra of A itself. Notice that if wethink of A as the free A-module on a basis element e1, then A⊗k is the free A-module on

ek1 = e1 ⊗ · · · ⊗ e1︸︷︷︸k times

.

We obtain the following lemma.

Lemma 9.10.2. The A-algebra homomorphism εe1 : A[X]→ TA(A) obtained by evalu-ating X at e1 is an isomorphism.

To develop the universal properties of tensor algebras more generally, let i : M →TA(M) be the canonical inclusion of M = M⊗1. Then the following proposition showsthat TA(M) is the free A-algebra on M .

Proposition 9.10.3. Let B be an A-algebra and let f : M → B be an A-module homo-morphism. Then there is a unique A-algebra homomorphism f : TA(M)→ B that makesthe following diagram commute.

M B

TA(M)

��f

��

��

�

i

�� f


Proof Just set f(m1 ⊗ · · · ⊗ mk) = f(m1) . . . f(mk). As the right-hand side is A-multilinear in m1, . . . ,mk, this specifies an A-module homomorphism, which is theneasily seen to be a ring homomorphism.

In particular, note that the elements of M = M⊗1 generate TA(M) as an A-algebra.We shall be able to construct A-algebras with other kinds of universal properties as

quotient rings of TA(M). Indeed, every A-algebra is a quotient of a tensor algebra.

Definition 9.10.4. Let M be an A-module and let a be the two-sided ideal of TA(M)generated by all elements of the form m1 ⊗m2 −m2 ⊗m1, with m1,m2 ∈M . Then thesymmetric algebra, SA(M), of M over A is the quotient ring TA(M)/a.

Example 9.10.5. Since TA(A) is commutative, it is easy to see that TA(A) ∼= SA(A) ∼=A[X].

Let ı : M → SA(M) be the composite of i : M → TA(M) with the canonical mapof the tensor algebra onto the symmetric algebra. Then the following proposition showsthat SA(M) is the free commutative A-algebra on M .

Proposition 9.10.6. Let B be a commutative A-algebra and let f : M → B be an A-module homomorphism. Then there is a unique A-algebra homomorphism f : SA(M)→B that makes the following diagram commute.

M B

SA(M)

��f

��

��

�

ı

�� f

Proof By Proposition 9.10.3, there is a unique A-algebra homomorphism f : TA(M)→B such that f ◦ i = f . Since B is commutative, f(m1 ⊗ m2 − m2 ⊗ m1) = 0 for allm1,m2 ∈ M . Thus, the ideal a is contained in the kernel of f , and hence f factorsuniquely through the canonical map from TA(M) to SA(M).

The next example, the exterior algebra, is easiest to understand in the context ofgraded algebras, so we shall develop these first.

Definitions 9.10.7. A graded A-module is an A-module M with an explicit direct sumdecomposition

M =∞⊕i=0

Mi

as an A-module. The elements of the submodule Mi are called the homogeneous elementsof degree i.

A graded homomorphism f : M → N between graded modules is one with theproperty that f(Mi) ⊂ Ni for all i ≥ 0.

A graded A-algebra, B, is a graded A-module which is an A-algebra (compatible withthe A-module structure) under a multiplication satisfying

1. 1 ∈ B0.

2. If b1 ∈ Bi and b2 ∈ Bj , then b1b2 ∈ Bi+j .


A graded A-algebra B is called skew commutative (or skew symmetric, or commuta-tive in the graded sense) if for b1 ∈ Bi and b2 ∈ Bj , we have

b1b2 = (−1)ijb2b1

Skew commutative algebras are important in algebraic topology and in homologicalalgebra, as the cohomology ring of a topological space is skew commutative, as are mostof the cohomology rings that arise purely algebraically. We shall see in Exercises 9.10.28that if 2 is a unit in A, then the exterior algebra on M is the free skew commutativealgebra on M , if we take M to be a graded module whose elements are all homogeneousof degree 1.

Examples 9.10.8.1. The tensor algebra TA(M) may be considered as a graded A-algebra with

(TA(M))i = M⊗i.

2. Specializing the preceding example to M = A gives an isomorph of the following:

The polynomial ring A[X] may be considered as a graded ring in which (A[X])i ={aXi | a ∈ A}. Thus, the homogeneous elements are the monomials.

3. Let X1, . . . , Xn be variables, and choose arbitrary non-negative integers ki, 1 ≤i ≤ n, for the degree of Xi. Then the polynomial ring A[X1, . . . , Xn] has a gradedalgebra structure in which the monomial aXr1

1 . . . Xrnn is a homogeneous element

of degree r1k1 + · · ·+ rnkn.

We would like to show that the symmetric algebras are graded algebras as well. Tosee this, we should define graded ideals.

Definitions 9.10.9. Let M be a graded A-module. Then a graded submodule, N , ofM is a direct sum

N =∞⊕i=0

Ni

where Ni is a submodule of Mi.If B is a graded A-algebra, then a graded ideal, a, of B is an ideal of B which is

graded as a submodule of B.

LetM be a graded A-module. Notice that a submoduleN ofM is a graded submoduleif and only if for each n ∈ N , the component of n lying in each Mi is also in N . (Here,we mean the component in the direct sum decomposition M =

⊕∞i=0Mi.)

Lemma 9.10.10. Let B be a graded A-algebra and let S ⊂ B be any subset consistingof homogeneous elements of B. Then the ideal of B generated by S is a homogeneousideal.

Of course, if a =⊕∞

i=0 ai is a graded ideal of B, then the graded module B/a =⊕∞i=0Bi/ai is a graded A-algebra. Since the elements m1 ⊗m2 −m1 ⊗m1 of the tensor

algebra are homogeneous of degree 2 in the standard grading, we obtain the followingcorollary.

Corollary 9.10.11. The symmetric algebra SA(M) has a natural grading in which theelements of M are homogeneous of degree 1.


Notice that if all the nonzero homogeneous elements of a graded algebra have evendegree, then the concepts of commutativity and skew commutativity coincide. Thus, ifwe set M⊗k to have homogeneous degree 2k in TA(M) for all k ≥ 0, then the resultantgrading on SA(M) gives an algebra that is skew commutative as well as commutative.

Definition 9.10.12. Let M be an A-module and let b ⊂ TA(M) be the two-sided idealgenerated by all elements of the form m ⊗m, with m ∈ M . Then the exterior algebraΛA(M) is the quotient ring TA(M)/b. We write ΛkA(M) for the image in ΛA(M) of M⊗k.

We write m1 ∧ · · · ∧ mk for the image in ΛkA(M) of the element m1 ⊗ · · · ⊗ mk ofTA(M).

If we grade TA(M) by setting M⊗i to be the homogeneous elements of degree i,then b is homogeneous, inducing a grading on ΛA(M) in which ΛiA(M) is the module ofhomogeneous elements of degree i. We shall take this as the standard grading on ΛA(M).

As above, the simplest example to work out is ΛA(A). Here, ae1 ⊗ ae1 is a mul-tiple of e1 ⊗ e1, so b is the ideal generated by e1 ⊗ e1. Utilizing the isomorphism ofLemma 9.10.2, which preserves grading, we see that ΛA(A) is isomorphic as a gradedalgebra to A[X]/(X2), which has basis 1, X as an A-module, with X2 = 0.

Lemma 9.10.13. Think of A as the free module on e1. Then Λ0A(A) and Λ1

A(A) arefree on 1 and e1, respectively, while ΛkA(A) = 0 for k > 1.

For general A-modules M , we have the following facts about ΛA(M).

Lemma 9.10.14. Let M be an A-module. Then the exterior algebra ΛA(M) is skewcommutative. Moreover, the square of any element of Λ2k+1

A (M) is zero, for all k ≥ 0.

Proof For m1,m2 ∈M , we have

(m1 +m2)2 = m21 +m1 ∧m2 +m2 ∧m1 +m2

2.

But the squares of elements of M are in b and hence are trivial in ΛA(M), and hencem1 ∧m2 = −m2 ∧m1 for all m1,m2 ∈M .

But an easy induction now shows that

(m1 ∧ · · · ∧mk) ∧ (m′1 ∧ · · · ∧m′

l) = (−1)kl(m′1 ∧ · · · ∧m′

l) ∧ (m1 ∧ · · · ∧mk).

The distributive law now shows ΛA(M) to be skew commutative.Finally, if α, β ∈ Λ2k+1

A (M), then the fact that they anticommute shows that (α +β)2 = α2 +β2. Clearly, any element of the form m1 ∧ · · · ∧m2k+1 squares to 0. Since theelements of this form generate Λ2k+1

A (M), the squaring operation on Λ2k+1A (M) must be

trivial.

Notice that there are no hypotheses about A in the preceding result. If 2 wereinvertible in A, then the statement regarding the squares of odd-degree elements wouldfollow from the skew commutativity. The point is that skew commutativity implies thatif x is homogeneous of odd degree, then x2 = −x2.

Lemma 9.10.15. Let A be a commutative ring in which 2 is invertible, and let B bea skew commutative graded A-algebra. Then the square of any homogeneous element ofodd degree is trivial.


Note that all we really need is that 2 fails to be a zero-divisor in B. Thus, this sortof analysis applies to situations where A is a domain of characteristic �= 2 and B is freeas an A-module.

We apply the results above to the study of universal properties of exterior algebras.Let M be an A-module. We may consider it to be a graded A-module in which allelements are homogeneous of degree 1. We call this the graded A-module in which M isconcentrated in degree 1.

We write ı : M → ΛA(M) for the composite of the natural inclusion of M = M⊗1 inTA(M) with the canonical map onto the quotient ring ΛA(M). Note that ı is a gradedmap from the above grading on M to the standard grading on ΛA(M). Also, since theideal b of TA(M) is generated by elements of degree 2, ı is an isomorphism of M ontoΛ1A(M). As in the proof of Proposition 9.10.6, the following proposition is immediate

from the universal property of the tensor algebra.

Proposition 9.10.16. Let M be an A-module, graded by concentration in degree 1 andlet B be a graded A-algebra. Let f : M → B be a graded A-module map with the propertythat f(m)2 = 0 for each m ∈M . Then there is a unique graded A-algebra homomorphismf : ΛA(M)→ B that makes the following diagram commute.

M B

ΛA(M)

��f

��

��

�

ı

�� f

Here, ΛA(M) is given the standard grading.If 2 is invertible in A and if B is skew commutative, then every graded A-algebra map

f : M → B has the above property. Thus, ΛA(M) is the free skew commutative A-algebraon M when 2 is a unit in A.

We wish to determine the structure of ΛA(An). One way to do this is via determinanttheory, as exterior algebras turn out to be a generalization of determinants.

But the argumentation turns out to be simpler if we develop the theory of skewcommutative algebras a bit more. By this method, we shall develop some more powerfulresults, as well as shed some light on determinant theory.

We first give an alternative formulation of the tensor product of graded algebras.

Definitions 9.10.17. Let M and N be graded A-modules. Then the preferred gradingon M ⊗A N is obtained by setting

(M ⊗A N)k =k⊕i=0

Mi ⊗A Nk−i.

Let B and C be graded A-algebras. Write B ⊗A C for the algebra structure onB ⊗A C obtained as follows: If b1, b2, c1, c2 are homogeneous elements, then

b1 ⊗ c1 · b2 ⊗ c2 = (−1)|b2|·|c1|b1b2 ⊗ c1c2,

where |b2| and |c1| stand for the degrees of b2 and c1, respectively.The standard terminology in homological algebra is to refer to B ⊗A C as the graded

tensor product of B and C. We shall follow this convention.


Warning: The graded tensor product is generally not isomorphic as an algebra to theusual algebra structure on B ⊗A C (the one from Proposition 9.6.1). The latter, whichis always commutative if B and C are commutative, can be thought of as the gradedtensor product if B and C are both concentrated in degree 0 (or in even degrees).

If we restrict attention to skew commutative A-algebras, the graded tensor producthas an important universal property. Note first that it is easy to see that if B and C areskew commutative, then so is B ⊗A C.

Proposition 9.10.18. The graded tensor product is the coproduct in the category ofskew commutative A-algebras. In other words, if B, C, and D are skew commutativeA-algebras and if f : B → D and g : C → D are graded A-algebra homomorphisms,then there is a unique graded A-algebra homomorphism h : B ⊗A C → D such that thefollowing diagram commutes.

B B ⊗A C C

D

��ι1

��

��

��

�

f��

h

�� ι2

��

��

��

g

Here, ι1(b) = b⊗ 1 and ι2(c) = 1⊗ c.Proof Since b⊗1 ·1⊗c = b⊗c, if there is a graded A-algebra homomorphism h makingthe diagram commute, we must have h(b⊗ c) = f(b)g(c). Since B ⊗A C is generated bythe elements of the form b⊗ c, the homomorphism h is uniquely defined by the diagram.

For the existence of h, note that setting h(b ⊗ c) = f(b)g(c) specifies a graded A-module homomorphism on B ⊗A C. And if b1, b2, c1, and c2 are homogeneous, then theskew commutativity of D shows that

h(b1 ⊗ c1 · b2 ⊗ c2) = h(b1 ⊗ c1) · h(b2 ⊗ c2).Since B ⊗A C is generated by elements b ⊗ c with b and c homogeneous, h is a homo-morphism of graded algebras.

For an A-module M , we have Λ0A(M) = A and Λ1

A(M) = M . Thus, if M and N areA-modules, then the homogeneous elements in degree 1 in ΛA(M) ⊗A ΛA(N) are givenby (M ⊗A A) ⊕ (A ⊗A N) ∼= M ⊕ N . This isomorphism gives the degree 1 part of theA-algebra isomorphism in the following proposition.

Proposition 9.10.19. Let M and N be A-modules. Then there is a natural isomor-phism of graded rings:

ΛA(M) ⊗A ΛA(N) ∼= ΛA(M ⊕N).

Proof Write α : (ΛA(M) ⊗A ΛA(N))1∼=−→M⊕N for the isomorphism described above.

Then Propositions 9.10.16 and 9.10.18 provide a unique graded A-algebra homomorphism

α : ΛA(M) ⊗A ΛA(N)→ ΛA(M ⊕N)

which restricts to α in degree 1, while Proposition 9.10.16 provides a unique gradedA-algebra homomorphism, β, in the opposite direction, restricting to α−1 in degree 1.

Thus, both α ◦ β and β ◦ α restrict to the identity in degree 1. Since both algebrasare generated by their elements of degree 1, β and α must be inverse isomorphisms.


Thus, given our analysis of ΛA(A), we are now able to compute ΛA(An).

Proposition 9.10.20. Let n ≥ 1. Then for 1 ≤ k ≤ n, ΛkA(An) is a free A-module ofrank

(nk

)and with basis given by any ordering of {ei1 ∧ · · · ∧ eik | i1 < · · · < ik}. Here,

e1, . . . , en is the canonical basis for An.For k > n, ΛkA(An) = 0, while Λ0

A(An) is, of course, A.

Proof Write An = Ae1⊕ · · · ⊕Aen. Then Proposition 9.10.19 gives an isomorphism ofgraded A-algebras

α : ΛA(Ae1) ⊗A · · · ⊗A ΛA(Aen)→ ΛA(An)

which carries ei1 ⊗ · · · ⊗ eik onto ei1 ∧ · · · ∧ eik . The result now follows easily fromLemma 9.10.13: Since Λ0

A(A) = A and ΛkA(A) = 0 for k > 1, the homogeneous terms ofdegree k in ΛA(Ae1) ⊗A · · · ⊗A ΛA(Aen) are given by the terms⊕

i1<···<ikΛ1A(Aei1)⊗A · · · ⊗A Λ1

A(Aeik).

We shall close this section with a discussion of Clifford algebras. Here, we shall applythe theory of graded algebras in a different context. The graded algebras we consideredpreviously were graded by the monoid N of non-negative integers under addition. Cliffordalgebras are graded over Z2.

Definition 9.10.21. A Z2-graded A-algebra, B, is an A-algebra with a direct sum de-composition B = B0 ⊕B1 as an A-module, such that

1. 1 ∈ B0.

2. If b1 ∈ Bi and b2 ∈ Bj , then b1b2 ∈ Bi+j , where the sum i+ j is taken mod 2.

A Z2-graded A-algebra B is skew commutative if for b1 ∈ Bi and b2 ∈ Bj , we have

b1b2 = (−1)ijb2b1.

Example 9.10.22. Let B be a graded A-algebra in the usual sense (with the gradingsover the non-negative integers). Then we may view B as a Z2-graded algebra as follows.If we write B′, for B with the induced Z2-grading, then we have

B′0 =

⊕k≥0

B2k and

B′1 =

⊕k≥0

B2k+1.

Note that if B is skew commutative in the usual sense, then B′ is a skew commutativeZ2-graded algebra.

Thus, Z2-gradings are strictly weaker than gradings over the non-negative integers.

Definition 9.10.23. Let x1, . . . , xn be a basis for a free A-module M . Then the Cliffordalgebra CliffA(x1, . . . , xn) is the quotient algebra TA(M)/c, where c is the two-sided idealgenerated by xi ⊗ xj + xj ⊗ xi for 1 ≤ i < j ≤ n, together with the elements x2

i + 1 for1 ≤ i ≤ n.


Here, the ideal is generated by homogeneous elements in the standard Z2-grading onTA(M), so as in the N-graded case, CliffA(x1, . . . , xn) inherits a Z2-grading from thaton the tensor algebra.

Example 9.10.24. Using the isomorphism of Lemma 9.10.2, we see that the Cliffordalgebra on a single generator is given by

CliffA(x) ∼= A[X]/(X2 + 1).

In particular, we obtain the complex numbers as C = CliffR(x).

Since the squares of the xi are nonzero in CliffA(x1, . . . , xn), the Clifford algebras arenot skew commutative, except in characteristic 2. Nevertheless, they may be computedin terms of a Z2-graded version of ⊗A.

Definition 9.10.25. Let B and C be Z2-graded A-algebras. Then B ⊗A C is a gradedalgebra structure on the A-module B ⊗A C, where the grading is given by

(B ⊗A C)0 = (B0 ⊗A C0)⊕ (B1 ⊗A C1) and(B ⊗A C)1 = (B1 ⊗A C0)⊕ (B0 ⊗A C1),

and the multiplication is given as follows: if b1, b2, c1, c2 are homogeneous elements, then

b1 ⊗ c1 · b2 ⊗ c2 = (−1)|b2|·|c1|b1b2 ⊗ c1c2,where |b2| and |c1| stand for the degrees of b2 and c1, respectively.

Since the algebras we care about here are not skew commutative, we have to be a bitmore careful about stating the universal property. The proof of the next proposition isleft to the reader.

Proposition 9.10.26. Let f : B → D and g : C → D be Z2-graded A-algebra ho-momorphisms such that for any pair of homogeneous elements b ∈ B and c ∈ C, wehave

f(b)g(c) = (−1)|b|·|c|g(c)f(b).

Then there is a unique Z2-graded A-algebra homomorphism h : B ⊗A C → D such thatthe following diagram commutes.

B B ⊗A C C

D

��ι1

��

��

��

�

f��

h

�� ι2

��

��

��

g

Here, ι1(b) = b⊗ 1 and ι2(c) = 1⊗ c.And that’s all we need for the proof of the following proposition.

Proposition 9.10.27. There is an isomorphism

α : CliffA(x1) ⊗A · · · ⊗A CliffA(xn)→ CliffA(x1, . . . , xn)

defined by α(xi) = xi for 1 ≤ i ≤ n. Thus,

{xi1 . . . xik | k ≥ 0, 1 ≤ i1 < · · · < ik ≤ n}gives a basis for CliffA(x1, . . . , xn) as an A-module. Here, the empty sequence, wherek = 0 gives the multiplicative identity element, 1.


Proof Proposition 9.10.26 shows that setting α(xi) = xi gives a well defined algebrahomomorphism, as stated. And if M = Ax1 ⊕ · · · ⊕ Axn, then setting β(xi) = xi givesa well defined A-algebra homomorphism

β : TA(M)→ CliffA(x1) ⊗A · · · ⊗A CliffA(xn).

Clearly, the ideal c lies in the kernel of β, so there is an A-algebra homomorphism

β : CliffA(x1, . . . , xn)→ CliffA(x1) ⊗A · · · ⊗A CliffA(xn)

with β(xi) = xi for i = 1, . . . , n. Since x1, . . . , xn generate both sides as A-algebras, αand β must be inverse isomorphisms.

Exercises 9.10.28.

1. Show that CliffR(x1, x2), the Clifford algebra on two generators over R, is isomor-phic to the division ring of quaternions, H.

2. Let M be an A-module and let B be a commutative A-algebra. Show that there isa B-algebra isomorphism

B ⊗A TA(M) ∼= TB(B ⊗AM).

Here, the algebra structure on the left is the ordinary tensor product of algebras.Note that if we give B the grading concentrated in degree 0 and use the standardgradings on the tensor algebras (in which M is concentrated in degree 1), then thealgebra structure on the left side agrees with that of B ⊗A TA(M), and we havean isomorphism of graded algebras.

3. Prove the analogues of the preceding problem where the tensor algebra is replacedby

(a) The symmetric algebra.

(b) The exterior algebra.

(c) The Clifford algebra, if M is the free A-module with basis x1, . . . , xn.

4. Let P be a finitely generated projective module over A of constant rank k. Showthat ΛiA(P ) = 0 for i > k.

‡ 5. Show that if M is any graded A-module, then the tensor algebra TA(M) has aunique grading as an A-algebra such that i : M → TA(M) is a graded map. Showthat TA(M), with this grading, is the free graded A-algebra on M .

6. Show that the grading on TA(M) given in the preceding problem induces gradedalgebra structures on both the symmetric and exterior algebras on M .

(a) Show that the symmetric algebra is the free skew commutative algebra on Mif all the nonzero homogeneous elements of M lie in even degree or if A hascharacteristic 2.

(b) Show that the exterior algebra is the free skew commutative algebra on M ifall the nonzero homogeneous elements of M lie in odd degree, provided that2 is a unit in A.


7. Let M be an arbitrary graded module over a ring A in which 2 is invertible. Whatis the free skew commutative algebra on M?

8. Show that SA(An) is isomorphic to the polynomial algebra in n variables over A.

9. Let B and C be graded A-algebras. Exhibit an explicit graded A-algebra isomor-phism from B ⊗A C to C ⊗A B.

10. Show that TA(An) is isomorphic to the monoid ring over A of the free monoid onn elements.

Chapter 10

Linear algebra

In the first part of this chapter, we develop the basic theory of linear algebra over com-mutative rings. The first three sections are devoted to the theory of traces and determi-nants. Then, we study the characteristic polynomial, giving a generalized version of theCayley–Hamilton Theorem.

In Sections 10.5 through 10.7, we study matrices over fields. Eigenvectors and eigen-values are introduced, and are used to study diagonalization. Then, Section 10.6 classifiesthe matrices over any field up to similarity, using rational canonical form. Section 10.7studies Jordan canonical form.

Then, we leave the realm of commutative rings, studying general linear groups overarbitrary rings. Section 10.8 studies Gauss elimination and gives generators of the groupGln(D) for a division ring D. Section 10.9 uses the material of Section 10.8 to define thealgebraic K-group K1(A) for a ring A.

Until we get to Section 10.8, we shall adopt the convention that all rings A arecommutative. And for the entirety of the chapter, we shall assume, unless otherwisestated, that the transformation induced by a matrix is induced by the left action ofmatrices on column vectors.

10.1 Traces

Recall that A is assumed commutative.

Definition 10.1.1. Let M = (aij) be an n × n matrix. By the trace of M , writtentr(M), we mean the sum of the diagonal entries in M . Thus, tr(M) =

∑ni=1 aii.

Recall from Lemma 7.1.16 that Mn(A) is an A-algebra via ι : A → Mn(A), whereι(a) = aIn, the matrix whose diagonal entries are all a’s and whose off-diagonal entriesare all 0’s. Thus, there is an induced A-module structure on Mn(A), where aM is thematrix product aIn ·M , for a ∈ A and M ∈Mn(A).

Lemma 10.1.2. The trace induces an A-module homomorphism tr : Mn(A)→ A.

The behavior of trace with respect to block sum is clear.

Lemma 10.1.3. Let M =(M ′ 00 M ′′

). Then tr(M) = tr(M ′) + tr(M ′′).

Lemma 10.1.4. Let M,M ′ ∈Mn(A). Then tr(MM ′) = tr(M ′M).

416

CHAPTER 10. LINEAR ALGEBRA 417

Proof Let M = (aij) and M ′ = (bij). Write MM ′ = (cij) and M ′M = (dij). Then

tr(MM ′) =n∑i=1

cii =n∑i=1

n∑j=1

aijbji,

while

tr(M ′M) =n∑i=1

dii =n∑i=1

n∑j=1

bijaji.

Thus, while cii �= dii, when we add up all the diagonal terms in MM ′ and in M ′M , weget the same result.

Proposition 10.1.5. Let M,M ′ ∈Mn(A) be similar matrices. Then tr(M) = tr(M ′).

Proof Similarity means that M ′ = M ′′M(M ′′)−1 for some invertible matrix M ′′ ∈Mn(A). But tr(M ′′M(M ′′)−1) = tr(M(M ′′)−1M ′′) = tr(M) by Lemma 10.1.4.

Recall that since A is commutative, the rank of a finitely generated free A-module isuniquely defined by Theorem 7.9.8. Let N be a free A-module of rank n. Recall fromCorollary 7.10.20 that if f : N → N is an A-module homomorphism, and if B and B′

are two different bases of N , then if M is the matrix of f with respect to B and if M ′ isthe matrix of f with respect to B′, then M and M ′ are similar.

Corollary 10.1.6. Let N be a finitely generated free A-module and let f : N → N bean A-module homomorphism. Let B and B′ be bases of N and let M be the matrix off with respect to B and let M ′ be the matrix of f with respect to B′. Then M and M ′

have the same trace.


Definition 10.1.7. Let N be a finitely generated free A-module and let f : N → N bean A-module homomorphism. Then the trace of f , tr(f), is defined to be the trace ofthe matrix of f with respect to any basis of N .

Recall that EndA(N) = HomA(N,N) is the endomorphism ring of N , with the prod-uct operation coming from composition of homomorphisms. By Lemma 7.10.15, if Nis free of rank n with basis B, then the transformation from EndA(N) to Mn(A) ob-tained by passage from a homomorphism f : N → N to its matrix with respect to Bis an isomorphism of rings. Explicitly, if B = x1, . . . , xn and if f(xj) =

∑ni=1 aijxi, for

1 ≤ j ≤ n, then the matrix of f with respect to B is just (aij).Since A is commutative, multiplication by a gives an A-module map ma : N → N

for any A-module N and any a ∈ A. Indeed, ma is easily seen to commute with eachf ∈ EndA(N), and we obtain an A-algebra structure μ : A → EndA(N) via μ(a) = ma

for all a ∈ A. The reader may note that the A-module structure on EndA(N) inducedby this A-algebra structure coincides with the A-module structures (all of them) given inLemma 9.7.3. Explicitly, for f ∈ EndA(N) and a ∈ A, the homomorphism af is definedby (af)(n) = f(an). Since multiplication by the matrix aIn induces multiplication by a,we obtain the following lemma.

Lemma 10.1.8. Let N be a free A-module of rank n. The isomorphism between EndA(N)and Mn(A) obtained from a basis B of N is an isomorphism of A-algebras.


Recall from Proposition 7.10.11 that, for an appropriate choice of basis, the matrixrepresenting a direct sum f1⊕ f2 of homomorphisms is the block sum of the matrices forf1 and f2. We obtain the following proposition.

Proposition 10.1.9. Let N be a finitely generated free A-module. Then tr : EndA(N)→A is an A-module homomorphism satisfying the following properties.

(1) tr(f ◦ g) = tr(g ◦ f), for f, g ∈ EndA(N).

(2) If f is a unit in EndA(N), then tr(f ◦ g ◦ f−1) = tr(g) for any g ∈ EndA(N).

(3) If N = N1 ⊕ N2, where N1 and N2 are finitely generated free modules, and iffi ∈ EndA(Ni) for i = 1, 2, then tr(f ⊕ g) = tr(f) + tr(g).

10.2 Multilinear alternating forms

Recall that A is a commutative ring.We develop some material needed to show the existence and basic properties of de-

terminants. Note that an n× n matrix may be viewed as an n-tuple of column vectors.Thus, we may identify Mn(A) with the product An × · · · ×An of n copies of An.

Definitions 10.2.1. Let N be an A-module. An n-form on N is any function

f : N × · · · ×N︸︷︷︸n times

→ A

from the product of n copies of N to A.An n-form f : N × · · · × N → A is multilinear if for each i ∈ {1, . . . , n} and each

collection x1, . . . , xi−1, xi+1, . . . , xn ∈ N , the function g : N → A given by

g(x) = f(x1, . . . , xi−1, x, xi+1, . . . , xn)

is an A-module homomorphism.1

A multilinear alternating n-form on N is a multilinear n-form f : N × · · · ×N → Awith the additional property that f(x1, . . . , xn) = 0 if xi = xj for some i �= j.

Our goal is to show that there is a unique multilinear alternating n-form

Δ : An × · · · ×An → A

with the property that Δ(e1, . . . , en) = 1. We will then define the determinant asfollows: If M is the n × n matrix whose i-th column is xi for 1 ≤ i ≤ n, thendetM = Δ(x1, . . . , xn). In the process of showing that Δ exists and is unique, weshall derive formulæ that will allow us to demonstrate the important properties of deter-minants.

Lemma 10.2.2. Let f : N ×· · ·×N → A be a multilinear alternating n-form. Then forany n-tuple (x1, . . . , xn) and any i < j, if (y1, . . . , yn) is obtained from (x1, . . . , xn) byexchanging the positions of xi and xj (i.e., yi = xj, yj = xi, and yk = xk for k �= i, j),then f(y1, . . . , yn) = −f(x1, . . . , xn). In words, if we exchange the positions of xi andxj, then the value of f changes sign.

1This is identical to the notion of A-multilinearity given in Definitions 9.4.21.


Proof Define a function of two variables, g : N ×N → A by

g(x, y) = f(x1, . . . , xi−1, x, xi+1, . . . , xj−1, y, xj+1, . . . , xn).

Then we are asked to show that g(y, x) = −g(x, y) for any x, y ∈ N .Notice that g itself is a multilinear alternating 2-form on N . Indeed, the general case

will follow if we verify the result for n = 2.Now,

0 = g(x+ y, x+ y) = g(x, x) + g(y, y) + g(x, y) + g(y, x),

where the first equality holds because g is alternating, and the second because g ismultilinear. Since g(x, x) = g(y, y) = 0, we obtain g(x, y) + g(y, x) = 0, as desired.

Recall that the sign of a permutation is given by a group homomorphism ε : Sn →{±1} characterized by the property that ε(τ) = −1 for each transposition τ .

Proposition 10.2.3. Suppose that there exists a multilinear alternating n-form

Δ : An × · · · ×An → A

with the property that Δ(e1, . . . , en) = 1. Then given x1, . . . , xn ∈ An, if we writexj =

∑ni=1 aijei, we obtain that

Δ(x1, . . . , xn) =∑σ∈Sn

ε(σ)aσ(1),1 . . . aσ(n),n.

Since the right-hand side is defined without any reference to Δ, we see that Δ, if it exists,is unique, and may be calculated by the stated formula.

Proof Since xj =∑ni=1 aijei for j = 1, . . . , n, the multilinearity of Δ gives

Δ(x1, . . . , xj) = Δ

(n∑

i1=1

ai11ei1 , . . . ,

n∑in=1

ainnein

)=

∑i1,...,in∈{1,...,n}

ai11 . . . ainnΔ(ei1 , . . . , ein).

Since Δ is alternating, unless the i1, . . . , in are all distinct, we must have Δ(ei1 , . . . , ein) =0. However, if the i1, . . . , in are all distinct, the function σ : {1, . . . , n} → {1, . . . , n}defined by σ(k) = ik is a permutation. Indeed, this describes a bijective correspondencebetween Sn and the collection of all n-tuples i1, . . . , in of distinct elements of {1, . . . , n}.Thus, we see that

Δ(x1, . . . , xn) =∑σ∈Sn

aσ(1),1 . . . aσ(n),nΔ(eσ(1), . . . , eσ(n)).

Thus, it suffices to show that if σ ∈ Sn, then Δ(eσ(1), . . . , eσ(n)) = ε(σ). If σ is atransposition, this follows from Lemma 10.2.2, together with the fact that Δ(e1, . . . , en) =1. But recall that any permutation is a product of transpositions, and that if σ is theproduct of k transpositions then ε(σ) = (−1)k. Thus, we can argue by induction on thenumber of transpositions needed to write σ as a product of transpositions.


Thus, suppose that σ = τσ′, where τ is a transposition, and where

Δ(eσ′(1), . . . , eσ′(n)) = ε(σ′)

by the induction hypothesis. Then ε(σ) = ε(τ)ε(σ′) = −ε(σ′), and it suffices to showthat

Δ(eσ(1), . . . , eσ(n)) = −Δ(eσ′(1), . . . , eσ′(n)).

But the sequence eσ(1), . . . , eσ(n) is obtained from eσ′(1), . . . , eσ′(n) by exchanging two ofits terms, because τ is a transposition. Thus, the result follows from Lemma 10.2.2.

The formula for Δ obtained in the preceding proposition is useful for a number ofthings, but it is hard to show that it gives a multilinear alternating form. Thus, weshall use a different formula to show that Δ exists. This alternative formula will alsohave applications. To state it, it is easiest (although not necessary) to use the languageand notation of matrices, rather than forms. Thus, we shall first formalize the matrixapproach to the problem.

Definition 10.2.4. The determinant for n × n matrices over A is a function det :Mn(A) → A which exists if and only if there is a multilinear alternating n-form Δ :An×· · ·×An → A with the property that Δ(e1, . . . , en) = 1. The determinant is definedas follows: if xi is the i-th column of M for 1 ≤ i ≤ n, then det(M) = Δ(x1, . . . , xn).

We shall often write detM for det(M).

Note that the columns, in order, of In are precisely e1, . . . , en. The following lemmais immediate.

Lemma 10.2.5. The n×n determinant, if it exists, is the unique function det : Mn(A)→A which is multilinear and alternating in the columns of the matrices and satisfiesdet In = 1.

The next corollary is now immediate from Proposition 10.2.3.

Corollary 10.2.6. Suppose there is an n × n determinant det : Mn(A) → A. Then ifM = (aij), we have

detM =∑σ∈Sn

ε(σ)aσ(1),1 . . . aσ(n),n.

This has a useful consequence. First, we need a definition.

Definition 10.2.7. LetM ∈Mn(A), withM = (aij). Then the transpose ofM , writtenM t, is the matrix whose ij-th entry is aji.

Thus, the transpose of M is obtained by reflecting the entries of M through the maindiagonal. Our determinant formula will now give the following.

Proposition 10.2.8. Suppose there is an n × n determinant det : Mn(A) → A. Thenfor any M ∈Mn(A), the determinants of M and its transpose are equal.

Proof Let M = (aij). Consider the expansion

detM =∑σ∈Sn

ε(σ)aσ(1),1 . . . aσ(n),n.


Note that for any permutation τ , the product aσ(1),1 . . . aσ(n),n is just a rearrangementof the product aσ(τ(1)),τ(1) . . . aσ(τ(n)),τ(n). Applying this with τ = σ−1, we get

aσ(1),1 . . . aσ(n),n = a1,σ−1(1) . . . an,σ−1(n).

But if we set M t = (bij), so that bij = aji, we see that this latter product is justbσ−1(1),1 . . . bσ−1(n),n.

Since ε : Sn → {±1} is a homomorphism, and since each element of {±1} is itsown inverse, we see that ε(σ) = ε(σ−1) for each σ ∈ Sn. Summing up the equalitiesaσ(1),1 . . . aσ(n),n = bσ−1(1),1 . . . bσ−1(n),n over all σ ∈ Sn, we get

detM =∑σ∈Sn

ε(σ−1)bσ−1(1),1 . . . bσ−1(n),n.

But since the passage from σ to σ−1 gives a bijection of Sn, this is just the expansion ofdet(M t).

Corollary 10.2.9. The n×n determinant, if it exists, is a multilinear alternating func-tion on the rows of a matrix, as well as the columns.

We shall now show by induction on n that an n× n determinant exists (and hence amultilinear alternating n-form Δ exists on An with Δ(e1, . . . , en) = 1). We first need adefinition.

Definition 10.2.10. Let M ∈ Mn(A). We write Mij for the (n − 1) × (n − 1) matrixobtained by deleting the i-th row and j-th column of M . We call it the ij-th (n− 1)×(n− 1) submatrix of M .

The determinant formula given next is called the expansion with respect to the i-throw.

Proposition 10.2.11. The n× n determinant det : Mn(A)→ A exists for all n ≥ 1. If1 ≤ i ≤ n, the determinant may be given by the following inductive formula: If M = (aij),then

detM =n∑j=1

(−1)i+jaij detMij ,

where detMij is the determinant of the ij-th (n− 1)× (n− 1) submatrix of M .

Proof It is easy to see that the 1 × 1 determinant is given by det(a) = a. Thus, byinduction, it suffices to show that if the (n − 1) × (n − 1) determinant exists, then theabove formula is multilinear and alternating in the columns, and carries In to 1.

First, we show multilinearity. Suppose that the columns of M , M ′, and M ′′ are allequal except for the k-th column, and that the k-th column of M is equal to the sum ofthe k-th columns of M ′ and M ′′. We want to show that if we define the n×n determinantby the above formula, then detM = detM ′ + detM ′′.

Note that if j �= k, then the columns of the (n − 1) × (n − 1) submatrices Mij , M ′ij

and M ′′ij are all equal except for the l-th column, where l = k if k < j and l = k − 1 if

k > j. Moreover, the l-th column of Mij is the sum of the l-th columns of M ′ij and M ′′

ij .


Thus, if j �= k, then detMij = detM ′ij +detM ′′

ij , as the (n−1)× (n−1) determinantis multilinear. Also note that Mik = M ′

ik = M ′′ik. Now write M = (aij), M ′ = (a′ij), and

M ′′ = (a′′ij), so that aij = a′ij = a′′ij for j �= k and aik = a′ik + a′′ik. We have

detM =n∑j=1

(−1)i+jaij detMij

=

⎛⎝∑j =k

(−1)i+jaij(det(M ′ij) + det(M ′′

ij))

⎞⎠+ (−1)i+k(a′ik + a′′ik) det(Mik).

By the remarks above, this is precisely detM ′ + detM ′′.Similarly, if M is obtained from M ′ by multiplying the k-th column by a, then

det(Mij) = adet(M ′ij) for j �= k by the inductive hypothesis. Since Mik = M ′

ik and sinceaik = aa′ik, a similar argument shows that detM = adetM ′.

Next, we need to show that det is an alternating function in the columns. Thus, if thek-th and k′-th columns of M are equal, we must show detM = 0. Note that if j �= k, k′,then there are two equal columns in Mij . Thus, det(Mij) = 0 by induction. Noting thataik = aik′ , we see that the expansion of detM with respect to the i-th row gives

detM = (−1)i+kaik det(Mik) + (−1)i+k′aik det(Mik′).

Assume that k < k′. If k′ = k + 1, then Mik = Mik′ , and the signs (−1)i+k and(−1)i+k

′are opposite. Thus, the two terms cancel and detM = 0. Alternatively, we

have k′ > k + 1, and Mik is obtained from Mik′ by permuting the k-th column of Mik′

past the (k+1)-st column through the (k′−1)-st column. In other words, Mik is obtainedfrom Mik′ by permuting the k-th column of Mik′ past k′ − k − 1 columns, so the signchanges k′−k−1 times, giving det(Mik) = (−1)k

′−k−1 det(Mik′). Inserting this into theabove formula gives detM = 0 as desired.

Finally, we need to show that det In = 1. Note first that if j �= i, then the k-thcolumn of (In)ij is 0, where k = i if i < j and k = i − 1 if i > j. Thus, det((In)ij) = 0by the multilinearity of the (n − 1) × (n − 1) determinant. The expansion now givesdet In = (−1)i+i det((In)ii). But it is easy to see that (In)ii = In−1, and the result nowfollows by induction.

Note that the ij-th (n − 1) × (n − 1) submatrix of M is the transpose of the ji-th(n− 1)× (n− 1) submatrix of M t. Since the determinant of a matrix and its transposeare equal, and since transposition reverses the roles of rows and columns, we obtain thefollowing corollary.

Corollary 10.2.12. For 1 ≤ j ≤ n, the determinant of an n × n matrix M = (aij)may be computed by the following expansion, called the expansion according to the j-thcolumn:

detM =n∑i=1

(−1)i+jaij detMij .

Given that we’ve established the existence of n × n determinants for all n over allcommutative rings, we can now ask about change of rings. The following is immediatefrom the expansion of the determinant in terms of permutations (Corollary 10.2.6).


Proposition 10.2.13. Let f : A → B be a ring homomorphism between commutativerings and let f∗ : Mn(A)→Mn(B) be the induced map (i.e., if M = (aij), then the ij-thcoordinate of f∗(M) is f(aij)). Then

Mn(A)f∗−−−−→ Mn(B)

det

⏐⏐1 det

⏐⏐1A

f−−−−→ B

commutes. Thus, if M ∈Mn(A), then det(f∗(M)) = f(detM).

Remarks 10.2.14. The connection between multilinear n-forms and tensor productsis as follows. Suppose given a multilinear n-form f : N × · · · × N → A. By Proposi-tion 9.4.23, there is a unique A-module homomorphism, f , from the tensor product of ncopies of N to A making the following diagram commute.

N × · · · ×N︸︷︷︸n times

A

N ⊗A · · · ⊗A N︸︷︷︸n times

��f

��

ι

�� f

Here, ι(x1, . . . , xn) = x1 ⊗ · · · ⊗ xn.Since ι is A-multilinear in the sense of Definitions 9.4.21, the composite g ◦ ι is a

multilinear n-form for any A-module homomorphism, g, from the n-fold tensor productof N with itself into A. We obtain a one-to-one correspondence between the multilinearn-forms on N and the A-module homomorphisms into A from the tensor product overA of n copies of N .

If the multilinear n-form, f , is alternating, then it is easy to see that there is a uniquefactorization of the homomorphism f : N ⊗A · · · ⊗A N → A as follows.

N ⊗A · · · ⊗A N A

ΛnA(N)��

π

��f�� f

Here, ΛnA(N) is the module of homogeneous elements of degree n in the exterior algebraΛA(N), and π is induced by the natural map from the tensor algebra TA(M) onto theexterior algebra.

But it’s easy to check that if f : ΛnA(N) → A is an A-module homomorphism, thensetting

f(x1, . . . , xn) = f(x1 ∧ · · · ∧ xn)gives a multilinear alternating n-form on N . Here, x1 ∧ · · · ∧xn is the standard notationfor the image of (x1, . . . , xn) under π ◦ ι. The end result is that there is a one-to-onecorrespondence between the multilinear alternating n-forms on N and the A-modulehomomorphisms from ΛnA(N) to A.

The case of greatest interest, of course, is for N = An. Here, Proposition 9.10.20shows that ΛnA(An) is the free A-module with basis e1 ∧ · · · ∧ en. Thus, if Δ : ΛnA(An)→


A is the unique A-module homomorphism that takes e1 ∧ · · · ∧ en to 1, then settingΔ(x1, . . . , xn) = Δ(x1 ∧ · · · ∧ xn) gives the unique multilinear alternating n-form whosevalue on (e1, . . . , en) is 1. We see that the existence and uniqueness of determinants is animmediate consequence of Proposition 9.10.20 and the universal properties of the tensorproduct and exterior algebra.

However, the uniqueness formula given in this section is useful, as are the expansionsof the determinant by rows and columns. And the easiest way to check the validity ofthese expansions is by the arguments given here: First, one shows that the expansionby rows is a multilinear alternating n-form carrying (e1, . . . , en) to 1. By uniqueness,it must give the determinant. Then one applies the uniqueness formula to show thatdetM = detM t. Thus, we have not wasted any effort here.

10.3 Properties of determinants

Determinants behave nicely with respect to matrix multiplication.

Proposition 10.3.1. Let M,M ′ ∈Mn(A). Then

det(MM ′) = (detM)(detM ′).

Proof The argument here is very similar to that in Proposition 10.2.3. Indeed, we shallmake use of the argument itself.

Recall from the preceding section that if Δ : An × · · · × An → A is the uniquemultilinear alternating n-form on An, and if M ′′ ∈ Mn(A) has columns y1, . . . , yn, thendetM ′′ is defined to be equal to Δ(y1, . . . , yn). Recall also that the i-th column of M ′′

is equal to the matrix product M ′′ei, where ei is the i-th canonical basis vector of An,which we have identified with the space of n× 1 column vectors.

Thus, det(MM ′) = Δ(MM ′e1, . . . ,MM ′en). Let us write M = (aij) and M ′ = (bij).Thus, M ′ej =

∑ni=1 bijei. Thus,

det(MM ′) = Δ

(n∑

i1=1

bi11Mei1 , . . . ,

n∑in=1

binnMein

)=

∑i1,...,in∈{1,...,n}

bi11 . . . binnΔ(Mei1 , . . . ,Mein).

Since Δ is alternating, Δ(Mei1 , . . . ,Mein) = 0 if ij = ik for some j �= k. As in the proofof Proposition 10.2.3, this implies that the nonzero summands above are in one-to-onecorrespondence with elements of Sn, so that we can make the identification

det(MM ′) =∑σ∈Sn

bσ(1),1 . . . bσ(n),nΔ(Meσ(1), . . . ,Meσ(n)).

Again we appeal to the proof of Proposition 10.2.3. The argument there which showsthat Δ(eσ(1), . . . , eσ(n)) = ε(σ)Δ(e1, . . . , en) goes over verbatim to this context, andshows that Δ(Meσ(1), . . . ,Meσ(n)) = ε(σ)Δ(Me1, . . . ,Men). But Δ(Me1, . . . ,Men) =detM . Thus, the previously displayed formula simplifies to

det(MM ′) =∑σ∈Sn

bσ(1),1 . . . bσ(n),nε(σ) detM

= detM∑σ∈Sn

ε(σ)bσ(1),1 . . . bσ(n),n

= (detM)(detM ′)


by the expansion for detM ′ of Proposition 10.2.3.

Since det In = 1, this says that det : Mn(A) → A is a monoid homomorphismbetween the multiplicative monoids of Mn(A) and A. As such, it must restrict to agroup homomorphism from the group of invertible elements (i.e., units) in Mn(A) to thegroup of units of A. Recall that the unit group of Mn(A) is Gln(A).

Corollary 10.3.2. The determinant restricts to a group homomorphism

det : Gln(A)→ A×.

Since the multiplicative monoid of A is abelian, we also have the following corollary.

Corollary 10.3.3. Let M,M ′ ∈Mn(A) be similar matrices. Then detM = detM ′.

We shall now investigate the relationship between determinants and invertibility.

Definition 10.3.4. Let M ∈ Mn(A). The adjoint matrix, Madj, of M is the matrixwhose ij-th entry is (−1)i+j detMji. Here, Mji is the (n − 1) × (n − 1) submatrixobtained by deleting the j-th row and i-th column of M .

As might be expected from the definition of the adjoint matrix, the expansions of thedeterminant in terms of rows and columns are important in the next proof.

Proposition 10.3.5. Let M ∈Mn(A) and let Madj be its adjoint matrix. Then

MMadj = MadjM = (detM)In.

Here, (detM)In is the matrix whose diagonal entries are all detM , and whose off-diagonal entries are all 0.

Proof Note that the ij-th entry of MMadj is∑nk=1 aik(−1)j+k detMjk. If i = j, this

is just the expansion of detM with respect to the i-th row, so the diagonal entries ofMMadj are equal to detM as desired. Thus, MMadj has the desired form provided wecan show that

∑nk=1(−1)j+kaik detMjk = 0 whenever i �= j. But indeed, the sum in

question is just the expansion with respect to the j-th row of the matrix obtained bythrowing away the j-th row of M and replacing it with a second copy of the i-th. Inparticular, it is the determinant of a matrix with two equal rows. Since the determinantof a matrix is equal to the determinant of its transpose, the determinant vanishes on amatrix with two equal rows. Thus, the off-diagonal entries of MMadj are all 0, as desired.

The argument that MadjM = (detM)In is analogous, using column expansions inplace of row expansions.

Our next corollary follows from the fact that M , Madj, and aIn commute with eachother for any a ∈ A.

Corollary 10.3.6. A matrix M ∈ Mn(A) is invertible if and only if detM is a unit ofA. If detM ∈ A×, then M−1 = ((detM)−1In)Madj.

Next, we consider the determinant of a block sum.

Proposition 10.3.7. Let M ′ ∈Mm(A), let M ′′ ∈Mk(A), and let M = M ′⊕M ′′. ThendetM = (detM ′)(detM ′′).


Proof We argue by induction on m, using the expansion of detM with respect to thefirst row.

If m = 1, let M ′ = (a). Then the first row of M is (a, 0, . . . , 0). Thus, in theexpansion of detM with respect to the first row, all terms but the first must vanish, anddetM = adetM11. But a = detM ′ and M11 = M ′′, so the result follows.

For m > 1, the first row of M is (a11, . . . , a1m, 0, . . . , 0), where (a11, . . . , a1m) is thefirst row of M ′. Thus,

detM = a11 detM11 − a12 detM12 + · · ·+ (−1)1+ma1m detM1m.

But for j ≤ m, we can write M1j as a block sum: M1j = M ′1j ⊕M ′′. Since M ′

1j is(m − 1) × (m − 1), our induction hypothesis gives detM1j = (detM ′

1j)(detM ′′), andhence

detM =m∑j=1

(−1)1+ja1j(detM ′1j)(detM ′′)

= (detM ′′)m∑j=1

(−1)1+ja1j(detM ′1j)

= (detM ′′)(detM ′)

The fact that matrices whose determinants are units are always invertible, togetherwith the product rule for determinants, shows that the set of n×nmatrices of determinant1 forms a subgroup of Gln(A).

Definition 10.3.8. We write Sln(A) for the group of n× n matrices over A of determi-nant 1. We call it the n-th special linear group of A.

Special linear groups tend to contain the hard part in terms of our understandings ofgeneral linear groups:

Proposition 10.3.9. For any commutative ring A and any n > 0, we have a split shortexact sequence of groups

0→ Sln(A) ⊂−→ Gln(A) det−−→ A× → 0.

Proof It suffices to construct a section for det : Gln(A) → A×. We define s : A× →Gln(A) by s(a) = (a)⊕ In−1. Proposition 10.3.7 shows that det s(a) = a, and the resultfollows.

Recall that if N is a finitely generated free A-module and if f ∈ EndA(N), then thematrices of f with respect to any two bases are similar.

Definition 10.3.10. Let N be a finitely generated free A-module and let f ∈ EndA(N).Then the determinant of f , det f , is the determinant of the matrix of f with respect toany basis of N .

LetN be a free module of rank n. Then a choice of basis for n gives a ring isomorphismfrom EndA(N) to Mn(A). Recall that the unit group of EndA(N) is called AutA(N).

Theorem 10.3.11. Let A be a commutative ring and let N be a finitely generated freeA-module. Then the determinant gives a function

det : EndA(N)→ A

with the following properties.


(1) det(f ◦ g) = (det f)(det g).

(2) An endomorphism f ∈ EndA(N) is an isomorphism if and only if det f ∈ A×.

(3) Restriction gives a group homomorphism det : AutA(N)→ A×.

(4) If f ∈ EndA(N) and g ∈ AutA(N), then det(gfg−1) = det f .

(5) If N = N1 ⊕ N2, where N1 and N2 are finitely generated free modules, and iffi ∈ EndA(Ni) for i = 1, 2, then det(f1 ⊕ f2) = (det f1)(det f2).

(6) Let f : A → B be a ring homomorphism, with B commutative. Then for g ∈EndA(N), the endomorphism 1B ⊗ g of B ⊗A N satisfies det(1B ⊗ g) = f(det g).

Proof All these assertions but the last should be clear from the results in this section.The last follows from Corollary 9.4.20.

Exercises 10.3.12.1. Let K be a field and let V be a finite dimensional vector space over K. Letf ∈ EndK(V ). Show that det f = 0 if and only if ker f �= 0.

2. Using rings of fractions, show that the preceding problem shows that if A is anintegral domain and if N is a finitely generated free A-module, then f ∈ EndA(N)has determinant 0 if and only if ker f �= 0.

3. Show that if A is not an integral domain, then there are endomorphisms of freemodules whose determinant is nonzero, but whose kernel is also nonzero.

4. Cramer’s Rule Let A be a commutative ring and let M ∈ Mn(A). Show thatfor y ∈ An, the i-th coordinate of Madjy is equal to the determinant of the matrixobtained by replacing the i-th column of M with y. Deduce that if detM ∈ A×,then the i-th coordinate of the solution of Mx = y is equal to (detM)−1 times thedeterminant of the matrix obtained by replacing the i-th column of M with y.

5. Show that if M is a diagonal matrix (i.e., the off-diagonal entries are all 0), thenthe determinant of M is the product of the diagonal entries.

† 6. Show that if M =(M ′ X0 M ′′

), then detM = (detM ′)(detM ′′). Show the same

for M =(M ′ 0Y M ′′

).

7. Write fM : An → An for the linear transformation induced by the matrix M ∈Mn(A). Show that the matrix M has the form

(M ′ X0 M ′′

)with M ′ ∈ Mk(A)

and M ′′ ∈ Mn−k(A) if and only if fM (Ak) ⊂ Ak, where Ak is the submodule ofAn generated by e1, . . . , ek.

8. Let N,N1, and N2 be free modules of finite rank, and suppose that the rows of thefollowing commutative diagram are exact.

0 �� N1⊂ ��

f1

��

Nπ ��

f

��

N2��

f2

��

0

0 �� N1⊂ �� N

π �� N2�� 0

Show that tr(f) = tr(f1) + tr(f2) and det f = det f1 · det f2.


9. Permutation Matrices. For σ ∈ Sn, let Mσ ∈ Mn(A) be the matrix whose i-thcolumn is eσ(i).

(a) Show that the passage from σ to Mσ gives an injective group homomorphismη : Sn → Gln(A): η(σ) = Mσ.

(b) Show that η induces an action of Sn on An under which each σ carries ei toeσ(i) for 1 ≤ i ≤ n.

(c) Show that detMσ = ε(σ).(d) Let xi be the i-th column of M ∈Mn(A) for i = 1, . . . , n. Show that the i-th

column of M ·Mσ is xσ(i) for all σ ∈ Sn and all i = 1, . . . , n. Compare this tothe action of Sn on (An)n (i.e., the matrices, when considered as n-tuples ofcolumns) given by Problem 23 of Exercises 3.3.23.

(e) Let yi be the i-th row of M ∈Mn(A) for i = 1, . . . , n. Show that the i-th rowof Mσ ·M is yσ−1(i).

10. The Vandermonde Determinant. Define Mn ∈Mn(Z[X1, . . . , Xn]) by

Mn =

⎛⎜⎜⎜⎝1 X1 X2

1 . . . Xn−11

1 X2 X22 . . . Xn−1

2...

1 Xn X2n . . . Xn−1

n

⎞⎟⎟⎟⎠ .

The element detMn ∈ Z[X1, . . . , Xn] is called the Vandermonde Determinant.

As shown in Proposition 7.3.23, we have ring isomorphisms

Z[X1, . . . , Xn] ∼= (Z[X1, . . . , Xj−1, Xj+1, . . . , Xn]) [Xj ]

for j = 1, . . . , n, where the isomorphism takes each Xi to Xi.

(a) Show that if i < j, then Xj −Xi divides detMn in Z[X1, . . . , Xn].(b) Show that detMn is divisible by

∏1≤i<j≤n(Xj − Xi). (Here, if n = 1, the

latter is just the empty product, which we set equal to 1.)(c) Show by induction on n that the Vandermonde Determinant is given by

detMn =∏

1≤i<j≤n(Xj −Xi).

For the inductive step, view both detMn and∏

1≤i<j≤n(Xj − Xi) as poly-nomials in Xn with coefficients in Z[X1, . . . , Xn−1]. Show that each one is apolynomial of degree n− 1 in Xn with leading coefficient detMn−1.

(d) Let K be a field and let x1, . . . , xn be distinct elements of K. Let y1, . . . , ynbe any collection of (not necessarily distinct) elements of K. Show that thereis a unique polynomial f(X) ∈ K[X] of degree n− 1 such that f(xi) = yi forall i.

11. Let P be a finitely generated projective module over A, and let α : P ⊕ Q∼=−→

An. Define ι : EndA(P ) → EndA(P ⊕Q) by ι(f) = f ⊕ 1Q and define α∗ :EndA(P ⊕Q)→ EndA(An) by α∗(g) = α ◦ g ◦ α−1. Define det : EndA(P )→ A tobe the composite

EndA(P ) ι−→ EndA(P ⊕Q) α∗−→ EndA(An) det−−→ A.


(a) Show that det : EndA(P ) → A is independent of the choice of isomorphismfrom P ⊕Q to An.

(b) Show that if we replace Q by Q⊕A and replace α by α⊕1A, then the functiondet : EndA(P )→ A doesn’t change.

(c) Deduce from Corollary 9.9.17 that if Q′ is any other finitely generated projec-tive such that P ⊕Q′ is free, then the above procedure applied to an isomor-phism P ⊕Q′ ∼= Am will produce exactly the same function det : EndA(P )→A.

(d) Show that det : EndA(P ) → A satisfies all the properties listed in Theo-rem 10.3.11.

12. Let f ∈ EndA(N), and let Λk(f) be the endomorphism of ΛkA(N) which restrictson generators to take each x1 ∧ · · · ∧ xk to f(x1) ∧ · · · ∧ f(xk).

Suppose that N is a free module of rank n. Thus, by Proposition 9.10.20, ΛnA(N)is a free module of rank 1.

(a) Deduce that Λn(f) must be induced by multiplication by some a ∈ A.(b) Let x1, . . . , xn be a basis for N and let M be the matrix of f with respect to

this basis. Show, via Proposition 10.2.3, that

f(x1) ∧ · · · ∧ f(xn) = detM · x1 ∧ · · · ∧ xn.(c) Use this to give a new proof that if M and M ′ are matrices for f with respect

to two different bases, then detM = detM ′. In particular, note that we couldhave defined the determinant of an endomorphism f , simply by stipulatingthat det f is the a of part (a) above. Note that if x1, . . . , xn is a basis for N ,then det f is characterized by the equality

f(x1) ∧ · · · ∧ f(xn) = det f · x1 ∧ · · · ∧ xn.(d) Use the properties of Λn(f) to give a quick proof that det(f ◦g) = det f ·det g.(e) Use the relationship between exterior algebras and direct sums to show that

det(f ⊕ g) = det f · det g, where f and g are endomorphisms of free modulesof rank n and k, respectively.

10.4 The characteristic polynomial

Let A be a commutative ring and let ι : A → A[X] be the natural inclusion. As iscustomary, we write ι(a) = a, and identify A as a subring of A[X]. Similarly, we considerMn(A) to be a subring of Mn(A[X]) via ι∗ : Mn(A)→ Mn(A[X]), and write M , ratherthan ι∗(M), for the image of M ∈Mn(A) under this inclusion.

Thus, if M ∈Mn(A), we may consider the element XIn −M ∈Mn(A[X]). We havethe determinant map det : Mn(A[X])→ A[X], and hence det(XIn−M) is a polynomial.

Definition 10.4.1. Let M ∈ Mn(A). Then the characteristic polynomial of M is thepolynomial chM (X) = det(XIn −M).

The characteristic polynomial turns out to be of greatest use in the analysis of ma-trices over a field. Note, however, that even if A is a field, the study of characteristicpolynomials will involve the use of determinants over the commutative ring A[X]. Thus,it is not enough for the purposes of field theory to study determinants over fields alone.


Lemma 10.4.2. Let M ∈Mn(A). Then chM (X) is a monic polynomial of degree n.

Proof Write M = (aij). Then (XIn −M) = (fij(X)), where

fij(X) ={ −aij if i �= jX − aii if i = j.

Thus, fij(X) has degree ≤ 0 if i �= j and has degree 1 if i = j. Now consider theexpansion of det(XIn−M) in terms of permutations. For every permutation other thanthe identity, at least two of the fσ(i),i must have degree ≤ 0. Thus, the product of termsassociated with each nonidentity permutation is a polynomial of degree < n − 1. Butthe term corresponding to the identity permutation is (X − a11) . . . (X − ann), a monicpolynomial of degree n. Thus, det(XIn−M) is also a monic polynomial of degree n.

We shall summarize some of the most basic properties of characteristic polynomials,and then specialize to some of the applications for matrices over a field.

Lemma 10.4.3. Let M and M ′ be similar matrices in Mn(A). Then chM (X) = chM ′(X).

Proof Suppose that M ′′M(M ′′)−1 = M ′. Then in Mn(A[X]), we have

M ′′(XIn −M)(M ′′)−1 = M ′′XIn(M ′′)−1 −M ′′M(M ′′)−1 = XIn −M ′,

since XIn is in the center of Mn(A[X]). Thus, XIn −M and XIn −M ′ are similar inMn(A[X]), and hence have the same determinant.

Thus, we may define the characteristic polynomial of an endomorphism.

Definition 10.4.4. Let N be a finitely generated free A-module and let f ∈ EndA(N).Then the characteristic polynomial chf (X) of f is equal to the characteristic polynomialof the matrix of f with respect to any chosen basis of N .

LetM ′ ∈Mn(A) andM ′′ ∈Mk(A). SinceXIn is a diagonal matrix, XIn−(M ′⊕M ′′)is the block sum of XIn −M ′ and XIn −M ′′. We may now apply Proposition 10.3.7:

Lemma 10.4.5. chM ′⊕M ′′(X) = chM ′(X) · chM ′′(X).

The roots of the characteristic polynomial have an important property.

Lemma 10.4.6. Let M ∈Mn(A). Then if we evaluate the characteristic polynomial ofM at a ∈ A, we get chM (a) = det(aIn−M), the determinant of the matrix (aIn−M) ∈Mn(A). In particular, a is a root of chM (X) if and only if det(aIn −M) = 0.

Proof Let ε : A[X] → A be the A-algebra homomorphism obtained by evaluating Xat a. Then Proposition 10.2.13 gives a commutative diagram

Mn(A[X]) Mn(A)

A[X] A

��det

��ε∗

��det

��ε

The result follows, as ε∗(XIn −M) = aIn −M .


An important result about the characteristic polynomial is what’s known as theCayley–Hamilton Theorem. We give a generalization of it here.

Theorem 10.4.7 (Generalized Cayley–Hamilton Theorem). Let N be a finitely gener-ated module over the commutative ring A and let f ∈ EndA(N). Let x1, . . . , xn be a setof generators for N over A, and suppose that f(xj) =

∑ni=1 aijxi. Write M = (aij) ∈

Mn(A). Then chM (X) is in the kernel of the evaluation map εf : A[X] → EndA(N)obtained by evaluating X at f .

In other words, if chM (X) = Xn + an−1Xn−1 + · · ·+ a0, then

chM (f) = fn + an−1fn−1 + · · ·+ a0 = 0

as an endomorphism of N .

Proof First note that sinceXIn−M t is the transpose ofXIn−M , chM (X) = chMt(X).Thus, we may use M t in place of M .

The subalgebra A[f ] of EndA(N) is the image of εf , and hence is commutative. Thus,Proposition 10.2.13 gives us a commutative diagram

Mn(A[X]) Mn(A[f ])

A[X] A[f ]

��εf ∗

��det

��det

��εf

.

Since chMt(f) = εf (chMt(X)), we see that chMt(f) is the determinant of the matrix(fIn−M t) ∈Mn(A[f ]). (This is, of course an abuse of notation, as the M t in fIn−M t

should be replaced by its image under the natural map Mn(A)→Mn(EndA(N)), whichhas not been assumed to be an embedding.)

Let Nn be the direct sum of n copies of N , thought of as column matrices with entriesin N . Then the A[f ]-module structure on N induces an Mn(A[f ])-module structure onNn by the usual formula for the product of an n× n matrix with a column vector.

Under this module structure, the formula used to define M shows that

(fIn −M t) ·

⎛⎜⎝ x1

...xn

⎞⎟⎠ = 0,

where x1, . . . , xn is the set of generators in the statement. Multiplying both sides on theleft by the adjoint matrix (fIn −M t)adj, we see that the diagonal matrix (chMt(f))In an-nihilates the column vector with entries x1, . . . , xn. But this says that the endomorphismchMt(f) carries each of x1, . . . , xn to 0. Since x1, . . . , xn generate N , chMt(f) = chM (f)is the trivial endomorphism of N .

Notice that if N is a free module with basis x1, . . . , xn then the matrix M in thestatement of the Generalized Cayley–Hamilton Theorem is just the matrix of f withrespect to the basis x1, . . . , xn. Thus, chM (X) is by definition equal to chf (X).

Corollary 10.4.8. Let N be a finitely generated free module over the commutative ringA and let f ∈ EndA(N). Then chf (f) = 0 in EndA(N).


Of course, if we start with a matrixM and let fM be the endomorphism of An inducedby M , then M is the matrix of fM with respect to the canonical basis. Moreover, thestandard A-algebra isomorphism from Mn(A) to EndA(An) carries M onto fM , so theclassical formulation of the Cayley–Hamilton Theorem is an immediate consequence:

Theorem 10.4.9 (Cayley–Hamilton Theorem). Let A be a commutative ring and letM ∈ Mn(A). Let α : A[X]→ Mn(A) be the A-algebra homomorphism with α(X) = M .Then the characteristic polynomial chM (X) is in the kernel of α.

Thus, chM (M) = 0, meaning that if chM (X) = Xn + an−1Xn−1 + · · · + a0, then

Mn + (an−1In)Mn−1 + · · ·+ (a0In) = 0 in Mn(A).

We now give a useful consequence of the Generalized Cayley–Hamilton Theorem.

Corollary 10.4.10. Let N be a finitely generated module over the commutative ring A,and let a be an ideal of A such that aN = N . Then there is an element a ∈ a such that1− a ∈ AnnA(N).

Proof Let x1, . . . , xn be a generating set for N . The elements of aN all have theform a1n1 + · · · + aknk for some k, where ai ∈ a and ni ∈ N for all i. Since the xigenerate N and since a is an ideal, distributivity allows us to write any such element asa′1x1 + · · ·+ a′nxn with a′i ∈ a for all i.

Since aN = N , we may write xj =∑ni=1 aijxi for all j, where aij ∈ a for all i, j.

Thus, the matrix M = (aij) is a matrix for the identity map, 1N , of N in the sense ofthe statement of the Generalized Cayley–Hamilton Theorem. As a result, we see thatchM (1N ) is the trivial endomorphism of N . But chM (1N ) is just multiplication by theelement chM (1) ∈ A. Thus, chM (1) lies in the annihilator of N .

Write chM (X) = Xn+∑n−1i=0 aiX

i. Since the entries ofM are all in a, Corollary 10.2.6shows that the coefficients ai all lie in a. So N is annihilated by 1− (−∑n−1

i=0 ai).

Exercises 10.4.11.1. Let M ∈Mn(A) and let chM (X) = Xn + an−1X

n−1 + · · ·+ a0. Show that an−1 =−tr(M).

† 2. Let M ∈ Mn(A) and let chM (X) = Xn + an−1Xn−1 + · · · + a0. Show that a0 =

(−1)n detM .

3. Use Corollary 10.4.10 to give a new proof of Nakayama’s Lemma when the ring Ais commutative.

† 4. Let a and b ideals in the integral domain A such that ab = b. Suppose that b isnonzero and finitely generated. Deduce that a = A.

10.5 Eigenvalues and eigenvectors

Here, we study matrices over a field, K.

Definitions 10.5.1. Let M ∈ Mn(K). We say that v ∈ Kn is an eigenvector of M ifMv = av for some a ∈ K. If v �= 0, this determines a uniquely, and we say that a is theeigenvalue, or characteristic root, associated to v.

The eigenspace for M of an element a ∈ K is the set of eigenvectors of M for whichMv = av.


More generally, if V is a vector space over K and if f ∈ EndK(V ), then we define theeigenvalues and eigenvectors of f analogously. The fact that eigenvalues are also calledcharacteristic roots is suggestive.

Lemma 10.5.2. Let M ∈Mn(K). Then a ∈ K is root of the characteristic polynomialchM (X) if and only if it is the eigenvalue associated to a nonzero eigenvector of M .

Moreover, the eigenspace of any a ∈ K with respect to M is a vector subspace ofKn. Specifically, the eigenspace of a is the kernel of the K-linear map obtained bymultiplication by the matrix aIn −M .

Proof By Lemma 10.4.6, a is a root of chM (X) if and only if det(aIn − M) = 0.But since K is a field, det(aIn − M) = 0 if and only if the K-linear map obtainedfrom multiplication by aIn − M fails to be invertible. However, a dimension count(Corollary 7.9.12) shows that every injective endomorphism of a finitely generated vectorspace is invertible. So det(aIn −M) = 0 if and only if the K-linear map obtained frommultiplication by aIn −M has a nontrivial kernel.

But (aIn −M)v = aInv −Mv = av −Mv, and hence (aIn −M)v = 0 if and only ifMv = av.

Recall that a diagonal matrix is a matrix whose off-diagonal entries are all 0.

Definition 10.5.3. A matrix M ∈Mn(K) is diagonalizable if M is similar to a diagonalmatrix.

It is valuable to be able to detect when a matrix is diagonalizable.

Proposition 10.5.4. A matrix M ∈ Mn(K) is diagonalizable if and only if there is abasis of Kn consisting of eigenvectors of M .

Proof Suppose first thatM is diagonalizable, so thatM ′M(M ′)−1 is a diagonal matrix,for some M ′ ∈ Gln(K). Recall from Corollary 7.10.10 that a matrix is invertible if andonly if its columns form a basis of Kn. Thus, the columns, (M ′)−1e1, . . . , (M ′)−1en forma basis of Kn. We claim that (M ′)−1ei is an eigenvector of M for each i.

To see this, note that if M ′′ = (aij) is a diagonal matrix, then M ′′ei = aiiei. Thus,the canonical basis vectors are eigenvectors for any diagonal matrix. In particular, foreach i, there is an ai ∈ K such that (M ′M(M ′)−1)ei = aiei. Multiplying both sides by(M ′)−1, we get M(M ′)−1ei = ai(M ′)−1ei, and hence (M ′)−1ei is an eigenvector for Mas claimed.

Conversely, suppose that Kn has a basis v1, . . . , vn consisting of eigenvectors of M .Thus, for each i there is an ai ∈ K such that Mvi = aivi. Let M ′ be the matrix whosei-th column is vi for 1 ≤ i ≤ n. Since its columns form a basis of Kn, M ′ is invertible.We claim that (M ′)−1MM ′ is a diagonal matrix.

To see this, note that

(M ′)−1MM ′ei = (M ′)−1Mvi = (M ′)−1aivi = aiei,

and hence the canonical basis vectors are eigenvectors for (M ′)−1MM ′. But this says thatthe i-th column of (M ′)−1MM ′ is aiei, and hence the off-diagonal entries of (M ′)−1MM ′

are all 0.

The reader who is approaching this material from a theoretical point of view maynot appreciate the full value of this last result and those related to it. The point is thatcharacteristic polynomials are reasonably easy to calculate. If we can then factor them,


we can use Gauss elimination to find bases for the kernels of the transformations inducedby the (aIn − M) as a ranges over the roots of the characteristic polynomial. If thedimensions of these eigenspaces add up to the size of the matrix, then one can show thatthe bases of these eigenspaces fit together to form a basis of Kn. Thus, modulo factoringthe characteristic polynomial, the determination of whether a matrix is diagonalizable,and the actual construction of a diagonalization, are totally algorithmic. And indeed,this sort of hands-on calculation can arise in actual research problems.

Exercises 10.5.5.1. LetM ∈Mn(K) and suppose that chM (X) = (X−a1) . . . (X−an), where a1, . . . , an

are n distinct elements of K. Show that M is diagonalizable.

2. What is the relationship between the eigenspaces of M and the eigenspaces ofM ′M(M ′)−1 for M ′ ∈ Gln(K)?

3. Show that if θ ∈ R is not a multiple of π, then the matrix

Rθ =(


)is not diagonalizable as an element of M2(R), but is diagonalizable when regardedas an element of M2(C). What are its complex eigenvalues?

4. Show that a block sum M1 ⊕M2 is diagonalizable if and only if each of M1 andM2 is diagonalizable.

5. Let M1 and M2 be elements of Mn(K) which commute with each other, and let Vbe the eigenspace of a ∈ K with respect to M1. Show that M2 preserves V (i.e.,M2v ∈ V for all v ∈ V ).

6. Suppose given a family {Mi | i ∈ I} of diagonalizable matrices in Mn(K) whichcommute with each other. Show that the Mi may be simultaneously diagonalized,in the sense that there is a single matrixM such thatMMiM

−1 is a diagonal matrixfor each i ∈ I. Conversely, show that any family of matrices that is simultaneouslydiagonalizable must commute with one another.

10.6 The classification of matrices

Let K be a field. We shall classify the elements of Mn(K) up to similarity. The key isthe Fundamental Theorem of Finitely Generated Modules over the P.I.D. K[X].

The connection between matrices and K[X]-modules is as follows. First off, Kn isan Mn(K)-module, where we identify Kn with the space of n × 1 column vectors, andMn(K) acts on Kn by matrix multiplication from the left. Note that if a ∈ K, then thematrix aIn acts on Kn as multiplication by a. In other words, the K-module structureon Kn induced by the Mn(K)-module structure together with the usual inclusion of Kin Mn(K) agrees with the original K-module structure on Kn.

Now let M ∈Mn(K). Then there is a K-algebra homomorphism α : K[X]→Mn(K)which takes X to M . And there is a K[X]-module structure on Kn obtained via α fromthe usual Mn(K)-module structure on Kn: for v ∈ Kn, Xv = Mv, and the action ofK ⊂ K[X] on Kn agrees with the usual action of K on Kn.

Conversely, let N be a K[X]-module such that the induced K-module structure on Nis n-dimensional. Because the elements a ∈ K commute with X in K[X], multiplication


by X induces a K-linear map from N to N . Thus, a choice of basis for N will provide amatrix M ∈Mn(K) corresponding to multiplication by X.

Thus, we can pass back and forth between n× n matrices and K[X] modules whoseunderlying K-vector space has dimension n. Of course, the passage from modules tomatrices involves a choice of basis. By varying this choice, we may vary the matrix upto similarity. Note that similarity is an equivalence relation on Mn(K). We call theresulting equivalence classes the similarity classes of n× n matrices.

Proposition 10.6.1. Let K be a field. Then the passage between matrices and K[X]-modules described above gives a one-to-one correspondence between the similarity classesof matrices in Mn(K) and the isomorphism classes of those K[X]-modules whose under-lying K-vector space has dimension n.

Proof Suppose that M,M ′ ∈ Mn(K) are similar: say M ′ = M ′′M(M ′′)−1 for someM ′′ ∈ Gln(K). Let N1 be the K[X]-module structure on Kn induced by M , and letN2 be the K[X]-module structure on Kn induced by M ′. Let fM ′′ : N1 → N2 be theK-linear map induced by M ′′ (i.e., fM ′′(v) = M ′′v for v ∈ N1 = Kn). Since M ′′ isinvertible, fM ′′ is bijective. We claim that fM ′′ is a K[X]-module isomorphism from N1

to N2.To see this, note that K[X] is generated as a ring by X and the elements of K. Thus,

it suffices to show that fM ′′(Xv) = XfM ′′(v) for all v ∈ N1 and that fM ′′(av) = afM ′′(v)for all v ∈ N1 and a ∈ K. The latter statement is immediate, as fM ′′ is a K-linear map.For the former, write fM , fM ′ for the linear transformations induced by multiplicationby M and M ′, respectively. Then the equation M ′ = M ′′M(M ′′)−1 implies that thefollowing diagram commutes.

N1 N2

N1 N2

��fM

��fM ′′

��fM ′

��fM ′′

Now on N1, multiplication by X is given by fM , and on N2 it is given by fM ′ . So thecommutativity of the diagram says precisely that fM ′′(Xv) = XfM ′′(v) for all v ∈ N1.

The passage from isomorphism classes of K[X]-modules whose underlying K-vectorspace has dimension n to similarity classes of matrices in Mn(K) is well defined becauseif f : N → N ′ is a K[X]-module isomorphism and if x1, . . . , xn is a basis for N , then thematrix for multiplication by X with respect to this basis is precisely equal to the matrixfor multiplication by X in N ′ with respect to the basis f(x1), . . . , f(xn).

If N is a K[X]-module whose underlying K-vector space has dimension n, then thematrix of X with respect to a basis of N produces a K[X]-module clearly isomorphic toN . Thus, to complete the proof, it suffices to show that if M,M ′ ∈ Mn(K) such thatthe K[X] module structures on Kn induced by M and M ′ are isomorphic, then M andM ′ are similar.

To see this, let N1 and N2 be the K[X] module structures on Kn induced by Mand M ′, respectively, and let f : N1 → N2 be a K[X]-module isomorphism. Thenf = fM ′′ for some M ′′ ∈ Mn(K). Moreover, since f is an isomorphism, M ′′ ∈ Gln(K).Reversing the steps in the earlier argument, the equation fM ′′(Xv) = XfM ′′(v) producesthe commutative diagram displayed above, which then gives the matrix equation M ′ =M ′′M(M ′′)−1.


Thus, we may apply the Fundamental Theorem of Finitely Generated Modules overa P.I.D. to classify matrices up to similarity.

Note that any K[X] module N whose underlying K-vector space is finite dimensionalmust be a torsion module: Any torsion-free summands would be isomorphic toK[X], andhence infinite dimensional over K. Thus, we may use the second form of the FundamentalTheorem (Theorem 8.9.20), which says that every torsion module over K[X] may bewritten uniquely as a direct sum

K[X]/((f1(X))r1)⊕ · · · ⊕K[X]/((fk(X))rk),

where f1(X), . . . , fk(X) are (not necessarily distinct) monic irreducible polynomials inK[X].

Next, recall from Proposition 7.7.35 that if f(X) ∈ K[X] has degree k, thenK[X]/(f(X))is a k-dimensional vector space over K, with a basis given by the images under the canon-ical map of 1, X, . . . ,Xk−1. We obtain the following proposition.

Proposition 10.6.2. Let M be a module over K[X] which has dimension n as a vectorspace over K. Then M has a unique (up to a reordering of the summands) decomposition

M ∼= K[X]/((f1(X))r1)⊕ · · · ⊕K[X]/((fk(X))rk),

where f1(X), . . . , fk(X) are (not necessarily distinct) monic irreducible polynomials inK[X], ri > 0 for all i, and

r1 deg f1 + · · ·+ rk deg fk = n.

The translation of this classification into matrices is called rational canonical form.To make this translation, we must study the matrices associated with the cyclic modulesK[X]/(f(X)). Thus, suppose that f(X) is any monic polynomial of degree m over Kand write x1, . . . , xm for the images (in order) of 1, X, . . . ,Xm−1 under the canonicalmap π : K[X]→ K[X]/(f(X)). We call this the standard basis of K[X]/(f(X)). Also,write f(X) = Xm + am−1X

m−1 + · · ·+ a0.Note that in the K[X]-module structure on K[X]/(f(X)), we have Xxi = xi+1 for

1 ≤ i < m. And Xxm is the image under the canonical map of Xm. Since f(X) is inthe kernel of the canonical map, this gives Xxm = −a0x1−· · ·−am−1xm. The resultingmatrix has a name.

Definition 10.6.3. Let f(X) = Xm + am−1Xm−1 + · · ·+ a0 be a monic polynomial of

degree m in K[X]. Then the rational companion matrix C(f) of f is the m×m matrixwhose i-th column is ei+1 for 1 ≤ i < m, and whose m-th column is⎛⎜⎝ −a0

...−am−1

⎞⎟⎠ .

We have shown the following.

Lemma 10.6.4. Let f(X) = Xm + am−1Xm−1 + · · · + a0 be a monic polynomial of

degree m in K[X]. Then the matrix obtained from the standard basis of K[X]/(f(X))of the transformation induced by multiplication by X is precisely the rational companionmatrix C(f).


Definition 10.6.5. We say that a matrix M is in rational canonical form if it is a blocksum

M = C(fr11 )⊕ · · · ⊕ C(frk

k )

of rational companion matrices, where f1(X), . . . , fk(X) are (not necessarily distinct)monic irreducible polynomials in K[X], and the exponents ri are all positive.

Propositions 10.6.1 and 10.6.2 now combine to give the classification of matrices upto similarity.

Theorem 10.6.6. (Classification of Matrices) Every similarity class of n × n matricesover a field K contains a unique (up to permutation of the blocks) matrix in rationalcanonical form.

The replacement of M by a matrix in its similarity class that’s in rational canonicalform is called “putting M in rational canonical form.” We call the resulting matrix therational canonical form of M .

We next consider the connection between rational canonical form and the character-istic polynomial.

Proposition 10.6.7. Let f(X) = Xm+am−1Xm−1 + · · ·+a0 be a monic polynomial of

degree m in K[X], and let C(f) be its rational companion matrix. Then the characteristicpolynomial of C(f) is precisely f(X).

Proof Write XIn−C(f) = (fij(X)) and consider the decomposition of its determinantin terms of permutations:

det(XIn − C(f)) =∑σ∈Sm

ε(σ)fσ(1),1(X) . . . fσ(m),m(X).

Now for i < m, the only nonzero terms of the form fji(X) are fii(X) = X and fi+1,i(X) =−1. So for i < m, the term corresponding to σ ∈ Sm vanishes unless σ(i) = i orσ(i) = i+ 1.

Note that since σ is a permutation, if σ(i) = i+1, then σ(i+1) �= i+1. In particular,if i < m is the smallest index for which σ(i) �= i, then we must have σ(j) = j + 1 fori ≤ j < m. Since σ fixes the indices less than i, this forces σ(m) = i.

Thus, for the term fσ(1),1(X) . . . fσ(m),m(X) to be nonzero, with σ not the identitypermutation, we must have σ equal to one of the cycles (i i+ 1 . . . m).

Now the term corresponding to the identity permutation is

Xm−1(X + am−1) = Xm + am−1Xm−1,

which are the leading two terms of f(X). And for σ = (i i + 1 . . . m), we haveε(σ) = (−1)m−i, while

fσ(j),j(X) =

⎧⎪⎨⎪⎩X if j < i

−1 if i ≤ j < m

ai−1 if j = m.

Putting this together, we see that the signs from the −1’s in the matrix cancel againstthe sign of σ, and we have

ε(σ)fσ(1),1(X) . . . fσ(m),m(X) = ai−1Xi−1,

the i − 1-st term in f . Adding this up over the displayed cycles σ, we get chC(f)(X) =f(X), as desired.


Corollary 10.6.8. Suppose given a matrix

M = C(fr11 )⊕ · · · ⊕ C(frk

k )

in rational canonical form. Then the characteristic polynomial of M is given by

chM (X) = (f1(X))r1 . . . (fk(X))rk .

There’s one situation in which the characteristic polynomial forces the rational canon-ical form.

Corollary 10.6.9. Let M be a matrix whose characteristic polynomial is square free, inthe sense that

chM (X) = f1(X) . . . fk(X)

where f1(X), . . . , fk(X) are distinct monic irreducible polynomials over K. Then therational canonical form for M is given by

C(f1)⊕ · · · ⊕ C(fk).

However, nonsimilar matrices can have the same characteristic polynomial. For in-stance, C((X − a)3), C((X − a)2) ⊕ C(X − a) and C(X − a) ⊕ C(X − a) ⊕ C(X − a)are all 3 × 3 matrices whose characteristic polynomial is (X − a)3. But all three lie indistinct similarity classes.

There is another invariant, which, together with the characteristic polynomial, willcharacterize these particular examples. Recall that the minimal polynomial, minM (X),of a matrix M ∈Mn(K) is the monic polynomial of least degree in the kernel of the eval-uation map εM : K[X]→Mn(K) which evaluates X at M . In particular, minM (X) gen-erates the kernel of εM , and hence the Cayley–Hamilton Theorem shows that minM (X)divides chM (X).

We can relate the minimal polynomial to the Fundamental Theorem of Finitely Gen-erated Modules over a P.I.D. as follows.

Lemma 10.6.10. Let M ∈ Mn(K), and let N be Kn with the K[X]-module structureinduced by M . Then minM (X) generates the annihilator of N over K[X].

Proof A polynomial f(X) acts on N by f(X)v = f(M)v. So f(X) annihilates N ifand only if the matrix f(M) induces the 0 transformation. But this occurs only whenf(M) is the 0 matrix, meaning that f(X) is in the kernel of εM .

But we may now apply our understanding of finitely generated torsion modules overK[X] in order to calculate the minimal polynomial of a matrix with a given rationalcanonical form.

Corollary 10.6.11. Suppose given a matrix

M = C(fr11 )⊕ · · · ⊕ C(frk

k )

in rational canonical form. Then the minimal polynomial of M is the least commonmultiple of the polynomials fri

i .


Proof The annihilator of a direct sum of modules is the intersection of the annihilatorsof the individual modules. In a P.I.D., the generator of an intersection of ideals is theleast common multiple of the generators of the individual ideals (Proposition 8.1.32).

Thus, if f1(X), . . . , fk(X) are the monic irreducible polynomials that divide the char-acteristic polynomial of M , and if si is the largest exponent such that the companionmatrix C(fsi

i ) appears in the rational canonical form of M for 1 ≤ i ≤ k, then

minM (X) = (f1(X))s1 . . . (fk(X))sk .

In particular, the minimal polynomials of C((X − a)3), C((X − a)2)⊕C(X − a), andC(X − a)⊕ C(X − a)⊕ C(X − a) are (X − a)3, (X − a)2, and X − a, respectively. Sothe minimal polynomial tells these three rational canonical forms apart.

Minimal polynomials also allow us to recognize diagonalizable matrices.

Corollary 10.6.12. Let M ∈Mn(K). Then M is diagonalizable if and only if

minM (X) = (X − a1) . . . (X − ak)for a1, . . . , ak distinct elements of K.

Proof If minM (X) has the stated form, then every rational companion block in therational canonical form of M has the form C(X−ai) = (ai) for some i. In particular, therational canonical form for M is a block sum of 1× 1 matrices, hence a diagonal matrix.

Conversely, a diagonal matrix is already in rational canonical form, with companionblocks C(X − a) = (a) for each diagonal entry a of the matrix. The least commonmultiple of the annihilators of the blocks then has the desired form.

But the minimal polynomial is primarily useful as a theoretical tool, as it’s difficultto calculate in practice.

It remains to discuss what happens to the rational canonical form when we pass froma given field to an extension field. The following can also be seen directly, using thestructure of the rational companion blocks.

Proposition 10.6.13. Let L be an extension field of K and let M ∈ Mn(K). Thenthe L[X]-module structure on Ln induced by the matrix M is isomorphic to the extendedmodule L[X]⊗K[X] K

n obtained from the K[X]-module structure on Kn induced by M .In particular, if Kn ∼= K[X]/(f1(X))⊕ · · · ⊕K[X]/(fk(X)), then

Ln ∼= L[X]/(f1(X))⊕ · · · ⊕ L[X]/(fk(X))

for the same polynomials f1(X), . . . , fk(X) ∈ K[X]. Thus, the rational canonical formfor M in Mn(L) may be found by determining the prime factorizations of the fi(X) inL[X] and applying the Chinese Remainder Theorem.

Proof Let f : Kn → Kn be the linear map induced by M . Then, as noted in Corol-lary 9.4.20, the effect of M on Ln may be identified with L⊗KKn 1⊗f−−→ L⊗KKn. Thus,if i : Kn → L⊗K Kn is the natural inclusion, then it is a K[X]-module homomorphism,where the K[X]-module structure on L ⊗K Kn is the restriction of the L[X]-modulestructure induced by M .

The universal property of extension of rings (Proposition 9.4.12) provides a uniqueL[X]-module homomorphism ı : L[X]⊗K[X] K

n → L⊗K Kn whose restriction to Kn isi. We claim that ı is an isomorphism.


To see this, let j : L → L[X] be the natural inclusion. Then there is a well definedhomomorphism j ⊗ 1 : L ⊗K Kn → L[X] ⊗K[X] K

n, which is easily seen to satisfyı ◦ (j ⊗ 1) = 1L⊗KKn . But if f(X) =

∑mi=0 αiX

i ∈ L[X], then for each v ∈ Kn, we have

f(X)⊗ v =m∑i=0

αi ⊗M iv

in L[X]⊗K[X] Kn. Thus, j ⊗ 1 is onto, and ı is an isomorphism.

We may now obtain the direct sum decomposition of Ln by applying the naturalisomorphisms B ⊗A A/a ∼= B/Ba for B a commutative A-algebra and a an ideal of A.

Exercises 10.6.14.1. What is the rational canonical form of the rotation matrix

Rθ =(


)?

Deduce from the classification theorem that if θ, φ ∈ [−π, π], then Rθ and Rφ aresimilar in M2(R) if and only if θ = ±φ.

2. What is the smallest n for which there exist two matricesM,M ′ ∈Mn(K) such thatchM (X) = chM ′(X) and minM (X) = minM ′(X) but M and M ′ are not similar?

3. Write Z3 = 〈T 〉. What is the rational canonical form of the transformation ofQ[Z3] induced by multiplication by T?

4. Write Z4 = 〈T 〉. What is the rational canonical form of the transformation ofQ[Z4] induced by multiplication by T? What do you get if you replace Q by C?

5. Write Z8 = 〈T 〉. What is the rational canonical form of the transformation ofQ[Z8] induced by multiplication by T? What do you get if you replace Q by R?What if you replace Q by C?

6. Let K be any field and let M ∈Mn(K). Show that M is similar to its transpose.

7. Let L be an extension field of K and let f1(X) and f2(X) be two monic irreducibleelements of K[X]. Show that f1 and f2 have no common prime factors in L[X].(Hint : How do f1 and f2 factor as polynomials over an algebraic closure of L?)

Deduce that if M,M ′ ∈Mn(K), then M and M ′ are similar in Mn(K) if and onlyif they are similar in Mn(L).

10.7 Jordan canonical form

Let K be a field. Suppose that the characteristic polynomial of M ∈ Mn(K) factors asa product of degree 1 polynomials in K[X]:

chM (X) = (X − a1)r1 . . . (X − ak)rk

where a1, . . . , ak are distinct elements of K. (This will always happen, for instance, ifthe field K is algebraically closed.) Then we shall give an alternative way of representingthe similarity class of M in Mn(K) by a canonical form.


In fact, we simply replace each rational companion matrix in the rational canonicalform of M by the matrix obtained for an alternate choice of basis for the summand inquestion. The matrix obtained from this alternative basis is called a Jordan block.

The key is the following.

Lemma 10.7.1. Let a ∈ K and write v1, . . . , vm for the images under the canonical mapK[X] → K[X]/((X − a)m) of 1, X − a, . . . , (X − a)m−1. Then v1, . . . , vm is a basis forK[X]/((X − a)m) over K.

Proof The elements v1, . . . , vm are all nonzero, as the polynomials (X − a)k with0 ≤ k < m have degree less than that of (X − a)m.

Notice that for 2 ≤ i ≤ m, vi = (X − a)i−1v1, and that (X − a)mv1 = 0. This isenough to show the linear independence of v1, . . . , vm: Let a1v1 + · · ·+ amvm = 0. If thecoefficients ai are not all 0, let i be the smallest index for which ai �= 0. Then

0 = (X − a)m−i(a1v1 + · · ·+ amvm) = aivm.

But vm �= 0, contradicting the assumption that the coefficients were not all 0.Since K[X]/((X − a)m) has dimension m over K, the result follows.

For 1 ≤ i < m, we have vi+1 = (X − a)vi, and hence Xvi = vi+1 + avi. And(X − a)vm = 0, so that Xvm = avm, and vm is an eigenvector for the transformationinduced by X, with eigenvalue a.

Definition 10.7.2. Let a ∈ K and let m ≥ 1. The m×m Jordan block with eigenvaluem is the matrix, J(a,m), whose i-th column is aei + ei+1 if 1 ≤ i < m and whose m-thcolumn is aem. Thus, the diagonal entries of J(a,m) are all a’s, and the off-diagonalentries are all 0 except for the entries just below the diagonal, which are all 1’s.

Thus,

J(a, 3) =

⎛⎝ a 0 01 a 00 1 a

⎞⎠ .

From the discussion above, we see that J(a,m) is the matrix for the transformationof K[X]/((X−a)m) induced by X, with respect to the basis v1, . . . , vm. The next lemmais immediate from Lemma 10.6.4.

Lemma 10.7.3. Let a ∈ K and let m ≥ 1. Then the Jordan block J(a,m) is similar tothe rational companion matrix C((X − a)m) in Mm(K).

In particular, for matrices whose characteristic polynomials factor as products ofdegree 1 terms, we’ll be able to translate back and forth between rational canonical formand a canonical form using Jordan blocks.

Definition 10.7.4. A matrix M ∈ Mn(K) is in Jordan canonical form if it is a blocksum of Jordan blocks:

M = J(a1, r1)⊕ · · · ⊕ J(ak, rk)

where a1, . . . , ak are (not necessarily distinct) elements of K and the exponents ri arepositive for 1 ≤ i ≤ k.


Theorem 10.6.6 and Lemma 10.7.3 now combine to give the following theorem.

Theorem 10.7.5. Let M ∈Mn(K) be a matrix whose characteristic polynomial factorsas a product of degree 1 polynomials over K. Then the similarity class of M in Mn(K)contains a unique (up to permutation of blocks) matrix in Jordan canonical form.

The replacement of M by a matrix in its similarity class that’s in Jordan canonicalform is called “putting M in Jordan canonical form.” We call the resulting matrix theJordan canonical form of M .

The Jordan blocks have certain features that make them useful. For simplicity ofnotation, we write kerM for the kernel of the transformation induced by a matrix M .

Lemma 10.7.6. Let a ∈ K and let M = J(a,m) for some m > 0. Then for 1 ≤ k ≤ m,a basis for ker(aIm −M)k is given by the canonical basis vectors em−k+1, . . . , em. Inparticular, the eigenspace of a with respect to M has dimension 1.

For b �= a, however, (bIm −M)k is invertible for all k.

Proof A glance at the matrix aIm −M shows that

(aIm −M)ei =

{−ei+1 if i < m

0 if i = m.

Induction on k then shows that

(aIm −M)kei =

{(−1)kei+k if i ≤ m− k0 if i > m− k.

The kernel of (aIm −M)k is now easily seen to be as stated.Since chM (X) = chC((X−a)k)(X) = (X − a)k, a is the only eigenvalue of M . So if

b �= a, then bIm −M invertible, and hence its powers are invertible. (Alternatively, it iseasy to see that det(bIm −M) = (b− a)m.)

This, in turn, allows us to calculate the Jordan canonical form of any matrix whosecharacteristic polynomial is a product of degree 1 polynomials. The next proposition isimmediate by adding up the dimensions of the kernels in the different Jordan blocks.

Proposition 10.7.7. Let M ∈Mn(K) such that chM (X) is a product of degree 1 poly-nomials in K[X]. Then for each eigenvalue a of M , the number of Jordan blocks witheigenvalue a occurring in the Jordan canonical form of M is equal to the dimension ofthe eigenspace of a with respect to M .

More generally, in the Jordan canonical form of M , the number of Jordan blocksJ(a,m) occurring with m ≥ k is equal to

dim ker (aIn −M)k − dim ker (aIn −M)k−1.

But this now says that if we can factor the characteristic polynomial as a product ofdegree 1 terms, then we may calculate the Jordan canonical form of a matrix algorith-mically. The point is that Gauss elimination gives an algorithm for calculating a basisfor the kernel of each (aIn −M)k.

Exercises 10.7.8.


1. A matrix M ∈ Gln(K) is said to be periodic if it has finite order in the groupGln(K). The period of M is its order.

Let K be an algebraically closed field and let M be a periodic matrix over K. Sup-pose that M has order m in Gln(K), where m is not divisible by the characteristicof K. Show that M is diagonalizable.

Deduce that such an M is similar to a matrix of the form (a1)⊕ · · · ⊕ (an), whereai is an m-th root of unity in K for each i, and the least common multiple of theorders of a1, . . . , an in K× is m.

2. Give an example of a periodic matrix over the algebraic closure of Zp that is notdiagonalizable.

‡ 3. Let G be a finite abelian group and let K be an algebraically closed field whosecharacteristic does not divide the order of G. Let ρ : G → Gln(K) be a grouphomomorphism. Show that there is a matrix M such that Mρ(g)M−1 is a diagonalmatrix for each g ∈ G.

Deduce that if |G| = m and if ζ is a primitive m-th root of unity in K, then thereare homomorphisms ρi : G → 〈ζ〉 ⊂ K× for i = 1 . . . , n, such that ρ is conjugateas a homomorphism into Gln(K) to the homomorphism that carries each g ∈ G tothe block sum (ρ1(g)) ⊕ · · · ⊕ (ρn(g)). (In the language of representation theory,this says that any finite dimensional representation of G over K is a direct sum ofone-dimensional representations.)

4. Let K be any field. Construct a homomorphism ρ : K → Gl2(K), from the additivegroup of K into Gl2(K), such that ρ(a) not diagonalizable if a �= 0.

5. Let M be a periodic matrix over a field K. Show that the minimal polynomialof M over K divides Xm − 1, where m is the period of M . Deduce that if thecharacteristic of K does not divide m and if K contains a primitive m-th root ofunity (i.e., an element with order m in K×), then M is diagonalizable over K.(The reader who has not read Chapter 11 may assume that K has characteristic0.)

10.8 Generators for matrix groups

Here, A is permitted to be any ring.A common technique for studying matrix groups is Gauss elimination. It is based on

row and column operations obtained by multiplying by the elementary matrices:

Definitions 10.8.1. Let x ∈ A and let i, j be distinct elements of {1, . . . , n}. We writeEij(x) for the n× n matrix whose diagonal entries are all 1 and whose only off-diagonalentry that is nonzero is the ij-th entry, which has value x. We shall refer to the matricesEij(x) as the elementary matrices.

If M is an n×k matrix, then the passage from M to Eij(x)·M is called an elementaryrow operation on M . If M is a k × n matrix, then the passage from M to M · Eij(x) iscalled an elementary column operation on M .

The elementary row and column operations behave as follows.

Proposition 10.8.2. Let x ∈ A and let i, j be distinct elements of {1, . . . , n}. LetM ∈Mn(A). Then the product Eij(x) ·M coincides with M in all rows but the i-th. The


i-th row of Eij(x) ·M is equal to the sum yi + xyj, where yi and yj are the i-th and j-throws of M , respectively. Here, we treat row vectors as left A-modules.

The product M ·Eij(x) coincides with M in all columns but the j-th. The j-th columnof M · Eij(x) is the sum xj + xix, where xj and xi are the j-th and i-th columns of M ,respectively. Here, we treat column vectors as right A-modules.

Proof Recall that the k-th row of a matrix M is the product ek ·M , where the canonicalbasis vector ek is regarded as a row vector. The result for Eij(x) ·M now follows fromthe fact that the k-th row of Eij(x) is ek if k �= i, and is ei + xej if k = i.

The case for M · Eij(x) is similar.

A special case of row operations allows us to determine a subgroup of Gln(A).

Corollary 10.8.3. For fixed i and j, we have

Eij(x) · Eij(y) = Eij(x+ y)

for all x, y ∈ A. Thus, each Eij(x) is invertible, with inverse Eij(−x), and the set{Eij(x) |x ∈ A} forms a subgroup of Gln(A) isomorphic to the additive group of A.

Recall that the commutator, [x, y], of two group elements is defined by [x, y] =xyx−1y−1, and that [x, y] = e if and only if x and y commute. The reader may ver-ify the next lemma by testing the transformations’ effect on the canonical basis vectors.

Lemma 10.8.4. If i′ �= j and i �= j′, then [Eij(x), Ei′j′(y)] = e. Here, x, y ∈ A, and, ofcourse, i �= j and i′ �= j′.

If i, j, and k are all distinct, then

[Eij(x), Ejk(y)] = Eik(xy).

The commutators [Eij(x), Eji(y)] are computable, but are less useful than the onesdisplayed above.

Definition 10.8.5. A square matrix M = (aij) is upper triangular if all the entriesbelow the main diagonal are 0 (i.e., if aij = 0 for i > j).

A square matrix M = (aij) is lower triangular if all the entries above the maindiagonal are 0 (i.e., if aij = 0 for i < j).

Elementary matrices are useful in understanding certain triangular matrices.

Lemma 10.8.6. Any upper or lower triangular matrix whose diagonal entries are all1’s is a product of elementary matrices.

Proof We shall treat the lower triangular case and leave the rest to the reader. Thus,suppose that M = (aij) is a lower triangular matrix in Mn(A) whose diagonal entriesare all 1’s. Let

Mj =n∏

i=j+1

Eij(aij).

Then the reader may verify that M = M1 . . .Mn−1.


In particular, matrices of the form(In 0X Ik

)or

(In Y0 Ik

)are products of ele-

mentary matrices, a fact that will allow us to show the following.

Lemma 10.8.7. Let M1,M2 ∈ Gln(A). Then there is a product, E, of elementarymatrices over A with the property that

E · (M1M2 ⊕ In) = M2 ⊕M1.

Proof Multiplication by(

In 0M−1

1 In

)takes M1M2 ⊕ In to

(M1M2 0M2 In

). Now

multiply by(In In −M1

0 In

), and we get

(M2 In −M1

M2 In

).

Multiplying this by(

In 0−In In

)gives

(M2 In −M1

0 M1

). Finally, we obtain M2⊕

M1 by multiplying this last matrix by(In In −M−1

1

0 In

).

Since Eij(x)−1 = Eij(−x), the inverse of a product of elementary matrices is also aproduct of elementary matrices. Thus, induction gives the following.

Corollary 10.8.8. Let M1, . . . ,Mk ∈ Gln(A). Then there is a product, E, of elementarymatrices with the property that

E · (M1 ⊕ · · · ⊕Mk) = (Mk . . .M1)⊕ I(k−1)n.

We shall make use of this for block sums of 1 × 1 matrices in the following. Thetechnique we shall use in the next proof is known as Gauss elimination.

Proposition 10.8.9. Let D be a division ring and let M ∈ Gln(D). Then there is aproduct, E, of elementary matrices over D, and an element a ∈ D such that

M = E · ((a)⊕ In−1) .

In particular, Gln(D) is generated by the elementary matrices, together with the matricesof the form (a)⊕ In−1, with a ∈ D×.

Proof By Corollary 10.8.8, it suffices to show that M may be reduced to a diagonalmatrix by a sequence of elementary row operations.

First, if the 11 entry of M is 0, choose any row whose first column entry is nonzero,and add it to the first (i.e., multiply M by E1i(1), where the i-th row has a nonzero entryin the first column). Thus, after applying an elementary row operation, if necessary, wemay assume that the 11 entry of our matrix is nonzero. If the resulting matrix is (aij),we may now multiply it by

∏ni=2Ei1(−ai1a−1

11 ), obtaining a matrix with no nonzerooff-diagonal entries in the first column.

Assume, inductively, that we’ve alteredM by a sequence of elementary row operationsso that the first k−1 columns have no nonzero off-diagonal entries. The diagonal entriesin these k − 1 columns must be nonzero, as otherwise the linear transformation inducedby our matrix will have a nontrivial kernel. Thus, the first k − 1 columns form a basisfor the (right) vector subspace of Dn generated by e1, . . . , ek−1. Thus, the k-th column


must have a nonzero entry below the (k−1)-st row; otherwise, the first k columns of ourmatrix will be linearly dependent.

Note that for i ≥ k, the first k− 1 entries of the i-th row are empty at this point. Soadding a multiple of the i-th row to any other row will not change the first k−1 columns.If the kk-th entry is 0, then multiply the matrix by Eki(1), where i > k is chosen sothat the ik-th entry is nonzero. Thus, without altering the first k − 1 columns, we mayassume that the kk-th entry is nonzero. Suppose that our matrix is now (bij). Multiplyit by ∏

1≤i≤ni =k

Eik(−bikb−1kk ),

and we obtain a matrix whose off-diagonal entries in the first k columns are all 0. Theresult now follows by induction on n.

The analogous result fails if D is replaced by some quite reasonable looking commu-tative rings. A method of analyzing the failure of this sort of result in a stable sense (i.e.,as n→∞) will be studied in Section 10.9.

Proposition 10.8.9 has a useful interrelation with determinant theory if we are workingover a field. Recall that the n-th special linear group, Sln(A), of a commutative ring Ais the group of matrices of determinant 1.

Proposition 10.8.10. Let K be a field and let n > 0. Then Sln(K) is the subgroup ofGln(K) generated by the elementary matrices.

Proof It is easy to see (e.g., by Problem 6 of Exercises 10.3.12) that the determinantof either an upper triangular matrix or a lower triangular matrix is given by the productof the diagonal terms. Thus, elementary matrices lie in Sln(K). But then, if M =E · ((a)⊕ In−1), with E a product of elementary matrices and a ∈ K, then detM = a.

We may now make use of our calculations of commutators in Lemma 10.8.4. We firstneed a definition.

Definition 10.8.11. A group G is perfect if the commutator subgroup [G,G] = G.

Proposition 10.8.12. Let K be a field. Then for n ≥ 3, the special linear group Sln(K)is perfect, and is the commutator subgroup of Gln(K).

Proof By Lemma 10.8.4, every elementary matrix is a commutator of elementary ma-trices, so Sln(K) is generated by commutators of elements of Sln(K). Thus, Sln(K) isperfect and is contained in the commutator subgroup of Gln(K).

But Sln(K) � Gln(K), and the quotient group Gln(K)/Sln(K) is the abelian groupK×. So Sln(K) contains the commutator subgroup of Gln(K) by Corollary 5.4.8.

Exercises 10.8.13.1. Let A be a Euclidean domain. Show that every element of Gln(A) may be written

as a product E · ((a) ⊕ In−1), where E is a product of elementary matrices anda ∈ A×. In particular, Sln(A) is generated by elementary matrices.

† 2. Show that Sl2(Z) is generated by a = E12(1) and b = E21(1). Show that x =ba−1b =

(0 −11 0

)has order 4, and y = b2a−1b =

(0 −11 −1

)order 3. Deduce that

Sl2(Z) is generated by torsion elements, despite the fact that it contains elementsof infinite order.


3. Show that Sln(Z) is generated by torsion elements for all n.

4. Let f : Zn → Zn be an injective homomorphism. Show that the cokernel of f is afinite group of order |det f |.

5. Let A be a commutative ring. Find the center of Sln(A).

10.9 K1

Once again, A can be any ring. Here, we study in a stable sense the failure of Gausselimination to solve matrix problems over general rings.

Stability here means allowing the size of our matrices to grow. Since we shall limitour interest to invertible matrices, we shall use the stability map

i : Gln(A)→ Gln+1(A)

given by i(M) = M ⊕ (1). Clearly, i is a group homomorphism.We can think of each Gln(A) as embedded in the set of infinite matrices over A by the

embedding that takes M ∈ Gln(A) to M ⊕ I∞. Under this identification, i : Gln(A) →Gln+1(A) is an inclusion of subsets. (Alternatively, we can just work abstractly in thedirect limit of the inclusions i.)

Definition 10.9.1. We set Gl(A) = Gl∞(A) to be the ascending union of the Gln(A):

Gl(A) =∞⋃n=1

Gln(A).

We call it the infinite general linear group of A.

The group structure we place on Gl(A) is, of course, the unique group structure thatagrees with the usual group structure on each Gln(A) ⊂ Gl(A).

Notice that if i �= j and if i and j are both ≤ n, then i : Gln(A)→ Gln+1(A) carriesthe n×n elementary matrix Eij(x) to the (n+1)× (n+1) elementary matrix Eij(x) forall x ∈ A. Thus, for each pair i, j of distinct positive integers, there is a uniquely definedelementary matrix Eij(x) ∈ Gl(A).

Definition 10.9.2. We write E(A) ⊂ Gl(A) for the subgroup of Gl(A) generated by theelementary matrices.

The next lemma is a key in understanding E(A).

Lemma 10.9.3. Let M ∈ Gln(A). Then M ⊕M−1 is a product of elementary matricesin Gl2n(A).

Proof By Lemma 10.8.7, there is a product of elementary matrices, E, in Gl2n(A),such that

M ⊕M−1 = E · (M−1M ⊕ In) = E.

Recall that a perfect group is one which is its own commutator subgroup.

Proposition 10.9.4. E(A) is a perfect group and is the commutator subgroup of Gl(A).


Proof Lemma 10.8.4 shows that E(A) is perfect and is contained in the commutatorsubgroup of Gl(A). It suffices to show that if M1,M2 ∈ Gl(A), then the commutator[M1,M2] = M1M2M

−11 M−1

2 is in E(A).Choose n large enough that M1 and M2 both lie in Gln(A). Then in Gl2n(A),

[M1,M2] = (M1 ⊕M−11 )(M2 ⊕M−1

2 ) · ((M2M1)−1 ⊕ (M2M1)).

The three matrices on the right are in E(A), by Lemma 10.9.3.

Elementary row operations, if conducted stably, may be used to replace a matrix byany element of its right coset under E(A) — so stably, Gauss elimination picks matrixrepresentatives for the elements of the quotient group Gl(A)/E(A). Thus, a calculationof this group will give us an indication of the inherent limitations on Gauss eliminationfor the study of matrices over A.

Definition 10.9.5. Let A be a ring. Then K1(A) = Gl(A)/E(A).

The groups K1(A) are sometimes called Whitehead groups, as they were first studiedby J.H.C. Whitehead. However, we shall reserve that term for a related group, to bedefined in Exercises 10.9.10, that was the primary motivation for Whitehead in developingthis theory.

The Whitehead groups in the latter sense are functors on groups, denoted Wh(G) fora group G. They arise in the s-Cobordism Theorem, which detects when an inclusionM ⊂W of manifolds is the inclusion ofM in its product with I. They also arise in surgerytheory, which is used to classify manifolds up to homeomorphism, diffeomorphism, etc.

The study of K1 continues the study of algebraic K-theory that we began with K0.We can think of K1 as giving a generalized notion of determinant theory, which workseven over noncommutative rings. The point is that K1(A) is the abelianization of Gl(A),so that similar matrices have the same value when viewed in K1(A).

Note that if f : A → B is a ring homomorphism, then the induced homomorphismf∗ : Gln(A) → Gln(B) carries elementary matrices to elementary matrices. Thus, thereis an induced map K1(f) : K1(A) → K1(B) making K1 a functor from rings to abeliangroups.

If A is commutative, note that i : Gln(A)→ Gln+1(A) preserves determinants. Thus,there is a homomorphism det : Gl(A) → A× that restricts on each Gln(A) to the usualdeterminant map.

Lemma 10.9.6. Let A be a commutative ring. Then there is a split short exact sequence

0→ Sl(A) ⊂−→ Gl(A) det−−→ A× → 0

where Sl(A) =⋃∞n=1 Sln(A) ⊂ Gl(A).

Proof The splitting comes from the fact that det : Gl1(A)→ A× is an isomorphism.

Since elementary matrices have determinant 1, E(A) ⊂ Sl(A).

Definition 10.9.7. Let A be a commutative ring. We write SK1(A) for the quotientgroup Sl(A)/E(A).

The following is then immediate from the above.


Proposition 10.9.8. Let A be a commutative ring. Then there is a split short exactsequence of abelian groups

0→ SK1(A) ⊂−→ K1(A) det−−→ A× → 0.

Thus, since the groups are abelian, there is an isomorphism

K1(A) ∼= SK1(A)⊕A×.

Calculations of K1 tend to be difficult. We shall not present any nontrivial calcula-tions here. The next result is immediate from Proposition 10.8.10.

Corollary 10.9.9. Let K be a field. Then SK1(K) = 0 and K1(K) = K×.

Exercises 10.9.10.1. Show that K1(A×B) ∼= K1(A)⊕K1(B).

2. Let G be a group. The 1×1 matrices {(±g) | g ∈ G} form a subgroup of Gl1(Z[G]),which we denote by ±G. We write π(±G) for the image of ±G in K1(Z[G]), andwrite

Wh(G) = K1(Z[G])/π(±G),

the Whitehead group of G.

Show that if G is abelian, there is a split short exact sequence

0→ SK1(Z[G])→Wh(G)→ Z[G]×/±G→ 0.

Chapter 11

Galois Theory

Galois Theory studies the algebraic extensions of a field k. One wishes to determinewhen two algebraic extensions are isomorphic by an isomorphism that is the identity onk, and to determine all such isomorphisms between them.

In particular, for a given extension field K of k, we would like to be able to calculateautk(K), the group of automorphisms (as a ring) of K that are the identity on k.

Notice that our interest here is in the ring homomorphisms between fields. It is cus-tomary to refer to these as homomorphisms of fields, and to implicitly assume, unlessspecifically stated to the contrary, that all mappings under discussion are homomor-phisms of fields. Thus, in this chapter, when we speak of embeddings between fieldsor automorphisms of a field, we shall assume that the mappings are homomorphisms offields.

Our primary interest will be in the finite extensions of k, as these are the extensionsabout which we can say the most. However, it is important to note that the study ofinfinite algebraic extensions is an area of ongoing research.

In Section 11.1, we show that if L is an algebraic closure of k, then every algebraicextension of k embeds in it. (We also show that any two algebraic closures of k areisomorphic by an isomorphism which restricts to the identity map on k.) Thus, ourfocus of interest is on the relationships between the subfields of the algebraic closure ofL.

If an extension of k is obtained by adjoining all roots of a given collection of poly-nomials, it is what’s known as a normal extension. Normal extensions K of k have thenice property that any two embeddings over k of K in an algebraic closure, L, of k havethe same image. Thus, there is a unique subfield of L that is isomorphic to K, and anytwo embeddings of K in L differ by an automorphism of K. Thus, in an essential way,the study of the homomorphisms out of K is determined by the calculation of autk(K).We develop the theory of normal extensions in Section 11.2, and then apply it to classifythe finite fields in Section 11.3.

The separable extensions are the easiest to analyze. These are the extensions K ofk with the property that the minimal polynomial over k of any element of K has norepeated roots. In particular, every algebraic extension is separable in characteristic 0.For a finite, separable extension, K, of k, the number of embeddings of K over k in analgebraic closure of k is equal to the degree of the extension. We develop the theory ofseparable extensions in Section 11.4.

Extensions that are normal and separable are known as Galois extensions. If K is aGalois extension of k, we shall write Gal(K/k) for the automorphism group autk(K).

450

CHAPTER 11. GALOIS THEORY 451

We call it the Galois group of K over k. In Section 11.5, we prove the FundamentalTheorem of Galois Theory, which studies the Galois group of a finite Galois extension.Among other things, it shows that the fixed points in K of the action of Gal(K/k) areprecisely k, and that the subfields of K containing k are precisely the fixed fields of thesubgroups of Gal(K/k).

Using the Fundamental Theorem of Galois Theory, we prove the Primitive ElementTheorem, showing that every finite separable extension is simple. Then, in Section 11.6,we prove the Fundamental Theorem of Algebra, that the complex numbers, C, are alge-braically closed.

In Section 11.7, we study cyclotomic extensions and characterize their Galois groups.In particular, we obtain a calculation of Gal(Q(ζn)/Q). We use this in Section 11.8 tostudy the splitting fields of polynomials of the form Xn − a. Galois groups for examplesof such extensions are computed in the text and exercises.

Sections 11.9 and 11.10 are concerned with Galois extensions with abelian Galoisgroups. We obtain a complete classification of the finite extensions of this sort, providedthat the base field has a primitive n-th root of unity, where n is an exponent for theGalois group.

In Section 11.11, we use our analysis of abelian extensions and of cyclotomic exten-sions to characterize the finite Galois extensions whose Galois group is solvable. As acorollary, we give a proof of Galois’ famous theorem that there is no formula for find-ing roots of polynomials of degree ≥ 5. We continue the discussion of such formulæ inSection 11.12.

Section 11.13 gives the Normal Basis Theorem, which shows that if K is a finiteGalois extension of k with Galois group G, then there is a basis for K as a k-vector spaceconsisting of the elements of a single orbit of the action of G on K. This shows that Kis free on one generator as a module over k[G].

Section 11.14 defines and studies the norm and trace functions, NK/k : K → k andtrK/k : K→ k for a finite extension K of k. These are extremely useful in studying thering of integers of a number field. We shall make use of them in Chapter 12.

11.1 Embeddings of Fields

Definition 11.1.1. Let K be an extension field of k. Then an intermediate field betweenk and K is a subfield of K containing k.

It is customary to indicate field extensions by vertical or upward slanting lines. Thus,the diagrams

K L

K1 K1 K2

k k

��

��

��

��

��

��

��

��

may be read as follows: The left-hand diagram indicates that K is an extension of kand that K1 is an intermediate field between them. The right-hand diagram gives anextension, L, of k, together with two intermediate fields, K1 and K2.


Definition 11.1.2. Let K and L be extensions of the field k. An embedding of K in Lover k is a homomorphism ν : K → L of fields such that ν restricts on k to its originalinclusion map into L. Pictorially, this gives

K L

k��

��

��ν

��

�

If K is an extension of k and ν : k → L is any homomorphism of fields, then anextension of ν to K is a homomorphism ν : K→ L whose restriction to k is ν. Here, wecould make a diagram as follows.

K L

k ν(k)

��ν

��ν

Recall that if f : A→ B is a ring homomorphism, then the induced homomorphismf∗ : A[X]→ B[X] is given by f∗

(∑ni=0 aiX

i)

=∑ni=0 f(ai)Xi.

Lemma 11.1.3. Let ν : k → L be an embedding of fields and let K = k(α) be a finitesimple extension of k. Let f(X) be the minimal polynomial of α over k. Then for anyroot β of ν∗(f) in L, there is a unique extension of ν to K that carries α to β. Indeed,the extension factors as a composite

k(α) ∼= k1(β) ⊂ L,

where k1 = ν(k).Moreover, if ν is an extension of ν to K, then ν(α) must be a root of ν∗(f) in L.

Thus, there is a one-to-one correspondence between the extensions of ν to K and theroots of ν∗(f) in L.

Proof Let ν be an extension of ν to K. Then, applying ν to both sides of the equationf(α) = 0, we see that (ν∗(f))(ν(α)) = 0, so ν(α) is a root of ν∗(f), as claimed.

Also, note that every element of K may be written as g(α) for some polynomialg(X) ∈ k[X]. As above, ν(g(α)) = (ν∗(g))(ν(α)), and hence ν is determined by its effecton α.

Finally, if β is a root of ν∗(f), we may construct a commutative diagram

k[X]/(f(X)) k1[X]/(ν∗(f))

k(α) k1(β)

��ν∗∼=

��εα ∼=

��εβ ∼=��ν

with ν(α) = β.

For algebraic extensions of the form k(α, β), the situation is more complicated, as thenatural method of analysis proceeds by stages: First extend ν over k(α), and then extendfurther to k(α, β). To get control over this procedure, we need to know the minimal


polynomial of β over k(α). Finding this can be a nontrivial task, even if we know theminimal polynomial of β over k. The point is that if ν : k→ L is a homomorphism, andif ν : k(α)→ L is an extension, then it may be the case that not every root of ν∗(minβ/k)is a root of ν∗(minβ/k(α)). And only the latter roots may be used to extend ν.

In practice, we shall be forced to confront such issues head on. But let us build upsome theory first.

Proposition 11.1.4. Let ν : k → L be an embedding of fields, where L is algebraicallyclosed, and let K be any algebraic extension of k. Then ν extends to an embedding ofK in L.

Proof If K is a finite extension of k, we may argue by induction on [K : k]. If α ∈ Kis not in k, then ν∗(minα/k) has a root in L because L is algebraically closed. Thus, νextends to ν : k(α)→ L by Lemma 11.1.3. But [K : k(α)] < [K : k], so ν extends to anembedding of K by induction.

If K is infinite over k, we use Zorn’s Lemma. We define a partially ordered set whoseelements consist of pairs (K′, ν′), where K′ is an intermediate field between k and K,and ν′ : K′ → L extends ν. Here, (K′, ν′) ≤ (K′′, ν′′) if K′ ⊂ K′′ and ν′′ extends ν′.

Here, if {(Ki, νi) | i ∈ I} is a totally ordered subset of this partially ordered set, letK′ =

⋃i∈I Ki. Then K′ is a subfield of K, and there is an embedding ν′ : K′ → L

defined by setting ν′(α) = νi(α), for any i such that α ∈ Ki. Now (K′, ν′) is an upperbound for the totally ordered subset in question, and hence the hypothesis of Zorn’sLemma is satisfied.

Thus, there is a maximal element (K′′, ν′′) in this partially ordered set. But then K′′

must equal K, as otherwise, we could extend ν′′ over a simple extension K′′(α) of K′′ inK, in which case (K′′, ν′′) would not be a maximal element in this set of extensions.

Corollary 11.1.5. Let ν : k1 → k2 be an isomorphism of fields and let Li be an algebraicclosure of ki for i = 1, 2. Then there is an embedding ν : L1 → L2 that extends ν:

L1 L2

k1 k2

��ν

��ν

Moreover, any such extension of ν gives an isomorphism from L1 to L2.In particular, if L1 and L2 are algebraic closures of a field k, then L1 and L2 are

isomorphic over k.

Proof Considering ν as an embedding into L2, Proposition 11.1.4 provides an extensionto an embedding ν : L1 → L2, as claimed.

Similarly, if μ : k2 → k1 is the inverse of the isomorphism ν, then μ provides anembedding of k2 in L1. Let μ : L2 → L1 extend μ.

Then (ν ◦ μ) : L2 → L2 is an embedding over k2. But since L2 is algebraic over k2,Proposition 8.2.19 shows that any embedding of L2 in itself over k2 is an automorphismof L2. So ν ◦ μ is surjective, and hence ν : L1 → L2 is surjective as well.


11.2 Normal Extensions

Definitions 11.2.1. Let f(X) be a polynomial over the field k and let K be an extensionfield of k. We say that f splits in K[X] (or splits in K) if

f(X) = a(X − α1)r1 . . . (X − αk)rk

in K[X], with a, α1, . . . , αk ∈ K and ri > 0 for i = 1, . . . , k.We say that K is a splitting field for f(X) if f splits as above in K[X] and K =

k(α1, . . . , αk).

Notice that since a is the leading coefficient of the right-hand side of the displayedequation, the fact that f(X) ∈ k[X] implies that a ∈ k. Of course, the elementsα1, . . . , αk are algebraic over k as they are roots of f(X).

Examples 11.2.2.

1. We’ve seen in Section 8.8 that Xn − 1 splits in Q(ζn)[X]. Since ζn is a root ofXn − 1, Q(ζn) is a splitting field for Xn − 1 over Q.

2. Let a > 0 in Q. Then {ζkn n√a | 0 ≤ k < n} gives n distinct roots of Xn − a. Thus,

Q( n√a, ζn) is a splitting field of Xn − a over Q.

3. The complex numbers C is a splitting field for X2 + 1 over R.

The next lemma is basic.

Lemma 11.2.3. Let K be an extension of k and suppose that f(X) ∈ k[X] splits inK[X] via

f(X) = a(X − α1)r1 . . . (X − αk)rk

with a, α1, . . . , αk ∈ K and with ri > 0 for i = 1, . . . , k.Let K1 be an intermediate field between k and K such that each of the roots αi lies in

K1. Then f(X) splits as above in K1[X] as well. In particular, the subfield k(α1, . . . , αk)of K is a splitting field of f .

Proof Both sides of the displayed equation lie in K1[X]. Since the natural map fromK1[X] to K[X] is injective, the result follows.

In an algebraically closed field, every polynomial of positive degree splits. Thus,Lemma 11.2.3 implies that splitting fields exist.

Corollary 11.2.4. Let L be an algebraically closed field containing k and suppose thatf(X) ∈ k[X] splits as

f(X) = a(X − α1)r1 . . . (X − αk)rk

in L[X], with a, αi ∈ L and ri > 0 for i = 1, . . . , k. Then K = k(α1, . . . , αk) is a splittingfield for f .

We shall treat the uniqueness of splitting fields in Proposition 11.2.8.The splitting field of a single polynomial is all we need think about if we’re concerned

only with finite extensions. But for infinite extensions, we need a generalization.


Definition 11.2.5. Let {fi(X) | i ∈ I} be any set of polynomials over the field k. ThenK is a splitting field for {fi(X) | i ∈ I} if

1. Each polynomial fi(X) with i ∈ I splits in K[X].

2. If {αj | j ∈ J} is the set of all roots in K of polynomials in {fi(X) | i ∈ I}, then

K = k(αj | j ∈ J).

Note that the splitting field for a finite set {f1(X), . . . , fk(X)} is just the splittingfield for the product f1(X) . . . fk(X).

The property of being a splitting field is one of the fundamental properties we willuse in our analysis of Galois theory. We abstract it as follows.

Definition 11.2.6. An extension K of k is a normal extension1 if K is the splitting fieldfor some family of polynomials in k[X].

Note that if K is a splitting field for the polynomials {fi(X) | i ∈ I} over k and ifK1 is any intermediate field between k and K, then K is also the splitting field for thepolynomials {fi(X) | i ∈ I} over K1.

Lemma 11.2.7. Let K be a normal extension of k and let K1 be an intermediate fieldbetween k and K. Then K is a normal extension of K1.

However, we shall see in Example 11.2.10 that an intermediate field between k and anormal extension K of k need not be normal over k.

As was done in Corollary 11.2.4 for a single polynomial, we may construct a splittingfield for any set of polynomials in k[X] as a subfield of any algebraic closure, L, of k: If{fi(X) | i ∈ I} is any set of polynomials over k, then each fi(X) splits in L[X], since Lis algebraically closed. Thus, if {αj | j ∈ J} is the set of all roots in L of the polynomialsin {fi(X) | i ∈ I}, then k(αj | j ∈ J) is a splitting field for {fi(X) | i ∈ I} over k. Wenow show that splitting fields are unique.

Proposition 11.2.8. Let {fi(X) | i ∈ I} be a set of polynomials over the field k, andlet K be a splitting field for {fi(X) | i ∈ I}. Let L be any extension field of k with theproperty that each fi(X) splits in L[X] and let {αj | j ∈ J} be the set of all roots in L ofthe polynomials in {fi(X) | i ∈ I}. Then every embedding of K in L over k has imageequal to k(αj | j ∈ J), and hence gives an isomorphism from K onto k(αj | j ∈ J).

Note that if L is an algebraically closed field containing k, then K embeds in L over kby Proposition 11.1.4. As a result, we see that any two splitting fields for {fi(X) | i ∈ I}are isomorphic over k.

Proof Let ν : K→ L be an embedding over k. Let {βj | j ∈ J ′} be the set of all rootsin K of polynomials in {fi(X) | i ∈ I}. Then each βj is a root of one of the irreduciblefactors, say p(X), of one of the fi in k[X]. By Lemma 11.1.3, ν(βj) must also be a rootof p(X), and hence ν(βj) ∈ {αj | j ∈ J}. Thus, ν restricts to a function

ν : {βj | j ∈ J ′} → {αj | j ∈ J}.

Since K = k(βj | j ∈ J ′), it suffices to show that this function is onto.

1Some authors, e.g., Emil Artin, have used the expression “normal extension” to describe what weshall call a Galois extension.


Note that for each irreducible factor, p(X), of one of the fi(X), ν carries the roots ofp in K into the set of roots of p in L. Since ν is injective, it suffices to show that p hasthe same number of roots in K as in L. But if

p(X) = a(X − βj1)r1 . . . (X − βjk)rk

in K[X], with a ∈ k, with βj1 , . . . βjk distinct and with ri > 0 for all i, then

p(X) = a(X − ν(βj1))r1 . . . (X − ν(βjk))rk

in L[X]. Since ν is injective, p has k distinct roots in L, just as it did in K.

We can now characterize normal extensions.

Proposition 11.2.9. Let K be an algebraic extension of k. Then the following condi-tions are equivalent.

1. K is a normal extension of k.

2. If L is an algebraic closure of K, then every embedding of K in L over k has imageK, and hence gives an automorphism of K.

3. For every element α ∈ K, the minimal polynomial of α over k splits in K[X].

Proof Let K be a normal extension of k. Since K is algebraic over k, so is any algebraicclosure, L, of K (Corollary 8.2.18), and hence L is an algebraic closure for k as well.Thus, if K is the splitting field for {fi(X) | i ∈ I}, Then Proposition 11.2.8 shows thatK is the subfield of L obtained by adjoining all the roots in L of the polynomials in{fi(X) | i ∈ I}, and that any embedding of K in L over k must have image K. Thus,the first condition implies the second.

Suppose that the second condition holds. Let L be an algebraic closure for K and letα ∈ K. Then the minimal polynomial, f(X), of α over k splits in L[X]. To show that itsplits in K[X], it suffices to show that every root of f(X) in L must lie in K.

But if β is a root of f(X) in L, Lemma 11.1.3 provides an isomorphism ν : k(α) →k(β) over k. By Proposition 11.1.4, ν extends to an embedding ν : K → L. But thesecond condition says that the image of ν must be K, and hence β must lie in K. So thesecond condition implies the third.

Now suppose that the third condition holds. Since K is algebraic over k, K =k(αi | i ∈ I) for some collection of elements αi ∈ K that are algebraic over k (Corol-lary 8.2.17). But the minimal polynomials of the αi over k must split in K[X], andhence K is the splitting field for the set of minimal polynomials of the αi over K. So thethird condition implies the first.

We may use Proposition 11.2.9 to show that if K is a normal extension of k, then anintermediate field between k and K need not be normal over k.

Example 11.2.10. Q( 4√

2) is an intermediate field between Q and Q( 4√

2, i). The min-imal polynomial of 4

√2 over Q is X4 − 2, which is irreducible over Q by Eisenstein’s

criterion. But X4 − 2 fails to split in Q( 4√

2)[X], as Q( 4√

2) is contained in the realnumbers, and hence does not contain the roots ±i 4

√2 of X4−2. So Q( 4

√2) is not normal

over Q. But Q( 4√

2, i) is normal over Q, as it is the splitting field of X4 − 2.

We now give another application of Proposition 11.2.9.


Corollary 11.2.11. Let K be an extension of k and let {Ki | i ∈ I} be a collection ofintermediate fields that are normal over k. Then ∩i∈IKi is a normal extension of k.

Proof Let α ∈ ∩i∈IKi and let f(X) be the minimal polynomial of α over k. Then fsplits in each Ki[X], and hence also in K[X]. In particular, the set of roots of f in eachKi must coincide, and hence each root of f in K must lie in ∩i∈IKi. Since f splits inK[X], Lemma 11.2.3 shows that it splits in (∩i∈IKi)[X]. The result now follows fromProposition 11.2.9.

In particular, let K be a normal extension of k and let K1 be an intermediate fieldbetween k and K. Then the collection of intermediate fields between K1 and K that arenormal over k is nonempty. Taking their intersection, we see that the following definitionmakes sense.

Definition 11.2.12. Let K be a normal extension of k and let K1 be an intermediatefield between k and K. Then the normal closure of K1 over k in K is the smallestintermediate field between K1 and K that is normal over k.

And normal closures may be constructed as follows.

Proposition 11.2.13. Let K be a normal extension of k and let α1, . . . , αn ∈ K. Letfi(X) be the minimal polynomial of αi for i = 1, . . . , n and let β1, . . . , βk be the collectionof all roots in K the polynomials f1, . . . , fn. Then k(β1, . . . , βk) is the normal closure ofk(α1, . . . , αn) over k in K.

Also, if f(X) = f1(X) . . . fn(X), then k(β1, . . . , βk) is a splitting field for f(X)over k.

Proof Write K1 = k(α1, . . . , αn) and K2 = k(β1, . . . , βk).Since K is a normal extension of k and since fi(X) is the minimal polynomial of

αi ∈ K over k, Proposition 11.2.9 shows that each fi(X) splits in K[X], and hence f(X)does also. Since β1, . . . , βk is the set of all roots of f(X) in K, Lemma 11.2.3 shows thatK2 is the splitting field for f over k, and hence is normal over k.

But any normal extension of k in K containing α1, . . . , αn must contain all the rootsof the fi(X), and hence must contain K2. Thus, K2 is the smallest normal extension ofk in K containing K1, and the result follows.

Since any finite extension of k may be written as k(α1, . . . , αn) for some collectionof elements α1, . . . , αn, and since the splitting field of a single polynomial is finite, weobtain the following corollary.

Corollary 11.2.14. Let K be a normal extension of k and let K1 an intermediate fieldthat is finite over k. Then the normal closure of K1 over k in K is a finite extensionof k.

The proof of Proposition 11.2.13 generalizes to give the following proposition.

Proposition 11.2.15. Let K be a normal extension of k and let {αi | i ∈ I} ⊂ K. Letfi(X) be the minimal polynomial of αi for i ∈ I. Then the normal closure of k(αi | i ∈ I)over k in K is a splitting field for {fi(X) | i ∈ I}.Exercises 11.2.16.

1. Show that every extension of degree 2 is normal.

2. Show that for all n > 2, there are extensions of Q of degree n that are not normal.


3. Show that normality of extensions is not transitive: Give an example of extensionsk ⊂ K1 ⊂ K such that K1 is normal over k, K is normal over K1, but K is notnormal over k.

4. Let K be a normal extension of k and let K1 be an intermediate field between kand K. Let L be any extension field of K. Show that the image of any embeddingof K1 in L over k must lie in K.

5. Let K be a normal extension of k and let K1 be an intermediate field between kand K. Show that every embedding of K1 in K extends to an automorphism of K.

6. Let K be an algebraic extension of k. Let L1 and L2 be two different extensionsof K that are both normal over k. Show that the normal closures of K over k inL1 and L2 are isomorphic over K.

11.3 Finite Fields

Let k be a finite field. Then k must have characteristic p for some prime p, and hence kis a finite extension of Zp. We shall show that for each r > 0 there is a unique field oforder pr.

Recall that for any commutative ring A of characteristic p, the Frobenius map ϕ :A→ A, defined by ϕ(a) = ap for all a ∈ A, is a ring homomorphism. We shall write ϕn

for the n-fold composite of ϕ with itself.

Proposition 11.3.1. Let k be a finite field of order q = pr and let K be the splittingfield of Xqn −X over k. Then [K : k] = n.

Proof Since qn ≡ 0 mod p, the formal derivative of Xqn − X is just 1. Thus, byLemma 8.7.7, Xqn −X has no repeated roots in K. Since Xqn −X splits in K, this saysthat Xqn −X has qn distinct roots in K. Write S ⊂ K for the set of roots of Xqn −Xin K.

We claim that S is a subfield of K. To see this, note that since qn = prn, we haveα ∈ S if and only if ϕrn(α) = α, with ϕ the Frobenius, as above. Since ϕ is a ringhomomorphism, the elements fixed by ϕrn are easily seen to form a subfield of K.

Thus, S is a field containing all the roots of Xqn − X. Since K is obtained from kby adjoining these roots, we must have K = S, and hence K has qn elements. But then[K : k] = n, as claimed.

Thus, for any positive n, a finite field has an extension of degree n. Our next resultshows that this extension is unique.

Proposition 11.3.2. For any finite field k and any positive integer n, any two exten-sions of k of degree n are isomorphic over k. Thus, k has a unique extension of degree n.

Proof We have already shown that if k has order q, then the splitting field of Xqn −Xis an extension of k of degree n. Thus, by our uniqueness result for splitting fields(Proposition 11.2.8), it suffices to show that any extension of k of degree n is a splittingfield for Xqn −X.

If K is an extension of degree n over k, then it has qn elements. Thus, the unit group,K×, of K has order qn − 1. So every element α ∈ K× has exponent qn − 1 there, andhence is a root of Xqn−1 − 1. But then α is also a root of

X · (Xqn−1 − 1) = Xqn −X.


Since 0 is also a root of Xqn −X, Xqn −X has qn distinct roots in K, and hence splitsin K[X]. Since K is generated by the roots of this polynomial, it must be its splittingfield over k.

So, applying the above with k = Zp, we obtain the following corollary.

Corollary 11.3.3. For each prime p and each positive integer n, there is a unique iso-morphism class of fields of order pn.

We next analyze the uniqueness properties of subfields.

Proposition 11.3.4. Let K be the field of order pn. Then if k is a subfield of K, kmust have order pr for some r dividing n.

Conversely, if r divides n, then there is a unique subfield of K of order pr, given by

k = {α ∈ K |αpr

= α}.Proof Since we’re in characteristic p, any subfield, k, of K has order pr for some r.And, if [K : k] = s, then (pr)s = pn, so r divides n.

If r divides n, then Proposition 11.3.1 shows that the field of order pr has an extensionof order pn. Since there’s only one field, here denoted K, of order pn, K must have asubfield of order pr.

Thus, suppose that k is any subfield of K of order pr. It suffices to show that

k = {α ∈ K |αpr

= α}.Of course, {α ∈ K |αpr

= α} is the set of roots of Xpr − X in K. And the proofof Proposition 11.3.2 shows that every element of k must be a root of Xpr −X, simplybecause k is a field with pr elements. Since Xpr −X may have at most pr roots in K,k must be the entire collection of such roots in K, and hence must coincide with thedisplayed subset.

Of course, there is much more of interest in the study of finite fields. For instance,if K is the field of order pn, and if α ∈ K is not contained in a proper subfield, thenK = Zp(α). We may then ask about the minimal polynomial of α over Zp. Of course,every irreducible polynomial of degree n over Zp occurs in this manner.

One source of elements α with K = Zp(α) is as follows. Since K× is finite, it is acyclic group by Corollary 7.3.14. We shall refer to the generators of K× as the primitive(pn − 1)-st roots of unity over Zp. If α is one such, then no proper subfield of K maycontain α, and hence K = Zp(α).

Exercises 11.3.5.1. Find the smallest number pn such that the field, K, of order pn has an element α

that is not a primitive (pn − 1)-st root of unity, but K = Zp(α), anyhow.

2. Show that X4 +X + 1 is irreducible over Z2. Find a primitive 15-th root of unityin the field K = Z2[X]/(X4 +X + 1).

3. Let p be a prime congruent to −1 mod 4. Show that X2 + 1 is irreducible inZp[X], and hence K = Zp[X]/(X2 + 1) is the field of order p2. Note that K has amultiplication similar to that of the complex numbers.

4. Find a primitive 24-th root of unity in K = Z5[X]/(X2 − 2).

5. What are the infinite subfields of the algebraic closure of Zp?


11.4 Separable Extensions

Definition 11.4.1. Let K be an extension of k. We say that α ∈ K is separable over kif α is algebraic over k and if α is not a repeated root of its minimal polynomial over k.

We say that an extension K of k is separable if every element of K is separable overk.

Proposition 8.7.8, together with our understanding of splitting fields, will give us abetter ideal of what separability means.

Proposition 11.4.2. Let K be an algebraic extension of k and let α ∈ K. Let f(X) bethe minimal polynomial of α over k. Then α is separable over k if and only if f(X) hasno repeated roots in its splitting field over k. In this case, if L is any extension of k inwhich f splits, then f(X) factors as

f(X) = (X − α1) . . . (X − αn)

in L[X], where α1, . . . , αn are distinct elements of L.All algebraic extensions in characteristic 0 are separable. In characteristic p �= 0, if α

is not separable over k, then, if L is any extension of k in which the minimal polynomialf(X) splits, we have a factorization

f(X) = (X − α1)pk

. . . (X − αn)pk

in L[X], where α1, . . . , αn are distinct elements of L and k > 0.

Proof Proposition 8.7.8 tells us that α is separable over k if and only if the formalderivative f ′(X) �= 0, in which case f(X) has no repeated roots. Thus, if f splits inL[X], it must split as stated.

Proposition 8.7.8 goes on to say that in characteristic 0, no irreducible polynomialhas repeated roots, while in characteristic p, any irreducible polynomial with repeatedroots has the form f(X) = h(Xpk

), where k > 0 and h(X) is an irreducible polynomialin k[X] with h′(X) �= 0.

Thus, h(X) is separable. Let L be an algebraically closed field containing k. Thenin L[X], we have

h(X) = (X − β1) . . . (X − βn)

for distinct elements β1, . . . , βn ∈ L.Since L is algebraically closed, we can find roots αi of Xpk−βi for i = 1, . . . , n. Thus,

in L[X], we have

Xpk − βi = Xpk − αpk

i = (X − αi)pk

,

where the last equality comes from the fact that the Frobenius map is a homomorphism.Putting this together, we see that

f(X) = h(Xpk

) = (X − α1)pk

. . . (X − αn)pk

in L[X], and hence also in k(α1, . . . , αn)[X]. The result now follows from the uniquenessof splitting fields (Proposition 11.2.8).


Since all algebraic extensions in characteristic 0 are separable, it might appear at firstthat the study of separability is mainly of interest in characteristic p. However, separabil-ity has important consequences whenever it holds. We shall explore these consequencesbelow.

Proposition 11.4.2 allows us to recast separability in terms of polynomials.

Definition 11.4.3. An irreducible polynomial over a field k is separable if it has norepeated roots in its splitting field.

More generally, we say that a polynomial f(X) ∈ k[X] is separable if each of itsirreducible factors in k[X] is separable.2

The next corollary is now immediate from Proposition 11.4.2.

Corollary 11.4.4. Let L be an extension of k and let α ∈ L be algebraic over k. Then αis separable over k if and only if the minimal polynomial, f(X), of α over k is separable.

Proposition 11.4.2 also gives the following.

Corollary 11.4.5. Let L be an extension field of k and let α ∈ L be separable over k.Let K be any intermediate field between k and L. Then α is separable over K.

Proof The minimal polynomial of α over K divides the minimal polynomial of α overk in K[X]. Since minα/k(X) has no repeated roots in its splitting field, the same musthold for minα/K.

Corollary 11.4.6. Let K be a separable extension of k and let K1 be an intermediatefield between k and K. Then K is separable over K1, and K1 is separable over k.

Proof K is separable over K1 by Corollary 11.4.5. K1 is separable over k becauseevery element of K is separable over k.

Here is another case in which separability is automatic.

Proposition 11.4.7. Every algebraic extension of a finite field is separable.

Proof We first show that every algebraic extension of Zp is separable. Thus, supposethat K is an algebraic extension of Zp, and let f(X) be the minimal polynomial of α ∈ Kover Zp. Then if α is not separable over Zp, Proposition 8.7.8 shows that f(X) = g(Xp)for some g(X) ∈ Zp[X]. But in Zp[X], Lemma 8.6.6 shows that g(Xp) = (g(X))p, whichis not irreducible. So α must have been separable.

Now suppose given a finite field k of characteristic p and an algebraic extension K ofk. Then k is a finite extension of Zp, and hence K is algebraic over Zp by Corollary 8.2.18.But we’ve just shown that K must be separable over Zp. Since k is an intermediate fieldbetween Zp and K, the result follows from Corollary 11.4.6.

In the case of finite extensions, there is another very useful way to look at separability.

Definition 11.4.8. Let K be a finite extension of k and let L be an algebraic closure fork. Then the separability degree of K over k, written [K : k]s, is the number of distinctembeddings of K over k into L.

2Warning: Some authors call a polynomial separable if it has no repeated roots in its splitting field.We have chosen the present definition for its role in Corollary 11.4.16.


Note that since any two algebraic closures of k are isomorphic over k, the calculationof [K : k]s is independent of the choice of algebraic closure.

Proposition 11.4.2 shows that if α is algebraic over k and f is its minimal polynomial,then there is a factorization of the form

f(X) = (X − α1)r . . . (X − αn)r

in any field in which f(X) splits, where α1, . . . , αn are distinct. Here, r = 1 if k hascharacteristic 0. In characteristic p, we have r = pk, where k = 0 if and only if f isseparable.

Lemma 11.4.9. Let K be a simple extension K = k(α) of k, with α algebraic over k,and suppose that the minimal polynomial f(X) of α over k splits as

f(X) = (X − α1)r . . . (X − αn)r

in its splitting field. Then [K : k]s = n, and hence

[K : k] = r · [K : k]s.

Thus, [K : k] = [K : k]s if and only if α is separable over k, and if α is inseparable overk, then

[K : k] = pk · [K : k]s,

where p is the characteristic of k and k > 0.

Proof Let L be an algebraic closure of K, and suppose that the factorization of f(X)above takes place in a splitting field that’s contained in L. Then Lemma 11.1.3 showsthere are exactly n embeddings of K into L over k, obtained by mapping α to α1, . . . , αn.Thus, [K : k]s = n. The rest follows from the above summary of the results in Proposi-tion 11.4.2.

Notice that since the embeddings of an extension k(α1, . . . , αn) over k are determinedby their effect on α1, . . . , αn, [K : k]s is finite for any finite extension K of k. The nextlemma will be useful for dealing with non-simple extensions.

Lemma 11.4.10. Let K be a finite extension of k and let ν : k→ k1 be an isomorphism.Let L1 be an algebraic closure for k1. Then [K : k]s is equal to the number of distinctextensions ν : K→ L1 of ν to K.

Proof Let L be an algebraic closure for k. Let ν : L → L1 be an embedding of L inL1 that extends ν:

L L1

k k1

��ν

��ν

Then ν is an isomorphism by Corollary 11.1.5.Let S be the set of embeddings of K into L1 extending ν and let T be the set

of embeddings of K into L over k. Then there’s a bijection ν∗ : T → S given byν∗(μ) = ν ◦ μ.


Corollary 11.4.11. Let K be a finite extension of k and let K1 be an intermediate fieldbetween k and K. Then

[K : k]s = [K : K1]s[K1 : k]s.

Proof Let L be an algebraic closure of k. Then each of the [K1 : k]s distinct embeddingsof K1 in L over k has [K : K1]s distinct extensions to K.

Now we can put things together.

Proposition 11.4.12. Let K be a finite extension of k. Then [K : k]s divides [K : k],and the two are equal if and only if K is a separable extension of k. If K is inseparableover k, we have

[K : k] = pr[K : k]s,

where p is the characteristic of k and r > 0.If K = k(α1, . . . , αk), then K is a separable extension of k if and only if α1, . . . , αk

are separable over k.

Proof Write K = k(α1, . . . , αk). Let K0 = k and let Ki = k(α1, . . . , αi) for 1 ≤ i ≤ k.Then

[K : k]s =k∏i=1

[Ki−1(αi) : Ki−1]s, while

[K : k] =k∏i=1

[Ki−1(αi) : Ki−1].

For i = 1, . . . , k, Lemma 11.4.9 tells us that [Ki : Ki−1]s divides [Ki : Ki−1], with thetwo being equal if and only if αi is separable over Ki−1. If αi is inseparable over Ki−1,then

[Ki : Ki−1] = pri · [Ki : Ki−1]s,

where p is the characteristic of k and ri > 0.By passage to products, we see that if αi is separable over Ki−1 for i = 1, . . . , k, then

[K : k]s = [K : k], and that otherwise

[K : k] = pr · [K : k]s,

where p is the characteristic of k and r > 0.Suppose that αi is separable over k for i = 1, . . . , n. Then Corollary 11.4.5 shows

that each αi is separable over Ki−1, and hence [K : k]s = [K : k] by the argument above.Thus, it suffices to show that if [K : k]s = [K : k], then K is a separable extension of

k. But if [K : k]s = [K : k] and if α ∈ K, then the multiplicativity formulæ for degreesand separability degrees show that

[k(α) : k]s = [k(α) : k] and [K : k(α)]s = [K : k(α)].

And the former equality shows that α is separable over k by Lemma 11.4.9.

Corollary 11.4.13. Let K be a finite extension of k and let K1 be an intermediate fieldbetween k and K. Then K is separable over k if and only if both K1 is separable over kand K is separable over K1.


Proof The multiplicativity formulæ for degrees and for separability degrees show that[K : k]s = [K : k] if and only if both [K : K1]s = [K : K1] and [K1 : k]s = [K1 : k].

We may now generalize this to infinite algebraic extensions.

Proposition 11.4.14. Let K be an algebraic extension of k and let K1 be an intermedi-ate field between k and K. Then K is separable over k if and only if both K1 is separableover k and K is separable over K1.

Proof By Corollary 11.4.6, it suffices to show that if K is separable over K1 and K1

is separable over k, then K is separable over k. Thus, let α ∈ K and let f(X) be theminimal polynomial of α over K1. Say f(X) =

∑ni=0 αiX

i with αi ∈ K1 for i = 0, . . . , nand with αn = 1. Then f(X) is also the minimal polynomial of α over the finite extensionk(α0, . . . , αn−1) of k.

Since α is not a repeated root of f(X), α is separable over k(α1, . . . , αn). By Propo-sition 11.4.12, k(α1, . . . , αn)(α) is separable over k(α1, . . . , αn). And any intermediatefield between k and K is separable over k. Since the extensions are now finite over k,the result follows from Corollary 11.4.13.

Corollary 11.4.15. Let K = k(αi | i ∈ I), where αi is separable over k for each i ∈ I.Then K is a separable extension of k.

Proof Each element α ∈ K is contained in a finite extension k(αi1 , . . . , αik) of K. Theresult now follows from Proposition 11.4.12.

The next corollary is now immediate from Corollary 11.4.4.

Corollary 11.4.16. A polynomial f(X) ∈ k[X] is separable if and only if the splittingfield of f is a separable extension of k. More generally, if {fi(X) | i ∈ I} is any collectionof separable elements of k[X], then the splitting field of {fi(X) | i ∈ I} is a separableextension of k.

Corollary 11.4.17. Let K be a normal extension of k and let K1 be an intermediatefield between k and K such that K1 is separable over k. Then the normal closure of K1

over k in K is a separable extension of k as well.

Proof Write K1 = k(αi | i ∈ I). Since αi is separable over k, its minimal polynomial,fi(X), over k is separable. By Proposition 11.2.15, the normal closure of K1 over k inK is a splitting field for {fi(X) | i ∈ I}, so the result follows from Corollary 11.4.16.

Proposition 11.4.12 shows that the separability degree of a finite extension divides itsdegree.

Definitions 11.4.18. Let K be a finite extension of k. Then the inseparability degree,[K : k]i, of K over k is defined by

[K : k]i = [K : k]/[K : k]s.

If [K : k]i = [K : k], we say that K is a purely inseparable extension of k.

We may now apply Proposition 11.4.12.

Corollary 11.4.19. Let k be a field of characteristic p �= 0 and let K be a finite extensionof k. Then [K : k]i is a power of p.


Exercises 11.4.20.1. Show that every algebraic extension of a perfect field is separable.

† 2. Let k be a field of characteristic p �= 0. Let K be an extension of k and let K1

be the collection of elements of K that are separable over k. Show that K1 is anintermediate field between k and K, and hence a separable extension of k. Weshall refer to K1 as the separable closure of k in K.

(a) Show that the separable closure of K1 in K is just K1.

(b) Let α ∈ K. Show that the minimal polynomial of α over K1 has the formXpr − β for some r ≥ 0 and some β ∈ K1.

(c) Show that K is a purely inseparable extension of K1 and that [K1 : k] =[K : k]s, while [K : K1] = [K : k]i.

11.5 Galois Theory

We shall be able to treat finite extensions of the following sort.

Definitions 11.5.1. An extension field K of k is a Galois extension if it is normal andseparable.

If K is a Galois extension of k, then the Galois group, Gal(K/k), of K over k is thegroup of automorphisms (as a field) of K over k.

The group structure on the Galois group, of course, is given by composition of auto-morphisms.

We’d like to be able to compute the Galois groups of finite Galois extensions. Thiscan require some detailed study of the extension in question. Indeed, it can be difficultto compute the degree of a particular Galois extension. Look at the work, for instance,that went into the calculation of [Q(ζn) : Q] in Section 8.8. However, we shall see thatthis calculation is the most difficult part of computing Gal(Q(ζn)/Q).

Here, we shall develop the general theory of Galois groups. Note that if K is a Galoisextension of k, then the Galois group, G = Gal(K/k), acts on K, making K a G-set.We wish to determine the orbit of an element α ∈ K under this action. Note that sinceK is a normal extension of k, the minimal polynomial of α over K splits in K[X].

Lemma 11.5.2. Let K be a Galois extension of k and let α ∈ K. Suppose that theminimal polynomial, f(X), of α over k splits as

f(X) = (X − α1) . . . (X − αk)

in K[X]. Then the orbit of α under the action of G = Gal(K/k) is precisely {α1, . . . , αk}.

Proof Let σ ∈ G. Since σ is an automorphism over k, σ∗ : K[X]→ K[X] carries f(X)to itself. Thus, by Lemma 11.1.3, σ(α) must be a root of f(X). Thus, it suffices to showthat for any root, αi, of f(X) there is an automorphism, σ, of K over k, with σ(α) = αi.

First, Lemma 11.1.3 provides an isomorphism σ0 : k(α)→ k(αi) over k with σ0(α) =αi. Regarding σ0 as an embedding into an algebraic closure, L, of K, we may extend itto an embedding σ : K→ L over k by Proposition 11.1.4. But K is a normal extensionof k, so any embedding of K in L has image K by Proposition 11.2.9. Thus, σ may beregarded as an automorphism of K.


In particular, if K is a Galois extension of k and if f(X) ∈ k[X] is a polynomialthat splits in K[X], then the set of roots of f in K form a sub-G-set of K. Note thatif K is the splitting field of f over k and if α1, . . . , αn are the roots of f in K, thenK = k(α1, . . . , αn), and hence any automorphism of K over k is determined by its effecton α1, . . . , αn.

Corollary 11.5.3. Let K be the splitting field over k of the separable polynomial f(X) ∈k[X] and let α1, . . . , αn be the roots of f in K. Then every element σ ∈ Gal(K/k) re-stricts to give a permutation of {α1, . . . , αn}. Moreover, the passage from elements ofthe Galois group to their restriction to {α1, . . . , αn} induces an injective group homo-morphism

ρ : Gal(K/k)→ Sn.

We shall make extensive use of the following notions.

Definitions 11.5.4. Let K be a Galois extension of k and let G = Gal(K/k). For eachsubgroup H ⊂ G, the fixed field of H, KH , is defined by

KH = {α ∈ K |σ(α) = α for all σ ∈ H}.

For an intermediate field, K1, between k and K, we define the subgroup GK1 associ-ated to K1 by

GK1 = {σ ∈ G |σ(α) = α for all α ∈ K1}.

Lemma 11.5.5. Let K be a Galois extension of k and let G = Gal(K/k). Then foreach subgroup H ⊂ G, KH is an intermediate field between K and k, and for eachintermediate field K1 between K and k, GK1 is a subgroup of G.

Lemma 11.5.6. Let K be a Galois extension of k and let G = Gal(K/k). Let K1 bean intermediate field between k and K. Then K is a Galois extension of K1 and thesubgroup GK1 of G associated to K1 is the Galois group Gal(K/K1) of K over K1.

Proof K is normal over K1 by Lemma 11.2.7, and is separable over K1 by Corol-lary 11.4.6. So K is a Galois extension of K1 as claimed.

Since k is a subfield of K1, every automorphism of K over K1 also fixes k. So we mayregard Gal(K/K1) as a subgroup of Gal(K/k): It is the subgroup consisting of thoseelements that fix K1. By definition, this subgroup is GK1 .

Note that Example 11.2.10 shows that in a Galois extension K of k, there may beintermediate fields not normal over k.

We may now apply Corollary 11.4.6.

Lemma 11.5.7. Let K be a Galois extension of k, and let K1 be an intermediate fieldbetween K and k. Then K1 is a Galois extension of k if and only if K1 is normalover k.

Thus, if an intermediate field K1 between k and K is normal over k, then we candiscuss the relationship between Gal(K/k) and Gal(K1/k).


Lemma 11.5.8. Let K be a Galois extension of k and let K1 be an intermediate fieldbetween k and K such that K1 is normal over k. Then every element of Gal(K/k)carries K1 onto itself, and hence restricts to an automorphism of K1 over k. Passagefrom automorphisms of K over k to their restrictions to K1 gives a group homomorphism,

ρ : Gal(K/k)→ Gal(K1/k).

Proof Let L be an algebraic closure of K. Since K is algebraic over K1, L is alsoan algebraic closure of K1. The restriction of any automorphism of K over k to K1

gives an embedding of K1 over k in K, and hence also in L. By Proposition 11.2.9, thisembedding must have image K1. The rest follows easily.

We shall refer to ρ : Gal(K/k)→ Gal(K1/k) as the restriction homomorphism.

Lemma 11.5.9. Let K be a finite Galois extension of k and let G = Gal(K/k). Thenthe order of G is given by

|G| = [K : k].

Proof Let L be an algebraic closure for K. Since K is a normal extension, everyembedding of K in L over k has image K, and hence gives an automorphism of K(Proposition 11.2.9). Thus, the order of G is equal to the number of distinct embeddingsof K in L over k, which is by definition the separability degree, [K : k]s of K over k.But since K is separable over k, [K : k]s = [K : k] (Proposition 11.4.12), and the resultfollows.

Lemma 11.5.10. Let G be a finite group of automorphisms of a field K and let k = KG,the fixed field of G. Then the degree of K over k is less than or equal to the order of G:

[K : k] ≤ |G|.Proof Let |G| = n and write G = {σ1, . . . , σn}. Let α1, . . . , αm be a basis for K as avector space over k. Supposing that m > n, we shall derive a contradiction.

We have n linear equations in m unknowns in K given by

σ1(α1)x1 + · · ·+ σ1(αm)xm = 0...

σn(α1)x1 + · · ·+ σn(αm)xm = 0.

Since m > n, this system has a nontrivial solution. Let β1, . . . , βm be a nontrivialsolution with the smallest possible number of nonzero entries. By a permutation of theindices of the α’s and β’s, if necessary, we may arrange that for some r > 0, we haveβi �= 0 for 1 ≤ i ≤ r, but βi = 0 for i > r. Thus, we obtain equations

σ1(α1)β1 + · · ·+ σ1(αr)βr = 0...

σn(α1)β1 + · · ·+ σn(αr)βr = 0.

Note that since the αi are all nonzero, we must have r > 1.Multiplying everything by β−1

r , we may assume that βr = 1. Also, since σ1 is an auto-morphism of K over k and since the αi are linearly independent over k, σ1(α1), . . . , σ1(αr)


are linearly independent over k, and hence at least one of the βi must lie outside of k.Permuting indices if necessary, we may assume that β1 is not in k. Since k is the fixedfield of the action of G on K, there must be an element τ ∈ G for which τ(β1) �= β1.

Now apply τ to both sides of each of the above equations. Since G is a group, theelements τσ1, . . . , τσn form a permutation of σ1, . . . , σn. So, permuting rows, we see thatτ(β1), . . . , τ(βr) is a solution to the last displayed set of linear equations. But then

τ(β1)− β1, . . . , τ(βr)− βris also a solution for these equations. Since τ(β1)− β1 �= 0, this is a nontrivial solution.But since βr = 1 and since τ is an automorphism of K, the r-th term is 0. Thus, weobtain a nontrivial solution to our original set of equations which has less than r nonzeroentries. This contradicts the minimality of r, and hence contradicts the assumption thatn < m.

We may now prove the Fundamental Theorem of Galois Theory.

Theorem 11.5.11. (Fundamental Theorem of Galois Theory) Let K be a finite Galoisextension of k and let G = Gal(K/k). Then there is a one-to-one correspondence betweenthe set of subgroups of G and the set of intermediate fields between k and K that associatesto each subgroup H of G the fixed field KH and associates to each intermediate field K1

between k and K the group GK1 = Gal(K/K1).A subgroup H of G is normal in G if and only if KH is a normal extension of k.

Moreover, if H � G, then the restriction map ρ : G → Gal(KH/k) is surjective, withkernel H, and hence Gal(KH/k) ∼= G/H.

Thus, if K1 is an intermediate field that is normal over k, we have a short exactsequence

1→ Gal(K/K1)⊂−→ Gal(K/k)

ρ−→ Gal(K1/k)→ 1

of groups.

Proof We first show that for a subgroup H ⊂ G, we have

H = Gal(K/KH).

Since H fixes KH , we have H ⊂ Gal(K/KH), so it suffices to show that |Gal(K/KH)| ≤|H|. But Lemma 11.5.9 gives |Gal(K/KH)| = [K : KH ] and Lemma 11.5.10 shows that[K : KH ] ≤ |H|, so H = Gal(K/KH) as claimed.

Thus, the one-to-one correspondence will follow if we show that

K1 = KGal(K/K1)

for each intermediate field K1. Clearly, K1 ⊂ KGal(K/K1), so it suffices to show that ifα ∈ K does not lie in K1, then there is an element σ ∈ Gal(K/K1) for which σ(α) �= α.

Now the minimal polynomial, minα/K1(X), of α over K1 has degree > 1, as α �∈K1. Moreover, since K is a normal extension of K1, minα/K1(X) splits in K[X], byProposition 11.2.9. Since K is separable over K1, minα/K1(X) has no repeated roots inK, by Proposition 11.4.2. Thus, the number of roots of minα/K1(X) in K is equal to itsdegree. In particular, minα/K1(X) has a root β ∈ K, with β �= α.

Let ν : K1(α) → K1(β) be the isomorphism over K1 that carries α to β. LetL be an algebraic closure of K. Then Proposition 11.1.4 provides an extension of ν


to an embedding ν : K → L. Since K is a normal extension of K1, any embeddingof K in L over K1 has image K (Proposition 11.2.9). Thus, ν may be viewed as anautomorphism of K, and hence an element of Gal(K/K1). Since ν(α) �= α, we haveshown that KGal(K/K1) = K1, as claimed.

Let H be a subgroup of G = Gal(K/k) and let σ ∈ G. Then it’s easy to see that thefixed field of σHσ−1 is given by

KσHσ−1= σ(KH).

By the one-to-one correspondence between the subgroups of G and the intermediate fieldsbetween k and K, we see that σ normalizes H if and only if σ(KH) = KH . Thus, H isnormal in G if and only if every element of G carries KH onto itself.

If KH is a normal extension of k, then every element of G carries KH onto itself byLemma 11.5.8. Conversely, suppose that every element of G carries KH onto itself andlet L be an algebraic closure of K. Since K is algebraic over KH , so is L, and hence Lis an algebraic closure of KH also. Thus, Proposition 11.2.9 shows that KH is a normalextension of k if every embedding of KH in L over k has image KH .

Thus, let ν : KH → L be an embedding over k. Then ν extends to an embeddingν : K → L. But K is normal over k, so ν gives an automorphism of K, and hence anelement of G. But our assumption on KH is that every element of G carries KH ontoitself, so the image of ν must be KH . Thus, KH is normal over k. In particular, we seethat H � G if and only if KH is a normal extension of k.

Suppose, now, that H � G. Let L be an algebraic closure of K. Then any automor-phism, σ, of KH over k may be considered to be an embedding of KH in L. Since Kis algebraic over KH , σ extends to an embedding of K in L, which must have image K,since K is a normal extension of k. In particular, σ extends to an automorphism of K,and hence σ is in the image of the restriction homomorphism ρ : G→ Gal(KH/k).

Thus, ρ is onto. But its kernel is the subgroup of G that fixes KH , which, by theone-to-one correspondence above, is precisely H.

Thus, if we’re given a finite Galois extension K of k, we can find all the intermediatefields by finding the subgroups of Gal(K/k) and determining their fixed fields. We shallcarry this out in some exercises in later sections. Meanwhile, here’s an example in whichwe already know the intermediate fields.

Example 11.5.12. Finite fields: Let K be the unique (Corollary 11.3.3) field with pn

elements. Then K is the splitting field of Xpn −X over Zp by Proposition 11.3.1, andhence is a normal extension of Zp. Proposition 11.4.7 shows K to be separable over Zp,so K is a Galois extension of Zp.

Now let ϕ : K → K be the Frobenius homomorphism. Since K is finite, ϕ is anautomorphism of K. As a ring homomorphism, it fixes 1, and hence fixes Zp, so ϕ ∈Gal(K/Zp). Since K is finite, K× is a cyclic group of order pn − 1, by Corollary 7.3.14.Let α be a generator of K×. Then for k < n,

ϕk(α) = αpk �= α,

so the order of ϕ in Gal(K/Zp) is at least n. But ϕn takes each element of K to its pn-thpower, and hence is the identity map, as each element of K is a root of Xpn −X.

Thus, ϕ has order n in Gal(K/Zp). But the order of Gal(K/Zp) is equal to [K : Zp],which is n, so Gal(K/Zp) is the cyclic group generated by ϕ.

Note that the subfields of K, as enumerated in Proposition 11.3.4, are given as thefixed fields of ϕr for integers r dividing n.


We may also make some immediate calculations using the discriminant. Recall fromCorollary 11.5.3 that the Galois group of the splitting field of a separable polynomial fembeds as a subgroup of the permutation group on the roots of f .

Corollary 11.5.13. Let K be the splitting field of the separable polynomial f(X) overk and let α1, . . . , αn be the roots of f in K. Let ρ : Gal(K/k) → Sn be the inclusioninduced by restricting an automorphism of K over k to its action on {α1, . . . , αn}.

Let

δ =∏i<j

(αj − αi) ∈ K.

Then an element, σ, of Gal(K/k) gives an even permutation of {α1, . . . , αn} if andonly if it fixes δ. Thus, ρ gives an embedding of Gal(K/k) into An if and only if thediscriminant Δ(f), which is the square of δ, has a square root in k.

Proof The use of δ to test whether ρ(σ) is even follows from Proposition 7.4.8. Inparticular, ρ(Gal(K/k)) is contained in An if and only if Gal(K/k) fixes δ. But theFundamental Theorem of Galois Theory says this happens if and only if δ ∈ k.

That δ is a square root of the discriminant Δ(f) follows from Corollary 7.4.13. SinceΔ(f) ∈ k, the result follows.

Note that the value of δ depends on the ordering of the roots of f . In particular, δ isonly determined up to sign.

The behavior of the discriminant is enough to characterize the behavior of splittingfields of separable cubics.

Corollary 11.5.14. Let K be the splitting field over k of the separable, irreducible cubicf(X). Then

1. If Δ(f) is a square in k, then Gal(K/k) = Z3.

2. If Δ(f) is not a square in k, then Gal(K/k) = S3.

Proof Let α be a root of f in K. Then f is the minimal polynomial of α, so k(α) hasdegree 3 over k. Thus, [K : k], which is the order of Gal(K/k), is divisible by 3.

Since f is irreducible and separable, it has three distinct roots, so Gal(K/k) embedsin S3. The only subgroups of S3 whose order is divisible by 3 are A3 = Z3 and S3 itself.The result now follows from Corollary 11.5.13.

The case of reducible cubics, of course, depends only on finding the number of rootsof the cubic that lie in k. We shall treat the issue of finding the roots of a cubic inSection 11.11.

Remarks 11.5.15. We’ve only given the explicit formula for the discriminant of a cubicof the form X3 + pX + q: In Problem 5 of Exercises 7.4.14, it is shown that

Δ(X3 + pX + q) = −4p3 − 27q2.

But the general case may be recovered from this by the method of completing the cube.Note that if f(X) = X3 + bX2 + cX + d, then setting g(X) = f(X − (b/3)) gives us acubic of the desired form. But if α1, α2, and α3 are the roots of f , then α1+b/3, α2+b/3,and α3 + b/3 are the roots of g. Thus, f and g have the same value for δ, and hence forΔ, as well.


Note that in all of the above, we started out with a Galois extension and studied thebehavior of its Galois group. Suppose, on the other hand, that we start out with a groupG of automorphisms of a field K and set k equal to the fixed field KG. What can we sayabout K as an extension of k?

First, we consider the induced action on polynomials. Suppose that G is a group ofautomorphisms of the ring A and that B = AG, the fixed ring of the action of G. Foreach σ ∈ G, we have an induced ring homomorphism σ∗ : A[X] → A[X], obtained byapplying σ to the coefficients of the polynomials. Passage from σ to σ∗ gives an actionof G on A[X]. The following lemma is immediate.

Lemma 11.5.16. Let G be a group of automorphisms of the ring A and let B = AG.Then the fixed ring of the induced G-action on A[X] is given by

A[X]G = B[X].

We now give a theorem of Artin.

Theorem 11.5.17. Let G be a finite group of automorphisms of the field K and letk = KG. Then K is a Galois extension of k, and Gal(K/k) = G.

Proof Let α ∈ K and let α1, . . . , αk be the orbit of α under the action of G, with, say,α = α1. Thus, {α1, . . . , αk} is the set of all distinct elements of K of the form σ(α) withσ ∈ G. In particular, {α1, . . . , αk} is the smallest sub-G-set of K containing α.

Write

f(X) =k∏i=1

(X − αi) ∈ K[X].

Since {α1, . . . , αk} is preserved by the action of G, we have σ∗(f(X)) = f(X) for allσ ∈ G, and hence f(X) ∈ K[X]G = k[X]. Thus, since α is a root of f , the minimalpolynomial of α over k must divide f(X).

It’s not hard to show that f is the minimal polynomial of α over k, but it’s unnecessaryfor our argument. Since f has no repeated roots, neither does the minimal polynomial,so α is separable over k. Also, the minimal polynomial of α splits in K[X]. Since α waschosen arbitrarily, K is a normal extension of k by Proposition 11.2.9.

Thus, K is a Galois extension of k, and G embeds in Gal(K/k). In particular, wehave

k = KG = KGal(K/k).

Since passage from subgroups of G to their fixed fields gives a one-to-one correspondencebetween the subgroups of Gal(K/k) and the intermediate fields between k and K, wemust have G = Gal(K/k).

The following application of this result plays an important role in understanding theunsolvability of polynomials of degree > 4. We shall come back to it later.

Example 11.5.18. Let k be a field. Then the symmetric group Sn acts on the fieldof rational functions k(X1, . . . , Xn) by permuting the variables. In particular, Sn is theGalois group of k(X1, . . . , Xn) over the fixed field k(X1, . . . , Xn)Sn .


Thus, every finite group is the Galois group of some field extension.Next, we give an application of the Fundamental Theorem. We shall prove the Primi-

tive Element Theorem: that every finite separable extension is simple. Recall, here, thatan extension K of k is simple if K = k(α) for some α ∈ K that is algebraic over k. Wefirst give a characterization of simple extensions.

Proposition 11.5.19. A finite extension K of k is simple if and only if there are onlyfinitely many intermediate fields between K and k.

Proof Suppose that K is simple, with K = k(α). Let K1 be an intermediate fieldbetween k and K and let g(X) be the minimal polynomial of α over K1, say

g(X) = Xk + αk−1Xk−1 + · · ·+ α0.

Let K2 = k(α0, . . . , αk−1) be the field obtained by adjoining the roots of g(X) to k.Then g(X) lies in K2[X] and has α as a root. Since K2 is a subfield of K1 and sinceg(X) is irreducible in K1[X], it must be irreducible in K2[X]. Thus, g(X) is the minimalpolynomial of α over K2. Since K is the result of adjoining α to any intermediate fieldbetween k and K, we obtain

[K : K2] = deg g(X) = [K : K1],

and hence K2 = K1.In particular, this says that K1 is obtained by adjoining the coefficients of g(X) to

k. Note that if f(X) is the minimal polynomial of α over k, then g(X) is a monic factorof f(X) in K[X]. Since f(X) has only finitely many monic factors in K[X], there canbe only finitely many intermediate fields.

For the converse, let K be a finite extension of k such that there are only finitelymany intermediate fields between k and K. If K is a finite field, then K× is cyclic, andhence K = k(α) for any generator α of K×. Thus, we shall assume that K, and hencealso k, is infinite.

Since [K : k] is finite, we have K = k(α1, . . . , αn) for some α1, . . . , αn ∈ K. Byinduction on n, it suffices to show that if K = k(α, β), then K = k(γ) for some γ.

Consider the subfields of the form k(α + aβ) for a ∈ k. Since there are only finitelymany intermediate fields between k and K and since k is infinite, there must be elementsa �= b ∈ k such that

k(α+ aβ) = k(α+ bβ) = K1 ⊂ K.

But then (α+ aβ)− (α+ bβ) = (a− b)β is in K1, and hence β and α are in K1. Thus,K1 = K, and we may take γ = α+ aβ.

The Primitive Element Theorem is now an easy application of the Fundamental The-orem.

Theorem 11.5.20. (Primitive Element Theorem) Every finite separable extension issimple.

Proof Let K be a finite separable extension of k and let L be the normal closure ofK over k in an algebraic closure of K. Then Corollary 11.4.17 shows L to be a Galoisextension of k. By Corollary 11.2.14, L is finite over k, and hence is a finite Galoisextension of k.


Since K is an intermediate field between k and L, it suffices to show that there areonly finitely many intermediate fields between k and L. But this is immediate fromthe one-to-one correspondence between the intermediate fields between k and L and thesubgroups of the Galois group Gal(L/k): Since the Galois group is finite, it has onlyfinitely many subgroups.

In computing Galois groups, it is often important to see how an extension is built upout of smaller extensions.

Definition 11.5.21. Suppose given a diagram

L

K1 K2

k

��

��

��

��

�

��

�

of field extensions. Then the compositum, K1K2, of K1 and K2 in L is the smallestsubfield of L containing both K1 and K2.

If L = K1K2, we say that L is a compositum of K1 and K2. However, it should benoted that extension fields K1 and K2 of k may have composita that are not isomorphicover k: The specific embeddings of K1 and K2 in the larger field are needed to determinethe compositum.

We can often deduce information about the compositum from information about K1

and K2. We shall discuss this further in the exercises.Note that in the abstract, neither K1 nor K2 is assumed to be algebraic over k.

Exercises 11.5.22.1. Compute the Galois groups over Q of the splitting fields of the following polyno-

mials.

(a) X3 − 3X + 1

(b) X3 − 4X + 2

(c) X3 +X2 − 2X − 1

2. Let K be a Galois extension of k. Show that an element α ∈ K is a primitiveelement for K over k (i.e., that K = k(α)) if and only if no non-identity elementσ ∈ Gal(K/k) satisfies σ(α) = α.

3. Let L be a compositum of K1 and K2 over k. Suppose that K1 is algebraic over k,with K1 = k(αi | i ∈ I). Show that L is algebraic over K2, with L = K2(αi | i ∈ I).

4. Let L be a compositum of K1 and K2 over k. Show that if any of the followingare descriptors of K1 as an extension of k, then the same descriptor holds for L asan extension of K2.

(a) normal

(b) separable

(c) Galois


(d) finite

(e) simple

† 5. Let L be a compositum of K1 and K2 over k, and suppose that K1 is a Galoisextension of k. Show that each element of Gal(L/K2) restricts on K1 to an el-ement of Gal(K1/k). Show that this restriction map provides an injective grouphomomorphism

ρ : Gal(L/K2)→ Gal(K1/k).

Deduce that if K1 is a finite Galois extension of k, then [L : K2] divides [K1 : k].

6. Let L be a compositum of K1 and K2 over k, and suppose that K1 is a finite Galoisextension of k.

(a) Suppose K1 ∩K2 = k, where the intersection is as subfields of L. Show that[L : K2] = [K1 : k]. (Hint : Let α be a primitive element for K1 over k andlet f(X) be the minimal polynomial of α over K2. Show that the coefficientsof f(X) must lie in K1 ∩K2, and hence f is also the minimal polynomial ofα over k.)

(b) Conversely, suppose that K1∩K2 properly contains k. Show that [L : K2] is aproper divisor of [K1 : k]. Thus, K1∩K2 = k if and only if [L : K2] = [K1 : k].

(c) Show that L is a quotient ring of K1 ⊗k K2, and that L ∼= K1 ⊗k K2 ifK1 ∩K2 = k.

7. Let K1 be the splitting field over k of f(X) ∈ k[X]. Let K2 be any extension ofk. Show that the splitting field of f(X) over K2 is a compositum of K1 and K2.

8. Let K1 and K2 be extension fields of k, and suppose that K1 is a finite normalextension of k. Show that any two composita of K1 and K2 over k are isomorphicover K2 (and hence over k).

9. Find extension fields K1 and K2 of Q with at least two different composita. (Hint :Look for an example where K1 = K2.)

10. Let K be the algebraic closure of Zp. Calculate Gal(K/Zp).

11. Give an example of a finite extension that is not simple.

11.6 The Fundamental Theorem of Algebra

We show that the complex numbers, C, are algebraically closed. We shall make use ofthe following application of the topology of the real line.

Lemma 11.6.1. Every odd degree polynomial over the real numbers, R, has a real root.

Proof Let f(X) ∈ R[X] be a polynomial of degree n, where n is odd. Divid-ing by the leading coefficient, if necessary, we may assume that f is monic, so thatf(X) = Xn + an−1X

n−1 + · · · + a0. We claim that if |x| is large enough, then |xn| >|an−1x

n−1 + · · ·+ a0|.To see this, note that if |x| > 1, then

|an−1xn−1 + · · ·+ a0| < (|an−1|+ · · ·+ |a0|)|xn−1|.


So if |x| > 1 is greater than |an−1| + · · · + |a0|, then |xn| is indeed greater than|an−1x

n−1 + · · ·+ a0|.In particular, if |x| is large enough, then f(x) has the same sign as xn, which, since n

is odd, has the same sign as x. Explicitly, if c is greater than both 1 and |an−1|+· · ·+|a0|,then f(−c) is negative and f(c) is positive, so the Intermediate Value Theorem showsthat f must have a root in the open interval (−c, c).Corollary 11.6.2. The field of real numbers, R, has no nontrivial extensions of odddegree.

Proof Let K be a nontrivial odd degree extension of R and let α be an element of Kthat’s not in R. Then [R(α) : R] divides [K : R], and hence is odd. But [R(α) : R] isthe degree of the minimal polynomial of α over R. Since every odd degree polynomialover R has a root in R, there are no irreducible elements of R[X] whose degree is anodd number greater than 1. So this is impossible.

Corollary 11.6.3. Let K be a finite extension of R. Then the degree of K over R mustbe a power of 2.

Proof Suppose, to the contrary, that there is an extension K of R whose degree overR is divisible by an odd prime. Let L be the normal closure of K over R in an algebraicclosure of K. Then L is finite over R by Corollary 11.2.14, and hence is a finite Galoisextension of R. By the multiplicativity formula for degrees, we see that [L : R] is divisibleby an odd prime.

Let G be the Galois group of L over R and let G2 be a 2-Sylow subgroup of G.Let K1 = KG2 , the fixed field of G2 in K. Then the Fundamental Theorem of GaloisTheory shows that G2 = Gal(K/K1), and hence the degree of K over K1 is |G2| byLemma 11.5.9. Thus, the multiplicativity formula for degrees gives

[K1 : R] = [G : G2],

the index of G2 in G. By the Second Sylow Theorem, [G : G2] is odd, and must begreater than 1, as |G| is divisible by an odd prime. In particular, K1 is a nontrivial odddegree extension of R, which is impossible by Corollary 11.6.2.

Of course, every extension of C is an extension of R.

Corollary 11.6.4. Let K be a finite extension of C. Then the degree of K over C mustbe a power of 2.

There’s only one further result needed to prove that C is algebraically closed.

Lemma 11.6.5. Every polynomial of degree 2 over C has a root in C.

Proof Completing the square gives

aX2 + bX + c = a(X +b

2a)2 + (c− b2

4a),

so a root is obtained by setting X + (b/2a) equal to any complex square root of (b2 −4ac)/4a2. In particular, it suffices to show that any complex number has a square rootin C.

The easiest way to see this is via the exponential function. The point is that forr, θ ∈ R with r ≥ 0, the complex number reiθ gives the point in the plane whose polar


coordinates are (r, θ). Thus, every complex number may be written in the form reiθ, andhence has

√reiθ/2 as a square root.

Or, explicitly, if x, y ∈ R with y �= 0, then the reader may check that a square rootfor x+ yi is given by

y√2(r − x) + i

√r − x

2

where r =√x2 + y2.

But every degree 2 extension is simple.

Corollary 11.6.6. There are no degree 2 extensions of C.

And now, we have the main theorem of this section.

Theorem 11.6.7. (Fundamental Theorem of Algebra) The field of complex numbers,C, is algebraically closed.

Proof It suffices to show that there are no irreducible elements of C[X] of degreegreater than 1. To see this, note that if f(X) is an irreducible polynomial of degree > 1,then the splitting field, K, of f(X) is a nontrivial extension of C.

By Corollary 11.6.4, [K : C] is a power of 2, and hence the Galois group Gal(K/C)is a 2-group. By Corollary 5.2.5, there is an index 2 subgroup, H, of Gal(K/C). Butthen KH is a degree 2 extension of C by the Fundamental Theorem of Galois Theory.Since C has no extensions of degree 2, the result follows.

Exercises 11.6.8.1. Show that every irreducible element of R[X] has either degree 1 or degree 2.

2. Show that the roots in C of an irreducible quadratic in R[X] are complex conjugatesof one another.

11.7 Cyclotomic Extensions

Definitions 11.7.1. Let K be a field. Then the n-th roots of unity in K are the rootsin K of the polynomial Xn − 1.

We say that ζ ∈ K is a primitive n-th root of unity if ζ has order n in K×.

Lemma 11.7.2. Let K be a field. Then the set of n-th roots of unity in K forms asubgroup of K×. Moreover, this subgroup is cyclic and contains all the elments of K×

with exponent n.In particular, K contains a primitive n-th root of unity if and only if there are n

distinct roots of Xn − 1 in K. In this case, any primitive n-th root of unity generatesthe group of n-th roots of unity in K.

Proof An element ζ ∈ K is an n-th root of unity if and only if ζn = 1. And this isequivalent to ζ having exponent n in K×.

The elements of exponent n in an abelian group always form a subgroup. So the n-throots of unity form a subgroup of K×. Since Xn − 1 has at most n roots, this subgroupis finite. And Corollary 7.3.14 shows that any finite subgroup of K× is cyclic.

Thus, the group of n-th roots of unity in K has order n if and only if K× has anelement of order n.


Examples 11.7.3.1. Every element x ∈ R× with |x| �= 1 has infinite order in R×. So the only roots of

unity in R are ±1, the primitive first root and primitive second root, respectively.

2. Let K be the finite field of order pn. Then K× is a cyclic group of order pn − 1.So K has primitive r-th roots for all r dividing pn − 1.

3. We’ve seen that ζn = e2πi/n is a primitive n-th root of unity in C. So the group ofn-th roots in C (or in any subfield of C containing Q(ζn)) is 〈ζn〉 = {ζkn | 0 ≤ k < n}.

For many purposes, it is useful to adjoin a primitive n-th root of unity to a field k.Note that if ζ is a primitive n-th root of unity in an extension field K of k, then k(ζ)contains n distinct roots of Xn − 1. Thus, Xn − 1 splits in k(ζ), and hence k(ζ) is asplitting field for Xn − 1 over k.

Conversely, if K is a splitting field for Xn − 1 over k, then K contains a primitiven-th root of unity if and only if Xn − 1 has n distinct roots in K. If K does contain aprimitive n-th root of unity, ζ, then the roots of Xn − 1 in K are the distinct powers ofζ. Since K is the splitting field of Xn − 1 over ζ, we see that K = k(ζ).

Corollary 11.7.4. Let k be a field. Then we may adjoin a primitive n-th root of unityto k if and only if Xn − 1 has n distinct roots in its splitting field over k. The result ofadjoining a primitive n-th root of unity to k, when possible, is always a splitting field forXn − 1.

The next thing to notice is that it isn’t always possible to adjoin a primitive n-throot to a field k. For instance, in Zp[X], we have Xp− 1 = (X − 1)p, which has only theroot 1 in any field of characteristic p. So it is impossible to adjoin a primitive p-th rootof unity to a field of characteristic p.

Proposition 11.7.5. We may adjoin a primitive n-th root to a field k if and only if nis not divisible by the characteristic of k.

Proof The formal derivative of Xn− 1 is nXn−1. If n is divisible by the characteristicof k, then nXn−1 = 0, and hence Lemma 8.7.7 tells us that Xn − 1 has repeated rootsin its splitting field. In particular, there cannot be n distinct roots of Xn − 1.

If n is not divisible by the characteristic of k, 0 is the only root of nXn−1 in anyextension of k. In particular, Xn − 1 and nXn−1 have no common roots in the splittingfield, K, of Xn − 1 over k. Lemma 8.7.7 now shows that Xn − 1 has no repeated rootsin K, so there must be n distinct roots there.

Definition 11.7.6. Let k be a field and let n be a positive integer not divisible by thecharacteristic of k. We shall refer to the splitting field of Xn−1 over k as the cyclotomicextension obtained by adjoining a primitive n-th root of unity to k.

Since the roots of Xn − 1 form a subgroup of K×, we may make an improvement onCorollary 11.5.3.

Proposition 11.7.7. Let K be the cyclotomic extension of k obtained by adjoining aprimitive n-th root of unity to K. Let ζ be a primitive n-th root of unity in K and write〈ζ〉 = {ζk | 0 ≤ k < n} for the group of n-th roots of unity in K. Write Aut(〈ζ〉) for thegroup of automorphisms of the cyclic group 〈ζ〉. Then each σ ∈ Gal(K/k) restricts on〈ζ〉 to an element of Aut(〈ζ〉). We obtain an injective group homomorphism

ρ : Gal(K/k)→ Aut(〈ζ〉).


Recall from Proposition 4.5.4 that the automorphism group of a cyclic group of ordern is isomorphic to Z×

n , the group of units of the ring Zn. In multiplicative notation, theautomorphism of 〈ζ〉 induced by a unit k ∈ Z×

n takes ζm to ζmk. The group structureon Z×

n is calculated in Section 4.5. Of crucial importance for our study here is the factthat Z×

n is abelian.Recall from Proposition 8.8.7 that [Q(ζn) : Q] = φ(n), where φ(n) is the order of Z×

n .Since ρ : Gal(Q(ζn)/Q)→ Aut(〈ζn〉) is injective, we obtain the following corollary.

Corollary 11.7.8. The Galois group of Q(ζn) over Q restricts isomorphically to theautomorphism group of the abelian group 〈ζn〉.

Recall from Corollary 7.11.10 that every field, K, of characteristic 0 has a unique,natural, Q-algebra structure. So it has a unique subfield isomorphic to Q, and everyautomorphism of K restricts to the identity on that subfield. Moreover, every subfieldof K must contain Q, so the collection of intermediate fields between Q and K coincideswith the collection of all subfields of K.

Example 11.7.9. We have Z×8 = {1, 3, 5,−1}, with a group structure isomorphic to

Z2 × Z2. Note that the automorphism of 〈ζ8〉 induced by −1 is induced by Galoisautomorphism obtained by restricting the complex conjugation map to Q(ζ8). The Galoisautomorphisms corresponding to 3 and 5 are induced by the standard isomorphisms overQ from Q(ζ8) to Q(ζ3

8 ) and Q(ζ58 ). These isomorphisms exist because ζ8, ζ3

8 , ζ58 , and ζ7

8

all have the same minimal polynomial, Φ8(X), over Q by Proposition 8.8.7.The subgroups of Z×

8 form the following lattice, with the downward lines representinginclusions of subgroups.

e

〈3〉〈5〉 {±1}

Z×8

��

��

�

��

��

�

��

��

��

��

With this, we can find all the subfields of Q(ζ8). Since each nontrivial subgroup ofZ×

8 has index 2, we can find its fixed field by finding a single element outside of Q thatit fixes.

In the case of the subgroup {±1}, we know that complex conjugation exchanges ζ8and ζ8, and hence fixes

ζ8 + ζ8 = 2 re ζ8 =√

2.

(Here, the last equality comes from the fact that ζ8 = 1√2

+ 1√2i.) So the fixed field of

{±1} is Q(√

2).The remaining fixed fields may be calculated similarly, using the known effect of the


subgroups in question on 〈ζ8〉. We obtain the following lattice of subfields.

Q(ζ8)

Q(i√

2) Q(i) Q(√

2)

Q

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Exercises 11.7.10.1. Let k be a field of characteristic p �= 0 and suppose that n is divisible by p. How

many roots does Xn − 1 have in its splitting field over k? Is the splitting field inquestion a cyclotomic extension of k?

2. Let K be a cyclotomic extension of k. Show that every intermediate field betweenk and K is a Galois extension of k.

3. What are the orders of the cyclotomic extensions obtained by adjoining a primitiven-th root of unity to Zp for all primes p ≤ 11 and all n prime to p with n ≤ 11?

4. Show that the fixed field of the restriction to Q(ζn) of complex conjugation isQ(ζn + ζn). Deduce that Q(ζn + ζn) = Q(ζn) ∩ R. What are the roots of theminimal polynomial of ζn + ζn over Q?

5. Find all the subfields of Q(ζ5).

6. What is the Galois group of Q(ζ7 + ζ7) over Q? What are the other subfieldsof Q(ζ7)?

7. What is the Galois group of Q(ζ9 + ζ9) over Q? What are the other subfieldsof Q(ζ9)?



10. Show that every cyclic group occurs as the Galois group of a Galois extension of Q.

11. Let k be a field and let n be an integer that is not divisible by the characteristic ofk. Since the cyclotomic polynomial Φn(X) has integer coefficients, we may regardit as an element of k[X]. Let K be the cyclotomic extension obtained by adjoininga primitive n-th root of unity to k. Show that K is the splitting field of Φn(X)over k and that the roots of Φn(X) in K are precisely the primitive n-th roots ofunity.

12. Show that Φpr (X) is irreducible over Qp for all r > 0. Deduce that if K is thecyclotomic extension obtained by adjoining a primitive pr-th root to Qp, thenGal(K/Qp) = Z×

pr . (Hint : Note that (X + 1)pr − 1 = Xpr

in Zp[X].)


13. Let n be an integer relatively prime to p and let K be the cyclotomic extension ofZp obtained by adjoining a primitive n-th root of unity. Show that Gal(K/Zp) isthe subgroup of Z×

n generated by p. What, then, is the degree of K over Zp?

14. Suppose that (n, p− 1) > 2. Show that Φn(X) is not irreducible over Qp.

11.8 n-th Roots

We shall study the Galois theory of the splitting fields of polynomials of the form Xn−a.Lemma 11.8.1. Let k be a field and let 0 �= a ∈ k. Let n be a positive integer notdivisible by the characteristic of k. Then the splitting field of Xn−a over k is a separableextension of k and has the form K = k(α, ζ), where α is a root of Xn − a and ζ is aprimitive n-th root of unity.

The set of roots in K of Xn − a is {αζk | 0 ≤ k < n}, and hence

Xn − a =n−1∏k=0

(X − αζk)

in K[X].

Proof Let L be an algebraic closure of K. Choose any root α of Xn − a in L and anyprimitive n-th root of unity, ζ ∈ L, and set K = k(α, ζ). Then {αζk | 0 ≤ k < n} gives ndistinct roots of Xn−a in K, and hence Xn−a factors as stated in K[X]. In particular,Xn − a has no repeated roots in K, and hence is a separable polynomial over k. SinceK is the splitting field of Xn − a over k, it is separable over k by Corollary 11.4.16.

The simplest special case of such extensions is when k already contains a primitiven-th root of unity.

Proposition 11.8.2. Let k be a field containing a primitive n-th root of unity, ζ. Let0 �= a ∈ k and let K be a splitting field for Xn − a over k. Then the Galois group of Kover k is a subgroup of Zn.

Explicitly, if α is any root of Xn − a in K, then for each σ ∈ Gal(K/k) there is aninteger s such that

σ(αζk) = αζk+s for 0 ≤ k < n.

The passage from σ to s gives an injective group homomorphism from Gal(K/k) to Zn.

Proof The roots ofXn−a are given by {αζk | 0 ≤ k < n}, so Corollary 11.5.3 shows thatrestricting elements of Gal(K/k) to their effect on {αζk | 0 ≤ k < n} gives an injectivegroup homomorphism from Gal(K/k) to the group S({αζk | 0 ≤ k < n}) of permutationson {αζk | 0 ≤ k < n}.

For σ ∈ Gal(K/k), let σ(α) = αζs. Since ζ ∈ k, it is fixed by σ, so

σ(αζk) = σ(α)σ(ζk) = αζsζk = αζk+s

for all values of k. In other words, the action of σ on the set {αζk | 0 ≤ k < n} coincideswith multiplication by ζs.

Define ϕ : Zn → S({αζk | 0 ≤ k < n}) by ϕ(s)(αζk) = αζk+s. Then ϕ is easilyseen to be an injective group homomorphism. Since the image of the restriction mapρ : Gal(K/k)→ S({αζk | 0 ≤ k < n}) lies in the image of ϕ, the result follows.


Thus, if k contains a primitive n-th root of unity and 0 �= a ∈ k, then the Galoisgroup over k of the splitting field of Xn − a is determined by the degree of this splittingfield over k. As usual, the degree is not always easy to compute.

For the general case of splitting fields of polynomials of the form Xn − a with n notdivisible by the characteristic of k, Lemma 11.8.1 shows that the splitting field, K, mustcontain a principal n-th root of unity, ζ. But then k(ζ) is a normal extension of k. Wemay apply Proposition 11.8.2 to describe the Galois group of K over k(ζ) and applyProposition 11.7.7 to describe the Galois group of k(ζ) over k. We can then put thesecalculations together using the Fundamental Theorem of Galois Theory.

Recall from Section 4.7 that if H and H ′ are groups, then an extension of H by H ′

is a group G, together with a surjective homomorphism f : G → H ′ with kernel H.In particular, the Fundamental Theorem of Galois Theory shows that if K is a Galoisextension of k and if K1 is an intermediate field normal over k, then the restrictionmap ρ : Gal(K/k) → Gal(K1/k) presents Gal(K/k) as an extension of Gal(K/K1) byGal(K1/k).

Recall from Lemma 4.7.9 that if H is abelian and if f : G → H ′ is an extension ofH by H ′, then there is an induced action of H ′ on H via automorphisms, αf : H ′ →Aut(H), defined as follows. For x ∈ H ′, write x = f(g). Then αf (x) acts on H byαf (x)(h) = ghg−1 for all h ∈ H. This action is an important invariant of the extension.Notice that if αf is not the trivial homomorphism, then G cannot be abelian.

Proposition 11.8.3. Let k be a field and let 0 �= a ∈ k. Let n be a positive integer notdivisible by the characteristic of k and let K be the splitting field of Xn − a over k.

Let ζ be a primitive n-th root of unity in K and let K1 = k(ζ). Write H =Gal(K/K1), H ′ = Gal(K1/k) and G = Gal(K/k). Then the restriction map ρ : G→ H ′

presents G as an extension of H by H ′.We have inclusion maps H ⊂ Zn and H ′ ⊂ Z×

n , and the action of H ′ on H inducedby the extension ρ coincides with the composite

H ′ ⊂ Z×n∼= Aut(Zn)

ε−→ Aut(H)

where ε is obtained by restricting an automorphism of Zn to its effect on H.

Proof Recall that a cyclic group has a unique subgroup of order d for every d thatdivides is order. Thus, every subgroup of a cyclic group is characteristic, so ε is welldefined (Lemma 3.7.8).

The only other point that isn’t covered by the discussion prior to the statement ofthis proposition is the verification that the action of H ′ on H is given by ε. Thus, let σ bean automorphism of K1 over k. Then there is an element r ∈ Z×

n such that σ(ζk) = ζrk

for all k.An extension of σ to an automorphism of K is determined by its effect on α. For our

purposes, all we need to know is that if σ extends σ, then σ(α) = αζs for some valueof s. Consequently, σ(αζk) = αζrk+s. It’s now easy to see that σ−1(αζk) = αζt(k−s),where t represents the inverse of r in Z×

n .We now wish to calculate the effect of conjugating an automorphism τ ∈ H =

Gal(K/K1) by σ. Here, Proposition 11.8.2 gives us an integer l such that τ(αζk) = αζk+l

for all k. And τ corresponds to the element l ∈ Zn under the identification of H with asubgroup of Zn.

A direct verification now shows that στσ−1 takes αζk to αζk+rl. This correspondsto the standard action of Z×

n on Zn, as claimed.

At this point, we need to consider some examples.


Example 11.8.4. We calculate the Galois group of the splitting field of X4 − 2 over Qand determine the intermediate fields. Here, we have a diagram

Q( 4√

2, i)

Q( 4√

2) Q(i)

Q

��

�2��

4

��

4 ��

��

2

where the numbers on the extension lines represent the degrees of the extensions. Thesedegrees are calculated as follows.

Here, X4− 2 is irreducible over Q by Eisenstein’s criterion, and hence is the minimalpolynomial of 4

√2 over Q. So [Q( 4

√2) : Q] = 4. Now Q( 4

√2) ⊂ R, so it doesn’t contain

i, and hence [Q( 4√

2, i) : Q( 4√

2)] = 2. The rest now follows from the multiplicativity ofdegrees, together with the fact that i has degree 2 over Q.

Now the fact that [Q( 4√

2, i) : Q(i)] = 4, shows that Gal(Q( 4√

2, i)/Q(i)) = Z4, byProposition 11.8.2. And the non-trivial element of Gal(Q(i)/Q) extends to the restrictionto Q( 4

√2, i) of the complex conjugation map. Since complex conjugation has order 2,

we see that the extension ρ : Gal(Q( 4√

2, i)/Q) → Gal(Q(i)/Q) splits. It is easy to seethat the resulting extension of Z4 by Z2 is precisely the dihedral group, D8, of order 8.Here, we can take the generator, b, of order 4 to act via b( 4

√2 ik) = 4

√2 ik+1, and take the

generator, a, of order 2 to act via a( 4√

2 ik) = 4√

2 i−k. The inverted lattice of subgroupsof D8 is given by

e

〈a〉〈ab2〉〈b2〉〈ab〉〈ab3〉

〈a, b2〉〈b〉〈ab, b2〉

D8

��

��

��

��

��

��

��

��

�

��

��

�

��

��

�

��

��

�

��

��

�

��

��

�


The associated lattice of subfields is given by

Q( 4√

2, i)

Q( 4√

2) Q(i 4√

2) Q(ζ8) Q(ζ84√

2) Q(ζ−18

4√

2)

Q(√

2) Q(i) Q(i√

2)

Q

��

��

��

��

��

��

��

��

��

��

��

��

��

��

�

��

��

��

��

�

��

��

��

��

��

��

��

�

��

��

��

��

�

To see this, note that ζ8 = 1√2

+ 1√2i is in Q( 4

√2, i). Note also that both ζ8

4√

2 and

ζ−18

4√

2 are roots of X4 + 2, and hence have degree 4 over Q by Eisenstein’s criterion.The rest follows by checking that the elements in question are left fixed by the appropriatesubgroups of D8.

The preceding example, while straightforward in itself, points to a potential difficultyin calculating the Galois groups of splitting fields of polynomials of the form Xn − a:It is sometimes more difficult than in the preceding example to compute the degree ofQ( n√a, ζn) over Q(ζn). For instance, consider the following related example.

Example 11.8.5. Consider the splitting field, Q( 8√

2, ζ8), of X8−2 over Q. Once again,X8 − 2 satisfies Eisenstein’s criterion, so Q( 8

√2) has degree 8 over Q. But ζ8 satisfies

the polynomial X2−√2X + 1 over Q( 8√

2), so the degrees of our extensions are given asfollows.

Q( 8√

2, ζ8)

Q( 8√

2) Q(ζ8)

Q

��2��

�� 4

��

8 ��

4

Thus, Gal(Q( 8√

2, ζ8)/Q) is an extension of Z4 by Z×8∼= Z2 × Z2. In this case, it turns

out that the extension does not split. We shall leave the further analysis of this exampleto the exercises.

Exercises 11.8.6.1. Show that X4 − 2 and X4 + 2 have the same splitting field over Q.

2. Find a primitive element for Q(i, 4√

2) over Q.


3. Show that the Galois group, G, of Q( 8√

2, ζ8) over Q contains an element of order8, and hence is an extension of Z8 by Z2. Compute the action of Z2 on Z8 inducedby this extension and determine whether G is a split extension of Z8 by Z2. (Hint :Show that the minimal polynomial of 8

√2 over Q(ζ8) is not invariant under the

action of the Galois group of Q(ζ8) over Q.)

4. Show that Gal(Q( 8√

2, ζ8)/Q) cannot be a split extension of Z4 by Z2 × Z2.

5. Find all the subfields of Q( 8√

2, ζ8).

6. Show that the splitting field of X8 + 2 over Q has degree 16 over Q. Calculate itsGalois group over Q. Determine whether it is isomorphic to the Galois group ofthe splitting field of X8 − 2 over Q.


2, ζ3).


2, ζ5).

9. Let 0 �= a ∈ Q and let K be the splitting field of Xn − a over Q. Suppose that Khas degree n over Q(ζn). Show that Gal(K/Q) is a split extension of Zn by Z×

n .

10. Let a be an integer divisible by a prime number q, but not by q2. Let p be anyprime. Show that the Galois group of Q( p

√a, ζp) over Q is a split extension of Zp

by Zp×.

11. Let a be an integer divisible by a prime number q, but not by q2. Let n = p1, . . . , pk,where p1, . . . , pk are distinct primes with the property that pi �≡ 1 mod pj for alli, j. Show that the Galois group of Q( n

√a, ζn) over Q is a split extension of Zn by

Zn×.

12. Let K be the splitting field over k of Xn − a, where n is not divisible by thecharacteristic of k, and 0 �= a ∈ k. Let ζ be a primitive n-th root of unity in K.Show that the extension

1→ Gal(K/k(ζ))→ Gal(K/k)ρ−→ Gal(k(ζ)/k)→ 1

splits whenever [K : k(ζ)] = [k(α) : k], where α is an arbitrarily chosen root ofXn − a in K. Show that this latter condition holds if and only if k(ζ) ∩ k(α) = k.

13. Let K be a field containing a primitive n-th root of unity and let 0 �= a ∈ K.Let α be a root of Xn − a in some extension field of K. Show that the minimalpolynomial of α over K has the form Xd − αd, where d is the smallest divisor of nwith the property that αd ∈ K. Show that in this case n/d is the largest divisor ofn with the property that K contains an n/d-th root of a.

14. Let a be a positive integer divisible by a prime number q but not by q2. Show thatthe splitting field of Xn − a over Q is a split extension of Zn by Z×

n if and only ifthe real k-th root, k

√a, of a does not lie in Q(ζn)∩R for any k > 1 that divides n.

15. Give the analogue of the preceding problem for a negative integer a.

16. Let Ka be the splitting field of X8 − a over Q. Calculate the degree of Ka over Qfor all positive integers a.

17. Show that Q(ζ5) ∩R = Q(√

5). What does this say about the splitting fields ofthe polynomials X10 − a for positive integers a? (Hint : Calculate the minimalpolynomial of ζ5 + ζ5 over Q.)


11.9 Cyclic Extensions

Definitions 11.9.1. We say that a Galois extension is cyclic if its Galois group is cyclic.Similarly, abelian extensions and solvable extensions will denote Galois extensions

whose Galois groups are abelian and solvable, respectively.We say that a Galois extension has exponent n if its Galois group has exponent n in

the sense of group theory (i.e., σn = 1 for each σ ∈ G.)

We shall first consider cyclic extensions whose degree is not divisible by the charac-teristic of k.

Recall from Proposition 11.8.2 that if k contains a primitive n-th root of unity andif 0 �= a ∈ k, then the splitting field of Xn− a over k is a cyclic extension of exponent n.

If k does not contain a primitive n-th root of unity, then there may be cyclic extensionsof exponent n over k that are not splitting fields of polynomials of the form Xn −a. For instance, the Galois group of Q(ζm) over Q is Z×

m, which has numerous cyclicquotient groups. And each one of these quotient groups is the Galois group over Q of anintermediate field between Q and Q(ζm) by the Fundamental Theorem of Galois Theory.By varying m we may show that Q has cyclic extensions of every order.

On the other hand, let K be the splitting field of Xn − a over Q with 0 �= a ∈ Q.Then a careful analysis of Proposition 11.8.3 shows that Gal(K/Q) cannot be cyclicunless either n = 2 or K = Q(ζn) for n = 2, 4, pr, or 2pr for some odd prime p.

This puts severe restrictions on the cyclic extensions of Q which arise as splittingfields of polynomials of the form Xn− a, so we see that there are many cyclic extensionsof Q that do not have this form.

We shall not attempt a systematic study of cyclic extensions of degree n in the casethat n is not divisible by the characteristic of k and yet k fails to contain a primitiven-th root of unity. But in the case that k does contain a primitive n-th root of unity, weshall give a converse to Proposition 11.8.2. Indeed, in this case, every cyclic extension ofexponent n over k is the splitting field of a polynomial of the form Xn − a.

We shall make use of Artin’s Linear Independence of Characters theorem.

Proposition 11.9.2. (Linear Independence of Characters) Let G be any group and K afield. Suppose given n distinct group homomorphisms, f1, . . . , fn, from G to K×, wheren ≥ 1. Then f1, . . . , fn are linearly independent over K as functions from G to K. Inother words, given any collection of elements a1, . . . , an ∈ K that are not all 0, there isa g ∈ G such that

a1f1(g) + · · ·+ anfn(g) �= 0.

Proof We argue by induction on n, with the result being trivial for n = 1. Assumingthe result true for n − 1, suppose we’re given coefficients a1, . . . , an ∈ K, not all 0,such that a1f1(g) + · · · + anfn(g) = 0 for all g ∈ G. By our induction hypothesis, wemay assume that all the coefficients ai are nonzero. So by multiplying both sides of theequation by a−1

1 , if necessary, we may assume that a1 = 1. We obtain

f1(g) + a2f2(g) + · · ·+ anfn(g) = 0

for all g ∈ G.Since the homomorphisms f1, . . . , fn are distinct, there is an x ∈ G such that f1(x) �=

fn(x). Substituting xg for g in the above equation and applying the fact that the fi arehomomorphisms, we obtain

f1(x)f1(g) + (a2f2(x))f2(g) + · · ·+ (anfn(x))fn(g) = 0


for all g ∈ G. Now multiply both sides of the equation by (f1(x))−1. Subtracting theresult from the first displayed equation, the terms f1(g) cancel and we are left with thefollowing:

a2[1− (f1(x))−1f2(x)]f2(g) + · · ·+ an[1− (f1(x))−1fn(x)]fn(g) = 0

for all g ∈ G. Since f1(x) �= fn(x), this is a nontrivial dependence relation betweenf2(x), . . . , fn(x), contradicting the inductive assumption.

Since the embeddings from a field K into a field L restrict to homomorphisms fromK× to L×, we obtain a proof of the following corollary, which was originally due toDedekind.

Corollary 11.9.3. Let σ1, . . . , σn be distinct embeddings from the field K into a field Land let β1, . . . , βn ∈ L be coefficients that are not all 0. Then there is an α ∈ K suchthat

β1σ1(α) + · · ·+ βnσn(α) �= 0.

Our most common use of this will be when σ1, . . . , σn are automorphisms of K itself.

Proposition 11.9.4. Let G be a group of automorphisms of K over k and let f : G→k× be a group homomorphism. Then there is an element α ∈ K such that

f(σ) =α

σ(α)

for all σ ∈ G.

Proof Consider the linear combination∑σ∈G f(σ) · σ. Since the coefficients f(σ) are

all nonzero, we can find an a ∈ K such that∑σ∈G f(σ)σ(a) �= 0. Set

α =∑σ∈G

f(σ)σ(a).

Note that each f(τ) lies in k, and hence is fixed by each σ ∈ G. Thus, if we apply σto each side of the equation, we get

σ(α) =∑τ∈G

f(τ)(στ)(a)

=∑στ∈G

(f(σ))−1f(στ)(στ)(a)

= (f(σ))−1 · α,where the second equality follows from the fact that f is a group homomorphism. Theresult now follows.

We may now give the promised converse to Proposition 11.8.2.

Proposition 11.9.5. Let k be a field containing a primitive n-th root of unity and letK be a cyclic extension of k of exponent n. Then K is the splitting field over k of apolynomial of the form Xn − a for some a ∈ k.


Proof Since k has a primitive n-th root of unity, the splitting field for Xd − a for anyd dividing n and any element a ∈ K has the form k(α) for any given single root α ofXd − a. In particular, if n = dk and if a ∈ k, then Xd − a and Xn − ak have the samesplitting field over k.

Thus, by induction, we may assume that K has degree n over k. Let σ be a generatorof the Galois group of K over k and let ζ ∈ k be a primitive n-th root of unity. Then thereis a homomorphism f : Gal(K/k) → k× such that f(σ) = ζ−1. By Proposition 11.9.4,there is an element α ∈ K such that f(σ) = α/σ(α), and hence σ(α) = ζ · α.

Thus, the orbit of α under the action of the Galois group is {ζkα | 0 ≤ k < n}, andhence the minimal polynomial of α over k is

g(X) =n−1∏k=0

(X − ζkα)

by Lemma 11.5.2. The constant term of g(X) is, up to sign, a power of ζ times αn. Sinceζ ∈ k, this implies that αn ∈ k. But then Xn − αn has exactly the same roots in K asg(X) does, and hence g(X) = Xn − αn.

Since the minimal polynomial of α over k has degree n, k(α) = K, and the resultfollows.

We now consider the case of cyclic extensions whose degree is divisible by the char-acteristic of k. We shall content ourselves with the study of the extensions whose degreeis equal to the characteristic of k, as this will be sufficient for a description of solvableextensions.

We will need a little preparation first.

Definition 11.9.6. Let K be a finite Galois extension of k and let G = Gal(K/k). Letα ∈ K. Then

TK/k(α) =∑σ∈G

σ(α).

The function TK/k is a kind of a trace. We shall explore traces more generally inSection 11.14.

Note that since we’re summing over all σ ∈ G, TK/k(α) is fixed by every element ofG. Thus, TK/k(α) ∈ k. Note that as a linear combination of automorphisms of K, TK/k

cannot be identically 0, by Corollary 11.9.3. The fact that elements of Gal(K/k) fix know gives the next lemma.

Lemma 11.9.7. Let K be a finite Galois extension of k. Then TK/k : K → k is anontrivial homomorphism of k-vector spaces.

With this, we can prove the following special case of an additive form of what’s knownas Hilbert’s Theorem 90. Note that every Galois extension of prime degree is cyclic.

Proposition 11.9.8. Let k be a field of characteristic p �= 0 and let K be a Galoisextension of k of degree p. Let σ be a generator of Gal(K/k). Then there is an α ∈ Ksuch that α− σ(α) = 1.

Proof Since TK/k : K → k is a nontrivial homomorphism of k-vector spaces, there isan element β ∈ K such that TK/k(β) = 1. Let

α = β + 2σ(β) + 3σ2(β) + · · ·+ (p− 1)σp−2(β).


Then

α− σ(α) = β + σ(β) + · · ·+ σp−2(β)− (p− 1)σp−1(β).

Since −(p− 1) = 1, this is precisely TK/k(β) = 1, as required.

This sets up a result of Artin and Schreier.

Proposition 11.9.9. Let k be a field of characteristic p �= 0. Then any Galois extensionof degree p over k is the splitting field of a polynomial of the form Xp−X−a with a ∈ k.

Conversely, if a ∈ k and if Xp −X − a has no roots in k, then its splitting field is aGalois extension of k of degree p.

Proof Let K be a Galois extension of degree p over k and let σ generate the Galoisgroup of K over k. Then Proposition 11.9.8 provides an element α ∈ K such thatα− σ(α) = 1. Thus, σ(α) = α− 1, so α is not in k, but

σ(αp − α) = (σ(α))p − σ(α) = (α− 1)p − (α− 1) = αp − α,

since passage to p-th powers is a homomorphism. Since σ generates the Galois group ofK over k, this implies that αp − α ∈ k.

Thus, α is a root of Xp −X − a, where a = αp − α ∈ k. In particular, the splittingfield of Xp −X − a is an intermediate field between k and K that properly contains k.Since K has degree p over k, this forces the splitting field of Xp − X − a to equal Kitself.

For the converse, let a ∈ k and suppose that f(X) = Xp −X − a has no roots in k.Since the formal derivative of Xp −X − a has no roots, the splitting field, K, of f is aseparable extension of k. Let α ∈ K be a root of f . Then (α+1)p− (α+1) = a, so α+1is also a root of f . By induction, the p distinct roots of f are α, α + 1, . . . , α + (p − 1).Thus, K = k(α).

Let σ be any non-identity element of Gal(K/k). Then σ(α) = α+ r for some r with1 ≤ r ≤ p − 1. But then induction shows that σk(α) = α + kr. Since r generates Zpadditively, all the roots of f are in the orbit of α under the action of the Galois group.Thus, f is the minimal polynomial of α over k by Lemma 11.5.2, and hence K has degreep over k.

Exercises 11.9.10.1. Do X2 − 2 and X4 − 4 have the same splitting field over Q?

2. Let K be a finite Galois extension of k. A function f : Gal(K/k)→ K× is called asolution to Noether’s equations if f(σ) · σ(f(τ)) = f(στ) for all σ, τ ∈ Gal(K/k).

(a) Show that f is a group homomorphism if and only if it takes value in k×.

(b) Show that a function f : Gal(K/k)→ K× is a solution to Noether’s equationsif and only if there is a nonzero element α ∈ K such that f(σ) = α/σ(α) forall σ ∈ Gal(K/k).

11.10 Kummer Theory

Here, we generalize the preceding section to study abelian extensions of exponent n overa field containing a primitive n-th root of unity.


Definition 11.10.1. Let k be a field containing a primitive n-th root of unity. A Kum-mer extension of k is the splitting field of a polynomial of the form

f(X) = (Xn − a1) . . . (Xn − ak),

for elements a1, . . . , ak ∈ k.

One direction of our study is easy.

Proposition 11.10.2. Let k be a field containing a primitive n-th root of unity and letK be a Kummer extension of k. In particular, let K be the splitting field over k of

f(X) = (Xn − a1) . . . (Xn − ak),

where a1, . . . , ak ∈ k. Then K is an abelian extension of k of exponent n.

Proof Let ζ be a primitive n-th root of unity in k and let G = Gal(K/k).Let Ki ⊂ K be the splitting field of Xn − ai over k and let Gi be the Galois group

of Ki over k. Then we have restriction maps ρi : G → Gi, obtained by restricting theaction of G to Ki. These assemble to give a homomorphism

ρ : G→k∏i=1

Gi

whose i-th component function is ρi. By Proposition 11.8.2, each Gi is a cyclic group ofexponent n. Since abelian groups of exponent n are closed under products, it suffices toshow that ρ is injective.

By Corollary 11.5.3, an element of G is determined by its effect on the roots of f .But every root of f(X) is a root of Xn − ai for some i, and hence lies in one of the Ki.Now any element of the kernel of ρ acts trivially on each Ki, and hence acts trivially onthe set of roots of f(X). So it must be the identity element by Corollary 11.5.3.

The converse here is very similar to the argument for the cyclic case. But we shallprove a little more.

First, recall that if A and B are abelian groups, then HomZ(A,B) denotes the set ofgroup homomorphisms from A to B, and that HomZ(A,B) itself has a natural abeliangroup structure given by addition of homomorphisms: If B is written in additive notation,then we have (f + g)(a) = f(a) + g(a) for all f, g ∈ HomZ(A,B) and a ∈ A.

Lemma 11.10.3. Let k be a field containing a primitive n-th root of unity and let G bea finite abelian group of exponent n. Then there is an isomorphism of groups between Gand HomZ(G,k×).

Proof Let ζ be a primitive n-th root of unity in k. Since any element of k× of exponentn is a root of Xn − 1, and since the roots of Xn − 1 in k× are precisely the powers ofζ, the image of any homomorphism from G to k× must lie in 〈ζ〉, the subgroup of k×

generated by ζ. Thus, we obtain an isomorphism

HomZ(G, 〈ζ〉) i∗−→∼= HomZ(G,k×).

Of course, 〈ζ〉 is isomorphic to Zn. For compatibility with earlier chapters, let uswrite G in additive notation. Thus, the Fundamental Theorem of Finite Abelian Groups


gives an isomorphism from G to a direct sum of cyclic groups (which were given to be ofprime power order in the first proof of the Fundamental Theorem) whose order dividesn:

G ∼= Zm1 ⊕ · · · ⊕ Zmk,

where mi divides n for i = 1, . . . , k.Proposition 9.7.7 now gives an isomorphism of groups

HomZ(G,Zn) ∼=k⊕i=1

HomZ(Zmi,Zn),

so the result follows if we show that HomZ(Zm,Zn) ∼= Zm whenever m divides n.Now Proposition 2.5.17 gives a bijection from HomZ(Zm,Zn) to the set of elements

of Zn with exponent m. This bijection takes a homomorphism f : Zm → Zn to f(1),the result of evaluating f at the generator 1. Since Zn is abelian, this bijection is easilyseen to be a group isomorphism from HomZ(Zm,Zn) to the subgroup of Zn consistingof those elements of exponent m.

Since m divides n, the elements of exponent m in Zn are precisely the elements ofthe cyclic subgroup of Zn of order m, so the result follows.

The converse to Proposition 11.10.2 is included in the next proposition.

Proposition 11.10.4. Let k be a field containing a primitive n-th root of unity and letK be an abelian extension of k of exponent n. Then K is a Kummer extension of k.

We may also express the Galois group of K over k as follows. Let G = Gal(K/k)and let r be the smallest positive integer which is an exponent for G. Let A ⊂ K× bethe collection of elements whose r-th powers lie in k. Write Ar and (k×)r for the setsof r-th powers of A and k×, respectively. Then A, Ar, and (k×)r are subgroups of K×,and there are group isomorphisms

G ∼= A/k× ∼= Ar/(k×)r.

Proof This time, we shall apply the classification of finite abelian groups given by theFundamental Theorem of Finitely Generated Modules over a P.I.D. Here, we obtain anisomorphism

η : G∼=−→ Zm1 ⊕ · · · ⊕ Zmk

,

where each mi divides mi−1 for i = 2, . . . , k. Thus, m1 is the smallest positive exponentfor G. By our hypothesis, m1 divides n. We shall write G itself in multiplicative form.

Write σi ∈ G for the element which corresponds under η to the image of the canonicalgenerator of the summand Zmi

under its standard inclusion in the direct sum. (Thus,each σi has order mi, and G is the internal direct product of the subgroups generatedby the σi.) Let ζ be a primitive n-th root of unity in k, and write ζm for the principalm-th root of unity ζn/m for all integers m dividing n.

Let fi : G→ k× be the homomorphism determined by

fi(σj) =

{1 if i �= j

ζ−1mi

if i = j.

Notice that up to sign, fi is the image of σi under the isomorphism fromG to HomZ(G,k×)that is given in Lemma 11.10.3. Thus, HomZ(G,k×) is generated by f1, . . . , fk.


By Proposition 11.9.4, we can find elements α1, . . . , αk ∈ K such that fi(σ) =αi/σ(αi) for all σ ∈ G and all i. Thus,

σi(αj) =

{αj if i �= j

ζmiαi if i = j.

In particular, the orbit of αi under G is {ζjmiαi | 0 ≤ j < mi}. Thus, as shown in

the proof of Proposition 11.9.5, αmii ∈ k, and the minimal polynomial of αi over k is

Xmi − ai, where ai = αmii .

Since k contains a primitive n-th root of unity, k(αi) contains all the roots of Xn −an/mi

i . Thus, k(α1, . . . , αk) is a Kummer extension of k, being the splitting field over kof

f(X) = (Xn − an/m11 ) . . . (Xn − an/mk

k ).

Note that the displayed calculation of σi(αj) shows that no non-identity element ofG = Gal(K/k) acts as the identity on k(α1, . . . , αk). Thus, the Galois group of K overk(α1, . . . , αk) is the trivial group, and hence K = k(α1, . . . , αk). Thus, K is a Kummerextension of k, as claimed.

We now consider the second part of the statement of this proposition. As observedabove, the smallest positive exponent for G is r = m1. Since passage to r-th powers is ahomomorphism in an abelian group with multiplicative notation, the subsets A, Ar, and(k×)r are easily seen to be subgroups of K×. Moreover, passage to r-th powers inducesa surjective homomorphism

μ : A/k× → Ar/(k×)r.

Now let α ∈ A and suppose that the class of α in A/k× lies in the kernel of μ. Thus,αr = ar for some a ∈ k×. But the r-th roots of ar all differ by multiplication by powersof ζr. Since ζr ∈ k, this forces α ∈ k×, and hence μ is an isomorphism.

Thus, it suffices to provide an isomorphism from A/k× to HomZ(G,k×). Let α ∈ A.Then αr ∈ k, and hence α is a root of the polynomial Xr−αr ∈ k[X]. Since the roots ofthis polynomial all differ from α by multiplication by a power of ζr, we see that α/σ(α)must be a power of ζr for all σ ∈ G.

In particular, if τ ∈ G, α/τ(α) is fixed by each element σ ∈ G. Thus,

(α/σ(α)) · (α/τ(α)) = (α/σ(α)) · σ (α/τ(α)) = α/στ(α).

Thus, there is a homomorphism fα : G→ k×, defined by setting fα(σ) = α/σ(α).Since the elements of G are automorphisms of K, the passage from α to fα is easily

seen to give a homomorphism γ : A→ HomZ(G,k×). And α ∈ A lies in the kernel of γif and only if σ(α) = α for all σ ∈ G, so the kernel of γ is precisely k×.

Thus, it suffices to show that γ is onto. But this is precisely the output of Propo-sition 11.9.4. To avoid making an elaboration on an earlier argument, simply considerthe generators fi of HomZ(G,k×) constructed above. We used Proposition 11.9.4 toconstruct elements αi ∈ K such that fi(σ) = αi/σ(αi) for all σ ∈ G. And the αi wereshown above to lie in A, and hence fi = γ(αi).

This shows that, with k, K, and A as in the statement of Proposition 11.10.4, then,with great redundancy, we may express K as the splitting field of {Xr − a | a ∈ Ar}.Thus, K is determined uniquely, up to isomorphism over k, by the subset Ar ⊂ k×. Thisgives one direction of the next corollary.


Corollary 11.10.5. Let k be a field containing a primitive r-th root of unity. Then thereis a one-to-one correspondence between the finite abelian extensions of k whose smallestpositive exponent is r and the subgroups B ⊂ k× such that (k×)r ⊂ B and B/(k×)r is afinite group whose smallest positive exponent is r.

Proof It suffices to show that if B is such a subgroup, and if K is the splitting field of{Xr − b | b ∈ B} over k, then there is a finite collection b1, . . . , bk ∈ B such that K is thesplitting field of {Xr − bi | 1 ≤ i ≤ k}.

To see this, suppose given a finite collection b1, . . . , bs ∈ B, and let K1 ⊂ K bethe splitting field of {Xr − bi | 1 ≤ i ≤ s}. If K1 = K, then we’re done. Otherwise,let B1 be the set of all elements of k that are r-th powers of elements of K×

1 . ThenProposition 11.10.4 shows that the index of (k×)r in B1 is strictly less than the index of(k×)r in B.

In particular, B1 �= B, and hence we can find an element bs+1 that is in B but not inB1. But then the splitting field of {Xr − bi | 1 ≤ i ≤ s+ 1} is strictly larger than K1, sothe result follows by induction on the index of (k×)r in B.

Exercises 11.10.6.1. Find fields k ⊂ K1 ⊂ K such that k contains a primitive n-th root of unity, K1

is a Kummer extension of k of exponent n, K is a Kummer extension of K1 ofexponent n, but Gal(K/k) is not abelian.

2. Let k be a field of characteristic p and let a1, . . . , ak ∈ k. Show that the splittingfield of

f(X) = (Xp −X − a1) . . . (Xp −X − ak)

over k is an abelian extension of exponent p.

3. Let k be a field of characteristic p and let K be a finite abelian extension of k ofexponent p. Show that K is the splitting field of a polynomial of the form

f(X) = (Xp −X − a1) . . . (Xp −X − ak),

with a1, . . . , ak ∈ k. (Hint : Consider the subgroups of Gal(K/k) of index p.)

4. Let k be a field of characteristic p and let K be a finite abelian extension of k ofexponent p. Let G = Gal(K/k), and write ψ : K → K for the homomorphismψ(α) = αp−α. Show that G, HomZ(G,Zp), (ψ−1(k))/k, and (ψ(ψ−1(k)))/(ψ(k))are all isomorphic.

5. Let k be a field of characteristic p and let ψ : k → k be given by ψ(a) = ap − a.Show that there is a one-to-one correspondence between the isomorphism classesover k of finite abelian extensions of k of exponent p and the additive subgroupsof k that contain ψ(k) as a subgroup of finite index.

6. Find extension fields Zp(X) ⊂ k ⊂ K such that k is an abelian extension of expo-nent p over Zp(X), K is an abelian extension of degree p over k, and Gal(K/Zp(X))is nonabelian.


11.11 Solvable Extensions

We can now address the issue of finding finding roots of polynomials as functions of thecoefficients.

The starting point, of course, is the quadratic formula. Over any field k of charac-teristic not equal to 2, the roots of the quadratic aX2 + bX + c have the form

−b−√b2 − 4ac2a

.

Here, we allow√b2 − 4ac to stand for either root of X2 − (b2 − 4ac) in our preferred

choice of splitting field for k. Indeed, in most fields (including C), there is no preferredchoice for a square root, so the quadratic formula should be taken as specifying the twosolutions in no particular order.

Now, we consider the cubic.

Example 11.11.1. Let k be a field whose characteristic is not equal to either 2 or 3.Consider the cubic g(X) = X3 + bX2 + cX + d. If b �= 0, we “complete the cube” bymaking the substitution f(X) = g(X−(b/3)). This gives us a cubic that does not involveX2, which we write as

f(X) = X3 + pX + q.

The solution to a cubic in this form is originally due to Cardano. Fix an algebraic closure,L, of k, in which all roots are to be taken, and let

Δ = −4p3 − 27q2.

As shown in in Problem 5 of Exercises 7.4.14, this is the discriminant of the cubic f(X).Choose a preferred square root,

√−3Δ, of −3Δ. We shall choose cube roots

α =3

√−q

2−√−3Δ

18and β =

3

√−q

2+√−3Δ

18

with some care. First, as the reader may verify, note that for any choices of α and β wehave (3αβ)3 = −p3.

Now if ζ is a primitive third root of unity in L, the roots α and β are determined upto multiplication by a power of ζ. In particular, for any choice of α, there is a uniquechoice of β for which 3αβ = −p. Having thus chosen β, we set γ = α+ β. Then

γ3 = α3 + β3 + 3αβ(α+ β)= −q − pγ,

and hence γ is a root of f . Note that the relationship between α and β gives

γ =3

√−q

2−√−3Δ

18− p

3

⎛⎝ 3

√−q

2−√−3Δ

18

⎞⎠−1

.

Here, if we start by choosing a square root of −3Δ, then the three different choices fora cube root of −q/2−√−3Δ/18 will all give roots of f via this formula.


It’s not hard to see that the three roots thus constructed are all distinct unless thediscriminant Δ = 0. But Corollary 7.4.13 shows that Δ(f) = 0 if and only if f hasrepeated roots. We leave it to the reader to verify that in this degenerate case, that theprocedure above produces all the roots of f in their correct multiplicities.

Of course, the roots of the original polynomial g(X) may be obtained from those off by subtracting b/3.

The procedure for solving a quartic is more complicated, but the result is that theroots may be expressed as algebraic functions of the coefficients.

Definition 11.11.2. An algebraic function of several variables is one obtained by com-posing the various operations of sums, differences, products, powers, quotients, and ex-traction of n-th roots.

Note that the only coefficients introduced by these operations will be integers. Weshall consider algebraic functions with more general coefficients in the next section.

Due to the presence of n-th roots, of course, algebraic “functions” are not necessarilywell defined as functions: Choices are made every time a root is extracted. But this isthe best we can ask for, as far as solving for the roots of polynomials over arbitrary fieldsis concerned.

Given that polynomials of degree ≤ 4 admit general solutions as algebraic functions(in characteristics not equal to 2 or 3), the thing that’s truly striking is that in degrees 5and above there exist polynomials over Z whose roots cannot individually be expressedas algebraic functions of the coefficients. So not only is there no general formula forsolving polynomials in degree ≥ 5, there are even examples (many of them, it turns out)which do not admit an individual formula of the desired sort for their solution.

It was Galois who first discovered this, by developing a theory of solvable extensions.An extension is solvable, of course, if it is a Galois extension whose Galois group is asolvable group. A concept more closely related to solving polynomials is that of solvabilityby radicals.

Definition 11.11.3. An extension k1 of k is said to be solvable by radicals if there isan extension K of k1 and a sequence of extensions

k = K0 ⊂ K1 ⊂ · · · ⊂ Km = K

such that each Ki is obtained from Ki−1 by adjoining a root of a polynomial Xni − aifor some ni > 0 and some ai ∈ Ki−1.

The next lemma is now clear.

Lemma 11.11.4. The splitting field of a polynomial f(X) ∈ k[X] is solvable by radicalsif and only if the roots of f may be expressed as algebraic functions on some collectionof elements in k.

In particular, if the roots of a polynomial f(X) ∈ k[X] may be written as algebraicfunctions of its coefficients, then the splitting field of f over k is solvable by radicals.

In characteristic 0, we shall see that a finite Galois extension is solvable by radicalsif and only if it is a solvable extension. In characteristic p �= 0, the situation is morecomplicated: A Galois extension of degree p is certainly solvable, but not necessarilyby radicals. Here, we shall relate solvable extensions to a different sequential solvabilitycriterion, which agrees with solvability by radicals in characteristic 0. We shall continuethe discussion of solvability by radicals in characteristic p in Exercises 11.11.10.


Definition 11.11.5. We say that the extension k1 of k is solvable by cyclic extensionsif there is an extension K of k1 and a sequence of extensions

k = K0 ⊂ K1 ⊂ · · · ⊂ Km = K

such that each extension Ki−1 ⊂ Ki has one of the following forms:

1. Ki is obtained from Ki−1 by adjoining a root of a polynomial of the form Xni−ai,where ni is not divisible by the characteristic of k and ai ∈ Ki−1.

2. Ki is obtained from Ki−1 by adjoining a root of a polynomial of the formXp−X−aifor some ai ∈ Ki−1. Here, p is the characteristic of k.

This concept is indeed identical to solvability by radicals when the fields have char-acteristic 0.

Lemma 11.11.6. Solvable extensions have the following closure properties.

1. Let K be a finite Galois extension of k. If K is solvable over k, then it is solvableover any intermediate field, and if K1 is an intermediate field that is normal overk, then Gal(K1/k) must be solvable as well.

Conversely, if K1 is an intermediate field such that K1 is solvable over k and Kis solvable over K1, then K must be solvable over k, as well.

2. Let K1 and K2 be finite solvable extensions of k and let L be a compositum of K1

and K2 over k. Then L is a solvable extension of k.

Proof The first assertion is immediate from the Fundamental Theorem of Galois Theoryand Corollary 5.5.10. For the second assertion, write Ki = k(αi) and let fi(X) be theminimal polynomial of αi over k. Then L = k(α1, α2) is the splitting field of f1(X)f2(X)over k, and hence is Galois over k. The restriction maps ρi : Gal(L/k) → Gal(Ki/k)provide a homomorphism

Gal(L/k)(ρ1,ρ2)−−−−→ Gal(K1/k)×Gal(K2/k).

Since an element of Gal(L/k) is determined by its effect on α1 and α2, (ρ1, ρ2) is injective.The result follows from Corollary 5.5.10.

The next proposition should come as no surprise.

Proposition 11.11.7. A finite Galois extension k1 of k is solvable by cyclic extensionsif and only if it is a solvable extension.

Proof Suppose that k1 is a finite solvable extension of k. Let n be the product of allthe prime divisors of [k1 : k] not equal to the characteristic of k, and let ζ be a primitiven-th root of unity in the algebraic closure of k1. Then k1 is contained in k1(ζ), and wehave inclusions

k ⊂ k(ζ) ⊂ k1(ζ).

Since k(ζ) is one of the permissible types of extension over k in the definition of solvabilityby cyclic extensions, it suffices to show that k1(ζ) is solvable by cyclic extensions overk(ζ).


Now k1(ζ) is a compositum of k1 and k(ζ) over k. By Problem 5 of Exercises 11.5.22,k1(ζ) is a Galois extension of k(ζ) and Gal(k1(ζ)/k(ζ)) is isomorphic to a subgroup ofGal(k1/k). By Corollary 5.5.7, this makes G = Gal(k1(ζ)/k(ζ)) a solvable group. Also,k(ζ) has a primitive q-th root of unity for every prime q that divides |G| and is not equalto the characteristic of k.

Write k1(ζ) = K. Since G is solvable, Proposition 5.5.12 provides a subnormal series

1 = Gm � Gm−1 � . . . � G0 = G,

where the successive quotient groups Gi−1/Gi are cyclic groups of prime order.Set Ki = KGi for i = 0, . . . ,m. Applying the Fundamental Theorem of Galois The-

ory to the extensions Ki−1 ⊂ Ki ⊂ K, we see that Ki is a Galois extension of Ki−1, withGalois group isomorphic to Gi−1/Gi, a cyclic group of prime order, q. The extensionKi−1 ⊂ Ki is now easily seen to fall into one of the two types in Definition 11.11.5,using Proposition 11.9.5 when q is not equal to the characteristic of k, and using Propo-sition 11.9.9 otherwise. Thus, K = k1(ζ) is solvable by cyclic extensions over K0 = k(ζ),as claimed.

Conversely, suppose k1 is solvable by cyclic extensions over k, via a sequence

k = K0 ⊂ K1 ⊂ · · · ⊂ Km = K.

Here, Ki = Ki−1(αi), where αi ∈ Ki is a root of a polynomial fi(X) ∈ Ki−1[X], andfi(X) may be written either as Xni − ai or as Xp −X − ai, depending on whether theextension Ki−1 ⊂ Ki satisfies the first or the second condition of Definition 11.11.5.

Since k1 ⊂ K, Lemma 11.11.6 shows that it suffices to show that K is contained in asolvable extension of k.

We first claim that each Ki is contained in a solvable extension of Ki−1: If fi(X) =Xp−X−ai, this is immediate from Proposition 11.9.9. Otherwise let K′

i be the splittingfield of fi over Ki−1. Then Proposition 11.8.3 shows that Gal(K′

i/Ki−1) is an extensionof an abelian group by an abelian group, and hence is solvable by Corollary 5.5.10.

Thus, it suffices to show that if K0 ⊂ K1 ⊂ K2 such that K1 is contained in a finitesolvable extension of K0 and K2 is contained in a finite solvable extension of K1, thenK2 is contained in a finite solvable extension of K0.

Of course, we may assume that K2 is solvable over K1. Let L1 be a finite extensionof K1 which is solvable over K0 and let L2 be the compositum of K2 and L1 over K1 inan algebraic closure L of K1. By Problem 5 of Exercises 11.5.22, L2 is Galois over L1,and Gal(L2/L1) is isomorphic to a subgroup of Gal(K2/K1), and hence is solvable.

Let L2 be the normal closure of L2 in L over K0. Then Gal(L2/K0) is an extensionof Gal(L2/L1) by Gal(L1/K0), so it suffices to show that Gal(L2/L1) is solvable.

Write L2 = L1(α) and let σ1, . . . , σk be the embeddings of L2 in L over K0. SinceL1 is normal over K0, σi(L1) = L1 for all i, and hence σi(L2) = L1(σi(α)). ThusL2 = L1(σ1(α), . . . , σk(α)), is the compositum of σ1(L2), . . . , σk(L2) over L1. Now σi(L2)is a solvable extension of σi(L1) = L1. Thus, L2 is a compositum of finite solvableextensions of L1, and hence is solvable by Lemma 11.11.6.

We shall now show that there are splitting fields over Q of polynomials of degree 5and higher that are not solvable by radicals.

Lemma 11.11.8. Let k be any subfield of R and let f(X) be an irreducible polynomialof degree 5 over k with three real roots and two non-real complex roots. Let K be thesplitting field of f over k. Then Gal(K/k) = S5, and hence K is not a solvable extensionof k.


Proof Let α be a root of f in K. Then f is its minimal polynomial, so k(α) has degree5 over k. Thus, [K : k] is divisible by 5.

We shall identify G = Gal(K/k) with its image in the permutation group of the set ofroots of f . Since the only elements of order 5 in S5 are 5-cycles, we see that G containsa 5-cycle, σ.

Since complex conjugation restricts to the identity on k ⊂ R, it must restrict to anautomorphism of the normal extension K of k. Since conjugation fixes the three realroots of f and must move the two non-real roots, it gives a transposition, τ , as an elementof G ⊂ S5.

Let us now number the roots of f to get an explicit isomorphism from the permutationgroup on the roots of f to S5. First, let’s write τ = (1 2). Next, note that since the orbitunder 〈σ〉 of any one root is the full set of roots of f , there is a power, say σk, of σ suchthat σk(1) = 2. Now, continue the numbering of the roots so that σk = (1 2 3 4 5). Thenconjugating τ by the powers of σk, we see that (2 3), (3 4) and (4 5) lie in G. Since anytransposition may be written as a composite of transpositions of adjacent numbers, Gmust contain all the transpositions, and hence all of S5.

Of course, A5, as a nonabelian simple group (Theorem 4.2.7), is not solvable (Lemma 5.5.3).Since subgroups of solvable groups are solvable, K is not solvable over k.

One of the interesting things about this study of solvability by radicals is that we canuse the information about finite extensions of Q to deduce information about solutionsof polynomials over algebraically closed fields, such as C.

Theorem 11.11.9. Let k be any field of characteristic 0 and let k ≥ 5. Then there isno general formula that gives the roots of every polynomial of degree k over k as algebraicfunctions of its coefficients.

Proof It is sufficient to show that there is a polynomial of degree k over Q ⊂ K whoseroots cannot be expressed as algebraic functions of its coefficients. Note that if f is apolynomial of degree 5 over Q with this property, then f(X) · (X − 1)k−5 gives such apolynomial in degree k. Thus, it suffices to show that there is a polynomial of degree 5over Q whose splitting field is not a solvable extension of Q.

By Lemma 11.11.8, it suffices to find an irreducible polynomial of degree 5 over Qwith exactly 3 real roots. We claim that f(X) = X5 − 4X + 2 is such a polynomial.Clearly, f is irreducible by Eisenstein’s criterion. We shall use elementary calculus toshow that it has the desired behavior with regard to roots.

Note that the derivative f ′(X) = 5X4−4 has two roots, at± 4√

4/5. Between these tworoots, f ′(X) is negative, and outside them, it is positive. Thus, f is strictly increasingon both (−∞,− 4

√4/5] and [ 4

√4/5,∞), and is strictly decreasing on [− 4

√4/5, 4

√4/5].

Clearly, f(−2) is negative and f(2) is positive, so it suffices by the Intermediate ValueTheorem to show that f(− 4

√4/5) is positive and f( 4

√4/5) is negative. But this follows

easily from the fact that 4/5 ≤ 4√

4/5 ≤ 1.

Exercises 11.11.10.1. Let p be a prime and let f be a polynomial of degree p over Q with exactly p− 2

real roots. Show that the Galois group of the splitting field of f over Q is Sp.

2. Let k be an algebraic extension of Zp. Show that every finite extension of k is bothsolvable and solvable by radicals.

3. Let K be a finite solvable extension of k in characteristic p �= 0, such that [K : k]is relatively prime to p. Show that K is solvable by radicals over k.


† 4. We can recover an analogue of Theorem 11.11.9 for some infinite fields in char-acteristic p, provided that we first generalize the Fundamental Theorem of GaloisTheory to the case of non-separable extensions. Here, if K is a normal extensionof k, we write autk(K) for the group of automorphisms (as a field) of K whichrestrict to the identity on k.

(a) Let K be a normal extension of k and let K1 be the separable closure of k inK. (See Problem 2 of Exercises 11.4.20.) Show that K1 is a Galois extensionof k and that restriction of automorphisms to K1 gives an isomorphism ofgroups,

ρ : autk(K)∼=−→ Gal(K1/k).

(b) Suppose given extensions k ⊂ K1 ⊂ K2, such that both K1 and K2 arenormal over k. Show that restriction of automorphisms from K2 to K1 givesa well defined restriction homomorphism ρ : autk(K2)→ autk(K1), inducinga short exact sequence

1→ autK1(K2)⊂−→ autk(K2)

ρ−→ autk(K1)→ 1.

(c) Let n = prs, where (p, s) = 1 and let a ∈ k. Let K be the splitting field ofXn− a over k. Show that the separable closure of k in K is the splitting fieldof Xs − a over k.

(d) Let K be a normal extension of k which is solvable by radicals. Show thatautk(K) is a solvable group.

We shall see below that there is a polynomial over the field of rational functionsZp(X1, . . . , Xn) whose splitting field has Galois group Sn. Thus, polynomials ofany degree ≥ 5 over a field whose transcendence degree over Zp is ≥ 5 do not admita general formula to solve for their roots as algebraic functions of their coefficients.

11.12 The General Equation

We’ve seen that there can be no general formula for finding the roots of the polynomialsof a given degree ≥ 5 as algebraic functions of the coefficients.

Recall that the algebraic functions we’ve been considering can be thought of as havinginteger coefficients. The question can be generalized as follows.

Definition 11.12.1. Let k be a field. Then a k-algebraic function is a function ofseveral variables, defined on an extension of k, obtained by composing the variousoperations of sums, differences, products, powers, quotients, extraction of n-th roots,and addition or multiplication by arbitrary elements (to be thought of as constants)of k.

Thus, a k-algebraic function is an algebraic function with coefficients in k.In a field of characteristic 0, Q-algebraic functions coincide with our earlier definition

of algebraic functions with integer coefficients. Similarly, in characteristic p, the Zp-algebraic functions coincide with the algebraic functions with integer coefficients. Thus,integer coefficients are the most general kind one could ask about for a formula thatwould apply to every field of a given characteristic.


But what about finding roots of polynomials over extensions of R? Every polynomialover R itself is solvable by radicals, because the roots lie in the radical extension C of R.So, abstractly, the roots are algebraic functions on elements of R. But this abstract facttells us absolutely nothing about the way in which the roots depend on the coefficientsof the polynomial.

So we might ask if there is a general solution for polynomials over any extensionfield of R, where the roots are given as R-algebraic functions of the coefficients of thepolynomial.

Here, we cannot point to particular polynomials over R itself for counterexamples,as the splitting fields are always solvable by radicals. The counterexamples will be givenby polynomials over extension fields of R.

Nor is there anything special about R here. We may work over an arbitrary field k.For any field k, the symmetric group Sn acts on the field of rational functions

k(X1, . . . , Xn) by permuting the variables. Write Si for the i-th elementary functionsi(X1, . . . , Xn) of the variables X1, . . . , X1. We’ve already seen (Proposition 7.4.7) thatthe fixed points of the restriction of the above action of Sn to the subring k[X1, . . . , Xn]is a polynomial algebra with generators S1, . . . , Sn. Thus, the field of rational func-tions k(S1, . . . , Sn) is contained in the fixed field (k(X1, . . . , Xn))Sn . We shall show thatthe fixed field is equal to k(S1, . . . , Sn), which then shows, via Theorem 11.5.17, thatk(X1, . . . , Xn) is a Galois extension of k(S1, . . . , Sn), with Galois group Sn.

The relevance of this to solving polynomials comes from Lemma 7.4.4, which showsthat in k(X1, . . . , Xn)[X], we have the factorization

Xn − S1Xn−1 + S2X

n−2 + · · ·+ (−1)nSn = (X −X1) . . . (X −Xn).

Thus, X1, . . . , Xn are the roots of the separable polynomial

fn(X) = Xn − S1Xn−1 + S2X

n−2 + · · ·+ (−1)nSn,

and hence k(X1, . . . , Xn) is the splitting field for fn(X) over k(S1, . . . , Sn). Clearly, thisextension is solvable by radicals if and only if there is a formula expressing the rootsX1, . . . , Xn as k-algebraic functions of S1, . . . , Sn.

For n ≤ 4, where such formulæ exist, they turn out to be universal, away fromcharacteristics ≤ n, though that is not implied by the setup here: There cannot be adirect comparison of extensions between the one studied here and the splitting field ofan arbitrary polynomial.

Nevertheless, if there is a universal formula for finding the roots as k-algebraic func-tions of the coefficients of the polynomial, then it will show up in the study of thisexample.

For n ≥ 5, we do not even need to verify that k(S1, . . . , Sn) is equal to the fixedfield of the Sn action to draw conclusions regarding solvability. Since Sn is the Galoisgroup of k(X1, . . . , Xn) over the fixed field of this action (Theorem 11.5.17), and sincek(S1, . . . , Sn) is a subfield of this fixed field, Sn must be a subgroup of the Galois groupof k(X1, . . . , Xn) over k(S1, . . . , Sn), and hence the Galois group cannot be solvable whenn ≥ 5.

Recall from Problem 4 of Exercises 11.11.10 that even in characteristic p �= 0, aGalois extension which is solvable by radicals must be solvable. Therefore, no solutionby radicals for the above polynomial fn(X) is possible for n ≥ 5.

Note that since S1, . . . , Sn are algebraically independent, k(S1, . . . , Sn) is isomorphicto k(X1, . . . , Xn). This shows that any field of transcendence degree ≥ 5 over Zp admitspolynomials of degree ≥ 5 whose roots cannot be written as k-algebraic functions of thecoefficients of the polynomial. We also obtain the next proposition.


Proposition 11.12.2. Let k be any field and let n ≥ 5. Then there is no general formulathat expresses the roots of polynomials over extension fields of k as k-algebraic functionsof the coefficients of the polynomials.

Proposition 11.12.3. Let k be any field, and let Sn act on the field of rational func-tions k(X1, . . . , Xn) by permuting the variables. Then the fixed field of this actionis k(S1, . . . , Sn), where Si is the i-th elementary symmetric function in the variablesX1, . . . , Xn. Thus, k(X1, . . . , Xn) is a Galois extension of k(S1, . . . , Sn) with Galoisgroup Sn.

Proof As noted above, since X1, . . . , Xn are the roots of the separable polynomialfn(X), k(X1, . . . , Xn) is the splitting field of fn(X) over k(S1, . . . , Sn). Thus, if G isthe Galois group of k(X1, . . . , Xn) over k(S1, . . . , Sn), then restricting the action of G to{X1, . . . , Xn} gives in embedding of G in the group of permutations on {X1, . . . , Xn}.

Since the action of Sn on k(X1, . . . , Xn) fixes k(S1, . . . , Sn), Sn embeds in G, re-specting the usual action of Sn on the variables. Thus, the inclusion of Sn in G is anisomorphism. The Fundamental Theorem of Galois Theory now shows that k(S1, . . . , Sn)must be the fixed field of the Galois group Sn.

11.13 Normal Bases

Definition 11.13.1. Let K be a finite Galois extension of k. Then a normal basis forK over k is an orbit G · α of the action of G = Gal(K/k) on K with the property thatthe elements of G · α form a basis for K as a vector space over k.

Thus, if G = {σ1, . . . , σn}, then a normal basis for K over k is a k-basis for K of theform σ1(α), . . . , σn(α) for some α ∈ K.

We shall show here that every finite Galois extension of an infinite field has a normalbasis.

So far, what we know is that every finite Galois extension has a primitive element. Itis easy to see that an element α ∈ K is a primitive element for K over k if and only if theorbit of α under the action of G = Gal(K/k) has |G| elements, and hence is isomorphicto G as a G-set.

Thus, if the orbit of α is a normal basis for K over k, then α is a primitive element.But the converse of this is false. For instance, i,−i fails to be a normal basis for Q(i)over Q.

Clearly, it will be useful to derive a condition under which a collection of elementsα1, . . . , αn will be a basis for K over k. We generalize things slightly.

Lemma 11.13.2. Let K be a finite separable extension of k and let L be the normalclosure of K over k in an algebraic closure of K. Let σ1, . . . , σn be the distinct embeddingsof K over k into L, and let α1, . . . , αn ∈ K. Then α1, . . . , αn is a basis for K as a vectorspace over k if and only if the matrix (σi(αj)) ∈Mn(L) is invertible.

Proof Since any embedding, σ, of K over k into the algebraic closure of K extendsto an embedding of the normal extension L, the image σ(K) must lie in L. Thus, byProposition 11.4.12, n, which is the number of distinct embeddings of K in L over k, isequal to the degree of K over k.

Let M = (σi(αj)). If M is not invertible, then its rows are linearly dependent overL. So we can find β1, . . . , βn ∈ L such that

β1σ1(αj) + · · ·+ βnσn(αj) = 0


for all j. Thus, the linear combination∑ni=1 βiσi vanishes on the k-linear subspace of

K generated by α1, . . . , αn. But Corollary 11.9.3 shows that∑ni=1 βiσi cannot be the

trivial homomorphism, so α1, . . . , αn cannot generate K as a vector space over k, andhence cannot form a basis for K over k.

Conversely, if α1, . . . , αn are not linearly independent, we can find coefficients a1, . . . , an ∈k such that

∑ni=1 aiαi = 0. But then

∑nj=1 ajσi(αj) = 0 for all i, and hence the columns

of M are not linearly independent.

Lemma 11.13.2, which makes use of Corollary 11.9.3 in its proof, may be used toprove the following proposition, which generalizes Corollary 11.9.3 in the cases of greatestinterest. We shall make use of Proposition 7.3.26, which shows that if K is an extensionof the infinite field k, then any polynomial f(X1, . . . , Xn) over K which vanishes on everyn-tuple (a1, . . . , an) ∈ kn must be the 0 polynomial.

Proposition 11.13.3. (Algebraic Independence of Characters) Let K be a finite sepa-rable extension of the infinite field k, and let L1 be the normal closure of K over k in analgebraic closure of K.

Let σ1, . . . , σn be the distinct embeddings of K into L1 over k. Then σ1, . . . , σnare algebraically independent over any extension field L of L1, in the sense that iff(X1, . . . , Xn) is a polynomial over L such that

f(σ1(α), . . . , σn(α)) = 0

for all α ∈ K, then f = 0.

Proof Let α1, . . . , αn be a basis for K as a vector space over k and consider the matrixM = (σi(αj)). Then Lemma 11.13.2 shows M to be invertible.

Write fM : Ln → Ln for the linear transformation induced by the matrix M . Then ifwe use the basis α1, . . . , αn to identify K with kn, we see that the function which takesα ∈ kn to f(σ1(α), . . . , σn(α)) is given by the restriction to kn of the composite f ◦ fM .Thus, our assumption on f is that f ◦ fM restricts to 0 on kn ⊂ Ln. But since fM islinear, its component functions are polynomials, so f ◦ fM = 0 by Proposition 7.3.26.But M is invertible, so f = f ◦ fM ◦ fM−1 = 0, as claimed.

Finally, we can prove the Normal Basis Theorem.

Theorem 11.13.4. (Normal Basis Theorem) Let K be a finite Galois extension of theinfinite field k. Then K has a normal basis over k.

Proof Let σ1, . . . , σn be the distinct elements of G = Gal(K/k), with σ1 equal tothe identity element. For each σ ∈ G, define a variable Xσ. To get an ordering of thevariables, identify Xσi

= Xi.Consider the matrix M over K[X1, . . . , Xn] whose ij-th entry is Xσiσ

−1j

. We claimthat M has nonzero determinant. To see this, make the substitution Xσ1 = 1 andXσi = 0 for i > 1. Under this substitution, M evaluates to the identity matrix. By thenaturality of determinants, detM must be nonzero.

The determinant, detM , lies in the ground ring K[X1, . . . , Xn], so we may write

detM = f(X1, . . . , Xn),

where f(X1, . . . , Xn) is a nonzero polynomial in n variables. By Proposition 11.13.3,there is an α ∈ K such that

f(σ1(α), . . . , σn(α)) �= 0.


By naturality of determinants, f(σ1(α), . . . , σn(α)) is the determinant of the matrixobtained by substituting σi(α) for each occurrence of Xσi in the matrix M . Thus,

f(σ1(α), . . . , σn(α)) = det(σiσ

−1j (α)

).

By Lemma 11.13.2, σ−11 (α), . . . , σ−1

n (α) forms a basis for K over k, which is clearly anormal basis, as desired.

Exercises 11.13.5.

1. Find a normal basis for Q(ζ8) over Q.

2. Let K be a finite Galois extension of k and write G = Gal(K/k). Show that theaction of G on K defines a k[G]-module structure on K that extends the usualk-module structure. Show that a normal basis of K over k induces an isomorphismof k[G]-modules from k[G] to K.

11.14 Norms and Traces

Let K be a finite extension of k and let α ∈ K. Write μα : K → K for the transformationinduced by multiplication by α: μα(β) = αβ. Note that μα is transformation of vectorspaces over K, and hence over k as well. We obtain a ring homomorphism,

μ : K→ Endk(K)

by setting μ(α) = μα.The transformation μα contains quite a bit of information about α over k. For in-

stance, if f(X) is a polynomial over k, then, as elements of Endk(K), we have μf(α) =f(μα). In particular, the minimal polynomial of α over k is equal to the minimal poly-nomial of μα as a k-linear transformation of K.

The easiest things to study about μα are its trace and determinant.

Definitions 11.14.1. The norm of K over k, denoted by NK/k, is given by the com-posite

Kμ−→ Endk(K) det−−→ k.

The trace of K over k, denoted by trK/k, is given by the composite

Kμ−→ Endk(K) tr−→ k.

Here, det and tr are defined as for any finite dimensional vector space over k. Thestandard properties of determinants and traces give us the following lemma.

Lemma 11.14.2. Let K be a finite extension of k. Then trK/k : K→ k is a homomor-phism of vector spaces over k, while NK/k : K→ k is a monoid homomorphism with theproperty that NK/k(α) = 0 if and only if α = 0.

If a ∈ k, then trK/k(a) = [K : k] · a and NK/k(a) = a[K:k].

It can be useful to express the norm and trace in Galois-theoretic terms.


Proposition 11.14.3. Let K be a finite extension of k. Let L be an algebraic closure ofK and let σ1, . . . , σk be the collection of all distinct embeddings of K in L over k. Thenfor α ∈ K we have

NK/k(α) = (σ1(α) . . . σk(α))[K:k]i andtrK/k(α) = [K : k]i · (σ1(α) + · · ·+ σk(α)) ,

where [K : k]i is the inseparability degree of K over k.In particular, if K is separable over k, then

NK/k(α) = σ1(α) . . . σk(α) andtrK/k(α) = σ1(α) + · · ·+ σk(α).

Proof Let f(X) = Xm + am−1Xm−1 + · · · + a0 be the minimal polynomial of α

over k. Then 1, α, . . . , αm−1 is a basis for k(α) over k. Let β1, . . . , βr be any basisfor K over k(α). Then the matrix for multiplication by α with respect to the basisβ1, β1α, . . . , β1α

m−1, . . . , βr, βrα, . . . , βrαm−1 is the block sum of r = [K : k(α)] copies

of the rational companion matrix, C(f), of f(X):

C(f) =

⎛⎜⎜⎜⎜⎜⎝0 0 . . . 0 −a0

1 0 . . . 0 −a1

. . .0 0 . . . 0 −am−2

0 0 . . . 1 −am−1

⎞⎟⎟⎟⎟⎟⎠ .

It is easy to see, using the decomposition of the determinant in terms of permuta-tions (or via Proposition 10.6.7 and Problem 2 of Exercises 10.4.11), that det(C(f)) =(−1)ma0. Thus,

NK/k(α) = ((−1)ma0)[K:k(α)]

.

Even more simply, we have

trK/k(α) = [K : k(α)] · (−am−1).

In Proposition 11.4.2, it is shown that

f(X) = (X − α1)[k(α):k]i . . . (X − αs)[k(α):k]i

in L[X], where α1, . . . , αs are all distinct, and s = [k(α) : k]s. Thus, there are s distinctembeddings, τ1, . . . , τs, of k(α) in L, characterized by τi(α) = αi. Thus,

am−1 = −[k(α) : k]i (τ1(α) + · · ·+ τs(α)) and

a0 = (−1)m (τ1(α) . . . τs(α))[k(α):k]i .

By Lemma 11.4.10, each τi has [K : k(α)]s distinct extensions to embeddings of K inL. Thus,

σ1(α) . . . σk(α) = (τ1(α) . . . τs(α))[K:k(α)]s andσ1(α) + · · ·+ σk(α) = [K : k(α)]s · (τ1(α) + · · ·+ τs(α)) .

The result now follows from the fact that [K : k]i = [K : k(α)]i · [k(α) : k]i.


The norm and trace satisfy an important transitivity property, a special case of whichappeared in the preceding proof.

Proposition 11.14.4. Let K be a finite extension of k and let L be a finite extensionof K. Then

NL/k = NK/k ◦NL/K andtrL/k = trK/k ◦ trL/K.

More generally, if V is a finite dimensional vector space over K and if f ∈ EndK(V ),then

detk(f) = NK/k(detK(f)) andtrk(f) = trK/k(trK(f)).

Here, for F = K or k, detF (f) and trF (f) stand for the determinant and trace, respec-tively, of f when regarded as a transformation of a vector space over F .

Proof Choose a basis v1, . . . , vn for V over K and a basis α1, . . . , αk for K over k. Forf ∈ EndK(V ), let M = (βij) be the matrix of f as a K-linear mapping, with respect tothe basis v1, . . . , vn.

We may also consider the matrix of f as a k-linear mapping with respect to the k-basis α1v1, . . . , αkv1, . . . , α1vn, . . . , αkvn. The matrix of f with respect to this basis is ablock matrix, where the blocks are the matrices of the k-linear transformations μβij

of K,with respect to the basis α1, . . . , αk. Thus, if φ : K→Mk(k) is the ring homomorphismthat carries α ∈ K to the matrix of μα with respect to α1, . . . , αk, then the matrix of fover k is the block matrix M = (φ(βij)).

Note that trK/k(α) = trk(φ(α)) and NK/k(α) = detk(φ(α)) for all α ∈ K. Thus, itsuffices to show that trk(M) = trk(φ(trK(M))), and detk(M) = detk(φ(detK(M))).

For traces, this is obvious: The trace of M is the sum of the traces of the diagonalblocks φ(β11), . . . , φ(βnn).

For determinants, note that the passage from M = (βij) to M = (φ(βij)) gives aring homomorphism from Mn(K) to Mnk(k). Clearly, M is invertible if and only if Mis invertible. In the non-invertible case, the determinants are 0, so it suffices to checkthe result on generators of Gln(K). By Proposition 10.8.9, Gln(K) is generated byelementary matrices and matrices of the form (α)⊕ In−1, with α ∈ K×.

If M is an elementary matrix, then M is either upper or lower triangular with only1’s on the diagonal, so detk(M) = 1. Since detK(M) = 1, the desired equality holds.

In the remaining case, M is diagonal and M is a block sum, and the result is easy tocheck.

Exercises 11.14.5.1. Let K be a finite extension of k and let α ∈ K. Show that the matrix of μα (with

respect to any k-basis of k) is diagonalizable over the algebraic closure of k if andonly if α is separable over k.

2. Let K be a finite inseparable extension of k. Show that trK/k is the trivial homo-morphism.

3. Calculate NC/R : C→ R and trC/R : C→ R explicitly.


4. Give explicit calculations of the norm and trace over Q of an extension of the formQ(√m), where m ∈ Z is not a perfect square.

5. Let K be a finite extension of k and let f(X) be the minimal polynomial of α ∈ Kover k. Let a ∈ k. Show that NK/k(a− α) = (f(a))[K:k(α)].

Chapter 12

Hereditary and SemisimpleRings

So far, we’ve been able to classify the finitely generated modules over P.I.D.s, and, viaGalois theory, to develop some useful tools toward an understanding of the category offields and the classification of particular types of extensions of a given field.

In general, we’d like to be able to classify particular types of rings, and to classify thefinitely generated modules, or even just the finitely generated projective modules, over aparticular ring.

Such results are hard to come by: Ring theory is incredibly rich and varied. Here, weshall look at generalizations of fields and P.I.D.s and obtain a few classification theorems.

As a generalization of the concept of field, we shall study the semisimple rings: thoserings with the property that every ideal of the ring is a direct summand of it. We shalldevelop a structure theory that will allow for a classification of these rings, modulo aclassification of division rings.

We shall also develop methods for classifying the modules over a semisimple ring.Given Maschke’s Theorem (Section 12.1) these give some of the fundamental techniquesin the study of the representation theory of finite groups.

The class of rings which simultaneously generalizes both P.I.D.s and semisimple ringsis the class of hereditary rings: those with the property that every ideal is a projectivemodule. The hereditary rings are much too broad a class of rings to permit any kindof useful classification results at this stage. Thus, we shall spend most of our timestudying the hereditary integral domains, which, it turns out, are precisely the ringsknown as Dedekind domains (though we shall introduce the latter under a more classicaldefinition).

Dedekind domains first came to light in the study of number fields. Here, a numberfield is a finite extension field of the rational numbers. Number fields have subrings,called their rings of integers, that behave somewhat like the integers do as a subring ofQ.

Rings of integers are fundamental in the study of number theory. In fact, the falseassumption that the ring of integers of a number field is a P.I.D. led to a false proof ofFermat’s Last Theorem. The circle of ideas surrounding this fact led to the first studiesof ideal theory.

The truth of the matter is that the ring of integers of a number field is a Dedekinddomain. As a result, the study of the ideal and module theory of Dedekind domains is

506

CHAPTER 12. HEREDITARY AND SEMISIMPLE RINGS 507

important for an understanding of number theory. In fact, the classification of the finitelygenerated projective modules over these rings of integers is fundamental in solving someof the basic questions in both algebra and topology.

In Section 12.1, we prove Maschke’s Theorem, which states that if G is a finite groupand K is a field whose characteristic does not divide the order of the group, then thegroup algebra K[G] is semisimple. We also characterize the finitely generated projectivemodules over a semisimple or hereditary ring.

In Sections 12.2 and 12.3, we develop the theory of semisimple rings. We show thata semisimple ring is a product of matrix rings over division rings, where there is onefactor in the product for each isomorphism class of simple left A-modules. We alsoshow that every finitely generated left A-module may be written uniquely as a directsum of simple modules. A complete determination of the ring and its finitely generatedmodules depends only on finding representatives for each isomorphism class of simpleleft A-modules.

Applying this to a group algebra K[G] that satisfies the hypothesis of Maschke’sTheorem results in a calculation of the representations of G over K. In the exercises, wedetermine the simple K[G] modules and construct the associated splittings of K[G] forsome familiar examples of groups and fields.

We also show that left semisimplicity and right semisimplicity are equivalent.Then, in Section 12.4, we give characterizations of both hereditary and semisimple

rings in terms of homological dimension.In Section 12.5, we give some basic material on hereditary rings. Then, in Section 12.6,

we define and characterize the Dedekind domains. We give both unstable and stable clas-sifications of the finitely generated projective modules over a Dedekind domain, modulothe calculation of the isomorphism classes of ideals of A. The isomorphism classes ofnonzero ideals form a group under an operation induced by the product of ideals. Thisgroup is known as the class group, Cl(A). Calculations of class groups are important innumerous branches of mathematics. We show here that K0(A) ∼= Cl(A).

Section 12.7 develops the theory of integral extensions and defines and studies the ringof integers, O(K), of a number field. By means of a new characterization of Dedekinddomains, we show that the rings of integers are Dedekind domains.

12.1 Maschke’s Theorem and Projectives

Definitions 12.1.1. A nonzero ring A is left hereditary if every left ideal is a projectiveA-module.

A nonzero ring A is left semisimple if each left ideal a of A is a direct summand of A.

Of course, semisimple rings are hereditary.

It is important to underline the distinction between a submodule N ⊂ M being adirect summand and merely being isomorphic to one. (The reader may want to takeanother look at Definition 9.8.4 and Lemma 9.8.5.) For instance, in a P.I.D., a proper,nonzero ideal (a) is isomorphic to the full ring A, but is not a direct summand of it, asthe sequence

0→ (a)→ A→ A/(a)→ 0

fails to split. The point is that A/(a) is a torsion module, and hence cannot embed in A.

Examples 12.1.2.


1. The simplest example of a hereditary ring is a P.I.D. Here, every ideal is free onone generator. However, as just noted, a P.I.D. is semisimple if and only if it is afield.

2. A product K × L of fields is easily seen to be semisimple: Its nonzero ideals areK × 0 and 0× L.

There are some interesting hereditary domains that are not P.I.D.s. Such domainsare what’s known as Dedekind domains, and figure prominently in number theory. Weshall study them in Sections 12.6 and 12.7.

Since a direct summand of a finitely generated module is finitely generated, we obtainthe following important observation.

Lemma 12.1.3. Left semisimple rings are left Noetherian.

In contrast, there are examples of hereditary rings that are not Noetherian.There is an alternative notion of semisimple rings that appears in the literature. We

give the definition here, but defer most of the discussion until later.

Definition 12.1.4. A nonzero ring is Jacobson semisimple if its Jacobson radical R(A) =0.

We shall see that a ring is left semisimple if and only if it is both Jacobson semisimpleand left Artinian. We shall also see that every left semisimple ring is right semisimple.Thus, in the presence of Jacobson semisimplicity, the left Artinian property implies theright Artinian property.

One important application of semisimple rings is to the study of the representationsof finite groups. For a finite group G, representation theory concerns the classificationof the finitely generated K[G]-modules, for a field K. The relevance of semisimple ringscomes from Maschke’s Theorem, which says that if G is a finite group and if K is afield whose characteristic doesn’t divide the order of G, then the group ring K[G] issemisimple.

A number of treatments of representation theory bypass this result in favor of a treat-ment depending on inner products in complex vector spaces. This approach only worksfor fields of characteristic 0, of course, and it gives some students the false impressionthat Maschke’s Theorem is something much deeper than it turns out to be. Thus, toillustrate how easy it is to prove Maschke’s Theorem, as well as to motivate the study ofsemisimple rings, we shall give it in this section. But first, we shall give the fundamentalresult regarding the finitely generated projective modules over hereditary and semisimplerings.

Proposition 12.1.5. Let A be a left hereditary ring. Then every submodule of a finitelygenerated projective left A-module is projective. In fact, it is isomorphic to a finite directsum of ideals.


If A is semisimple, then every submodule of a finitely generated projective module isa direct summand of it.

Proof Since every finitely generated projective module is a direct summand of a freemodule, it suffices to show that every left submodule M ⊂ An is isomorphic to a finitedirect sum of ideals, if A is hereditary, and is additionally a direct summand of An if Ais semisimple. We argue by induction on n, with the case n = 1 being immediate fromthe definitions.

Let π : An → A be the projection onto the last factor. Then π(M) = a, a left ideal ofA. Let K be the kernel of π : M → a and identify the kernel of π : An → A with An−1.Then we have a commutative diagram

0 0 0

0 K M a 0

0 An−1 An A 0

��

��

��

��

��

i

��π

��

j

��

��

k

�� π ��

where the straight lines are all exact.By the inductive hypothesis, K is isomorphic to a finite direct sum of ideals. Since

a is projective, the upper short exact sequence splits, and hence M ∼= K ⊕ a is alsoisomorphic to a finite direct sum of ideals.

Let P be projective and let N ⊂ P . By Proposition 9.8.6 and the definition, we seethat

0→ N⊂−→ P → P/N → 0

splits if and only if P/N is projective. So Lemma 9.8.5 shows that N is a direct summandof P if and only if P/N is projective. Thus, for the result at hand, if A is semisimpleand M ⊂ An, it suffices to show that the cokernel of j : M ⊂ An is projective.

Write C ′, C, and C ′′, respectively, for the cokernels of i, j, and k in the above diagram.Then the Snake Lemma (Problem 2 of Exercises 7.7.52) shows that we may complete theabove diagram to


0 0 0

0 K M a 0

0 An−1 An A 0

0 C ′ C C ′′ 0

0 0 0

��

��

��

��

��

i

��π

��

j

��

��

k

��

��

��π

��

��

��

��

��

��

��

��

��

so that all straight lines are exact.The semisimplicity of A shows k to be the inclusion of a direct summand, while i

may be assumed to be so by induction. Thus, C ′ and C ′′ are projective. But the shortexact sequence of C’s then splits, and hence C is projective, too.

We shall now give the proof of Maschke’s Theorem. We begin with two lemmas.

Lemma 12.1.6. Let G be a finite group and let A be a commutative ring. Let M andN be A[G]-modules. Then there is an action of G on HomA(M,N) given by

(g · f)(m) = gf(g−1m)

for g ∈ G, f ∈ HomA(M,N), and m ∈M .Each g ∈ G acts via an A-module homomorphism on HomA(M,N), and the fixed

points HomA(M,N)G are precisely the A[G]-module homomorphisms, HomA[G](M,N).

Proof We treat the last assertion, leaving the rest to the reader.The point is that A[G] is generated by A and G. Thus, if f is in HomA(M,N), then

it already commutes with the actions of A on M and N , so it lies in HomA[G](M,N) ifand only if f(gm) = gf(m) for all g ∈ G and m ∈ M . But multiplying both sides byg−1, this says that (g−1 · f)(m) = f(m) for all g and m.

Since passage from g to g−1 is a bijection on G, this is equivalent to f lying inHomA(M,N)G.

The hypothesis in Maschke’s Theorem that the characteristic of K does not dividethe order of G will allow us to use the following technique.

Lemma 12.1.7. Let A be a commutative ring and let G be a finite group whose order,|G|, maps to a unit under the unique ring homomorphism Z → A. Then for any twoA[G]-modules, there is a retraction R : HomA(M,N) → HomA[G](M,N) of A-modulesgiven by

R(f) =1|G|

∑g∈G

(g · f), so that (R(f))(m) =1|G|

∑g∈G

gf(g−1m)


for f ∈ HomA(M,N) and m ∈M .

Proof Looked at in terms of group actions, the result is nearly immediate. It is alsouseful to compute it by hand. If f ∈ HomA[G](M,N), then

(R(f))(m) =1|G|

∑g∈G

gf(g−1m) =1|G|

∑g∈G

f(m) =1|G| |G|f(m)

for all m ∈M , so R(f) = f .For an arbitrary f ∈ HomA(M,N) and x ∈ G, we have

(R(f))(xm) =1|G|

∑g∈G

gf(g−1xm)

=1|G|

∑g∈G

x(x−1g)f((x−1g)−1m)

= x1|G|

∑x−1g∈G

(x−1g)f((x−1g)−1m)

= x(R(f))(m).

So R(f) commutes with the action of G. As noted above, because it also commutes withthe action of A, R(f) is an A[G]-module homomorphism.

The reader may easily verify that R is an A-module homomorphism between the Homgroups.

The trick of adding things up over the elements of the group and then dividing bythe order of G is an averaging technique akin to integration. In fact, the exact sameargument applies, using the Haar integral for the averaging process, in the case where Mand N are the spaces associated to representations of a compact Lie group. One obtains aresult precisely analogous to the one we shall derive below for finitely generated modulesover the group algebra of a finite group: Every finite dimensional representation over Ror C of a compact lie group is a direct sum of irreducible representations. But we aregetting ahead of the presentation here, so let us now prove Maschke’s Theorem.

Theorem 12.1.8. (Maschke’s Theorem) Let G be a finite group and let K be a fieldwhose characteristic does not divide the order of G. Then K[G] is semisimple.

In fact, if N is a K[G]-submodule of any K[G]-module M , then N is a direct summandof M as a K[G] module.

Proof The semisimplicity of K[G] follows from the statement of the second paragraphby taking M = K[G]. In fact, the second paragraph also follows from the semisimplicityof K[G], as we shall see in Proposition 12.4.1.

Thus, suppose given an inclusion i : N ⊂ M of K[G]-modules. By Lemma 9.8.5, itsuffices to produce a retraction for i, i.e., a K[G]-module homomorphism f : M → Nthat restricts to the identity map on N .

Of course, any inclusion of vector spaces over K admits such a retraction as vectorspaces, by Corollary 7.9.15. So let f : M → N be a homomorphism of K-vector spacesthat restricts to the identity map on N . Since the characteristic of K does not dividethe order of G, we may apply Lemma 12.1.7, producing a K[G]-module homomorphismR(f) : M → N with

(R(f))(m) =1|G|

∑g∈G

gf(g−1m)


for all m ∈ M . Applying this to n ∈ N , we see that f(g−1n) = g−1n since f is aretraction, so (R(f))(n) = (1/|G|)|G|n = n, so R(f) is a retraction of K[G]-modules.

Exercises 12.1.9.1. Show that the product of a finite set of hereditary rings is hereditary, and that the

product of a finite set of semisimple rings is semisimple.

2. Write Z2 = 〈T 〉 = {1, T} as a multiplicative group. Show that Z2[Z2] = Z2[〈T 〉]is not semisimple by showing that the ideal generated by 1 + T is not a directsummand. Deduce that the hypothesis that the characteristic of K does not dividethe order of G is necessary in Maschke’s Theorem.

3. Write Z2 = 〈T 〉 and let K be a field of characteristic �= 2. Let a be the idealof K[Z2] generated by 1 + T . Write down an explicit direct sum decompositionK[Z2] ∼= a ⊕ b. Note that b is one-dimensional as a vector space over K. Whateffect does T have on a nonzero element of b?

4. A projection operator on an A-module M is an A-module homomorphism f : M →M such that f ◦ f = f . Show that a submodule N ⊂ M is a direct summand ifand only if it is the image of a projection operator. Show that the complementarysummand may be taken to be the image of 1M − f , which is also a projectionoperator. Deduce that a finitely generated module is projective if and only if it isthe image of a projection operator on An for some n.

5. An element x ∈ A is idempotent if x2 = x. Show that a left ideal a ⊂ A is adirect summand of A if and only if it is the principal left ideal generated by anidempotent in A.

6. Show that A splits as a product of rings if and only if there is an idempotentelement other than 0 or 1 in the center of A.

12.2 Semisimple Rings

For semisimple rings, Proposition 12.1.5 has strong implications.

Corollary 12.2.1. Let A be a left semisimple ring. Then every finitely generated leftA-module is projective.

Proof Let M be a finitely generated left A-module. Then there is a surjection f :An →M , and hence an exact sequence

0→ K⊂−→ An

f−→M → 0.

Proposition 12.1.5 now shows the inclusion ofK in An to be that of a direct summand,and hence the exact sequence splits. Thus, M is isomorphic to a direct summand of An,and hence is projective.

Of course, we’d now like to develop techniques to classify the finitely generated mod-ules over a semisimple ring. Indeed, by the end of this section, we shall obtain a completeclassification, modulo the identification of the isomorphism classes of minimal nonzeroleft ideals. Note that A need not be semisimple in the next definition.

Definition 12.2.2. An A-module M is semisimple if each of its submodules is a directsummand of it.


Clearly, a left semisimple ring is a ring that is semisimple as a left module over itself.Note that if N ′ ⊂ N ⊂M are inclusions of submodules and if N ′ is a direct summand

of M , then it must also be a direct summand of N : A retraction r : M → N ′ of theinclusion of N ′ restricts to a retraction r : N → N ′.

Lemma 12.2.3. A submodule of a semisimple module is semisimple.

Recall that a nonzero A-module M is simple, or irreducible, if it has no submodulesother than 0 and itself. We shall use the next lemma as part of a characterization ofsemisimple modules.

Lemma 12.2.4. Suppose that the A-module M is generated by simple submodules in thesense that there is a family {Mi | i ∈ I} of simple submodules of M such that

∑i∈IMi =

M . Then there is a subset J ⊂ I such that M is the internal direct sum of the Mj forj ∈ J .

Proof Consider the set, S, of all subsets J ⊂ I with the property that the sum∑j∈JMj

is direct in the sense that the map⊕

j∈JMj → M induced by the inclusions of the Mj

is injective. We claim that Zorn’s Lemma shows that S has a maximal element, J .To see this, note that for any J ⊂ I, if the natural map

⊕j∈JMj → M fails to be

injective, then there is a finite subset j1, . . . , jk of J such that the natural map

Mj1 ⊕ · · · ⊕Mjk →M

fails to inject. But if J is the union of a totally ordered subset F ⊂ S, then any finitesubset j1, . . . , jk of J must lie in one of the elements of F , so this cannot happen. Thus,F is bounded above in S, and the hypothesis of Zorn’s Lemma is satisfied.

Let J be a maximal element of S. We claim that⊕

j∈JMj → M is onto. If not,then since M is generated by the submodules Mi, there must be an i such that Mi is notcontained in

∑j∈JMj . But then, since Mi is simple, we must have Mi ∩

∑j∈JMj = 0.

Thus, if we take J ′ = J ∪{i}, the sum∑j∈J′ Mj is easily seen to be direct, contradicting

the maximality of J . Thus, M is the internal direct sum of the submodules Mj forj ∈ J .

Lemma 12.2.5. A nonzero semisimple module, M , contains a simple module.

Proof Let 0 �= m ∈M . If Am is contained in every nonzero submodule of M , then Amis simple, and we’re done. Otherwise, let N be maximal in the set of submodules of Mthat do not contain m. Since M is semisimple, it is an internal direct sum M = N +N ′.We claim that N ′ is simple.

Suppose it isn’t. By Lemma 12.2.3, it is semisimple, so it may be written as theinternal direct sum of two nonzero submodules: N ′ = P + Q. Thus, M is an internaldirect sumM = N+P+Q. But the maximality ofN shows thatmmust lie in bothN+Pand N +Q, contradicting the statement that the summation N + P +Q is direct.

We can now characterize semisimple modules.

Proposition 12.2.6. Let M be a nonzero A-module. Then the following conditions areequivalent.

1. M is a direct sum of simple modules.

2. M is a sum of simple submodules.


3. M is semisimple.

Proof The first condition obviously implies the second, and the second implies the firstby Lemma 12.2.4.

We shall show that the second condition implies the third. Thus, suppose that M =∑i∈IMi, where each Mi ⊂M is simple. Let N ⊂M be any submodule. We shall show

that N is a direct summand of M .Note that Mi∩N is a submodule of the simple module Mi, and hence either Mi ⊂ N

or Mi ∩N = 0. Since the Mi generate M , if N �= M , then there must be some indices isuch that Mi ∩N = 0.

Thus, suppose N �= M , and put a partial ordering by inclusion on the collection ofsubsets J ⊂ I with the property that N ∩∑

j∈JMj = 0. Then Zorn’s Lemma showsthat there is a maximal subset with this property. We claim that if J is a maximal suchsubset, then M is the internal direct sum of N and

∑j∈JMj .

To see this, since N ∩∑j∈JMj = 0, it suffices to show that N +

∑j∈JMj = M .

But the maximality of J shows that if i �∈ J , then N ∩ (Mi +∑j∈JMj) �= 0, and hence

Mi ∩ (N +∑j∈JMj) �= 0. By the simplicity of Mi, we see that Mi ⊂ N +

∑j∈JMj .

Since the Mi generate M , N +∑j∈JMj = M , and hence the second condition implies

the third.We now show that the third condition implies the second. Thus, suppose that M

is semisimple. Let {Mi | i ∈ I} be the set of all simple submodules of M and let N =∑i∈IMi. If N �= M , then M is an internal direct sum M = N +N ′, with N ′ �= 0. By

Lemmas 12.2.3 and and 12.2.5, N ′ contains a simple module M ′, which must be equalto one of the Mi, as those are all the simple submodules of M . But then, M ′ ⊂ N ∩N ′,which contradicts the directness of the summation M = N +N ′. Thus, M = N , and Mis generated by simple modules.

When we apply Proposition 12.2.6 to A itself, we get a nice bonus.

Corollary 12.2.7. A left semisimple ring, A, is left Artinian, and is a finite direct sumof simple left ideals. In addition, the Jacobson radical, R(A) = 0, so that A is Jacobsonsemisimple.

Proof Proposition 12.2.6 shows that A is a possibly infinite direct sum of simple leftideals. But a ring cannot be an infinite direct sum of ideals, as 1 will have only finitelymany nonzero components with respect to this decomposition, and any multiple of 1 willinvolve only those summands. Thus, A is a finite direct sum of simple left ideals.

Clearly, simple modules are Artinian, and by Lemma 7.8.7, any finite direct sum ofArtinian modules is Artinian, so A is a left Artinian ring.

Write A as an internal direct sum of simple left ideals: A = a1 + · · · + an. LetNi =

∑j =i aj . Then A/Ni is isomorphic to ai as an A-module. Since the submodules of

A/Ni are in one-to-one correspondence with the submodules of A containing Ni, the Nimust be maximal left ideals. But ∩ni=1Ni = 0, so R(A) = 0, as claimed.

We can also use Proposition 12.2.6 to recognize the semisimplicity of a given ring.

Corollary 12.2.8. Let D be a division ring and let n > 0. Then Mn(D) is semisimple.

Proof Let ai ⊂ Mn(D) be the set of matrices whose nonzero entries all lie in the i-thcolumn. Then D is the direct sum of the ideals ai, each of which is isomorphic to Dn,the space of column vectors.


For any nonzero vector v ∈ Dn, there is a unit M ∈Mn(D)× with Me1 = v. So anyMn(D)-submodule of Dn that contains v must contain e1, as well as any other nonzeroelement of Dn, so Dn is a simple module over Mn(D).

We can now give a complete classification of the finitely generated modules over asemisimple ring, modulo the determination of the isomorphism classes as A-modules ofthe simple left ideals. The next lemma will play an important role.

Lemma 12.2.9. (Schur’s Lemma) Let M be a simple module over the ring A. Then theendomorphism ring EndA(M) is a division ring.

If N is another simple module and if M and N are not isomorphic as A-modules,then HomA(M,N) = 0.

Proof It suffices in both cases to show that any nonzero homomorphism between simplemodules M and N is an isomorphism (and hence a unit in EndA(M) if M = N).

Thus, let f : M → N be a homomorphism of simple modules. If f �= 0, then thekernel of f is not M . But since M is simple, the only other possibility is ker f = 0.Similarly, since N is simple, the image of f is either 0 or N , so if f is nontrivial, it isonto.

Corollary 12.2.10. A left semisimple ring A has only finitely many isomorphism classesof simple modules, each of which is isomorphic to a simple left ideal. In consequence, ev-ery semisimple A-module is isomorphic to a direct sum of ideals, and hence is projective.

Proof Let M be a simple A-module and let f : A → M be a nontrivial A-modulehomomorphism. By the simplicity of M , f is onto. Thus, M is finitely generated, andhence projective by Corollary 12.2.1. Thus, f admits a section, and hence M is isomor-phic to a direct summand of A, and hence to a simple left ideal. By Proposition 12.2.6,we see that semisimple A-modules are isomorphic to direct sums of simple left ideals.

Corollary 12.2.7 gives an A-module isomorphism

A ∼= a1 ⊕ · · · ⊕ ak

from A to a direct sum of simple left ideals. By Schur’s Lemma, if a is a simple left ideal ofA, then at least one of the composites a ⊂ A proj−−→ ai must be an isomorphism: Otherwise,the inclusion a ⊂ A is the zero map. Thus, there are finitely many isomorphism classesof simple left ideals.

In some cases, we obtain a decomposition A ∼= a1 ⊕ · · · ⊕ ak of A as a direct sum ofsimple left ideals that are all isomorphic to one another as A-modules. For example, theproof of Corollary 12.2.8 shows that if D is a division ring, then Mn(D) is a direct sumof simple left ideals each of which is isomorphic to the space of column vectors, Dn. Inthis case, the above argument shows that every simple left ideal of Mn(D) is isomorphicto Dn. Thus, Mn(D) is an example of a simple ring:

Definition 12.2.11. A simple ring is a semisimple ring with only one isomorphism classof simple left ideals.

Summarizing the discussion of matrix rings, we have:

Corollary 12.2.12. Let D be a division ring. Then Mn(D) is a simple ring for n ≥ 1.


Corollary 12.2.13. Let a and b be simple left ideals in the ring A and let M be a leftA-module that is a sum of simple submodules all of which are isomorphic as modules tob.

If a and b are not isomorphic as A-modules, then any A-module homomorphismf : a→M is trivial, a ⊂ Ann(M), and hence aM = 0.

Alternatively, if A is semisimple and a and b are isomorphic as A-modules, thenaM = M .

Proof First note that by Lemma 12.2.4, M is a direct sum M =⊕

i∈IMi, where eachMi is isomorphic to b.

Suppose, then, that a and b are not isomorphic and let f : a → M be an A-modulehomomorphism. Then if πi : M →Mi is the projection onto the i-th summand, Schur’sLemma shows that πi ◦ f = 0. Since a map into a direct sum is determined by itscoordinate functions, f = 0.

Now let m ∈ M . Then there is an A-module homomorphism f : a → M given byf(a) = am. In particular, if a is not isomorphic to b, f = 0, and hence a ⊂ Ann(m).Since this is true for all m ∈M , a ⊂ Ann(M).

Now suppose that a and b are isomorphic and that A is semisimple. We have aM =⊕i∈I aMi, so it suffices to show that aMi = Mi for all i. The hypothesis now is that

Mi is isomorphic to a, so let f : a → Mi be an isomorphism. Since A is semisimple, ais a direct summand of A, so f extends to an A-module homomorphism f : A → Mi.Since f is an A-module homomorphism out of A, there is an element m ∈Mi such thatf(x) = xm for all x ∈ A. Since f restricts to an isomorphism from a to Mi, am = Mi,so aMi = Mi, and hence aM = M , as claimed.

In particular, if M is a sum of copies of a particular simple left ideal, a, then everysimple submodule of M must be isomorphic to a.

We can now split a semisimple ring into a product of simple rings, one for eachisomorphism class of simple left ideals. We shall give a slicker proof of the followingproposition in Section 12.3. We include it now, as a hands-on argument in instructive.

Proposition 12.2.14. Let A be a semisimple ring and let a1, . . . , ak be representativesof the distinct isomorphism classes of simple left ideals in A. Let bi ⊂ A be the sum ofall the left ideals of A isomorphic to ai. Then bi is a two-sided ideal for each i, and A isthe internal direct sum

A = b1 ⊕ · · · ⊕ bk.

Also, there are simple rings Bi for 1 ≤ i ≤ k and an isomorphism of rings

f : A∼=−→ B1 × · · · ×Bk

such that f restricts to an isomorphism fi : bi∼=−→ Bi of A-modules for all i. In particular,

fi(ai) represents the single isomorphism class of simple left ideals of Bi.

Proof By Proposition 12.2.6, A is a sum of simple left ideals, so A = b1 + · · · + bk.Since the simple left ideals generate A, bi will be a two-sided ideal if and only if bic ⊂ bifor every simple left ideal c.

If c ∼= ai, then c ⊂ bi by the definition of the latter. Since c is a left ideal, bic ⊂ c ⊂ bi,as desired.

Suppose, then, that c �∼= ai. Since b is generated by ideals isomorphic to ai, bc = 0 byCorollary 12.2.13. Thus bi is a two-sided ideal.


We claim now by induction on i that each sum b1 + · · ·+bi is direct. This is certainlytrue for i = 1. Assuming the result for i− 1, write

a = (b1 + · · ·+ bi−1) ∩ bi.

Then the result can only be false for i if a �= 0. But a ⊂ bi, so if a �= 0, then its simpleleft ideals must all be isomorphic to ai by Corollary 12.2.13. So another application ofCorollary 12.2.13 shows that aia = a.

But a ⊂ b1 + · · ·+ bi−1, so Corollary 12.2.13 also shows that a1a = 0. Thus, A is thedirect sum of the bi, as claimed.

Clearly, bibj = 0 for i �= j. (This follows simply because bi and bj are two-sidedideals with bi ∩ bj = 0.) Thus, suppose that a = a1 + · · ·+ ak and b = b1 + · · ·+ bk arearbitrary elements of A, with ai, bi ∈ bi for all i. Then ab = a1b1 + · · ·+ akbk.

Thus, if we write 1 = e1 + · · · + ek with ei ∈ bi for all i, then each ei acts as a two-sided identity if we restrict the multiplication to the elements of bi alone. Thus, there isa ring Bi whose elements and whose operations are those of bi, and whose multiplicativeidentity element is ei. Clearly, A is isomorphic as a ring to the product of the Bi.

Note that the A-module structure on bi pulls back from the obvious Bi-module struc-ture, so ai is a simple Bi module, and Bi is a sum of copies of ai. Thus, Bi is a simplering, as claimed.

The next lemma is very easy, but may not be part of everyone’s developed intuition.

Lemma 12.2.15. Let M be a left module over the ring A and let Mn be the direct sumof n copies of M . Then EndA(Mn) is isomorphic to the ring of n × n matrices overEndA(M).

In particular, if M is a simple A-module and if D = EndA(M), then EndA(Mn) isisomorphic to the ring of n× n matrices over the division ring D.

Proof If f : Mn →Mn is an A-module homomorphism, then the matrix associated tof has ij-th entry given by the composite

Mιj−→Mn f−→Mn πi−→M,

where ιj is the inclusion of the j-th summand of Mn and πi is the projection onto thei-th summand.

The matrices then act on the left of Mn in the usual manner, treating the n-tuplesin Mn as column vectors with entries in M .

When A is commutative, the preceding argument is a straightforward generalizationof the proof that the endomorphism ring of An is isomorphic to Mn(A). For noncommu-tative rings, the representation of EndA(An) that’s given by Lemma 12.2.15 is differentfrom the one we’re used to. The point is that the evaluation map

ε : EndA(A)∼=−→ A

that takes f to f(1) is an anti-homomorphism: ε(fg) = ε(g) · ε(f), so that EndA(A) isisomorphic to the opposite ring, Aop, of A. Thus, Lemma 12.2.15 gives an isomorphismfrom EndA(An) to Mn(Aop).

However, the presence of opposite rings here should come as no surprise: The morecustomary method of representing the transformations of left free modules by right matrixmultiplication induces an isomorphism from EndA(An) to the opposite ring of Mn(A),


as discussed in Proposition 7.10.22. (Indeed, some algebraists write homomorphisms ofleft modules on the right and use the opposite multiplication as the default in discussingcomposites and endomorphism rings.) The situation is resolved by the next lemma.

Lemma 12.2.16. Passage from a matrix to its transpose induces an isomorphism

Mn(Aop) ∼= Mn(A)op.

We shall now classify the finitely generated modules over a semisimple ring.

Theorem 12.2.17. Let A be a semisimple ring and let a1, . . . , ak represent the distinctisomorphism classes of simple left ideals of A (and hence of simple modules, by Corol-lary 12.2.10). Let M be a finitely generated module over A. Then there is a uniquelydetermined sequence r1, . . . , rk of non-negative integers with the property that

M ∼= ar11 ⊕ · · · ⊕ ark

k .

Proof Corollary 12.2.1 shows that every finitely generated A-module is projective, soProposition 12.1.5 shows that it is isomorphic to a finite direct sum of ideals. Lemma 12.2.3shows that a nonzero ideal is semisimple as an A-module, and hence is a direct sum ofsimple ideals. Since semisimple rings are Noetherian, the direct sum must be finite.

Thus, it suffices to show that if

ar11 ⊕ · · · ⊕ ark

k∼= as11 ⊕ · · · ⊕ ask

k ,

then ri = si for all i.Now Corollary 12.2.13 shows that if we multiply the two modules on the left by the

ideal ai, we get an isomorphism arii

∼=−→ asii . Thus, we can argue one ideal at a time. In

particular, it suffices to show that if a is any simple left ideal of A and if

f : ar∼=−→ as,

then r = s.We shall apply the technique of proof of Lemma 12.2.15. Let D = EndA(a) and let

Mf be the s× r matrix over D whose ij-th entry is the composite

aιj−→ ar

f−→ asπi−→ a.

Then it is easy to see that f is induced by matrix multiplication by Mf .Similarly, f−1 is represented by an r × s matrix, Mf−1 , with entries in D. Since the

passage from A-module homomorphisms to matrices is easily seen to be bijective, the factthat f and f−1 are inverse isomorphisms shows that MfMf−1 = Is and Mf−1Mf = Ir.Since the rank of a D-module is well defined, this implies r = s.

This implies the following, weaker result.

Corollary 12.2.18. Let A be a semisimple ring and let a1, . . . , ak be representatives ofthe isomorphism classes of the simple left ideals of A. Then K0(A) is the free abeliangroup on the classes [a1], . . . , [ak].


Proof Theorem 12.2.17 shows that the monoid, FP (A), of isomorphism classes offinitely generated projectives over A is isomorphic to the submonoid of the free abeliangroup on [a1], . . . , [ak] consisting of the sums

r1[a1] + · · ·+ rk[ak]

whose coefficients ri are all non-negative. (In other words, FP (A) is the free abelianmonoid on [a1], . . . , [ak].)

The universal property of the Grothendieck construction gives a homomorphism fromK0(A) to the free abelian group on [a1], . . . , [ak] that is easily seen to provide an inverseto the obvious map in the other direction.

We can use Lemma 12.2.15 to give a version of what’s known as Wedderburn’s The-orem. We will give another version in Section 12.3.

Theorem 12.2.19. Let A be a simple ring, with simple left ideal a. Let D = EndA(a).Then A is isomorphic as a ring to Mn(Dop), where n is the number of copies of a thatappear in an A-module isomorphism A ∼= an.

Thus, a ring A is simple if and only if it is isomorphic to Mn(D) for some n ≥ 1 anda division ring D.

Proof If A ∼= an, then Lemma 12.2.15 gives us an isomorphism

Aop ∼= EndA(A) ∼= EndA(an) ∼= Mn(D).

Lemma 12.2.16 now gives an isomorphism A ∼= Mn(Dop). Here, D is a division ring bySchur’s Lemma, hence so is Dop. Conversely, Corollary 12.2.12 shows that Mn(D) issimple for any division ring D.

Now let A be semisimple, and let a1, . . . , ak represent the isomorphism classes ofsimple left ideals in A. Then Proposition 12.2.14 produces simple rings Bi for 1 ≤ i ≤ ksuch that A ∼= B1× · · ·×Bk as a ring, and such that ai is isomorphic as an A-module tothe simple left ideal a′i of Bi. Note that since the map A → Bi induced by the productprojection is surjective, pulling back the action gives an isomorphism

EndBi(a′i)

∼=−→ EndA(ai).

Corollary 12.2.20. Let A be a semisimple ring and let a1, . . . , ak represent the isomor-phism classes of simple left ideals in A. Let Di = EndA(ai) and let ni be the exponentof ai occurring in an A-module isomorphism

A ∼= an11 ⊕ · · · ⊕ ank

k .

Then there is a ring isomorphism

A ∼= Mn1(D1op)× · · · ×Mnk

(Dkop).

Since finite products of semisimple rings are semisimple, we see that a ring is semisim-ple if and only if it is isomorphic to a finite product of matrix rings over division rings.

Proof The only thing missing from the discussion prior to the statement is the iden-tification of the integers ni. But this follows from the decomposition A ∼= b1 ⊕ · · · ⊕ bkof A as a direct sum of two-sided ideals that’s given in Proposition 12.2.14. Each bi isisomorphic to Bi as an A-module, and is a direct sum of ideals isomorphic to ai. So thenumber of copies of ai that appear in a direct sum decomposition as above for A is equalto the number of copies of a′i that occur in such a decomposition of Bi.


The entire theory of this section goes through, of course, for right modules. The endresult is the same: products of matrix rings over division rings. Such matrix rings areboth left and right simple.

Corollary 12.2.21. A ring is left semisimple if and only if it is right semisimple and isleft simple if and only if it is right simple.


Exercises 12.2.22.1. Show that a semisimple ring is commutative if and only if it is a product of fields.

Show that the decomposition of a commutative semisimple ring A as a direct sumof simple ideals has the form

A ∼= a1 ⊕ · · · ⊕ ak,

where a1, . . . , ak are simple left ideals all lying in distinct isomorphism classes.

2. Corollary 8.8.9 gives an explicit isomorphism from Q[Zn] to a product of fields.Give analogous decompositions for R[Zn] and C[Zn].

3. What are the idempotents in Q[Z2r ]?

4. Let A be any ring and let M be a finitely generated semisimple A-module. Supposethat

M ∼= Mr11 ⊕ · · · ⊕Mrk

k ,

where M1, . . . ,Mk are mutually nonisomorphic simple A-modules. Show that

EndA(M) ∼= Mr1(D1)× · · · ×Mrk(Dk),

where Di = EndA(Mi).

5. Give an example of an Artinian ring A with a simple left ideal a such that aa = 0.Deduce from Corollary 12.2.13 that A cannot be semisimple.

12.3 Jacobson Semisimplicity

Recall that a nonzero ring is Jacobson semisimple if the Jacobson radical R(A) = 0 andis Jacobson simple if it has no two-sided ideals other than 0 and A.

Recall from Corollary 12.2.7 that every semisimple ring is Jacobson semisimple, aswell as being Artinian.

Lemma 12.3.1. A simple ring A is Jacobson simple.

Proof Let b be a nonzero two-sided ideal of A. By Lemmas 12.2.3 and 12.2.5, there isa simple left ideal a ⊂ b. As A is simple, it is a sum of simple left ideals isomorphic toa, and hence aA = A by Corollary 12.2.13. Since b is a two-sided ideal, aA ⊂ b, and theresult follows.

Jacobson semisimplicity is a very weak condition all by itself. The next exampleshows, among other things, that Jacobson semisimple rings need not be Artinian.

Example 12.3.2. Let A be a P.I.D. with infinitely many prime ideals. Then uniquefactorization shows that the intersection of all the maximal ideals of A is 0, so A isJacobson semisimple. Note that each of the quotient modules A/(p) with (p) prime is asimple module, so that A has infinitely many isomorphism classes of simple modules.

We shall make use of the following concept, which could have been given much earlier.

Definition 12.3.3. An A-module M is faithful if AnnA(M) = 0.


Since the annihilator of a module is a two-sided ideal, every nonzero module over aJacobson simple ring is faithful. And every nonzero ring admits a simple module, as, ifm is a maximal left ideal in the ring A, then Lemma 9.1.4 shows A/m to be a simplemodule. The next lemma now follows from Lemma 9.1.2, which shows every proper leftideal to be contained in a maximal one.

Lemma 12.3.4. A Jacobson simple ring admits a faithful simple module.

We’ve seen that an A-module structure on an abelian group M is equivalent to spec-ifying a ring homomorphism ϕ : A → EndZ(M). Here, for a ∈ A, the associatedhomomorphism ϕ(a) : M →M is given by ϕ(a)(m) = am.

If we examine this more carefully, we can get additional information. First, notethat if M is an A-module, then EndA(M) is a subring of EndZ(M), and that M is anEndA(M)-module via f ·m = f(m) for f ∈ EndA(M) and m ∈M .

Now write B = EndA(M), and note that for a ∈ A, ϕ(a) is a B-module homomor-phism. We obtain the following lemma.

Lemma 12.3.5. Let M be an A-module and let B = EndA(M). Then there is a canon-ical ring homomorphism ϕ : A → EndB(M) given by ϕ(a)(m) = am for all a ∈ A andm ∈M . In particular, the kernel of ϕ is AnnA(M).

Recall from Schur’s Lemma that if M is a simple module over the ring A, thenEndA(M) is a division ring. The next lemma now follows from Lemma 12.3.5.

Lemma 12.3.6. Let M be a faithful simple module over the ring A, and let D =EndA(M). Then the canonical map ϕ : A→ EndD(M) gives an embedding of A into theendomorphism ring of the vector space M over the division ring D.

We shall show that if A is Artinian and M is a faithful simple A-module, then Mis finite dimensional as a vector space over D and the map ϕ : A → EndD(M) is anisomorphism. Since EndD(M) is isomorphic to the ring of n×n matrices over a divisionring, this will show that A is simple.

We shall prove this using Jacobson’s Density Theorem (Theorem 12.3.8). The nextlemma is a key step. Note that M here is semisimple, rather than simple.

Lemma 12.3.7. Let M be a semisimple A-module and let B = EndA(M). Then forany f ∈ EndB(M) and any nonzero m ∈M , there is an a ∈ A with

f(m) = am.

Proof This would be immediate if M were simple, as then Am = M . More generally,Am is a direct summand of M as an A-module, so there is an A-module homomorphismr : M → Am that is a retraction of M onto Am, i.e., if i : Am → M is the inclusion,then r ◦ i is the identity on Am.

Now i ◦ r : M → M is an A-module homomorphism, hence an element of B =EndA(M). Thus, f commutes with the action of i ◦ r, so

f(am) = f((i ◦ r)(am)) = (i ◦ r)f(am)

for all a ∈ A, and hence f(Am) ⊂ Am. But this says f(m) ∈ Am, and the resultfollows.

Theorem 12.3.8. (Jacobson Density Theorem) Let M be a simple A-module and letD = EndA(M). Let f ∈ EndD(M). Then for any finite subset m1, . . . ,mn of M , thereis an a ∈ A such that f(mi) = ami for all i.


Proof Recall from Lemma 12.2.15 that EndA(Mn) is isomorphic to the ring of n × nmatrices Mn(D), which acts by matrix multiplication on Mn, where elements of Mn aretreated as n× 1 column vectors with entries in M .

Write fn : Mn → Mn for the direct sum of n copies of f : Written horizontally,fn(x1, . . . , xn) = (f(x1), . . . , f(xn)) for each element (x1, . . . , xn) ∈ Mn. Since theaction of each d ∈ D commutes with that of f , it is easy to see that the action of amatrix (dij) ∈Mn(D) on Mn commutes with the action of fn.

In other words, if we setB = EndA(Mn), then fn ∈ EndB(Mn). Applying Lemma 12.3.7to the element (m1, . . . ,mn) ∈Mn, we see there’s an element a ∈ A such that a(m1, . . . ,mn) =fn(m1, . . . ,mn), i.e., (am1, . . . , amn) = (f(m1), . . . , f(mn)).

Corollary 12.3.9. Suppose that the ring A has a faithful simple module M and letD = EndA(M). Then A is left Artinian if and only if M is finite dimensional as avector space over D.

If M is an n-dimensional vector space over D, then the canonical map ϕ : A →EndD(M) is an isomorphism. Since EndD(M) is isomorphic to Mn(Dop), we obtainthat A is a simple ring. Moreover, A ∼= Mn as an A-module.

Proof Lemma 12.3.6 shows that ϕ is injective. Suppose that M is a finite dimensionalD-module, with basis m1, . . . ,mn. Then the Jacobson Density Theorem shows that forany f ∈ EndD(M), there is an element a ∈ A such that f(mi) = ami for all i. Sincem1, . . . ,mn generate M over D, this says that f agrees everywhere with multiplicationby a, and hence f = ϕ(a). Thus, ϕ : A→ EndD(M) is an isomorphism.

Now Proposition 7.10.22 gives an isomorphism from EndD(M) toMn(D)op, and henceto Mn(Dop) by Lemma 12.2.16. Thus, A is left Artinian and simple. Now, Mn(Dop) isa direct sum of n copies of its simple left ideal, so the same is true for A. The simple leftideal of A is isomorphic to M by Corollary 12.2.10, so A ∼= Mn.

Conversely, suppose that M is an infinite dimensional vector space over D, and letm1,m2, . . . ,mn, . . . be an infinite subset of M that is linearly independent over D. Thenfor each n > 1, there is an element of EndD(M) that vanishes on m1, . . . ,mn−1 and isnonzero on mn.

By the Jacobson Density Theorem, there is an a ∈ A that annihilates each ofm1, . . . ,mn−1, but does not annihilate mn. Thus, the inclusion

AnnA(m1, . . . ,mn−1) ⊃ AnnA(m1, . . . ,mn)

is proper. Here, the annihilator of a finite set is taken to be the intersection of theannihilators of its elements.

In particular, we have constructed an infinite descending chain of proper inclusionsof left ideals of A, and hence A cannot be left Artinian if M is infinite dimensional overD.

We obtain a characterization of simple rings.

Theorem 12.3.10. (Wedderburn Theorem) Let A be a ring. Then the following condi-tions are equivalent.

1. A is Jacobson simple and left Artinian.

2. A is Jacobson simple and right Artinian.

3. A is left (resp. right) Artinian and has a faithful simple left (resp. right) module.


4. A is simple.

5. A is isomorphic to the ring of n× n matrices over a division ring.

Proof Of course, we’ve already seen in Theorem 12.2.19 that a ring is left simple ifand only if it is right simple and that the last two conditions are equivalent. Thatwill also come out of what we’ve done in this section, together with the fact, fromCorollary 12.2.12, that the ring of n × n matrices over a division ring is both left andright simple.

Now simple rings are Jacobson simple by Lemma 12.3.1 and Artinian by Corol-lary 12.2.7, so the fourth condition for left simple rings implies the first.

Lemma 12.3.4 shows that the first condition implies the left version of the third,which implies the fifth by Corollary 12.3.9.

The right side implications are analogous and the result follows.

We now consider the Jacobson semisimple rings.

Lemma 12.3.11. Let A be a left Artinian Jacobson semisimple ring. Then there is afinite collection M1, . . . ,Mk of simple left A-modules such that

k⋂i=1

Ann(Mi) = 0.

Proof Let {Mi | i ∈ I} be a set of representatives for the isomorphism classes of simpleleft A-modules. (The isomorphism classes form a set because every simple left A-moduleis isomorphic to A/m for some maximal left ideal m of A.) Then Proposition 9.1.7 showsthat

R(A) =⋂i∈I

Ann(Mi).

Since A is left Artinian, we claim that R(A) is the intersection of a finite collection ofthe Ann(Mi). To see this, note that the Artinian property shows the collection of finiteintersections of the ideals Ann(Mi) has a minimal element. But if a =

⋂kj=1 Ann(Mij )

is a minimal such intersection, then a ∩ Ann(Mi) = a for all i. But then a ⊂ Ann(Mi)for all i, and hence a = R(A).

Since A is Jacobson semisimple, R(A) = 0, and the result follows.

The key step is the following proposition.

Proposition 12.3.12. Let A be a left Artinian Jacobson semisimple ring. Let M1, . . . ,Mk

be pairwise non-isomorphic simple left A-modules such that

k⋂i=1

Ann(Mi) = 0.

Suppose also that this intersection is irredundant in the sense that for no proper subcol-lection of M1, . . . ,Mk will the intersection of the annihilators be 0.

Let Di = EndA(Mi). Then each Mi is a finite dimensional vector space over thedivision ring Di, and we have a ring isomorphism

A ∼= Mn1(D1op)× · · · ×Mnk

(Dkop),


where ni is the dimension of Mi as a vector space over Di. We obtain an A-moduleisomorphism

A ∼= Mn11 ⊕ · · · ⊕Mnk

k .

In particular, every left Artinian Jacobson semisimple ring is semisimple.

Proof Note that eachMi is a faithful simple module overA/Ann(Mi). SinceA/Ann(Mi)is left Artinian, Theorem 12.3.10 shows it to be a simple ring. Note that EndA(Mi) =EndA/Ann(Mi)(Mi), so Theorem 12.3.10 shows that Mi is finite dimensional over Di, andgives us a ring isomorphism

A/Ann(Mi) ∼= Mni(Dopi ).

Corollary 12.3.9 gives an A-module isomorphism A/Ann(Mi) ∼= Mnii .

Since A/Ann(Mi) is a simple ring, it has no two-sided ideals other than 0 and itself.Thus, Ann(Mi) is a maximal two-sided ideal of A. By the irredundancy of the intersectionof the annihilators of the Mi, we see that if i �= j, Ann(Mi) �= Ann(Mj), and henceAnn(Mi) + Ann(Mj) = A. Thus, since the intersection of the Ann(Mi) is 0, the ChineseRemainder Theorem gives us a ring isomorphism

A ∼= A/Ann(M1)× · · · ×A/Ann(Mk).

The next proposition is now clear from Corollary 12.2.7, Lemma 12.3.11, Proposi-tion 12.3.12, Corollary 12.2.8, and the fact that a finite product of semisimple rings issemisimple.

Theorem 12.3.13. (Semisimple Wedderburn Theorem) Let A be a ring. Then the fol-lowing conditions are equivalent.

1. A is Jacobson semisimple and left Artinian.

2. A is Jacobson semisimple and right Artinian.

3. A is semisimple.

4. A is isomorphic to a finite product of finite dimensional matrix rings over divisionrings.

Let K be a field and let A be a K-algebra that is semisimple as a ring and finitedimensional as a vector space over K. Then any finitely generated A-module M is afinitely generated K-vector space, and there is a natural embedding of EndA(M) as aK-subalgebra of EndK(M). Thus, if M is simple, then D = EndA(M) is a divisionalgebra over K that is finite dimensional as a K-vector space.

In practice, to obtain a decomposition of such an A as product of matrix rings, it isoften convenient to hunt for simple modules Mi, calculate both Di and dimDi(Mi), andcontinue this process until the resulting product of matrix rings has the same dimensionas a K-vector space as A.

Exercises 12.3.14.1. Show that Jacobson semisimplicity is not preserved by localization.


2. For any ring A, show that A/R(A) is Jacobson semisimple.

3. Let G be a finite group and let K be an algebraically closed field whose char-acteristic doesn’t divide |G|. Let M be a simple module over K[G]. Show thatEndK[G](M) = K. Deduce that K[G] contains exactly dimK(M) copies of M inits decomposition as a direct sum of simple modules.

4. Let G be a finite group and K a field. Then a K[G]-module structure whose un-derlying K-vector space is Kn corresponds to a K-algebra homomorphism K[G]→Mn(K). Show that this induces a one-to-one correspondence between those K[G]modules whose underlying K-vector spaces have dimension n and the conjugacyclasses of group homomorphisms ρ : G → Gln(K). (Such homomorphisms ρ arecalled n-dimensional representations of G over K.)

Deduce that there is a one-to-one correspondence between the isomorphism classesof K[G]-modules of dimension 1 over K (which are a priori simple) and the set ofgroup homomorphisms from G to K×.

5. Deduce from Problem 3 of Exercises 10.7.8 that if G is a finite abelian group andK is an algebraically closed field whose characteristic does not divide |G|, then thesimple K[G] modules are precisely those with dimension 1 as vector spaces over K.

6. Write Q[D2n], R[D2n], and C[D2n] as products of matrix rings over division rings.

7. Write R[Q8] as a product of matrix rings over division rings.

12.4 Homological Dimension

We can also characterize the infinitely generated modules over a semisimple ring.

Proposition 12.4.1. Let A be a left semisimple ring. Then every left A-module isisomorphic to a direct sum of simple left ideals, and hence is projective, as well as beinga semisimple module over A.

Proof By Corollary 12.2.10, it suffices to show that every A-module is semisimple.Let M be an A-module and let N be a submodule. We define a partially ordered set

whose elements are submodules N ′ containing N , together with a retraction r : N ′ → Nfor the inclusion of N in N ′. The partial ordering is given by setting (N ′, r′) ≤ (N ′′, r′′)if N ′ ⊂ N ′′ and if r′′ restricts to r′ on N ′.

Zorn’s Lemma shows that there exists a maximal element (N ′, r′) of this set. Weclaim that N ′ = M , and hence N is a direct summand of M . This will complete theproof, as N was an arbitrary submodule of M .

Thus, suppose that m ∈M does not lie in N ′ and let fm : A→M be the A-modulehomomorphism that takes 1 to m. Let a = f−1

m (N ′). Then A is an internal direct sumA = a+b. Since ker fm ⊂ a, fm exhibits Am as the internal direct sum (Am∩N ′)+fm(b).As a result, the sum N ′ + fm(b) is direct. Now set r′′ : N ′ + fm(b) → N to be equalto r′ on N ′ and to be 0 on fm(b). This contradicts the maximality of (N ′, r′), and theresult follows.

This now gives a characterization of semisimple rings.

Corollary 12.4.2. Let A be a ring. Then the following conditions are equivalent.


1. A is left semisimple.

2. Every left A-module is projective.

3. Every finitely generated left A-module is projective.

Proof All that’s left to show is that the third condition implies the first. But if a is aleft ideal, then A/a is finitely generated, and hence the sequence

0→ a→ A→ A/a→ 0

splits. So a is a direct summand of A.

In view of Corollary 12.4.2, semisimple rings may be characterized in terms of thefollowing notion.

Definition 12.4.3. We say that the ring A has global left homological dimension ≤ nif for each left A-module M there is an exact sequence

0→ Pn → · · · → P0 →M → 0

where Pi is projective for all i.

We may now apply Corollary 12.4.2.

Corollary 12.4.4. A ring is left semisimple if and only if it has global left homologicaldimension 0.

Being hereditary is also equivalent to a homological condition. To show this, we shallgeneralize Proposition 12.1.5 to the case of infinitely generated modules. The followingproposition is due to Irving Kaplansky.

Proposition 12.4.5. Let F be a left free module over the left hereditary ring A. Thenany submodule M ⊂ F is isomorphic to a direct sum of ideals, and hence is projective.

Proof Let {ei | i ∈ I} be a basis for F . We shall put a well-ordering on I. This is apartial ordering with the property that any subset J ⊂ I has a smallest element, i.e.,an element j ∈ J such that j ≤ k for all k ∈ J . We may do this by the Well-OrderingPrinciple, which is equivalent to the Axiom of Choice.

For i ∈ I, let Fi be the submodule generated by {ej | j ≤ i} and let πi : Fi → A beinduced by the projection onto the summand generated by ei (i.e., πi(ej) = 0 for j �= iand πi(ei) = 1).

Write Mi = M ∩ Fi, and let ai = πi(Mi). Then we have a short exact sequence

0→ Ki →Miπ′

i−→ ai → 0.

Here π′i is the restriction of πi to Mi and Ki is the kernel of π′

i. Since A is hereditary, aiis projective, and the sequence splits. Write Ni ⊂Mi for the image of ai under a sectionfor π′

i. We claim that M is the internal direct sum of {Ni | i ∈ I}.To see this, suppose first that n1 + · · · + nk = 0, with nj ∈ Nij for j = 1, . . . , k and

with i1 < · · · < ik. Then

0 = πik(n1 + · · ·+ nk) = πik(nk).


Since πik : Nik∼=−→ aik , we must have nk = 0. Inductively, ni = 0 for all i.

Thus, it suffices to show that the Ni generate M . We argue by contradiction. Leti ∈ I be the smallest element such that Mi is not contained in

∑j∈I Nj , and let m ∈Mi

be an element not in∑j∈I Nj . Let si : ai → Ni be the inverse of the isomorphism

πi : Ni → ai. Write m′ = m− si(πi(m)).Then m′ is not in

∑j∈I Nj . And πi(m′) = 0. Thus, m′ lies in the submodule of F

generated by {ej | j < i}. Since m′ is a finite linear combination of these ej , it must liein Fj ∩M = Mj for some j < i, contradicting the minimality of i.

Since P.I.D.s are hereditary and their ideals are all free, we obtain the followingcorollary.

Corollary 12.4.6. Every projective module over a P.I.D. is free.

Lemma 12.4.7. (Schanuel’s Lemma) Suppose given exact sequences

0→ Ki−→ P

f−→M → 0 and 0→ Li−→ Q

f−→M → 0

of A-modules, with P and Q projective. Then K ⊕Q ∼= L⊕ P .

Proof We shall make use of the pullback construction that’s discussed in the proof ofProposition 9.8.2: P ×M Q = {(x, y) ∈ P ×Q | f(x) = g(y)}. We obtain a commutativediagram whose straight lines are exact:

0 0

N2 L

0 N1 P ×M Q Q

0 K P M 0

0

��

��

��f

��

ı

��

i

��

��

g

��ı ��f

��

g

��

g

�� i ��f ��

��

Here, f(x, y) = y and g(x, y) = x, while f and g are obtained by restricting f and g,respectively.

We now make some claims the reader should check. We claim that in any suchpullback diagram, the maps f and g are isomorphisms. Also, since f is surjective, so isf , and since g is surjective, so is g.


Since f is surjective and Q is projective, we have a split short exact sequence

0→ N1ı−→ P ×M Q

f−→ Q→ 0.

Since g is an isomorphism, K ⊕Q ∼= P ×M Q.Since g is surjective and P is projective, we have a split short exact sequence

0→ N2j−→ P ×M A

g−→ P → 0.

Since f is an isomorphism, L⊕ P ∼= P ×M Q.

We may now characterize hereditary rings in terms of homological dimension.

Proposition 12.4.8. A ring A is left hereditary if and only if it has global left homo-logical dimension ≤ 1.

Proof If A is hereditary and M is a left A-module, let f : F →M be surjective, whereF is free. Then we have a short exact sequence

0→ K → Ff−→M → 0.

By Proposition 12.4.5, K is projective. Since M was arbitrary, we see that A has globalleft homological dimension ≤ 1.

Conversely, suppose that A has global left homological dimension ≤ 1 and let a be aleft ideal of A. Then our hypothesis provides an exact sequence

0→ P1 → P0 → A/a→ 0,

with P0 and P1 projective. Of course, 0 → a → A → Aa → 0 is exact, so Schanuel’sLemma gives an isomorphism a ⊕ P0

∼= P1 ⊕ A. Thus, a is projective, and hence A ishereditary.

12.5 Hereditary Rings

Over a P.I.D., Proposition 12.1.5 shows that every finitely generated projective moduleis free. However, it is definitely not the case that every finitely generated projectivemodule over an arbitrary hereditary domain is free. For instance, as we shall see inSection 12.7, the ring of integers (to be defined there) in a finite extension of Q is ahereditary Noetherian domain. However, these rings of integers turn out very rarely tobe P.I.D.s. So the following proposition will give a recognition principle for non-free idealsin these rings, and hence a source of examples of finitely generated projective modulesthat are not free.

Proposition 12.5.1. Let a be a nonzero ideal in the commutative ring A. Then a isfree as an A-module if and only if it is free of rank 1, in which case it is a principal ideal(x) for some x ∈ A that is not a zero-divisor.

In consequence, if A is a commutative ring in which every ideal is free, then A is aP.I.D.


Proof If a is free on more than one generator, then picking a pair of the basis elementsinduces an injective homomorphism from A2 into a. But a is a submodule of A, so weget an injection from A2 into A, contradicting Theorem 9.5.20.

Thus, if an ideal a of A is free as an A-module, it must have rank 1, so there’s anisomorphism f : A

∼=−→ a. And if x = f(1), we have a = (x). Since f is injective, x is nota zero-divisor.

If every ideal is free, then every ideal is a principal, and it suffices to show that A isan integral domain. But if x ∈ A is a zero-divisor, then AnnA((x)) �= 0, so (x) cannotbe free.

Recall that a discrete valuation ring is a Euclidean domain which is local, and hencehas at most one prime element, up to equivalence of primes.

Corollary 12.5.2. A Noetherian hereditary local ring is a discrete valuation ring.

Proof The hereditary property says that every ideal is projective, and the Noetherianproperty that every ideal is finitely generated. But finitely generated projectives over alocal ring are free by Proposition 9.8.14, so Proposition 12.5.1 shows that our ring is aP.I.D.

A P.I.D. which is not a field is local if and only if it has exactly one prime element(up to equivalence of primes). And Problem 9 of Exercises 8.1.37 shows that such a ringis Euclidean, and hence a discrete valuation ring.

We now give another consequence of Proposition 12.5.1. Recall from Proposition 9.8.16that every finitely generated projective module P over an integral domain A has constantrank, meaning that the rank of the free Ap-module Pp is independent of the choice ofprime ideal p of A.

Corollary 12.5.3. Let A be an integral domain and let a be a nonzero ideal of A thatis finitely generated and projective as an A-module. Then a has constant rank 1.

Proof Let p be a prime ideal of A. Because A is a domain, ap is a nonzero ideal of Ap.Since ap is finitely generated and free over Ap, the result follows from Proposition 12.5.1.

Note the stipulation in the preceding that A be a domain is necessary.

Corollary 12.5.4. Let A be a Noetherian, hereditary commutative ring. Then everylocalization of A is a discrete valuation ring.

Proof Recall that every ideal of Ap has the form ap for an ideal a of A. Since A ishereditary and Noetherian, a is finitely generated and projective, so ap = Ap ⊗A a is afree Ap-module by Proposition 9.8.14. Thus, Ap is a P.I.D. by Proposition 12.5.1. SinceAp is local, Corollary 12.5.2 suffices.

Exercises 12.5.5.1. Show that a Noetherian hereditary commutative ring has Krull dimension ≤ 1.


12.6 Dedekind Domains

As we shall show in Theorem 12.6.8, a Dedekind domain is precisely a hereditary integraldomain. However, the definition we shall give is more classical, and is interesting fromthe point of view of the development of ideal theory.

The point is that a number field, meaning a finite extension of the rational numbers,Q, has a subring, called its ring of integers, which behaves somewhat like the integersdo as a subring of Q. At one time, it was assumed that the ring of integers of a numberfield would be a P.I.D. Indeed, a false proof of Fermat’s Last Theorem was given on thebasis of this false assumption.

The truth of the matter is that the ring of integers of a number field is a Dedekinddomain. We shall show this in Section 12.7. And in a Dedekind domain, the idealsbehave in the same way that the principal ideals do in a U.F.D. Indeed, ideal theorybegan with this observation. Ideals were originally thought of as idealized numbers.

Moving from elements to ideals, we shall make an analogy between prime elementsand maximal ideals and an analogy between irreducible elements and prime ideals.

Here, the operation on ideals that concerns us is the product of ideals:

ab = {a1b1 + · · ·+ akbk | k ≥ 1, ai ∈ a, and bi ∈ b for i = 1, . . . , k}Recall from Lemma 7.6.9 that if p, a, and b are ideals in the commutative ring A,

with p prime, then ab ⊂ p implies that either a ⊂ p or b ⊂ p. Since ab ⊂ a∩b, we obtainthe following lemma, which displays the analogy between prime ideals and irreducibleelements.

Lemma 12.6.1. Let p be a prime ideal in the commutative ring A. Suppose given idealsa and b with p = ab. Then either p = a or p = b.

Definition 12.6.2. A Dedekind domain is an integral domain in which every proper,nonzero ideal is a product of maximal ideals.

One result comes immediately from Lemma 12.6.1. Recall that the Krull dimensionof a commutative ring A is the number of inclusion maps in the longest possible chain ofproper inclusions

p0 ⊂ · · · ⊂ pn

of prime ideals of A.

Proposition 12.6.3. Every nonzero prime ideal in a Dedekind domain is maximal.Thus, a Dedekind domain that is not a field has Krull dimension 1.

Proof Let p be a nonzero prime ideal in the Dedekind domain A. Then we may write

p = m1 . . .mk

for (not necessarily distinct) maximal ideals m1, . . . ,mk. But an induction using Lemma 12.6.1now shows that p = mi for some i.

The following observation is important.

Lemma 12.6.4. Let A be an integral domain with fraction field K. Let a be a nonzeroideal of A and let f : a → A be an A-module homomorphism. Then there is an elementα ∈ K such that f is the restriction to a of the A-module homomorphism from K toitself that’s induced by multiplication by α. In fact, for any nonzero element a ∈ a, wehave f(a)/a = α.


Proof Let 0 �= a ∈ a. Then for x ∈ a, we have

af(x) = f(ax) = f(a)x,

so f(x) = (f(a)/a) · x as an element of K.

It is useful to establish as quickly as possible that the Dedekind domains are thehereditary integral domains. A key idea is the notion of invertible ideals.

Definition 12.6.5. A nonzero ideal a in an integral domain is invertible if there is anonzero ideal b such that ab is principal.

The next proposition displays the relevance of invertibility to the hereditary property.

Proposition 12.6.6. Let A be an integral domain. A nonzero ideal a of A is invertibleif and only if it is projective as an A-module. If a is an invertible ideal, it is also finitelygenerated.

Proof First suppose that a is invertible, with ab = (a). Then we can write a1b1 + · · ·+akbk = a, with ai ∈ a and bi ∈ b for i = 1, . . . , n. Let f : Ak → a be the A-modulehomomorphism that takes ei to ai for 1 ≤ i ≤ k. We shall construct a section for f ,showing that a is a finitely generated projective module over A.

Let b ∈ b. Since ab = (a), we see that multiplication by b/a carries a into A. Thus,there is a well defined homomorphism s : a→ Ak with

s(x) =(b1x

a, . . . ,

bkx

a

)for x ∈ a. Since a1b1 + · · ·+ akbk = a, we see that s is a section for f , as claimed.

Conversely, suppose that a is projective. Let f : F → a be a surjective homomor-phism, where F is the free A-module with basis {ei | i ∈ I}. Let s : a → F be a sectionof f . Since the direct sum embeds in the direct product, Lemma 12.6.4 shows there areelements αi ∈ K for each i ∈ I such that s(x) =

∑i∈I(αix)ei for each x ∈ a, where K is

the field of fractions of A. But the sum must be finite, so all but finitely many of the αimust be 0.

In particular, we see that f restricts to a surjection on a finitely generated freesubmodule of F , so we may assume that F is the finitely generated free module withbasis e1, . . . , en, and that s(x) = (α1x)e1 + · · ·+ (αnx)en for all x ∈ a.

Let f(ei) = ai. Because s is a section of f , we have

x = (α1a1)x+ · · ·+ (αnan)x

for all x ∈ a, and hence α1a1 + · · · + αnan = 1. Let a be a common denominator forα1, . . . , αn, and set bi = aαi ∈ A for 1 ≤ i ≤ n. Recall that multiplication by αi carriesa into A. Thus, multiplication by bi carries a into (a). Note that a1b1 + · · ·+ anbn = a.Thus, if we set b to be the ideal generated by b1, . . . , bn, we have ab = (a), and hence ais invertible.

Thus, projective ideals in an integral domain are finitely generated.

Corollary 12.6.7. A hereditary integral domain is Noetherian.


In the study of Dedekind domains, it is often convenient to work with a generalizednotion of ideals. IfA is a domain with fraction fieldK, we may consider the A-submodulesof K. If a and b are A-submodules of K, we define their product, ab, as if they wereideals:

ab = {α1β1 + · · ·+ αkβk | k ≥ 1, αi ∈ a, and βi ∈ b for i = 1, . . . , k}.

This product is easily seen to be associative and commutative and to satisfy the distribu-tive law with respect to the usual sum operation on submodules.

Certain of the A-submodules of K are generally known as fractional ideals. Weshall confine our formal treatment of fractional ideals to the case where A is a Dedekinddomain, which we treat in Exercises 12.6.23. We now give a characterization of Dedekinddomains.

Theorem 12.6.8. Let A be an integral domain. Then the following conditions are equiv-alent.

1. For any inclusion a ⊂ b of ideals in A, there is an ideal c such that a = bc.

2. A is hereditary.

3. A is a Dedekind domain.

Proof The first condition implies the second, as if a is a nonzero ideal of A and if0 �= a ∈ a, then (a) ⊂ a, so there is an ideal b with ab = (a). Thus, every nonzero idealof A is invertible, and hence A is hereditary.

The second condition implies the first as follows. Let a ⊂ b with a �= 0. Since A ishereditary, b is invertible. Say bc = (x). Then, as A-submodules of the fraction field ofA, we have (1/x)c ·b = A. Since a ⊂ b, (1/x)c ·a is an A-submodule of A, hence an ideal.Write (1/x)c · a = d. But then bd = Aa = a, and hence the first condition holds.

Thus, the first two conditions are equivalent. Now assume the third, so that A is aDedekind domain. The result is obvious if A is a field, so we assume it is not. We firstclaim that every maximal ideal of a is invertible. To see this, let m be a maximal ideal,and let 0 �= a ∈ m. Then (a) is a product of maximal ideals, say (a) = m1 . . .mk. Butthen m1 . . .mk ⊂ m. By Lemma 7.6.9, we must have mi ⊂ m for some i. Since mi andm are maximal, this gives m = mi. But mi is a factor if the principal ideal (a), so m isinvertible.

Clearly, a product of invertible ideals is invertible. Since every proper, nonzero idealin a Dedekind domain is a product of maximal ideals, every such ideal is invertible. Thus,Dedekind domains are hereditary.

It suffices to show that the first two conditions imply the third. Thus, let a be aproper, nonzero ideal in the hereditary domain A. If a is maximal, there’s nothing toshow. Otherwise, a ⊂ m1, with m1 maximal. By condition 1, we have a = m1a1 forsome ideal a1. By Problem 4 of Exercises 10.4.11 (which is an immediate consequence ofCorollary 10.4.10), the inclusion a ⊂ a1 is proper.

We continue inductively, writing a = m1 . . .mnan for n ≥ 1, until we arrive at thepoint where an is maximal. Since

a ⊂ a1 ⊂ · · · ⊂ an

are proper inclusions, an must be maximal for some finite n, since hereditary domainsare Noetherian.


The first thing that we should establish now is the uniqueness of the factorizationof ideals in a Dedekind domain. The hereditary property makes this easier. The nextlemma is obvious if the ideal a is principal. But the result extends to invertible ideals,as they are factors of principal ideals.

Lemma 12.6.9. Let a be an invertible ideal in a domain A, and let b and c be idealssuch that ab = ac. Then b = c.

Proposition 12.6.10. Let A be a Dedekind domain. Then any proper, nonzero idealhas a unique factorization of the form

a = mr11 . . .mrk

k

where m1, . . . ,mk are distinct maximal ideals and the exponents ri are positive for i =1, . . . , k. Here, uniqueness is up to a reordering of the factors mr1

1 , . . . ,mrk

k .

Proof Suppose given two such decompositions

mr11 . . .mrk

k = ns11 . . . nsl

l .

Then the left-hand side shows that the ideal in question is contained in m1. So Lemma 7.6.9shows that ni ⊂ m1 for some i. Renumbering the n’s we may assume that n1 ⊂ m1, butsince both are maximal, this gives n1 = m1.

By Lemma 12.6.9, this implies that

mr1−11 . . .mrk

k = ns1−11 . . . nsl

l ,

and the result follows by induction.

The first condition of Theorem 12.6.8 shows that if a and b nonzero ideals of aDedekind domain, then a ⊂ b if and only if a = bc for some ideal c of A. Uniquefactorization now gives a corollary.

Corollary 12.6.11. Let A be a Dedekind domain and let a = mr11 . . .mrk

k , where m1, . . . ,mk

are distinct maximal ideals of A, and each ri is positive. Then the ideals of A that containa are precisely those of the form ms1

1 . . .msk

k , where 0 ≤ si ≤ ri for all i.

We wish now to learn more about the structure of modules over a Dedekind domain.

Proposition 12.6.12. Let M be a finitely generated module over a Dedekind domain.Then M is projective if and only if it is torsion-free.

Proof By Proposition 12.1.5, every finitely generated projective module over a Dedekinddomain is a direct sum of ideals, and hence is torsion-free.

Conversely, let M be a finitely generated torsion-free A-module, with A a Dedekinddomain. Then the canonical map η : M →M(0) is injective, whereM(0) is the localizationof M at (0). Thus, M is a finitely generated A-submodule of a finitely generated vectorspace over the fraction field, K, of A.

Let e1, . . . , en be a K-basis for M(0). Since M is finitely generated, we can clear thedenominators of the coefficients with respect to this basis of the generators of M , pro-ducing a nonzero a ∈ A such that aM lies in the free A-module generated by e1, . . . , en.In particular, M is isomorphic to a submodule of An, and hence is projective by Propo-sition 12.1.5.


Note, then, that any finitely generated submodule of a torsion-free A-module is pro-jective, and hence flat. Thus, we obtain the next corollary from the proof given inCorollary 9.5.12.

Corollary 12.6.13. Let A be a Dedekind domain. Then an A-module is flat if and onlyif it is torsion-free. In particular, every finitely generated flat module is projective.

More information about the ideals will be useful if we are to get a better understandingof the classification of the finitely generated projective modules over a Dedekind domain.The decomposition as a product of maximal ideals has a great deal of redundancy asfar as describing the isomorphism classes of the ideals as A-modules. After all, if Ais a P.I.D., then, independent of this decomposition, any two ideals are isomorphic asmodules.

Lemma 12.6.14. Let a, b, c, and d be ideals of the domain A, and suppose that a ∼= cand b ∼= d as A-modules. Then ab ∼= cd.

Proof Let K be the fraction field of A. Lemma 12.6.4 gives elements α, β ∈ K withαa = c and βb = d. So αβab = cd.

Definition 12.6.15. Let A be a Dedekind domain. The class group, Cl(A), is theabelian group whose elements are the isomorphism classes, [a], of nonzero ideals of A,and whose multiplication is given by

[a][b] = [ab].

Of course, [A] is the identity element of Cl(A), and the inverse of [a] is given by [b],where ab is principal.

Class groups are quite difficult to compute. However, one thing is clear.

Lemma 12.6.16. Let A be a Dedekind domain. Then Cl(A) is the trivial group if andonly if A is a P.I.D.

Next, we shall show that any ideal of a Dedekind domain is generated by at most twoelements. Recall that a principal ideal ring is a commutative ring in which every ideal isprincipal.

Proposition 12.6.17. Let a be a proper, nonzero ideal in the Dedekind domain A. Thenthe quotient ring A/a is a principal ideal ring.

Proof Write a = mr11 . . .mrk

k , with m1, . . . ,mk distinct maximal ideals of A, and ri > 0for all i. Then Corollary 12.6.11 shows that the ideals of A/a are those of the form b/a,where b = ms1

1 . . .msk

k with 0 ≤ si ≤ ri for all i.For such a b, suppose we find b ∈ A such that b ∈ msi

i but b �∈ msi+1i for all i. Then

a calculation of exponents shows that (b) + a ⊂ b, but (b) + a cannot lie in any smallerideal containing a. Thus, (b) + a = b, and b/a = (b).

To find such a b, choose, for each i, a bi ∈ msii such that bi �∈ msi+1

i . Since mi isthe only maximal ideal containing msi+1

i , we have msi+1i + m

sj+1j = A for i �= j. By the

Chinese Remainder Theorem, there is a b ∈ A such that b ≡ bi mod msi+1i for all i. This

is our desired b.

Corollary 12.6.18. Let a be a non-principal ideal in the Dedekind domain A, and leta be any nonzero element in a. Then there is an element b ∈ a such that a is the idealgenerated by a and b.


Proof A/(a) is a principal ideal ring. Just set b equal to any element of a such thata/(a) = (b).

We now wish to classify the projective modules over a Dedekind domain. The nextlemma is basic.

Lemma 12.6.19. Let A be any commutative ring and let a and b be ideals of A whichare relatively prime in the sense that a + b = A. Then ab = a ∩ b.

Proof Clearly, ab ⊂ a∩b. But if 1 = a+ b with a ∈ a and b ∈ b, then for any c ∈ a∩b,we have c = ac+ cb ∈ ab.

Proposition 12.6.20. Let a and b be ideals in the Dedekind domain A. Then

a⊕ b ∼= A⊕ ab.

Proof First assume that a and b are relatively prime. Then, as the reader may check,there’s an exact sequence

0→ a ∩ bψ−→ a⊕ b

ϕ−→ A→ 0,

where ψ(x) = (x,−x), and ϕ(a, b) = a+ b. Since A is free, the sequence splits, giving anisomorphism a⊕ b ∼= A⊕ (a ∩ b), and the result follows from Lemma 12.6.19.

Thus, for an arbitrary pair of ideals a and b, it suffices, by Lemma 12.6.14, to find anideal a′ that is isomorphic to a as an A-module, such that a′ and b are relatively prime.

Let c be an ideal such that ac is principal. Say ac = (a). Applying Proposition 12.6.17to the ideal c/bc of A/bc, we see there’s an element c ∈ c such that c = bc + (c).

Multiplying both sides by a, we get

(a) = ac = abc + ca

= ab + ca

We can then divide both sides by a. We see that a′ = (c/a)a is an ideal of A isomorphicto a as a module, and b + a′ = A.

We are now able to completely classify the finitely generated projective modules overa Dedekind domain modulo an understanding of the isomorphism classes of ideals. Thus,we reduce the classification of the projective modules to the calculation of the class groupCl(A). The next theorem is due to Steinitz.

Theorem 12.6.21. Let A be a Dedekind domain. Suppose given nonzero ideals ai, 1 ≤i ≤ n and bi, 1 ≤ i ≤ k, of A. Then there is an A-module isomorphism

a1 ⊕ · · · ⊕ an ∼= b1 ⊕ · · · ⊕ bk

if and only if n = k and the product ideals a1 . . . an and b1 . . . bk are isomorphic asmodules.

Proof By Proposition 12.6.20,

a1 ⊕ · · · ⊕ an ∼= An−1 ⊕ (a1 . . . an),

and a similar isomorphism holds for the b’s. So if n = k and a1 . . . an ∼= b1 . . . bk, thenthe direct sum of the a’s is isomorphic to the direct sum of the b’s as claimed.


Conversely, suppose given an isomorphism

f : a1 ⊕ · · · ⊕ an∼=−→ b1 ⊕ · · · ⊕ bk.

We first claim that n = k. To see this, note that if c is any nonzero ideal of A andif K is the field of fractions of A, then the localization, K ⊗A c, of c at (0) is a one-dimensional vector space over K by Corollary 12.5.3. Thus, f ⊗ 1K is an isomorphismfrom an n-dimensional vector space over K to an k-dimensional one, so n = k.

Next, note that by Lemma 12.6.4, each composite

ajιj−→ a1 ⊕ · · · ⊕ an

f−→ b1 ⊕ · · · ⊕ bnπi−→ bi

is induced by multiplication by an element αij ∈ K. Here, ιj is the canonical inclusion inthe direct sum and πi is the projection onto the i-th factor. Thus, if M = (αij) ∈Mn(K),then, if we write the n-tuples in a1 ⊕ · · · ⊕ an vertically, we see that

f

⎛⎜⎝ a1

...an

⎞⎟⎠ = M ·

⎛⎜⎝ a1

...an

⎞⎟⎠ .

Note that f−1 is also represented by a matrix in Mn(K), which, by the naturality of thepassage between mappings and matrices, together with the fact that each of the abovedirect sums contains a K-basis for Kn, must be M−1.

We claim that multiplication by detM carries a1 . . . an into b1 . . . bn, and that mul-tiplication by det(M−1) carries b1 . . . bn into a1 . . . an. Since detM and det(M−1) areinverse elements of K (Corollary 10.3.2), this implies that multiplication by detM givesan isomorphism from a1 . . . an to b1 . . . bn, and the proof will be complete.

Let M ′ =

⎛⎜⎜⎜⎝a1 0 . . . 00 a2 . . . 0

. . .0 0 . . . an

⎞⎟⎟⎟⎠ be the diagonal matrix whose entries are a1, . . . , an,

where ai ∈ ai for i = 1, . . . , n. Then the ij-th entry of MM ′ lies in bi for all i, j. Thus,by the formula of Corollary 10.2.6, we see that det(MM ′) ∈ b1 . . . bn. But

det(MM ′) = (detM) · (detM ′) = (detM) · a1 . . . an,

by Propositions 10.3.1 and 10.3.7. Since elements of the form a1 . . . an generate a1 . . . an,

detM · a1 . . . an ⊂ b1 . . . bn,

as claimed.The same proof shows that multiplication by det(M−1) carries b1 . . . bn into a1 . . . an,

so the result follows.

And this gives the following, weaker result.

Corollary 12.6.22. Let A be a Dedekind domain. Then the reduced K-group K0(A) isisomorphic to the class group Cl(A). Thus, we get an isomorphism

K0(A) ∼= Z⊕ Cl(A).


Proof Let K be the fraction field of A. Then we can identify K0(A) with the kernel ofthe natural map K0(A) → K0(K). In particular, the elements of K0(A) are the formaldifferences [P ]− [Q] where the K-ranks of P and Q are equal. Note that if Q⊕Q′ ∼= An

is free, then

[P ]− [Q] = [P ⊕Q′]− [An],

so every element of K0(A) may be written as [P ′]− [An] for some P ′ such that K ⊗A P ′

is an n-dimensional vector space over K.Now Proposition 12.6.20 shows that if P is isomorphic to the direct sum a1⊕· · ·⊕an

of ideals, then

[P ]− [An] = [(a1 . . . an)⊕An−1]− [An] = [a1 . . . an]− [A].

It is now easy to see that there is a surjective group homomorphism ϕ : Cl(A)→ K0(A)given by

ϕ([a]) = [a]− [A].

By Lemma 9.9.8, [a] ∈ kerϕ if and only if there is an isomorphism a⊕An ∼= A⊕Anfor n sufficiently large. But Theorem 12.6.21 says this is impossible unless a ∼= A, inwhich case [a] is the trivial element of Cl(A).

Exercises 12.6.23.1. Show that a Dedekind domain is a U.F.D. if and only if it is a P.I.D.

2. Let A be a Dedekind domain with fraction field K. A fractional ideal of A is definedto be a finitely generated A-submodule of K.

(a) Show that an A-submodule i of A is a fractional ideal if and only if ai ⊂ A forsome a ∈ A. Deduce that every fractional ideal is isomorphic as an A-moduleto an ideal of A.

(b) Show that the nonzero fractional ideals of A form an abelian group I(A) underthe product operation for A-submodules of K.

(c) Show that there is an exact sequence

0→ A× → K× → I(A)p−→ Cl(A)→ 0.

Here K× maps onto the principal fractional ideals in I(A), and for any frac-tional ideal i, p(i) = [a] for any ideal a of A that is isomorphic to i as anA-module.

3. Let a and b be ideals of the Dedekind domain A. Show that the natural map

a⊗A b→ ab

localizes to be an isomorphism at each prime ideal of A, and hence is an isomor-phism of A-modules.

(a) Deduce that if we consider K0(A) to be the kernel of K0(A) to K0(K), andhence an ideal of K0(A), then the product of any two elements of K0(A) inthe ring structure of K0(A) is 0.


(b) Show that the elements of K0(A) that map to the multiplicative identity ofK0(K) form a subgroup of K0(A)× isomorphic to Cl(A). (This group of unitsin K0(A) is known as the Picard group, Pic(A), of A.)

4. Let a1, . . . , ak be nonzero ideals of the Dedekind domain A. Show that ΛkA(a1 ⊕· · · ⊕ ak) ∼= a1 . . . ak. Use this to give a shorter proof of Theorem 12.6.21.

12.7 Integral Dependence

Definitions 12.7.1. Let A ⊂ B be an inclusion of commutative rings. We say thatb ∈ B is integral over A if it is a root of a monic polynomial with coefficients in A.

We say that B itself is integral over A if every element of B is integral over A. Wealso say, in this situation, that B is integrally dependent on A.

We can characterize integral elements as follows. Recall that an A-module M isfaithful if AnnA(M) = 0.

Proposition 12.7.2. Let A ⊂ B be an inclusion of commutative rings and let b ∈ B.Then the following conditions are equivalent.

1. b is integral over A.

2. A[b] is finitely generated as an A-module.

3. There is a subring C ⊂ B, containing both A and b, such that C is finitely generatedas an A-module.

4. There is a faithful A[b]-module M that is finitely generated as an A-module.

Proof Suppose that b is a root of the monic polynomial f(X) ∈ A[X]. Then A[b] isa quotient of A[X]/(f(X)), which is a free module of rank deg f over A by Proposi-tion 7.7.35. Thus, the first condition implies the second.

Clearly, the second condition implies the third, and the third implies the fourth bytaking M = C. Thus, assume given an A[b] module M as in the fourth statement. Letx1, . . . , xn generate M over A. Then we may write bxj =

∑ni=1 aijxi for each j, where

aij ∈ A for all i, j.Let M ′ = (aij) ∈ Mn(A), and let fb ∈ EndA(M) be induced by multiplication by b.

Then the Generalized Cayley–Hamilton Theorem shows that chM ′(fb) = 0 in EndA(M).But M is a faithful A[b]-module, and hence the natural map A[b]→ EndA(M) (i.e., theA-algebra homomorphism that takes b to fb) is injective. Thus, chM ′(b) = 0 in B, andhence b is a root of the monic polynomial chM ′(X) ∈ A[X].

Clearly, if A ⊂ B ⊂ C and if c ∈ C is integral over A, then it is integral over B.Thus, an induction on the second condition above gives the next corollary.

Corollary 12.7.3. Let A ⊂ B be an inclusion of commutative rings and let b1, . . . , bn ∈B be integral over A. Then A[b1, . . . , bn] is finitely generated as an A-module.

Definitions 12.7.4. Let A ⊂ B be an inclusion of commutative rings. Then the integralclosure of A in B is the set of elements in B that are integral over A. Note that byCorollary 12.7.3 and Proposition 12.7.2, the integral closure of A in B is a subring of Bcontaining A.

We say that A is integrally closed in B if the integral closure of A in B is just A itself.An integral domain A is said to be an integrally closed domain if it is integrally closed

in its field of fractions.


We’ve already seen a number of examples of integrally closed domains.

Lemma 12.7.5. A unique factorization domain is integrally closed.

Proof Let A be a U.F.D. with fraction field K, and suppose that α ∈ K is integral overA. Write α = a/b, with a and b relatively prime, and let f(X) = Xk+ak−1X

k−1+· · ·+a0

be a monic polynomial over A with f(α) = 0. Then clearing denominators, we get

ak + ak−1ak−1b+ · · ·+ a0b

k = 0.

The result is that b must divide ak. Since a and b are relatively prime, this forces b tobe a unit, and hence α ∈ A.

Corollary 12.7.6. Suppose given inclusions A ⊂ B ⊂ C of commutative rings, suchthat B is integral over A. Then any element of C that is integral over B is also integralover A.

Proof Let f(X) = Xk + bk−1Xk−1 + · · ·+ b0 be a monic polynomial over B which has

c is a root. Then A[b1, . . . , bn] is finitely generated as an A-module. Since c is integralover A[b1, . . . , bn], A[b1, . . . , bn, c] is finitely generated as an A-module as well.

We now give an important special case of integral closures.

Proposition 12.7.7. Let A be a domain with fraction field K, and let L be an algebraicextension of K. Let B be the integral closure of A in L. Then the field of fractions of Bis equal to L. In fact, for each α ∈ L, there is an a ∈ A such that aα ∈ B. Also, B isintegrally closed in L, and hence B is an integrally closed domain.

Proof Let α ∈ L. Since L is algebraic over K, α is a root of a polynomial withcoefficients inK. In fact, by clearing denominators, we see that α is a root of a polynomialf(X) = anX

n + · · ·+ a0 with coefficients in A.Set g(X) = Xn+

∑n−1i=0 a

n−i−1n aiX

i. Then g(anα) = an−1n f(α) = 0, so anα is integral

over A.It suffices to show that B is integrally closed. But if α ∈ L is integral over B, then it

is integral over A by Corollary 12.7.6, and hence lies in B.

Lemma 12.7.8. Let A be an integrally closed domain with fraction field K. Let L be afinite extension of K, and let B be the integral closure of A in L. Then the norm andtrace behave as follows:

NL/K(B) ⊂ A and trL/K(B) ⊂ A.

Proof Let L′ be any extension of K and let σ : L → L′ be an embedding over K.Then if b ∈ L is a root of the monic polynomial f(X) ∈ A[X], σ(b) is also a root of f .By Proposition 11.14.3, both NL/K(b) and trL/K(b) are integral over A for each b ∈ B.Since these elements lie in K, the result follows from the fact that A is integrally closed.

Lemma 12.7.9. Let A be an integrally closed domain with fraction field K. Let L be afinite extension of K, and let B be the integral closure of A in L. Let b ∈ B. Then the(monic) minimal polynomial, f(X), of b over K lies in A[X].


Proof Let L be an algebraic closure of L. Then we have a factorization

f(X) = (X − b1)r1 . . . (X − bk)rk

in L[X]. For each i, there is an embedding of L in L that carries b to bi, so each bi mustbe integral over A.

The coefficients of f(X) are polynomials in the roots bi, and hence are integral overA. Since A is integrally closed and the coefficients lie in K, the result follows.

Corollary 12.7.10. Let A be an integrally closed domain with fraction field K. Let Lbe an extension field of K, and suppose that α ∈ L is integral over A. Then A[α] is afree A-module of rank [K(α) : K].

Recall that a number field is a finite extension of the rational numbers, Q.

Definitions 12.7.11. Let K be a number field, or, more generally, an algebraic exten-sion of Q. Then the integral closure of Z in K is known as the ring of integers of K, andis denoted by O(K).

The elements of O(K) are known as the algebraic integers in K.

As an example, we calculate the ring of integers of a quadratic extension of Q. Recallfrom Problem 8 of Exercises 8.2.24 that every quadratic extension of Q has the formQ(√n), where n is a square-free integer.

Proposition 12.7.12. Let n �= 1 be a square-free integer. Then

O(Q(√n)) =

{Z[√n] if n �≡ 1 mod 4

Z[

1+√n

2

]if n ≡ 1 mod 4.

Proof Write K = Q(√n) and let α = a + b

√n with a, b ∈ Q. Since the nontrivial

element of Gal(K/Q) takes√n to −√n, Proposition 11.14.3 gives

trK/Q(α)=(a+ b√n) + (a− b√n)=2a and

NK/Q(α)= (a+ b√n) · (a− b√n) =a2 − nb2.

Note that if b �= 0, then the minimal polynomial of α over Q is X2 − trK/Q(α)X +NK/Q(α). This, together with Lemma 12.7.8, shows that α ∈ O(K) if and only if 2a ∈ Zand a2 − nb2 ∈ Z.

Suppose α ∈ O(K). Then a = c/2 with c ∈ Z. Recall that n is square-free. Thus,the norm formula now gives b = d/2 with d ∈ Z, and further gives c2 − nd2 ≡ 0 mod 4.If c is even, then so is d, so that a, b ∈ Z, giving α ∈ Z[

√n]. If c is odd, then so is d,

and hence c2 ≡ d2 ≡ 1 mod 4. But this forces n ≡ 1 mod 4. Thus, if n �≡ 1 mod 4, thenO(K) = Z[

√n], as claimed.

If n ≡ 1 mod 4 and c and d are odd, then α = (c + d√n)/2 is easily seen to have

integral norm and trace, and hence to lie in O(K). Note that since 1 and√n lie in

Z[(1 +√n)/2], it is easy to see that (c+ d

√n)/2 also lies in Z[(1 +

√n)/2] if c and d are

odd. The result follows.

Note that negative values of n (including n = −1) are included in the above.Rings of integers of number fields are the central object of study in the field of alge-

braic number theory. There are many unanswered questions about them. Of particularinterest to many researchers is the study of the rings of integers of the cyclotomic ex-tensions of Q. It is a fact that O(Q(ζn)) = Z[ζn] for all integers n. We shall prove thispresently if n is prime.


Lemma 12.7.13. Let a be the principal ideal of O(Q(ζp)) generated by 1 − ζp, for aprime number, p. Then a ∩ Z = (p) and trQ(ζp)/Q(a) ⊂ (p).

Proof Since Φp(X) = 1 +X + · · ·+Xp−1, we see that

p = Φp(1) = (1− ζp) . . . (1− ζp−1p )

= NQ(ζp)/Q(1− ζp).Since NQ(ζp)/Q : O(Q(ζp)) → Z is a homomorphism of multiplicative monoids, we seethat 1−ζp cannot be a unit in O(Q(ζp)). Thus, the principal ideal a = (1−ζp) is proper,and hence a ∩ Z is a proper ideal of Z containing p. Since (p) is maximal, a ∩ Z = (p).

For k > 1, we have

1− ζkp = (1− ζp) · (1 + ζp + · · ·+ ζk−1p )

in O(Q(ζp)), so that 1 − ζkp ∈ a. Thus, if σ ∈ Gal(Q(ζp)/Q), then σ(1 − ζp) ∈ a. So ifα ∈ O(Q(ζp)), Proposition 11.14.3 gives

trQ(ζp)/Q(α(1− ζp)) =∑

σ∈Gal(Q(ζp)/Q)

σ(α)σ(1− ζp)

∈ a ∩ Z = (p).

Proposition 12.7.14. For any prime p, O(Q(ζp)) = Z[ζp].

Proof For p = 2, the result is trivial, so assume that p is odd. Let α = a0 + a1ζp +· · · + ap−2ζ

p−2p be an element of O(Q(ζp)), with ai ∈ Q for all i. We shall show that

ai ∈ Z for all i.Note first that

α(1− ζp) = a0(1− ζp) + a1(ζp − ζ2p) + · · ·+ ap−2(ζp−2

p − ζp−1p ).

By Proposition 11.14.3, if (k, p) = 1, we have

trQ(ζp)/Q(ζkp ) = ζp + ζ2p + · · ·+ ζp−1

p

= −1,

since ζp is a root of 1 + X + · · · + Xp−1. Thus, if 1 ≤ k ≤ p − 2, ζkp − ζk+1p lies in

the kernel of trQ(ζp)/Q. And since trQ(ζp)/Q(1) = [Q(ζp) : Q] = p − 1, we see thattrQ(ζp)/Q(1− ζp) = p. Thus,

trQ(ζp)/Q(α(1− ζp)) = trQ(ζp)/Q(a0(1− ζp))= a0 · p,

By Lemma 12.7.13, trQ(ζp)/Q(α(1− ζp)) ∈ (p) ⊂ Z, so a0 ∈ Z.Now, ζ−1

p = ζp−1p ∈ O(Q(ζp)), so ζ−1

p (α− a0) is in O(Q(ζp)). By the argument givenfor a0, we see that a1 ∈ Z. By induction, ai ∈ Z for all i.

We shall see presently that the ring of integers in a number field is a Dedekind domain.It behooves us to see how the concepts of integral dependence and of Dedekind domainsinterrelate. First, we shall show that Dedekind domains are integrally closed.


Lemma 12.7.15. Let A ⊂ B be an inclusion of commutative rings and let C be theintegral closure of A in B. Let S be a multiplicative subset of A. Then S−1C is theintegral closure of S−1A in S−1B.

Proof The integral closure of S−1A in S−1B is a subring of S−1B containing bothS−1A and the image of C under the canonical map η : B → S−1B. Thus, it mustcontain S−1C.

Now suppose that b/s is integral over S−1A. Since s lies in A, we see that η(b) mustalso lie in the integral closure of S−1A. Recall from Problem 9 of Exercises 7.11.27 thatwe may assume 1 ∈ S, so that η(b) = b/1.

It suffices to show that there’s an element of the form bt in C with t ∈ S, as b/s =(bt)/(st). Suppose, then, that b/1 is a root of

f(X) = Xn +an−1

sn−1Xn−1 + · · ·+ a0

s0,

where ai ∈ A and si ∈ S for all i. Now set t = s0 . . . sn−1 and set ti =∏j =i sj . Then

bt/1 is a root of

g(X) = Xn + an−1tn−1Xn−1 + · · ·+ aitit

n−i−1Xi + · · ·+ a0t0tn−1,

which we may think of as a polynomial with coefficients in A. So we are done if η isinjective. In the general case, g(bt) gives an element of A that lies in the kernel of η, sothat u · g(bt) = 0 for some u ∈ S. But if we write g(X) = Xn +

∑n−1i=0 a

′iX

i, we see thatbtu is a root of Xn +

∑n−1i=0 a

′iun−iXi, and the result follows.

Proposition 12.7.16. Let A be an integral domain. Then the following conditions areequivalent.

1. A is integrally closed.

2. Ap is integrally closed for all primes p of A.

3. Am is integrally closed for all maximal ideals m of A.

Proof If K is the fraction field, then K = Kp for each prime p. Thus, the fact that thefirst condition implies the second is immediate from Lemma 12.7.15, while the secondcondition immediately implies the third.

Suppose the third condition holds, and let C be the integral closure of A in the fractionfield K. By Lemma 12.7.15, the third condition implies that the inclusion Am ⊂ Cm issurjective for each maximal ideal m of A. By Corollary 7.11.26, this implies that theinclusion A ⊂ C is also surjective, and hence A is integrally closed.

Corollary 12.5.4 shows that every localization of a Dedekind domain is a discretevaluation ring. So Lemma 12.7.5 shows that every localization of a Dedekind domain isintegrally closed. Thus, Proposition 12.7.16 gives the following corollary.

Corollary 12.7.17. Dedekind domains are integrally closed.

We next wish to examine the relationship between integral dependence and Krulldimension.

Lemma 12.7.18. Let A ⊂ B be an inclusion of domains such that B is integral over A.Let b be any nonzero ideal of B. Then b ∩A �= 0.


Proof Let b be a nonzero element of b and let f(X) = Xn +∑n−1i=0 aiX

i be a monicpolynomial over A with f(b) = 0. Note that if a0 = 0, then f is divisible by X. Since Bis a domain, the result of this division will still have b as a root. By induction, we canfind a monic polynomial over A with a nonzero constant term with b as a root. But theconstant term is then easily seen to be an element of b ∩A.

Recall that for any ring homomorphism f : A → B between commutative rings andany prime ideal p of B, f−1(p) is a prime ideal of A.

Proposition 12.7.19. Let A ⊂ B be an inclusion of commutative rings such that B isintegral over A. Let

q0 ⊂ q1 ⊂ · · · ⊂ qn

be a sequence of proper inclusions of prime ideals of B. Then the inclusions

(q0 ∩A) ⊂ (q1 ∩A) ⊂ · · · ⊂ (qn ∩A)

are also proper.Thus, the Krull dimension of B is less than or equal to the Krull dimension of A.

Proof It suffices to assume that n = 1 above. Thus, suppose we have a proper inclusionq0 ⊂ q1 of prime ideals of B. Suppose, then, that q0 ∩A = q1 ∩A = p.

Then we have an inclusion of domains A/p ⊂ B/q0. And the prime ideal q1 = q1/q0

of the latter has the property that q1 ∩ A/p = 0. But B/q0 is integral over A/p, so thiscontradicts Lemma 12.7.18.

We wish to show the opposite implication with regard to the Krull dimensions of Aand B. Here is a start.

Lemma 12.7.20. Let A ⊂ B be an inclusion of domains such that B is integral over A.Then B is a field if and only if A is a field.

Proof One direction is immediate from Proposition 12.7.19, as a domain is a field ifand only if it has Krull dimension 0. Thus, if A is a field, so is B.

Conversely, suppose that B is a field, and let 0 �= a ∈ A. Then a−1 lies in B, andhence is integral over A. We obtain an equation

a−n = −an−1a−(n−1) − · · · − a0

with coefficients in A. Multiplying both sides of this equation by an−1, we get

a−1 = −an−1 − an−2a− · · · − a0an−1,

so a−1 lies in A. Since a was arbitrary, A is a field.

Corollary 12.7.21. Let A ⊂ B be an inclusion of commutative rings such that B isintegral over A. Let m be a maximal ideal of B. Then m ∩A is a maximal ideal of A.

Proof Since m ∩A is prime in A, we have an inclusion A/(m ∩A) ⊂ B/m of domains,where B/m is integral over A/(m∩A). Since B/m is a field, A/(m∩A) must be also.

Proposition 12.7.22. Let A ⊂ B be an inclusion of commutative rings such that B isintegral over A. Let p be a prime ideal of A. Then there is a prime ideal q of B withq ∩A = p.


Proof We have a commutative diagram

A B

Ap Bp

��

η

��⊂

��

η

��⊂.

By Lemma 12.7.15, Bp is integral over Ap. Thus, if m is a maximal ideal of Bp, Corol-lary 12.7.21 shows that m ∩ Ap must be the unique maximal ideal, pp, of Ap. Sincep = η−1(pp), the result follows from the commutativity of the square.

We can now prove what’s known as the Going Up Theorem of Cohen and Seidenberg.

Theorem 12.7.23. (Going Up Theorem) Let A ⊂ B be an inclusion of commutativerings such that B is integral over A. Suppose given a sequence of inclusions

p0 ⊂ · · · ⊂ pn

of prime ideals of A, together with a “lift” of the beginning part of the sequence to asequence of prime ideals of B: Thus, we are given a sequence

q0 ⊂ · · · ⊂ qk

of prime ideals of B, with k < n, such that qi ∩A = pi.Then we may extend this to a lift of the entire sequence, producing prime ideals qi of

B for i = k + 1, . . . , n such that qi ∩A = pi for all i, and so that

q0 ⊂ · · · ⊂ qn.

Proof An easy induction shows that we may assume that n = 1 and k = 0. Now passto the inclusion A/p0 ⊂ B/q0 of domains, and apply Proposition 12.7.22 to the primeideal p1/p0 of A/p0.

Corollary 12.7.24. Let A ⊂ B be an inclusion of commutative rings such that B isintegral over A. Then the Krull dimensions of A and B are equal in the sense that ifeither is finite, so is the other, and the two dimensions then agree.

Proof By Proposition 12.7.19, it suffices to show that if A has a chain of properinclusions of prime ideals of length n, then so does B. But this is immediate fromProposition 12.7.22 and the Going Up Theorem.

We shall now give a characterization of Dedekind domains to the effect that theyare the Noetherian, integrally closed domains of Krull dimension ≤ 1. First, we shallreexamine the notion of invertibility.

Definition 12.7.25. Let A be an integral domain with fraction field K and let a be anonzero ideal of A. We write a−1 for the A-submodule of K given by

a−1 = {α ∈ K |αa ⊂ A}.Note that for a ∈ a, we have aa−1 ⊂ A. Thus, a−1 is isomorphic as an A-module

to an ideal of A. This property constitutes one definition for a fractional ideal over ageneral domain.

Recall that there is a product operation on the A-submodules of K, generalizing theproduct of ideals. The inverse A-submodule of an ideal a satisfies the following importantproperty.


Lemma 12.7.26. Let a be a nonzero ideal of the domain A. Then a is invertible if andonly if aa−1 = A.

Proof Let 0 �= a ∈ a. Then aa−1 = (a)a−1 is an ideal of A. If aa−1 = A, we havea · (aa−1) = (a), so a is invertible.

Conversely, if a is invertible, there is an ideal b such that ab = (a) for some a ∈ A.But then a · ((1/a)b) = A, so (1/a)b ⊂ a−1, and A = a · ((1/a)b) ⊂ aa−1 ⊂ A, soaa−1 = A.

Recall from Proposition 7.6.16 that Noetherian rings possess a stronger maximalityprinciple than is found in general rings: If S is any nonempty family of ideals in aNoetherian ring, then there are maximal elements in S. This permits a sort of downwardinduction on ideals in a Noetherian ring. To show that a given property holds, onemay argue by contradiction, choosing a maximal element in the set of ideals in whichthe property doesn’t hold, and deriving a contradiction. Here is an illustration of thistechnique.

Lemma 12.7.27. Let A be a Noetherian commutative ring. Then every ideal of A con-tains a product of prime ideals.

If A is a Noetherian domain but not a field, then every nonzero ideal of A contains aproduct of nonzero prime ideals.

Proof We shall treat the statement in the first paragraph. The proof of the other issimilar.

Let a be maximal in the collection of ideals that do not contain a product of primeideals. Then a is neither A nor a prime ideal of A. In particular, we can find a, b ∈ Asuch that neither a nor b lies in a, but ab lies in a. Thus, a + (a) and a + (b) strictlycontain a.

By the maximality of a, this says that both a + (a) and a + (b) contain products ofprime ideals. But (a+(a))(a+(b)) ⊂ a, so a must also contain a product of prime ideals.Thus, the set of ideals that do not contain a product of primes must be empty.

Another example of Noetherian induction will be useful in our characterization ofDedekind domains.

Proposition 12.7.28. Let A be a Noetherian domain such that every maximal ideal ofA is invertible. Then A is a Dedekind domain.

Proof We claim that every proper, nonzero ideal is a product of maximal ideals. Sup-posing this false, let a be a maximal element in the set of proper, nonzero ideals thatare not products of maximal ideals. Then a cannot be maximal, and hence must becontained in a maximal ideal m.

Our hypothesis says that m is invertible, and hence m−1m = A by Lemma 12.7.26.Since a ⊂ m, m−1a ⊂ m−1m, and hence is an ideal of A. Since A ⊂ m−1, a ⊂ m−1a.

Note now that m−1a properly contains a, as otherwise ma = a, and hence Problem 4of Exercises 10.4.11 shows that a = 0.

Also, m−1a �= A, as otherwise, a = m. So the maximality of a shows that m−1a is aproduct of maximal ideals. But then a = m · m−1a is also a product of maximal ideals,and we obtain our desired contradiction.

Recall that Dedekind domains are Noetherian and integrally closed. Thus, the nextlemma will allow us to close in on our new characterization of Dedekind domains.


Lemma 12.7.29. Let A be a Noetherian, integrally closed domain that is not a field.Let m be a maximal ideal of A that is not invertible. Then the inverse fractional idealsatisfies m−1 = A.

Proof From the definition of m−1 we see that A ⊂ m−1 and m−1m ⊂ A. Thus, we haveinclusions

m ⊂ m−1m ⊂ Aof ideals. Since m is maximal, one of these inclusions is the identity, and since m is notinvertible, we must have m = m−1m.

Let α ∈ m−1. Since m−1m = m, we must have αm ⊂ m, and hence m is an A[α]-module. Since A is Noetherian, m is finitely generated as an A-module. Since everynonzero A[α]-submodule of the fraction field K is faithful, Proposition 12.7.2 showsthat α is integral over A. Since A is integrally closed, this says α ∈ A, and the resultfollows.

We can now give our new characterization of Dedekind domains.

Theorem 12.7.30. A domain is a Dedekind domain if and only if it is Noetherian,integrally closed, and has Krull dimension ≤ 1.

Proof That Dedekind domains have these three properties is given in Corollary 12.6.7(via Theorem 12.6.8), Corollary 12.7.17, and Proposition 12.6.3.

For the converse, Lemma 12.7.29 and Proposition 12.7.28 show that if A is a Noethe-rian, integrally closed domain of Krull dimension 1, then A is Dedekind provided we canshow that for any maximal ideal m of A, there is an element α of the inverse fractionalideal m−1 that does not lie in A. (To show the existence of such an α, all we shall needis that A is a Noetherian domain of Krull dimension 1.)

Let a ∈ m with (a) �= m. Since A is Noetherian, Lemma 12.7.27 shows that (a) con-tains a product of nonzero prime ideals, which are maximal, since A has Krull dimension1. Let r be the smallest integer such that (a) contains a product m1 . . .mr of r maximalideals. Since (a) �= m, r > 1. Since

m1 . . .mr ⊂ (a) ⊂ m,

one of the mi must lie in m because m is prime. Since both mi and m are maximal, weobtain that m = mi. For simplicity, assume i = 1.

By the minimality of r, we can find b ∈ m2 . . .mr such that b �∈ (a). Now

bm ⊂ m1 . . .mr ⊂ (a),

so (b/a)m ⊂ A. Since b �∈ (a), a doesn’t divide b, and hence α = (b/a) is not in A. Butα ∈ m−1, and the result follows.

We can now address the question of rings of integers.

Theorem 12.7.31. Let A be a Dedekind domain with fraction field K. Let L be a finiteseparable extension field of K, and let B be the integral closure of A in L. Then B is aDedekind domain with fraction field L.

In addition, B is finitely generated and projective as an A-module, with Ap ⊗A Bbeing a free Ap-module of rank [L : K] for each prime ideal p of A. In particular, if A isa P.I.D., then B is a free A-module of rank [L : K].

If m′ is a maximal ideal of B, then m = m′ ∩A is a maximal ideal of A, and B/m′ isa finite extension of A/m, with [B/m′ : A/m] ≤ [L : K].


Proof B is integrally closed by Proposition 12.7.7, and has Krull dimension ≤ 1 byProposition 12.7.19. Once we show that B is a finitely generated A-module, we will knowthat B is Noetherian, at which point Theorem 12.7.30 will show that B is Dedekind, asclaimed. Thus, it suffices to verify the assertions of the second and third paragraphs ofthe statement.

Let L be an algebraic closure of L and let σ1, . . . , σn be all the distinct embeddings(as a field) of L in L that restrict to the identity on K. Since L is separable over K,n = [L : K] by Proposition 11.4.12.

Recall from Proposition 11.14.3 that the trace function trL/K : L → K of L over Kis given by

trL/K(α) = σ1(α) + · · ·+ σn(α),

because L is separable over K. Since the σi are ring homomorphisms, trL/K is easilyseen to be a homomorphism of vector spaces over K. By Lemma 12.7.8, it restricts toan A-module homomorphism, trL/K : B → A.

Now define

tr : L→ HomK(L,K)

by tr(α)(β) = trL/K(αβ). We claim that tr is an isomorphism of K-vector spaces. To seethis, since both vector spaces have dimension n, it suffices to show that tr is injective.Thus, we must show that for each nonzero α in L, there exists a β ∈ L such thattrL/K(αβ) �= 0. Thus, since L is a field, it suffices to show that trL/K : L → K is notthe trivial homomorphism. Since trL/K is a sum of distinct embeddings, this followsimmediately from Corollary 11.9.3.

Recall from Proposition 12.7.7 that for any α ∈ L there is an a ∈ A such thataα ∈ B. In particular, beginning with any K-basis of L, we can construct a new basiswhose elements all lie in B. Let b1, . . . , bn be such a basis, and define f : L→ Kn by

f(α) = (trL/K(b1α), . . . , trL/K(bnα)).

Since b1, . . . , bn is a basis for L over K, any element in the kernel of f must also lie inthe kernel of tr. Thus, f is an isomorphism of K-vector spaces.

But trL/K maps B into A, so f is easily seen to restrict to an embedding of A-modules, f : B → An. Thus, since A is hereditary and Noetherian, B is a finitelygenerated projective A-module by Proposition 12.1.5. In fact, B is isomorphic as anA-module to a direct sum a1 ⊕ · · · ⊕ ak of ideals of A.

If B is the direct sum of k nonzero ideals, then Ap ⊗A B is easily seen to have rankk over Ap for each prime ideal p of A. But Proposition 12.7.7 shows that every elementof L has the form b/a for b in B and a ∈ A, so that L = K ⊗A B. Thus, k = n.

If m′ is a maximal ideal of B, then m = m′ ∩A is maximal in A by Corollary 12.7.21.Write S ⊂ A for the complement of m in A. We’ve just seen that S−1B is a free S−1A-module of rank [L : K], so it suffices to show that the natural maps A/m→ S−1A/S−1mand B/m′ → S−1B/S−1m′ are isomorphisms. The former is an isomorphism by Propo-sition 7.11.24, and the latter is an isomorphism by a similar argument.

Without the assumption of separability, we shall show the following.

Theorem 12.7.32. Let A be a Dedekind domain with fraction field K. Let L be a finiteextension of K and let B be the integral closure of A in L. Then B is a Dedekind domain.


Proof Let L0 be the separable closure of K in L. (See Problem 2 of Exercises 11.4.20.)Then L0 is separable over K, and L is purely inseparable over L0. Let C be the integralclosure of A in L0. Then Theorem 12.7.31 shows C to be a Dedekind domain. Since Bis the integral closure of C in L, we may assume that K = L0 and that L is a purelyinseparable extension of K.

By Problem 2 of Exercises 11.4.20, the minimal polynomial of any element β ∈ Lover K has the form Xpe − a for some a ∈ K. By Lemma 12.7.9, we have a ∈ A if andonly if β ∈ B. Since the degrees pe must all divide [L : K], there is an integer r suchthat βp

r ∈ K for all β ∈ L, and β ∈ B if and only if βpr ∈ A.

Let L be an algebraic closure of L and let ϕ : L→ L be the Frobenius homomorphism.Let K = (ϕr)−1(K) and let A = (ϕr)−1(A). Then the above shows that L ⊂ K, andB = L ∩ A. Since ϕ is a homomorphism of fields, A is isomorphic to A, and hence is aDedekind domain.

By Theorem 12.6.8 and Proposition 12.6.6, it suffices to show that every nonzero idealb of B is invertible. Since A is Dedekind, we know that Ab is invertible as an ideal of A.By Lemma 12.7.26, this means that if we set

M = {γ ∈ K | γ · Ab ⊂ A},

then M · Ab = A. This is easily seen to imply that there are elements γi ∈M and bi ∈ bsuch that

∑ki=1 γibi = 1.

Applying ϕr, we obtain that∑ki=1 δibi = 1, where δi = γp

r

i bpr−1i . By Lemma 12.7.26,

it suffices to show that each δi ∈ L and that δib ⊂ B.Since ϕr(γi) ∈ K and bi ∈ b, δi ∈ L, as desired. The result now follows since if b ∈ b,

δib = (γibi)pr−1(γib) lies in A ∩ L = B.

Exercises 12.7.33.1. Show that the assertion of Lemma 12.7.18 becomes false if we remove the assump-

tion that A and B are domains.

2. Let n be a square-free integer. What is the integral closure of Z[1/2] in Q(√n)?

3. Let A be a Dedekind domain and let B be the integral closure of A in a finiteextension of its field of fractions. Let p be a nonzero prime ideal of A. Show thatBp is a proper ideal of B. Suppose Bp = qr11 . . . qrk

k , where q1, . . . , qk are distinctprime ideals of B and each ri > 0. (We say that p ramifies in B if ri > 1 for somei.) Find all ideals b of B with the property that b ∩A = p.

4. Let A be an integrally closed domain and let B be the integral closure of A in afinite extension, L, of its field of fractions, K. Show that b ∈ B× if and only ifNL/K(b) ∈ A×.

5. Let p be a prime and let r > 0. Let a be the principal ideal of Z[ζpr ] generated by1 − ζpr . Show that a(p−1)pr−1

= (p), the principal ideal of Z[ζpr ] generated by p.Deduce (unless p = 2 and r = 1) that p ramifies in O(Q(ζpr )).

6. Let n > 1 be an integer with more than one prime divisor. Show that 1 − ζn is aunit in Z[ζn].

Bibliography

Atiyah, M.F., and Macdonald, I.G., Introduction to Commutative Algebra, Addison–Wesley, Massachusetts, 1969.

Bass, H., Algebraic K-theory , Benjamin, New York, 1968.

Bass, H. Ed., Algebraic K-Theory I , Lecture Notes in Mathematics 341, Springer–Verlag, Germany, 1973.

Brown, K.S., Cohomology of Groups, Springer–Verlag, Germany, 1982.

Cartan, H., and Eilenberg, S., Homological Algebra, Princeton University Press,New Jersey, 1956.

Coxeter, H.S.M., and Moser, W.O., Generators and Relations for Discrete Groups,Springer–Verlag, Germany, 1964.

Curtis, C., and Reiner, I., Representation Theory of Finite Groups and AssociativeAlgebras, Interscience Publishers, New York, 1962.

Gorenstein, D., Finite Simple Groups, Plenum Press, New York, 1982.

Hall, M., The Theory of Groups, Macmillan, New York, 1959.

Hilton, P.J., and Stammbach, U., A Course in Homological Algebra, Springer–Verlag, Germany, 1971.

Mac Lane, S., Homology , Springer–Verlag, Germany, 1967.

Milnor, J., Introduction to Algebraic K-theory , Annals of Mathematics Studies,Princeton University Press, New Jersey, 1971.

Mumford, D., Introduction to Algebraic Geometry , Notes, Department of Mathe-matics, Harvard University, Massachusetts, 1967.

Rotman, J.J., An Introduction to the Theory of Groups, Wm. C. Brown Publishers,Iowa, 1988.

Samuel, P., Algebraic Theory of Numbers, Hermann, France, 1972.

550


Serre, J.-P., Corps locaux , Publications de l’Institut do Mathematique deo l’Universitede Nancago VIII, Hermann, France, 1968.

Serre, J.-P., Linear Representations of Finite Groups, Springer–Verlag, Germany,1977.

Washington, L.C., Introduction to Cyclotomic Fields, Springer–Verlag, Germany,1982.

Weil, A., Basic Number Theory , Springer–Verlag, Germany, 1967.

Zariski, O., and Samuel, P., Commutative Algebra, Vols. I and II, Van Nostrand,New Jersey, 1958 and 1960.

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