Galois theory, commutative algebra, with applications to ...

Galois theory, commutative algebra, withapplications to finite fields

Preface

This is the written notes for the course entitled “Galois theory and commu-tative algebra” given in the fall of 2004. However, this notes is not faithful tothe one given in class as I found it necessary to rearrange some topics. Forexample, ring of algebraic integers was mentioned both in Galois theory andDedekind domain during the semester and in this notes only one chapter isdevoted to it. Another effort I am trying to make is putting more details tothe examples and exercise.

The purpose of the course is to build a bridge between basic algebra corecourses and advanced topics such as coding theory, algebraic number theory,and algebraic geometry. Hence we assume that the students have some basicbackground on ring theory and module theory.

The notes consists of three parts: Galois theory, introduction to commu-tative algebra, and applications to coding theory.

In the first part, Galois theory, we emphasize on the fundamental theoryof Galois group, and some basic properties of fields such as normality, separa-bility, and certain structures of their Galois group such as cyclic extensions.Throughout the first part, one of the main themes is finite field. We are alsohoping to cover some basic properties for function fields with characteristicgreater than 0. These are motivated by the fact that finite field is one of themost extensively used mathematical tools in other sciences and technology.On the other hand, function fields with characteristic greater 0 have attract-ing growing attention as they are used in various applications such as codingtheory. We would like to point out that we omit some very classical topicsin Galois theory such as ruler and compass constructions. It is because thatthe audience might have learned these topics from basic abstract algebra. Ifnot, they can be found in most books contained Galois theory such as theone by Jacobson [Jac85] or the one by Hungerford [Hun80].

The topics to be considered in part two, introduction to commutative al-

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4

gebra, besides some reviews, consists of primary decomposition, chain condi-tion, Noetherian and Artinian rings, discrete valuations rings, and Dedekinddomains. Hopefully they serve a preparation for learning algebraic numbertheory and algebraic curves which can be used for constructing geometriccodes, mainly Goppa codes.

A Dedekind domain is an commutative domain in which every ideal isfinitely generated, every nonzero prime ideal is a maximal ideal, and whichis integrally closed in its fraction field.

In another description, an commutative domain R is a Dedekind domainif and only if the localization of R at each prime ideal p of R is a discretevaluation ring.

Some examples of Dedekind domains are the ring of integers, the polyno-mial rings F [x] in one variable over any field F , and any other principal idealdomain. Not all Dedekind domains are principal ideal domains however. Themost important examples of Dedekind domains, and historically the motivat-ing ones, arise from algebraic number fields that we have considered before.

The study of Dedekind domains was initiated by Dedekind, who intro-duced the notion of ideal in a ring in the hopes of compensating for the fail-ure of unique factorization into primes in rings of algebraic integers. Whilenot all Dedekind domains are unique factorization domains, they all have aproperty closed to unique factorization: primary decomposition. This meansevery ideal can be uniquely factored as a product of primary ideals.

There is another example of Dedekind domains generalizing the examplethat F [x] being a Dedekind domain. Let C be any smooth algebraic curveover an algebraically closed field F . Then the ring of functions of C over Fis also a Dedekind domain. We use that here as a motivation for learningelliptic curves, commutative algebra, and algebraic geometry. See the firsttwo chapter of [Sil86] for a more detailed description of algebraic curves.

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In last part on coding theory consists of student presentations

• James Fiedler: Hamming codes and hat game

• Mehmed Dagli: BCH codes

• Tim Zick: Perfect codes

• Theodore Rice: Greedy codes and games

• Key One Chung: Goppa codes

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Basic notations

• Z: the ring of rational integers;

• Q: The field of rational numbers;

• F, E, K: fields;

• F [X]: the smallest ring containing both the field F and the set X;

• F (X): the smallest field containing both the field F and the set X; orthe field of fractions of F [X];

• | · |: cardinality or order;

• OK: the ring of integers of the field K;

• Fpn : a finite field of characteristic pn;

• F : an algebraic closure of the field F ;

• E/F : the field E is an extension of the field F ;

• [E : F ]: degree of the field extension;

• Gal(E/F ): The Galois group of the extension E/F ;

• InvG: the field of invariants of the automorphism group G;

• TEF (u): trace of the element u ∈ E;

• NEF (u): norm of the element u ∈ E;

• R,A: rings;

• M : module;

• l(M): length of the module M ;

• a, b, c: ideals;

• r(a), or√

a: the radical of the ideal a;

• p: prime ideals;

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• q: primary ideal;

• M: maximal ideal;

• N: the Nilradical of the ring;

• R: the Jacobson radical of the ring;

• S−1A: ring of fractions with respect to the multiplicative set S;

• Ap: ring A localized at the prime ideal p;

• dim A: the dimension of ring A;

• Ann(x): the annihilating ideal of element x in A;

• A1k: the affine line over the field k;

• Spec(A): the spectrum of the ring A;

Contents

I Galois theory 13

1 Field extension 15

1.1 Field extension and examples . . . . . . . . . . . . . . . . . . 15

1.2 Simple extension . . . . . . . . . . . . . . . . . . . . . . . . . 20

2 Galois groups of field extensions 23

2.1 Field isomorphism . . . . . . . . . . . . . . . . . . . . . . . . 23

2.2 Splitting fields of a polynomial . . . . . . . . . . . . . . . . . . 24

2.3 Multiple roots . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4 Algebraic closure . . . . . . . . . . . . . . . . . . . . . . . . . 28

2.5 Galois groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

2.6 Galois extension . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.7 Fundamental theorem of Galois theory . . . . . . . . . . . . . 33

3 Separability 37

3.1 Purely inseparable extensions . . . . . . . . . . . . . . . . . . 37

3.2 Separable extensions . . . . . . . . . . . . . . . . . . . . . . . 39

4 Cyclic extensions 45

4.1 Trace and Norm . . . . . . . . . . . . . . . . . . . . . . . . . . 45

4.2 Cyclic extension . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5 Finite fields 49

5.1 Properties of finite fields . . . . . . . . . . . . . . . . . . . . . 49

5.2 Irreducible polynomials over Fp . . . . . . . . . . . . . . . . . 51

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10 CONTENTS

II Commutative algebra 53

6 Rings and ideals-Review 55

6.1 Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6.2 Local rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

6.3 Nilradical and Jacobson radical . . . . . . . . . . . . . . . . . 56

6.4 Rings and modules of fractions . . . . . . . . . . . . . . . . . 57

6.5 Nakayama’s lemma . . . . . . . . . . . . . . . . . . . . . . . . 58

6.6 Integral dependency . . . . . . . . . . . . . . . . . . . . . . . 59

6.7 Integrally closed domains . . . . . . . . . . . . . . . . . . . . . 61

7 Primary decomposition 63

7.1 Primary ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.2 Primary decomposition . . . . . . . . . . . . . . . . . . . . . 64

7.3 Minimal primary decomposition . . . . . . . . . . . . . . . . 64

8 Chain condition 67

8.1 Chain condition . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8.2 Basic properties of Noetherian and Artinian rings . . . . . . . 68

9 Noetherian rings 73

9.1 Properties of Noetherian rings . . . . . . . . . . . . . . . . . . 73

9.2 Primary decomposition for Noetherian rings . . . . . . . . . . 76

10 Artinian Rings 79

10.1 Further properties of Artinian rings . . . . . . . . . . . . . . . 79

10.2 Local Artinian rings . . . . . . . . . . . . . . . . . . . . . . . 80

11 Discrete valuation ring and Dedekind domains 83

11.1 Discrete valuation rings . . . . . . . . . . . . . . . . . . . . . . 83

11.2 Dedekind domains . . . . . . . . . . . . . . . . . . . . . . . . 85

12 Algebraic integers 87

12.1 Algebraic integers form a ring and a Z-module . . . . . . . . . 87

12.2 Ring of integers of a number field is a Dedekind domain . . . . 89

12.3 Primary decomposition for imaginary quadratic number fields 89

CONTENTS 11

13 Algebraic curves 9113.1 Affine varieties and affine curves . . . . . . . . . . . . . . . . . 9113.2 An example -Elliptic curve . . . . . . . . . . . . . . . . . . . . 9213.3 An example of singular algebraic curve . . . . . . . . . . . . . 94

12 CONTENTS

Part I

Galois theory

13

Chapter 1

Field extension

1.1 Field extension and examples

Definition 1.1.1. A field is a commutative division ring. The smallest nat-ural number p such that summation of p copies of 1 yields 0 is called thecharacteristic of the field. Since a field contains no zero divisor, then sucha p, if it exits, must be a prime integer. If there is no such natural numberexists, the field is said to have characteristic 0. Give a field F , we use |F | todenote the cardinality of F .

Example 1.1.1. Let Q be the set of all rational numbers, then Q is a field.

Example 1.1.2. Z/5Z = {0, 1, 2, 3, 4} is a field. In particular, a + b = a + b,and ab = ab. Given a ∈ Z/NZ such that a 6= 0. Pick any representative afor a in Z. It is easy to see a is coprime to 5. Then there exist two integerss and t such that as + 5t = 1. Modulo 5, we have as = 1 and (a)−1 = s.

In general, given any prime number p, the ring Z/pZ is a field with pelements. We denote such a field by Fp. Such a field is called a prime finitefield.

Example 1.1.3. Z/10Z is not a field as 2 is a zero divisor.

Example 1.1.4. Let i be a primitive 4th root of unity such that i2 = −1. Theset {a + bi | a, b ∈ Q} is a vector space over Q of dimension 2. This set isa field. We only need to check every non-zero element is a unit. Given anya, b ∈ Q which are not simultaneously zero, then (a + bi) a−bi

a2+b2= 1. Hence

a + bi is a unit. This field is denoted by Q(i).

15

16 CHAPTER 1. FIELD EXTENSION

Example 1.1.5. Let i as above, then the set {a + bi | a, b ∈ F5} is not a field.It is because (2 + i)(2− i) = 0.

Exercise 1.1.1. Let S be an element such that S2 = −3. Prove that {a +bS|a, b ∈ F5} is a field. It is a field with 52 elements, denoted F5(S) or F52

1.

Remark 1.1.1. The difference between these two cases is that x2 = −1 hastwo solutions in F5. However, there is no solution of x2 = −3 in F5.

Definition 1.1.2. A field F is called an extension (field) of the field K if Fcontains K as a subfield. If F is an extension of K, their relation is denotedby F/K. The degree of the field extension of F over K, denoted by [F : K]is the dimension of F as a K-vector space. When [F : K] < ∞, F is calleda finite extension of K.

Example 1.1.6. The field Q(i) is a degree 2 extension of the field Q.

Example 1.1.7. Any finite field F must have characteristic p greater than 0.Hence F must contains Fp as a subfield.

Example 1.1.8. The field F52 is a degree 2 extension of F5.

Remark 1.1.2. Don’t confuse the ring Z/25Z and the field F52 . Z/25Z is noteven an integral domain.

Proposition 1.1.1. Let F be a finite field with characteristic p. Then thecardinality of F is a power of p.

Proof. Since Fp is a subfield of F and hence |F| = p[F:Fp].

Definition 1.1.3. Let R be a commutative integral domain with units, theset

{ab| a, b ∈ R, b 6= 0

}is a field. it is called the field of rational functions

of R.

Let F be a field and x an indeterminant. Then the polynomial ring F [x]is an integral domain. The quotient field of this ring, denoted by F (x), is

the set {f(x)g(x)|f(x), g(x) ∈ F [x], g(x) 6= 0}.

Exercise 1.1.2. Let F be a field and x an indeterminant. Check that F [x] isan integral domain and a principal ideal domain (P.I.D.).

Example 1.1.9. The field Q(x) is an extension of Q as every rational numberis a constant function. [Q(x) : Q] = ∞.

1We will see all fields with 25 elements are isomorphic

1.1. FIELD EXTENSION AND EXAMPLES 17

The ring Q[x] is a principle ideal domain. The polynomial x2 + 1 isirreducible over Q. Hence the ideal (x2 + 1) is a prime ideal and hence amaximal one.

Exercise 1.1.3. The quotient Q[x]/(x2 + 1) is isomorphic to Q(i).

Theorem 1.1.1. Let F be a field extension of E and let E be a field extensionof K. Then F is an extension of K and [F : K] := [F : E][E : K].[F : K] < ∞ if and only if [F : E] < ∞ and [E : K] < ∞.

Proof. When [F : K] < ∞, then F is an [F : E] dimensional vector spaceover E and E is an [E : K] dimensional vector space over K. Hence F is an[F : K] = [F : E][E : K] dimensional vector space over K.

Exercise 1.1.4. Let F :={a + 3

√2b + 3

√4c | a, b, c ∈ Q(i)

}. Prove that F is

a field and [F : Q] = 6.

If F is an extension of the field K and X is a set in F . Then we let K(X)(resp. K[X]) denote the intersection of all sub fields(subrings) of F whichcontain X

⋃K. To be more explicitly, if X is a finite set {u1, u2, ..., un} then

we write F (X) = F (u1, ..., un) (resp. F [X] = F [u1, ..., un]). If X = {u},then X(u) is said to be a simple extension of K. It is easy to see that theorder of the elements ui does not matter. Moreover, F (u1, ..., un−1)(un) =F (u1, u2, ..., un) (resp. F [u1, ..., un−1][un] = F [u1, u2, ..., un]).

Example 1.1.10. Let X = {√−3,

√−2}. Please describe F5(

√−3,

√−2) =

F5(√−3)(

√−2) = F52(

√−2). Now we would like to check whether

√−2

is in the field F52 or not. This is equivalent to say we would like to knowwhether x2+2 = 0 has any solution in F52 . We can check ±2

√−3 are the two

solutions of this equation. Hence F5(√−3,

√−2) = F5(

√−3) = F5(

√−2).

Theorem 1.1.2. If F is an extension field of K, u, ui ∈ F , and X ⊂ F ,then

1. the subring K[u] consists of all elements of the form f(u) where f is apolynomial with coefficients in K.

2. the subring K[u1, ..., um] consists of all elements of the form g(u1, ..., um)where g ∈ K[x1, ..., xn].

3. the subring K[X] consists of all elements of the form h(u1, ..., un) whereui ∈ X, n a positive integer, and h ∈ K[x1, ..., xn].


4. the subfield K(u) (resp. K(u1, ..., um), K(X)) is the quotient field ofthe ring K[u] (resp. K(u1, ..., um), K[X]).

Proof. We will only show the proof of 1 and it can be generalized to provestatement 2, 3, and 4.

1. All elements of the form f(u) where f(x) ∈ K[x] form a ring, denotedby R′. Since K[u] is a ring contains K ∩ {u}, hence it contains R′. SinceK ∩{u} ⊂ R′, and K[u] is the smallest ring contains K ∩{u}, hence K[u] ⊆R′. So R′ = K[u].

Exercise 1.1.5. Let X = {i}. We have described before Q(i) (resp. F5(√−3))

as a vector space over Q (resp. F5). Please describe the rings Q[i] andF5[√−3]. How about the field Q(i,

√2) and the ring Q[i,

√2]?

If L and M are subfields of a field F , the composite of L and M in F ,denoted by LM is the subfield generated by the set L

⋃M .

Example 1.1.11. The composite of Q(i) and Q(√

2) is the field Q(i,√

2) =Q(√

2 + i).

Definition 1.1.4. Let F be an extension field of K. An element u of Fis said to be algebraic over K provided that u is a root of some nonzeropolynomial f ∈ K[x]. An irreducible monic polynomial in K[x] of degreen ≥ 1 uniquely determined by the conditions that f(u) = 0 is called theminimal polynomial of u.

If u is not a root of any nonzero f ∈ K[x], u is said to be transcendentalover K. F is said to be an algebraic extension of K is every elements of Fis an algebraic extension over K. F is said to be a transcendental extensionextension if at least one element of F is transcendental over K.

Remark 1.1.3. It is a nontrivial fact that π, e ∈ R are transcendental over Q.For proofs, please refer to [Jac85]

Theorem 1.1.3. Let F be an extension field of K and E the set of allelements of F which are algebraic over K. Then E is a subfield of F .

Proof. Let u and v in F be algebraic over K, then K(u − v) ⊂ K(u, v) =K(u)(v) and if v 6= 0 K(uv−1) ⊂ K(u, v). Hence u + v and uv are alsoalgebraic over K. Hence the set of all elements of F which are algebraic overK form a subfield.

1.1. FIELD EXTENSION AND EXAMPLES 19

Theorem 1.1.4. If F is an extension field of K and u ∈ F is transcendentalover K, then the field K(u) is isomorphic to K(x) where x is an indetermi-nant.

Proof. The map f(u) 7→ f(x) gives the isomorphism.

Theorem 1.1.5. If F is an extension field of K and u ∈ F is algebraic overK, then

1. K(u) = K[u].

2. K(u) ∼= K[x]/(f) where f is the minimal polynomial of u. Then g(u) =0 if and only if f divides g.

3. [K(u) : K] = n.

4. {1, u, ..., un−1} is a basis of K(u) over K.

Proof. Let φ : K[x] → K[u] be the map sending g(x) to g(u). The kernel isgenerated by a single element f since K[x] is a P.I.D. Hence if there is anyother polynomial g such that g(u) = 0. Then g ∈ Kerφ. Hence f |g. Nowwe claim {1, u, ..., un−1} are linearly independent over K. Since K[u] is anintegral domain, and K[x]/(f) ∼= K[u], hence (f) is a prime ideal and hencea maximal ideal. So K[u] is a field.

The field K(u) is the quotient field of K[u]. So K[u] ⊂ K(u). Onthe other hand, K(u) is the smallest field which contains K and u. HenceK(u) ⊂ K[u]. Hence K(u) = K[u].

Corollary 1.1.1. Let f(x) be a monic irreducible polynomial in F [x]. Let ube any solution of f(x) = 0. Then F (u) is isomorphic to F [x]/(f(x)).

Exercise 1.1.6. How many degree 2 irreducible polynomials in F2. How aboutdegree n then?

Exercise 1.1.7. f(x) = x3 +x2− 1 is irreducible in F5. Let u be a root of theequation f(x) = 0. Please write u4 +1 as a linearly combination of {1, u, u2}over F5. Please find the inverse of u2 + 1 in F5(u).

Exercise 1.1.8. Let E = F (u) where u is algebraic of odd degree. Show thatE = F (u2).

Proof. F (u2) is a subfield of F (u). Hence [F (u) : F (u2)][F (u2) : F ] =[F (u) : F ]. Hence [F (u2) : F (u)] is either 1 or 2 and [F (u) : F ] is odd. Hence[F (u2) : F (u)] = 1 and [F (u2) : F ] = [F (u) : F ].


1.2 Simple extension

Theorem 1.2.1. If F is a finite extension of the field K, then F is finitelygenerated and algebraic over K.

Proof. Suppose [F : K] = n < ∞. Let {u1, ..., un} be a basis of the n-dimensional vector space F over K. Then F = K(u1, ..., un). Given anyu ∈ F , 1, u, ..., un is linearly dependent over K. Hence u is algebraic overK.

Exercise 1.2.1. Let E be an extension field of the field F such that i) [E :F ] < ∞, and ii) for any two subfields E1 and E2 containing F , either E1 ⊃ E1

or E2 ⊃ E1. Show that E has a primitive element over F .

Proof. We prove it by contradiction. Suppose E is not an simple extensionover F , then for any u ∈ E, F (u) ( E. Now we fix the choice for u suchthat the index [F (u) : F ] obtains the maximum among all choice u ∈ E.is Pick v ∈ E such that v /∈ F (u). Then we have either F (v) ⊆ F (u) orF (u) ⊆ F (v) by the assumption. The first circumstance won’t happen asv /∈ F (u) and the second circumstance implies [F (v) : F ] > [F (u) : F ] whichcontradicts the choice of u. Hence E is a simple extension.

We will like to method a corollary of this exercise. However, to prove itwe will need more properties of finite fields that we will obtain later.

Corollary 1.2.1. Any finite field is a simple extension over its prime field.

Example 1.2.1. In the field C, Q(i) and Q(√

2) are isomorphic as vector spaces(both are 2-dimensional Q spaces), but not as fields. To prove Q(i,

√2) =

Q(i +√

2).

Proof. As vector spaces, they are both two dimensional over Q. Hence theyare isomorphic. But as fields, if there is a field isomorphism φ : Q(i) →Q(√

2) then the image of φ(i) = a+b√

2 satisfies that (a+b√

2)2 = −1. Thisis impossible. Hence as fields they are not isomorphic. To prove the secondstatement, we first see that

(x− (i +√

2))(x− (i−√

2))(x− (−i +√

2))(x− (−i−√

2)) = x4− 2x2 + 9.

Hence i +√

2 is a root of the over Q irreducible polynomial x4− 2x2 + 9. Bytheorem 1.1.5, [Q(i+

√2) : Q] = 4. Hence (i+

√2)+(i−

√2) = 2i ∈ Q(i+

√2).

Similarly,√

2 ∈ Q(i +√

2). Hence Q(i,√

2) = Q(i +√

2).

1.2. SIMPLE EXTENSION 21

The next theorem indeed holds for any field which contains only infinitelymany elements. But we will state and prove it for a specified case only.

If F is a finite extension of the field Q, then F is called an algebraicnumber field, or simply a number field.

Example 1.2.2. The fields Q(i), Q(√

2, i) are algebraic number field.

Theorem 1.2.2. Let F be an algebraic number field. Then F is a simpleextension of Q. In other words, there exists an element u ∈ F such thatF = Q(u).

Proof. Since [F : Q] < ∞, F is finitely generated over Q. By induction,it is sufficient to prove given any two u1, u2 ∈ F , there exists an elementu ∈ F such that K(u1, u2) = K(u). Let f, g ∈ Q[x] be two relativelyprime polynomials. Suppose α1, ..., αn and β1, ..., βm are the roots of thesetwo polynomials respectively. Since they are relatively prime, they have norepeated roots. Choose t ∈ Q such that ai + tβj 6= α1 + tβ1 for all j 6= 1 andall i. Let u = α1 + tβ1. Consider f(u − tx). It is easy to see the solutionsare u − tx = αi. I.e. x = (α1 + tβ1 − αi)/t. Hence the common roots ofg(x) and f(u − tx) is β1. On the other hand, the greatest common divisorof f(u − tx) and g(x), which is x − β1 , is a polynomial in Q(u)[x] by theEuclidean algorithm. Hence β1 ∈ Q(u) and α1 = u − tβ1 ∈ Q(u). HenceQ(u) ⊂ Q(α1, β1) ⊂ Q(u).

Chapter 2

Galois groups of fieldextensions

2.1 Field isomorphism

Suppose E is an extension of K and F is an extension of field L, and σ : K →L is an isomorphism of fields. We will consider here under what condition σcan be extended to an isomorphism from E to F?

Theorem 2.1.1. Let σ : K → L be an isomorphism of fields, u an elementof some extension field of K and v an element of some extension field of L.Assume either

1. u is transcendental over K and v is transcendental over L; or

2. u is a root of an irreducible polynomial f ∈ K[x] and v is a root ofσf ∈ L[x].

then σ extends to an isomorphism of fields K(u) ∼= L(v) which maps u to v.

Proof. In case 1, it is obvious as K(u) ∼= K[x].In case 2, since σ : K → L is an isomorphism, the induced map

K[x]/(f) → L(x)/(σf)

g + (f) 7→ σ(g) + (σf)

is also an isomorphism when f is irreducible.

23

24 CHAPTER 2. GALOIS GROUPS OF FIELD EXTENSIONS

Corollary 2.1.1. Let E and F each be extension fields of K and let u ∈E and v ∈ F be algebraic over K. Then u and v are roots of the sameirreducible polynomials f ∈ K[x] if and only if there is an isomorphism offield K(u) ∼= K(v) which sends u onto v and is the identity on K.

Proof. “⇒ ” follows from theorem 2.1.1 when σ = 1K .“⇐” Let σ : K(u) ∼= K(v) be the isomorphism. Suppose the minimal

irreducible polynomial satisfy by u is f =∑

0≤i≤n aixi. Then σaiu

i = aivi =

0. Hence v is also a root of the irreducible polynomial f .

Theorem 2.1.2. If K is a field and f ∈ K[x] polynomial of degree n, thenthere exists a simple extension field F = K(u) of K such that

1. u ∈ F is a root of f ;

2. [K(u) : K] ≤ n, with equality holding iff it is irreducible in K[x];

3. if f is irreducible in K[x], then K(u) is unique up to an isomorphismwhich is the identity on K.

4. the number of such an extension is the same as the number of distinctroots of f .

Proof. Let u be a root of f , then [K(u) : K] ≤ n as 1, u, · · · , un are linearlydependent. If f is irreducible, iff 1, u, · · · , un−1 are linearly independent,hence [K(u) : K] = n. The last two claims follow from the previous corollary.

This theorem explains why we later need to consider multiple roots andsimple roots separately.

2.2 Splitting fields of a polynomial

Definition 2.2.1. Let F be a field, f(x) a monic polynomial in F [x]. Thenan extension E/F is called a splitting field over F of f(x) if

1. f(x) = (x− r1)...(x− rn) in E[x] and

2. E = F (r1, ..., rn), that is E is generated by the roots of f(x).

2.2. SPLITTING FIELDS OF A POLYNOMIAL 25

Example 2.2.1. Let f = x3 − 1 and ω a primitive cubic root of one. Thenthe splitting field of f over Q is Q(ω).

Example 2.2.2. Let f = x3 − 2. The the splitting field of f over Q contains3√

2, 3√

2ω, where ω = −1−√

32

. Hence the splitting field of f is Q( 3√

2, ω)

Example 2.2.3. Let F = F2, and f = x3 + x + 1. f is irreducible overF2. Put r1 = x + (f(x)) ∈ F [x]/(f) so F (r1) is a field and x3 + x + 1 =(x+r1)(x

2+ax+b) ∈ F (r1)[x]. Moreover, we can check that a = r1, b = 1+r21.

The element of F (r1) can be written as c+dr1 +er21, c, d, e ∈ F . Substituting

these in x2 + r1x + 1 + r21 we get (r2

1)2 + r1(r

21) + 1 + r2

1 = r41 + r3

1 + 1 + r21 = 0

as since r31 = r1 + 1 and r4

1 = r21 + r1. Hence E = F (r1) is the splitting field

of x3 + x + 1.

Example 2.2.4. Let F = Q and let f = xp − 1. The degree of its splittingfield over Q is p− 1.

Proof. Let µp = e2πi/p, then in Q(µp),

xp =

p∏i=1

(x− µip).

Theorem 2.2.1. Any monic polynomial f(x) ∈ F [x] of positive degree hasa splitting field E/F .

Proof. We prove it by induction. Suppose the statement is true for anypolynomial of degree < n. Let f be a monic irreducible polynomial of F [x].Let u be a solution of f , then let E ′ = F (u) be an intermediate field. Then,f factors over E ′ as several irreducible components, and each component hasdegree less than n. By our assumption, each component has a splitting fieldthen the composition of splitting fields obtained above is a splitting field forf .

Theorem 2.2.2. Let f(x) be a polynomial over a field F . Splitting fields forf(x) are F -isomorphic.

Proof. We use Proposition 1.1.5.


2.3 Multiple roots

We factor f(x) in its splitting field as

f(x) = (x− r1)k1 · · · (x− rs)

ks

so that ki > 0 and ri 6= rj if i 6= j. We call ki the multiplicity of the root ri.If ki = 1 then ri is called a simple root.

Lemma 2.3.1. Let f(x) be a monic polynomial of positive degree in F [x].Then all the roots of f in any splitting field E/F are simple if and only ifgcd(f, f ′) has degree 0.

Proof. Let f(x) = (x− r1)k1 · · · (x− rs)

ks , then

f ′ =s∑

i=1

ki(x− r1)k1 · · · (x− ri)

ki−1 · · · (x− rs)ks ∈ F [x]

and(x− r1)

k1−1 · · · (x− rs)ks−1 | gcd(f, f ′)

So gcd(f, f ′) has degree 0 iff ki = 1 for i = 1, ..., s.

Corollary 2.3.1. If f(x) is an irreducible polynomial with degree at least 1,then gcd(f, f ′) has degree greater than 0 iff f ′ = 0.

It is easy to see that every irreducible polynomial over a characteristic 0field has only simple roots.

Corollary 2.3.2. Let F be a field with characteristic p > 0. Let f(x) bean irreducible polynomial with degree at least 1. Then f ′ = 0 iff f(x) =∑n

i=1 aixpi.

In a characteristic p case (odd or even), we have

(x + y)pe

= xpe

+ ype

(2.1)

(2.2)

Exercise 2.3.1. Determine a splitting filed over Fp of xpe − 1 for any naturalnumber e.

Solution: The splitting field is Fp itself as xpe − 1 = (x− 1)pe.

2.3. MULTIPLE ROOTS 27

Lemma 2.3.2. If F has characteristic p > 0 and a ∈ F , then xp−a is eitherirreducible in F [x] or it is a pth power in F [x].

Proof. If xp − a is not irreducible, then gcd(f, f ′) = f . Suppose (xp − a) =g(x)h(x) where g(x) is monic of degree k and 1 ≤ k ≤ p − 1. Let E be asplitting field over F of xp − a and let b ∈ E be a root of g. Then bp = a.Hence xp − a = xp − bp = (x − b)p = g(x)h(x). Hence g(x) = (x − b)k andbk ∈ F . Since k and p are coprime, there exist u and v such that uk+vp = 1,hence b = (bk)u(bp)v ∈ F .

Example 2.3.1. Let F = Fp(t) be a function field, we claim that t is nota pth power in this field. If yes, we have t = (f(t)/(gt))p where f(t) =a0 + a1t + · · ·+ ant

n and g(t) = b0 + ...bmtm. Hence we have

(bp0 + b1t

p + · · · bpmtmp)t = ap

0 + ap1t

p + ...apnt

np

However, 1, t, t2, ... are independent over Fp. Hence all bi = 0, impossible.Now let f(x) = xp − t, it is irreducible over Fp. Let E be the splitting fieldof f . Then f has multiple roots as f ′(x) = 0.

Definition 2.3.1. A polynomial in F [x] is separable if its irreducible factorshave distinct roots. A field is said to be perfect if every irreducible polynomialin F [x] is separable .

By Lemma 2.3.1, any field with characteristic 0 is perfect. We will nowdiscuss when a field with characteristic p is perfect.

Theorem 2.3.1. A field F of characteristic p 6= 0 is perfect if and only ifF = F p, the field of pth powers of the elements of F .

Proof. If F 6= F p, then there is an a ∈ F − F p. Then f(x) = xp − a isirreducible and have multiple roots. Hence F is not perfect. If let f(x) be aninseparable irreducible polynomial in F [x]. Then (f(x), f ′(x)) 6= 1 and thisimplies that f(x) = a0 + apx

p + a2px2p + · · · and one of these ai is not a pth

root. For otherwise, f(x) will be a pth power, contrary to the irreducibilityof f(x). Hence F 6= F p.

Corollary 2.3.3. Every finite field F is perfect.


Proof. Since xp = yp implies (x− y)p, the map

F → Fp

x 7→ xp

is one to one and hence is also an isomorphism.

Exercise 2.3.2. Let F be imperfect of characteristic p. Show that xpe − a isirreducible if a /∈ F p.

Proof. If xpe − a = (x − α)pe= f(x)g(x) with deg f > 0, deg g > 0. I.e.

f(x) = (x − α)n1 ∈ F [x]. Hence αn1 ∈ F and αpe−n1 ∈ F . Let pi = (pe, n1)then i < n. By the Euclidean algorithm we know αpi ∈ F . Hence αpe−1 ∈ F .I.e. a ∈ F p.

2.4 Algebraic closure

Theorem 2.4.1. The following conditions on a field are equivalent.

1. Every nonconstant polynomial f ∈ F [x] has a root in F ;

2. every nonconstant polynomial f ∈ F [x] splits over F ; every irreduciblepolynomial in F [x] has degree 1;

3. there is no algebraic extension field of F (except itself);

4. there exists a subfield K of F such that F is algebraic over K and everypolynomial in K[x] splits in F [x].

A field that satisfies the equivalent conditions of theorem 2.4.1 is said tobe algebraically closed.

Example 2.4.1. The field of complex numbers is algebraically closed as everynonconstant polynomial f(x) ∈ C[x] has at least a root in C.

Theorem 2.4.2. If F is an extension field of K, then the following condi-tions are equivalent.

1. F is algebraic over K and F is algebraically closed;

2. F is a splitting field over K of the set of all polynomials in K[x].

2.5. GALOIS GROUPS 29

Any extension field F of a field K that satisfies the equivalent conditionsof theorem 2.4.2 is called an algebraic closure of K.

Theorem 2.4.3. Every field K has an algebraic closure. Any two algebraicclosure of K are K-isomorphic.

Proof. C.f. Theorem3.6 Page 259 of Hungerford [Hun80].

Example 2.4.2. Let Q denote an algebraic closure of Q, then Q ( C as bythe above theorem Q does not contain any transcendental elements over Q.

2.5 Galois groups

Let E be an extension of a field F and let G be the set of automorphisms of Ewhich fix F ; that is the set of automorphisms η of E such that η(a) = a for alla ∈ F . We also called such an automorphism of E to be an F -automorphismof E. The set of F -automorphisms form a group G and is called the Galoisgroup of E over F , denoted by Gal(E/F ).

Example 2.5.1. Let E = F (u) where u2 = a ∈ F and a is not a square.The {1, u} is a basis for E over F . Gal(E/F ) =< σ > is of order 2 whereσ(a + bu) = a− bu, a, b ∈ F .

Exercise 2.5.1. Find the Galois group Gal(Q(√

2,√

3)/Q).

Solution: We have discussed in section 1.2 that Q(√

2,√

3) is a simple ex-tension. Explicitly, E = Q(

√2,√

3) = Q(√

2+√

3). The minimal polynomialof√

2 +√

3 is given by

(x−√

2 +√

3)(x−√

2 +√

3)(x−√

2 +√

3)(x−√

2 +√

3) = x4 − 10x2 + 1

There are 4 Q-automorphisms of E which are specified by

σ1(√

2 +√

3) =√

2 +√

3, σ2(√

2 +√

3) =√

2−√

3,

σ3(√

2 +√

3) = −√

2 +√

3, σ4(√

2 +√

3) = −√

2−√

3.

Indeed the Galois group Gal(E/Q) =< σ2, σ3 >= Z2 ⊕ Z2.

Example 2.5.2. Let F be imperfect of characteristic p and let a ∈ F − F p.Then exercise 2.3.2 says f(x) = xp − a is irreducible. Moreover

xp − a = (x− u)p.

Let E be a splitting field of f(x). So the Galois group Gal(E/F ) = 1.


Example 2.5.3. Let F be a field and let E = F (t) where t is transcendentalover E. Any automorphism σ will send t to another generator of the fieldextension u of E/F . Moreover, u is a generator of E/F is and only if

u =at + b

ct + d, ad− bc 6= 0.

Hence Gal(E/F ) = PGL2(F ) where PGL2(F ) is the group of invertible 2×2matrices with entries in F modulo nonzero scalar matrices.

Exercise 2.5.2. Let E be a splitting field of x3 − 2 over Q. Prove that theGalois group Gal(E/F ) is isomorphic to S3.

Now let G be any group of automorphisms of a field E. Let

InvG = {a ∈ E | g(a) = a, g ∈ G.}

I.e. InvG are the set of elements of E which are not moved by any g ∈ G.Then from

g(a + b) = g(a) + g(b), g(ab) = g(a)g(b), g(1) = 1, g(a−1) = g(a)−1,

we see that InvG is a subfield of E, called the G-invariants or G-fixed subfieldof E.

Given two automorphism groups G1 and G2 of E and two subfields F1 andF2 of E, according to the definitions above, it is easy to verify the followingproperties:

1. If G1 ⊃ G2, then InvG1 ⊂ InvG2;

2. F1 ⊃ F2, then Gal(E/F1) ⊂ Gal(E/F2);

3. Inv(Gal(E/F )) ⊃ F ;

4. Gal(E/InvG) ⊃ G.

Theorem 2.1.2 implies

Lemma 2.5.1. Let E/F be a splitting field of a separable polynomial con-tained in F [x]. Then |(Gal(E/F )| = [E : F ].

2.5. GALOIS GROUPS 31

Lemma 2.5.2 (Artin’s Lemma). Let G be a finite group of automorphismsof a field E and let F = InvG. Then

[E : F ] ≤ |G|.

Proof. Let n = |G|. We only need to show that any m(> n) elements ofE are linearly dependent over F . Let G = {1, g2, ..., gn} and let u1, ..., um,be m(> n) elements of E. Then by linear algebra we know the system of nequations

m∑i=1

gi(uj)xi = 0 (2.3)

have nontrivial solutions. Suppose among all of them, (a1, ..., am) has theleast number of nonzero entries. Without loss of generality, we may assumea1 = 1. Now we claim all ai ∈ F . If all ai ∈ InvG then it is done. If not,we suppose ai /∈ InvG. We may assume gk(a2) 6= a2. Now we applied gk tothe system (2.3) to get

∑mj=1(gkgi)(uj)gk(bj) = 0. Since (gkg1, ..., gkgn) is a

permutation of (g1, ..., gn), we know (1, gk(b2), ..., gk(bm)) is also a solution of(2.3). Hence (0, b2 − gk(b2), ..., bm − gk(bm)) is also a nontrivial solution (asthe second entry a2 − gk(a2) 6= 0) with even fewer nonzero entries than theoriginal one.

Exercise 2.5.3. Let E = Fp(t) where t is transcendental over Fp. Let σ(t) =t + 1 and G =< σ >. Determine F = Inv(G) and show that [E : F ] = p.

Proof. Since the characteristic is p, so σp = id. Hence |G| = p. By Artin’slemma [E : InvG] ≤ p. Every f(t) ∈ Fp[t] satisfies f(t) = f(t + 1) impliesthat f(t)−f(0) is a multiply of g(t) = t(t−1) · · · (t−p+1). Such a polynomialg(t) is a minimal polynomial which is invariant under G. By induction, f(t)is a polynomial of g. It follows that InvG = Fp(g).

Now we show 1, t, ..., tp−1 is linearly independent over InvG. If not thereare ai(g) ∈ Fp[g] such that

∑p−1i=1 ai(g)ti = 0. We may assume that not every

ai(g) is a multiple of g (after dividing by a high enough power of g). Then∑p−1i=1 ai(g)ti mod g = 0. After modulo by g, the left hand side is a degree

at most p − 1 polynomial in t which is won’t equal to zero, the right handside. Hence [E : InvG] ≥ p. Hence [E : InvG] = p.

Definition 2.5.1. An extension E/F is called normal (algebraic) extensionif every irreducible polynomial in F [x] which has a root in E is a product oflinear factors in E[x].


Example 2.5.4. Let be a root of x3 − 2. Then Q(u)/Q is not a normalextension. (As the splitting field of x3 − 2 has degree 6 over Q).

Exercise 2.5.4. Let F be a field of characteristic p, a an element of F not ofthe form bp− b, b ∈ F . Determine the Galois group over F of a splitting fieldof xp − x− a.

Solution: Let f(x) = xp−x−a. Then f ′ = −1 6= 0. Hence xp−x−a onlyhas simple roots. By the assumption on a, xp−x− a does not have any rootover F . On the other hand, if u is a root of f(x), then f(u + 1) = (u + 1)p−(u+1)− a = up−u− a = 0. So f(x) = (x−u)(x−u− 1) · · · (x−u− p+1).Hence E = F (u) is a splitting field for f(x). Gal(E/F ) =< σ > whereσ(u) = u + 1.

2.6 Galois extension

Theorem 2.6.1. Let E be an extension field of a field F . Then the followingconditions on E/F are equivalent.

1. E is a splitting field over F of a separable polynomial f(x).

2. F = InvG for some finite group of automorphisms of E.

3. E is finite dimensional normal and separable over F .

Moreover, if E and F are as in (1) and G = Gal(E/F ) then F = InvG andif G and F are as in (2), then G = Gal(E/F ).

Proof. (1) ⇒ (2). Let G = Gal(E/F ) and F ′ = InvG. Then F ′ is a subfieldof E containing F . Hence E is a splitting field over F ′ of f(x) as well asover F and G = Gal(E/F ′). Hence by lemma 2.5.1 we know [F ′ : F ] = 1. Itfollows that F = Inv(Gal(E/F ))

(2) ⇒ (3). By Artin’s lemma we have [E : F ] ≤ |G|, so E is of finitedimensional over F . Let f(x) be an irreducible polynomial in F [x] havinga root r in E. Let {r1 = 1, r2, ..., rm} be the orbit of r under the action ofthe group G. Thus this set is the set of distinct elements of the form g(r).Hence if ξ ∈ G, then the set ((ξ(r1), ..., ξ(rm)) is a permutation of (r1, ..., rm).We have f(r) = 0 which implies f(ri) = 0. Then f is divisible by x − ri,i = 1, ...,m. Hence f is divisible by

∏mi=1(x− ri). Since ξg(x) = g(x) for any

ξ ∈ G, so the coefficients of g(x) are G-invariant, i.e. g(x) ∈ F [x]. Hence

2.7. FUNDAMENTAL THEOREM OF GALOIS THEORY 33

f(x) = g(x) (as f is irreducible) is a product of distinct linear factors inE[x]. It follows E is separable and normal over F and (3) holds.

(3) ⇒ (1). We assume E = F (r1, ..., rk) where ri are all algebraic over Fwith minimal polynomial fi(x). Then E is the splitting field of f =

∏i fi(x)

and f is separable.Moreover, we know by Artin’s lemma that [E : F ] ≤ |G| and Gal(E/F ) =

[E : F ]. Since G ⊂ Gal(E/F ) and we know G = Gal(E/F ).

Definition 2.6.1. If an extension E/F satisfies one (and hence all) of theabove equivalent conditions, then E is said to be Galois over F .

Exercise 2.6.1. Let E = Q(u) where u is a root of f(x) = x3+x2−2x−1 = 0.Verify that u′ = u2 − 2 is also a root of f(x) = 0, Determine Gal(E/Q) andshow that E is normal over Q.

2.7 Fundamental theorem of Galois theory

Theorem 2.7.1 (Fundamental theorem of Galois theory). Let E be aGalois extension field of a field F . Let G = Gal(E/F ). Let Γ = {H}, theset of subgroups of G, and Σ, the set of intermediate fields between E andF . The maps H 7→ InvH, K 7→ Gal(E/K), H ∈ Γ, K ∈ Σ are inverses andso are bijective. Moreover, we have the following

1. H1 ⊃ H2 ⇔ InvH1 ⊂ InvH2;

2. H = [E : InvH], [G : H] = [InvH : F ];

3. H is normal in G ⇔ InvH is normal over F . In this case

Gal(InvH/F ) ∼= G/H.

Proof. The bijection between this two sets Γ and Σ follows from theorem2.6.1. Claim 1) follows directly from the definition of InvH.

2) Since H = Gal(E/InvH). #H = #Gal(E/InvH) ≤ [E : InvH] ≤#H. Hence #H = #Gal(E/InvH). It follows that [G : H] = #G/#H =[E : F ]/[E : InvH] = [InvH : F ].

3). If H ∈ Γ and K = InvH is the corresponding invariants. Thenfor any γ ∈ G, InvγHγ−1 = γ(K). Hence H is normal then for anyγ ∈ G, γ(InvH) = InvH. Hence every γ ∈ G maps InvH to itself.


So we consider γ|K=InvH which gives the restriction homomorphism fromGal(E/F ) to Gal(InvH/F ). The kernel of this map are those elements inG which are identity on K, by the correspondence, they are in H. HenceG/H ∼= Gal(InvH/F ). The image G is a group of automorphisms in InvHand InvG = F . Hence theorem 2.6.1 implies that K/Inv(G) is a normalextension. Conversely, if K is normal over F . Let a ∈ K and let f(x) bethe minimal polynomial of a over F . Then over K, f(x) =

∏(x− ai), where

a1 = a. If γ ∈ G then f(γ(a)) = 0 implies that γ(a) = ai for some ai. Theγ(a) ∈ K. I.e. γ(K) ⊂ K, so H is normal in G.

Example 2.7.1. Let f = x17 − 1 ∈ Q[x]. Let E be a splitting field of f(x).Determine the Galois correspondence for this case.

Solution: Let µ17 = e2πi/17, then

f(x) = x17 − 1 =17∏i=1

(x− µi17).

Let σ(µ)17 = µ317, then G = Gal(E/Q) is cyclic and is generated by σi. The

order |Gal(E/Q)| = 16.Since G cyclic, it is easy to determine the set of all subgroups of G as

H = {< σ >, < σ2 >, < σ4 >, < σ8 >, < σ16 > }.Each subgroup Ga =< σa > is also cyclic and hence

InvGa =

16/a∑i=1

σai(x) |x ∈ E

= Q(

16/a∑i=1

σai(µ17)).

Let x1 =∑16/a

i=1 σai(µ17) = µ17 + µ217 + · · ·+ µ16

17 ∈ Q.

Let x2 =∑16/2

i=1 σa2(µ17) = µ217 + µ4

17 + µ617 + · · ·+ µ16

17.

Let x4 =∑16/4

i=1 σa4(µ17) = µ417 + µ8

17 + µ1217 + µ16

17.

Let x8 =∑16/8

i=1 σa8(µ17) = µ817 + µ16

17.Then Inv < σi >= Q(xi).Hence there are 3 intermediate fields between Q and E. Explicitly

Q = Q(x1) ⊂ Q(x2) ⊂ Q(x4) ⊂ Q(x8) ⊂ Q(µ17) = E.

2.7. FUNDAMENTAL THEOREM OF GALOIS THEORY 35

Exercise 2.7.1. Let K = F2 and let F = F2(t). Describe the separable closureand purely inseparable closure of F over K.

Solution: By example 2.5.3 we know

G = Gal(F/K) = PGL(2, F2) =< σ1, σ2 >∼= S3,

where σ1(t) = 1/t, σ2(t) = t + 1. The group G has 6 subgroups, which are

Γ = {{id}, < (σ1) >,< σ2 >,< σ1σ2σ1 >,< σ1σ2 >,G}.

Inv{id} = F2(t)

Inv < (σ1) > = F2(t2 + 1

t)

Inv < (σ2) > = F2(t(t + 1))

Inv < σ1σ2σ1 > = F2(t2

t + 1)

Inv < σ1σ2 > = F2(1 + t + t3

t(t + 1))

InvG = F2

Chapter 3

Separability

3.1 Purely inseparable extensions

Definition 3.1.1. Let F be an extension field of K. An algebraic elementu ∈ F is purely inseparable over K if its irreducible polynomial f in K[x]factors in F [x] as f = (x− u)m. F is a purely inseparable extension of K ifevery elements of F is purely inseparable over K.

Example 3.1.1. Let K = Fp(t) and y2 = t. Then the minimal polynomial ofy over K is X2 − t. The element y is separable over K is Char(K) = p 6= 2.Char(K) = 2, y is purely inseparable over K.

Theorem 3.1.1. Let F be an extension field of K. Then u ∈ F is bothseparable and purely inseparable over K if and only if u ∈ K.

Proof. If u ∈ K, then the minimal irreducible polynomial of uover K isx− u. Hence u is both separable and purely inseparable. It is easy to provethe converse by definitions.

Lemma 3.1.1. Let F be an extension field of K with charK = p 6= 0. Ifu ∈ F is algebraic over K, then upn

is separable over K for some n ≥ 0.

Proof. We use induction. Let f(x) be the irreducible polynomial of u overK. If deg f = 1 or f has no repeated roots, then u is separable. Otherwise,we have f ′ = 0, hence f is a polynomial in xp. Therefore, up is algebraic ofdegree less than u over K. Hence we can use induction (up)pm

is separableover K for some m ≥ 0.

37

38 CHAPTER 3. SEPARABILITY

Theorem 3.1.2. If F is an algebraic extension field of a field K of charac-teristic p 6= 0, then the following statements are equivalent.

1. F is purely inseparable over K;

2. the irreducible polynomial of any u ∈ F is of the form xpn − a ∈ K[x];if u ∈ F , then upn ∈ K for some n ≥ 0;

3. If u ∈ F , then upn ∈ K for some n ≥ 0;

4. the only elements of F which are separable over K are the elements ofK itself;

5. F is generated over K by a set of purely inseparable elements.

Proof. 1)⇒ 2). Given any u ∈ F , the irreducible polynomial is f = (x−u)m.f ′ = 0 and hence m = prm′ and (p, m′) = 1. Hence (xpr − upr

)m ∈ K.Since the coefficient of Xpr(m′−1) is upr

. If m′ 6= 1, then upr ∈ K andf = (xpr − upr

)m′is not irreducible. Hence m′ = 1.

2)⇒ 3)⇒ 4)⇒ 5) trivial, 4)⇒ 3) (implies by Lemma 3.1.1).3)⇒ 1) If u ∈ F , then upn

K for some n ≥ 0. Then g(x) = xpn − upn ∈K[x] satisfies by u. Hence the minimal polynomial of u is a divisor of g(x),and u is purely inseparable.

5)⇒ 3) Suppose ui is purely inseparable and by 3) we know upni

i ∈ K. If

u = uiuj then upni+nj ∈ K. If u = aui+buj, a, b ∈ K, then (aui+buj)pni+nj

=

apni+njupni+nj

i + bpni+njupni+nj

j ∈ K. By induction, we know every element inF is purely inseparable

Corollary 3.1.1. If F is a finite dimensional purely inseparable extensionfield of K and charK = p 6= 0, then [F : K] = pn for some n ≥ 0.

Proof. Suppose F = K(u1, · · · , un) and by the above theorem, all ui arepurely inseparable over K. Then [K(ui) : K] = pei for some i ≥ 0. The thecomposition of K(ui) also has p-power degree over K.

Lemma 3.1.2. Let E be an extension of the field F . If u is separable (resp.purely inseparable) over F , then u is separable (resp. purely inseparable) overE.

3.2. SEPARABLE EXTENSIONS 39

Proof. Let f(x) be the monic minimal polynomial of u over F and let f ′ bethe monic minimal polynomial of u over E. Then f ′|f . If u is separable overF , then f has only simple roots; and so if f ′. If u is purely inseparable overF , then by Theorem 3.1.2, f has only one single root; and so is f ′.

Exercise 3.1.1. If u ∈ F is separable over K and v ∈ F is purely inseparableover K, then K(u, v) = K(u + v). If uv 6= 0, then K(u, v) = K(uv).

Proof. We only need to consider characteristic p > 0 case.By theorem 3.1.2, there is an large enough n such that vpn ∈ K. Hence

(u + v)pn= upn

+ vpn(resp. (uv)pn

= upnvpn

. ) So upn ∈ K(u + v) (resp.upn ∈ K(uv)). By Lemma 3.1.2, u is also separable over K(upn

). We considerxpn − upn ∈ K(upn

)[x] and xpn − upn= (x − u)pn ∈ K(upn

)[x]. Hence uis both separable and purely inseparable over K(u(pn). By Theorem 3.1.1,u ∈ K(upn

) = K(u+v) (resp. u ∈ K(upn) = K(uv)). It follows v ∈ K(u+v)

(resp. v ∈ K(uv)).

Exercise 3.1.2. If charK = p 6= 0 and p - [F : K] < ∞, then F is separableover K.

3.2 Separable extensions

Lemma 3.2.1. If F is an extension field of K, X is a subset of F suchthat F = K(X) and every element of X is separable over K, then F is aseparable extension of K.

Proof. If v ∈ F , there exists u1, ...un ∈ X such that v ∈ K(u1, ..., un). I.e.Let fi be the irreducible polynomials of ui. The v is in the splitting fieldof the polynomial f =

∏fi which is separable. Hence E is normal and

separable. So v is separable.

Theorem 3.2.1. Let F be an algebraic extension of K, S the set of allelements of F which are separable over K, and P the set of all elements ofF which are purely inseparable over K.

1. S is a separable extension field of K

2. F is purely inseparable over S

3. P is a purely inseparable extension over K.


4. P ∩ S = K.

5. F is separable over P if and only if F = SP .

6. If F is normal over K, then S is Galois over K, F is Galois over Pand Gal(F/K) ∼= Gal(F/P ) ∼= Gal(S/K).

Proof. 1) follows from lemma 3.2.1.2) Let u ∈ F , the upn ∈ S for some n, by property 3 of theorem 3.1.2 we

know F is purely inseparable over K.3) use same argument as that for 5) ⇒ 3) of theorem 3.1.2.4) theorem 3.1.15) If F is separable over P , then F is separable over PS. Since F is purely

inseparable over S, hence F is purely inseparable over PS. Hence F = PS.If F = PS = P (S) is separable over P by lemma 3.2.1.

6) Let G = Gal(F/K). InvG ⊃ P . Let u ∈ InvG, let f ∈ K[x] bethe minimal polynomial of u. Since F is normal, then all roots of f ∈ F .Suppose there is another root v 6= u of f . Then σ : u 7→ v induces an ele-ment in Gal(F/K) which will be fix u. Hence InvG ⊂ P and InvG = P .By Galois theorem, we know F/P = F/InvG is Galois and G = Gal(F/P ).Now since every σ ∈ Gal(F/K) = Gal(F/P ) must sending separable el-ements to separable elements. Hence σ → σ|S defines a homomorphismθ : Gal(F/P ) → Gal(S/K). Since F is normal over S (as F is purely insep-arable over S), hence θ surjective. Since F/P is Galois, hence theta is one-to-one. Hence Gal(S/K) = Gal(F/P ). Now if u ∈ S is fixed by Gal(S/K)then it is also in P which is fixed by Gal(F/K). Hence u ∈ S ∩ P = K.

We will call P (resp. S) the purely inseparable (resp. separable) closureof F over K.

Corollary 3.2.1. If F is separable over E and E is separable over K, thenF is separable over K.

Proof. Let S be the maximal separable extension of K contain in F , thenE ⊂ S. F is purely inseparable over S. Since F is separable over E, it isseparable over S. Hence F = S.

Corollary 3.2.2. Let F be an algebraic extension of K and charK = p 6= 0.If F is separable over K, then F = KF pn

for each n ≥ 1. If [F : K] < ∞and F = KF p, then F is separable over K. In particular, u ∈ F is separableover K iff K(up) = K(u).


F

��

��

p.i.,normal

��111

1111

1111

1111

PS

p.i. !!BBB

BBBB

B

sep.}}{{

{{{{

{{

P

p.i.,normal !!CCCC

CCCC

S

sep.}}||

||||

||

K

Figure 3.1: Separable and purely inseparable closures

Proof. We first prove the last claim. If u is separable over K, K(u)/K(up) isseparable as K(u)/K is separable. K(u)/K(up) is purely inseparable as it isgenerated by u which purely inseparable over F (up). Hence K(u) = K(up).

If K(u) = K(up), then by induction K(u) = K(upn). Now for largeenough n, we have upn

being separable over K. Hence u is separable over K.

Now we prove the first statement.

It follows naturally that if F/K is separable, then F = KF = KF pnfor

all n ≥ 1.

If [F : K] < ∞, and F = KF p. Iteratively , F = KF pn= S. Hence F is

separable over K.

Definition 3.2.1. Let F/K be an algebraic extension and S be the largestsubfield of F separable over K. Then [S : K] is called the separable degree ofF over K, denoted by [F : K]s. The dimension [F : S], denoted by [F : K]i,is called the inseparable degree of F over K.

[F : K]i = 1, or [F : K]s = [F : K] iff F is separable over K. [F : K]s = 1or [F : K]i = [F : K] iff F is purely inseparable over K. [F : K] = [F :K]s[F : K]i.

Lemma 3.2.2. Let F be an extension field of E and E/K an extension andN/K is a normal extension containing F . If r is the cardinality of distinct E-monomorphism F → N and t is the cardinality of distinct K-monomorphismE → N , then rt is the cardinality of distinct K-monomorphism F → N .


Proof. Suppose r, t < ∞. Let τ1, ..., τr be all the distinct E-monomorphismF → N and σ1, ..., σt be the distinct monomorphism E → N . Each σi ex-tends to a K-automorphism of N . Each composition σiτj is a K-monomorphismof F → N . If σiτj = σaτb, then τ−1

a σiτj = τb. Hence σ−1a σi|E = id.Hence

a = i. Since σi is injective, hence σiτj = σiτb implies τj = τb. So all σjτj

K-monomorphism are distinct. ON the other hand, let σ : F → N be aK-monomorphism. Then σ|E = σi for some i. Then σ−1

i σ|E = id. Henceσ−1

i σ = τj for some j. So σ = σiτj.

Proposition 3.2.1. Let F/K be a finite extension and N/K be normal whichcontains F . Then the number of distinct K-monomorphisms of F → N is[F : K]s.

Proof. Let S be the maximal separable extension of K contained in F . EveryK-monomorphism S → N extends to a K-automorphism of N and by restric-tion a K-monomorphism F → N . We only need to worry charK = p 6= 0and suppose τ, σ are two K-monomorphism F → N such that τ |S = σ|S. Ifu ∈ F , then upn ∈ S, hence τ(upn

) − σ(upn) = (σ(u) − τ(u))pn

= 0 hencetau(u) = σ(u). So σ|S = τ |S. So it suffice to prove when F/K is separable.

We proceed by induction on n = [F : K] = [F : K]s. The case n = 1is trivial. Choose Choose u ∈ F − K, if [K(u) : K] < n, then we applyK(u) = E to the above lemma. If [K(u) : K] = n, then F = K(u) and[F : K] is the degree of the irreducible polynomial f ∈ K[x] of u. Every K-monomorphism σ : F → N is completely determined by v = σ(u) minimalpolynomial (separable) of u over K is n. Since v is a root of f , there areat most [F : K] of them. Since f splits completely in N is separable, hencethere are exactly [F : K] of them.

Corollary 3.2.3. If F/E and E/K, then [F : E]s[E : K]s = [F : K]s and[F : E]i[E : K]i = [F : K]i.

Proof. It follows from lemma 3.2.2 and prop 3.2.1.

Corollary 3.2.4. Let f ∈ K[x] be an irreducible monic polynomial over K,F a splitting field of f over K and u1 a root of f . Then

1. every root of f has multiplicity [K(u1) : K]i so that in F (x),

f = ((x− u1)...(x− un))[K(ui):K]i .

2. uK(u1):K]i1 is separable over K.


Proof. Assume charK = p 6= 0. 1) for σi : u1 7→ ui is a K-isomorphism ofF . Hence (x − u1)

r1 ...(x − rn)rn = f = σf = (x − σ(u1))r1 ...(x − σ(un))rn .

Since ui are distinct, by the unique factorization in K[x] we know ri = r areall the same and hence [K : F ] = deff = nr. Since there are n distinctK-monomorphism K(u1) → F and hence [K(u1) : K]s = n. It follows[K(u1) : K]i = r.

Since r is a power of p, uri are all distinct, then (x−ur

1)...(x−urn) ∈ K[x]

satisfied by ur1, and hence it is separable over K.

Chapter 4

Cyclic extensions

4.1 Trace and Norm

Definition 4.1.1. Let F be a finite extension of K and K an algebraicclosure of K containing F . Let σ1, ..., σr be all the distinct K-monomorphismF → K. If u ∈ F , the norm of u, denoted by NF

K(u) is defined by

NFK(u) = (σ1(u) · · ·σr(u))[f :K]i .

The trace of u, denoted by T FK (u) is the element

T FK (u) = [f : K]i(σ1(u) + · · ·+ σr(u)).

By definition, we can track that NFK(u), T F

K (u) ∈ K.

Theorem 4.1.1. If F is a finite dimensional Galois extension of K and

Gal(F/K) = {σ1, ·, σn},

then for u ∈ F

NFK(u) = σ1(u) · · ·σr(u);

T FK (u) = σ1(u) + · · ·+ σr(u).

Proof. When F/K is Galois, then F is separable over K. Hence [F : K]i =1.

Theorem 4.1.2. Let F/K be a finite extension. Then for all u, v ∈ F :

45

46 CHAPTER 4. CYCLIC EXTENSIONS

1. NFK(u)NF

K(v) = NFK(uv) and T F

K (u) + T FK (v) = T F

K (u + v);

2. if u ∈ K, then NFK(u) = u[F :K] and T F

K (u) = [F : K]u;

3. NFK(u) and T F

K (u) are elements of K. More precisely

NFK(u) = ((−1)na0)

[F :K(u)] ∈ K, T FK (u) = −[F : K(u)]an−1 ∈ K,

where f = xn +an−1xn−1 + · · ·+a0 ∈ K[x] is the irreducible polynomial

of u;

4. if E is an intermediate field, then

NEK(NF

E (u)) = NFK(u), TE

K (T FE (u)) = T F

K (u).

4.2 Cyclic extension

Definition 4.2.1. Let S be a nonempty set of automorphisms of a fieldF . S is linearly independent provided that for any a1, · · · , an ∈ F andσ1, · · · , σn ∈ S:

a1σ1(u) + · · · anσn(u) = 0,∀u ∈ F

then ai = 0 for every i.

Lemma 4.2.1. If S is a set of distinct automorphisms of a field F , then Sis linearly independent.

Proof. If not there exist nonzero ai ∈ F and distinct σi ∈ F such that

a1σ1(u) + · · · anσn(u) = 0,∀u ∈ F, (4.1)

such an n > 1. Among all choose one with n minimal. Since σ1 and σ2 aredistinct, there exists v ∈ F with σ1(v) 6= σ2(v). Apply (4.1) to uv we have

a1σ1(u)σ1(v) + · · ·+ anσn(u)σn(v) = 0. (4.2)

Multiply (4.1) by σ(v) gives:

a1σ1(u)σ1(v) + · · ·+ anσn(u)σ1(v) = 0. (4.3)

The difference of (4.4) and (4.2) is a relation

a1[σ2 − σ1(v)]σ2(u) + · · ·+ an[σn(v)− σ1(v)σn(u) = 0,∀u ∈ F. (4.4)

Not all the coefficients are zero and this contradicts the minimality of n.

4.2. CYCLIC EXTENSION 47

Definition 4.2.2. Let E/F is said to be cyclic (resp. abelian) if F/K isalgebraic and Galois and Gal(F/F ) is a cyclic (resp. abelian)group.

Example 4.2.1. Any finite field with characteristic p is a cyclic extension overFp.

Theorem 4.2.1. Let F be a cyclic extension field of K of degree n, σ agenerator of Gal(F/K). Then

1. T FK (u) = 0 if and only if u = v − σ(v) for some v ∈ F .

2. (Hilbert’s theorem 90) NFK(u) = 1 if and only if u = vσ(v)−1 for some

nonzero v ∈ F

Proof. T (u) = u + σu + · · ·+ σn−1u and Tu = u(σu) · · · (σn−1u).1) It is easy to prove that when u = v − σv then Tu = 0. Now we prove

the converse. Suppose Tu = 0. Since σ, · · · , σn are linearly independent,there exists a z ∈ F such that Tz 6= 0. Let w = Tz−1z. Then Tw = 1. Nowlet

v = uw+(u+σu)(σw)+(u+σu+σ2u)(σ2w)+· · ·+(u+σu+· · ·+σn−2u)(σn−2w),

then we can check v − σv = u.2) It is also easy to verify that if u = σ(v)−1v then Tu = 1. Conversely,

if N(u) = 1, then there exists y ∈ F such that the element v given by

v = uy + (uσu)σy + · · ·+ (uσu · · ·σn−2u)σn−2y + (uσu · · ·σun−1)σn−1y

is nonzero. The last summand is indeed N(u)σn−1y = σn−1y. We can verifythat u−1v = σv. Hence u = vσv−1.

Proposition 4.2.1. Let F/K be a cyclic extension of degree n = mpt where0 6= CharK = p and (m, p) = 1. Then there exists a chain of intermediatefields F ⊃ E0 ⊃ E1 ⊃ · · · ⊃ Et = K such that F/E0 is cyclic of degree mand for each 0 ≤ i ≤ t, Ei−1 is a cyclic extension of Ei of degree p.

Proposition 4.2.2. Let K be a field of characteristic p 6= 0. F is a cyclicextension field of K of degree p if and only if F is a splitting field over Kof an irreducible polynomial of the form xp − x − a ∈ K[x]. In this caseF = K(u) where u is any root of xp − x− a.

48 CHAPTER 4. CYCLIC EXTENSIONS

Proof. If F/K is a cyclic extension of degree p, let Gal(F/K) =< σ >. Letu ∈ F −K, then f(x) = (x − u) · · · (x − σp−1u) ∈ K[x] is irreducible. NowT F

K (1) = [F : K]1 = p = 0. Hence 1 = v − σv. Now σu = u + 1. Henceσup = (u + 1)p = up + 1 = up − u + σu. Henceσup − u = up − u = a ∈ K.Now let u = −v, it is easy to see u /∈ K. σu = −σv = 1− v = 1 + u, henceσiu = i + u. The irreducible polynomial for u is f(x) = (x − u)(x − 1 −u) · · · (x−p−1−u). Now f(x+u) = x(x−1) · · · (x−p−1) = xp−x. Hencef(x) = (x− u)p − (x− u) = xp − up − x + u = xp − x + a.

Conversely, we assume that F = K(u) where u is a root of an irreduciblepolynomial of the form f(x) = xp− x− a. σ(u) = u + 1 is an automorphismon F whose degree is p. Let G =< σ >. Then InvG = K. On the otherhand since f(x) = (x−u)(x− (u+1)) · (x− (u+p−1)) is irreducible and p isa prime number, we know deg f = p won’t be a multiple of char(K). Henceall the p roots of f are distinct. Hence [F : K] = p by Theorem 2.1.2. So Fis a cyclic extension of K of degree p.

Chapter 5

Finite fields

5.1 Properties of finite fields

Any finite extension of the prime field Fp is called a finite field.

Lemma 5.1.1. If F is a finite extension over Fp, then F has pn elementsfor some exponent n.

Proof. Suppose [F : Fp] = n, then as a vector space F , it is n dimensionalover Fp.

Lemma 5.1.2. For any natural number n, there is a finite field with pn

elements.

Proof. Let f(x) = xpn − x ∈ Fp[x]. Now let F be the splitting field of f(x).Now f(x)′ = −1 6= 0, then f(x) has distinct roots. We are going to proveall these roots form a field. It is easy to see of apn

= a and bpn= b, then

(a− b)pn= apn − bpn

= a− b and (ab−1)pn= ab−1. Since f has degree pn and

all roots are distinct, then all these roots for a field of cardinality pn.

Lemma 5.1.3. Any two fields of the same number of elements are isomor-phic.

Proof. Given a field F with pn elements, we consider F× the nonzero ele-ments. Since F is a field, f× form an abelian group of order pn. Hence forany a ∈ F , apn−1 = 1,i.e. apn

= a for every a ∈ F . Hence F is a splitting fieldof xpn − x over Fp. It follows that any two such fields are isomorphic.

49

50 CHAPTER 5. FINITE FIELDS

Lemma 5.1.4. Given any finite field F with pn elements. Then σ : a 7→ ap

is an automorphism of F . Gal(F/Fp) is cyclic generated by σ or order n.

Proof. 1) Now give two elements x and y in E we know (x ± y)p = xp ± yp

and (xy)p = xpyp and σ(x−1) = x−p = (σ(x))−1.2) σi(x) = σi−1(xp) = xpi

. Since any a in F with pn elements satisfyingapn

= a, we know σn = id. Hence the order of σ is exactly n. Now [F :Fp] = n is normal separable, hence #Gal(F/Fp) = [F : Fp] and it followsGal(F/Fp) =< σ >.

Corollary 5.1.1. The subgroup of Fpn are all Fpm where m|n.

Proof. We apply the fundamental theorem of Galois theorem.Since Gal(Fpn/Fp) =< σ >. Hence Γ the set of all subgroups of Gal(Fpn/Fp)

are {< σm >} for some natural number m dividing n. Now Inv(< σm >) =Fpm and Gal(Fpn/Fpm) =< σm > of order n/m.

More over since each element in Γ is normal, we have

Gal(Fpn/Fp)/Gal(Fpn/Fpm) = Gal(Fpm/Fp).

Exercise 5.1.1. Let F be a finite field and f(x) an irreducible polynomialin F [x]. Show that if the roots r1, ..., rn of f(x) are suitably order then theGalois group Gf consisting of the powers of (12 · · ·n).

Proof. Suppose F is a field of pm elements. Let u be a root of f(x), then[F (u) : F ] = n and upm

/∈ F . Now we claim xpmis also a root of f since

σm(f(u)) = f(σm(u)) = f(upm

).

is the splitting field of the polynomial f as F (u) is finite. Similarly σ2m(u) =

up2mis also a roof of f(x). We label all the roots as u, upm

, up2m, ..., up(n−1)m

,then Gal(E/F ) acts on the roots as powers of (12 · · ·n).

Given a prime finite field Fp, let Fp be an algebraic closure of Fp, whichas defined in section 2.4.

Theorem 5.1.1. Gal(Fp/Fp) =< σp >∼= Z.

Proof. It follows from Lemma 5.1.4.

5.2. IRREDUCIBLE POLYNOMIALS OVER FP 51

5.2 Irreducible polynomials over Fp

Example 5.2.1. There are 2 degree 1 irreducible polynomials over F2, namelyx and x+1. Moreover x(x−1) = x2−x. There is only 1 degree 2 irreduciblepolynomial over F2, namely x2+x+1. Moreover x(x−1)(x2+x+1) = x4−x.

Exercise 5.2.1. Find the number of degree d irreducible polynomials in Fp.

Sol: Let N(d) denote the number of irreducible polynomials over Fp.Consider the factorization of xpn − x = 0 which is the splitting polynomialof Fpn . Since Fpn

is separable, so all irreducible factors of xpn − x have onlysimple roots. We know there has no irreducible factor with degree largerthan n since every root is in Fpn

. Hence we get

n∑n=1

N(n)n = pn.

So N(n) are given by a recursive formula. Exact formula for N(n) is availableby using some basis function in number theory.

52 CHAPTER 5. FINITE FIELDS

Part II

Commutative algebra

53

Chapter 6

Rings and ideals-Review

In this section, in stead to give a full treatment of rings, modules, and integraldomains, we will only recall only some special topics relevant to our discussionin the next few sections. For full details of the basic ring or module theory,please refer to either [Jac85], [Hun80], or [AM69].

6.1 Ideals

Any ring A will be commutative with unit 1 unless specified. Let a be anideal of A, then φ : A → A/a is a surjective ring homomorphism.

Proposition 6.1.1. There is a one-to-one order-preserving correspondencebetween the ideals b of A which contain a and the ideals b of A/a, given byb = φ−1(b).

Let f : A → B be a ring homomorphism.

If a is an idea in A, then f(a) is not necessary an ideal in B. We definethe extension ae of a to e the ideal Bf(a) generated by f(a) in B.

Let b be an ideal in B, then f−1(b) is an ideal in A. This ideal is calledthe contraction bc of b. Let b be a prime ideal in B, then f−1(b) is a primeideal in A. But if b is maximal in B, f−1(b) may not be maximal.

Example 6.1.1. A = Z, B = Q, a = 0. Consider the embedding f : Z → Q.The ideal 0 is maximal in Q, however 0c = 0 is not maximal in Z.

Theorem 6.1.1. Every ring A 6= 0 has at least one maximal ideal.

55

56 CHAPTER 6. RINGS AND IDEALS-REVIEW

6.2 Local rings

A ring with exactly one maximal ideal M is called a local ring. The fieldk = A/M is called the residue field of A.

Example 6.2.1. Z/pnZ for any prime p and any natural number n is a localring with residue field Fp.

Example 6.2.2. Any field will be a local ring with 0 being the only maximalideal. The residue field of a field is itself.

Proposition 6.2.1. 1. Let A be a ring with M 6= (1) be an ideal of Asuch that every x ∈ A−M is a unit. Then A is a local ring and M isits maximal ideal.

2. Let A be a ring and M a maximal ideal of A, such that every elementof 1 + M is a unit in A, then A is a local ring.

Proof. 1. Every ideal 6= (1) consists of non-units, hence is contained in M.Hence M is the only maximal ideal of A.

2. Let x ∈ A− x, since M is maximal, the the ideal generated by x andM is (1), hence xy + m = 1 for some y ∈ A, m ∈ M. Now xy=1-m is a unit.Hence it follows from 1. A is local.

6.3 Nilradical and Jacobson radical

Proposition 6.3.1. The set N of all nilpotent elements in a ring A is anideal, and A/N has no nilpotent element 6= 0.

Proof. Since A is commutative, if x is nilpotent, then ax for any a ∈ A isnilpotent. If x, y ∈ N such that xn = ym = 0 then (x + y)n+m+1 = 0. HenceR is an ideal. Now consider a + N for a /∈ N. If it is nilpotent, then an ∈ N

for some n, so a itself is nilpotent.

The ideal N is called the nilradical of A.

Example 6.3.1. A = Z/9Z, N = {0, 3, 6} = (3).

Example 6.3.2. A = Z/12Z, N = {0, 6}.

Proposition 6.3.2. The nilradical N is the intersection of all the primeideals of A.

6.4. RINGS AND MODULES OF FRACTIONS 57

Proof. Let N′ be the intersection of all prime ideals of A. If f ∈ A isnilpotent, then for any prime ideal a fn = 0 ∈ a, for some n > 0. Hencef ∈ a. So N ⊆ N′. If f /∈ N, then let Σ be the set of ideals a with theproperty n > 0 ⇒ fn /∈ a. 0 ∈ Σ is nonempty. The Σ has a maximalelement, called b. If x, y /∈ b, then b + (x) and b + (y) strictly contain bhence not in σ; hence fm ∈ b + (x), fn ∈ b + (y) for some n, m > 0. Thenfm+n ∈ b + (xy) hence b + (xy) is not in Σ and therefor xy /∈ b. Hence b isprime. It is a contradiction.

Definition 6.3.1. Let a be an ideal of A, the radical of a, denoted by√a = {f ∈ A|f r ∈ a for some r > 0}. The set

√a is an ideal itself.

Example 6.3.3. (x2) is an ideal of k[x], then√

(x2) = (x). (x, y2) is an ideal

of k[x, y], then√

(x, y2) = (x, y).

Definition 6.3.2. The Jacobson radical R of A is the intersection of allmaximal ideals of A.

Example 6.3.4. In Z, maximal ideals are (p) for primes p. Then the Jacobsonradical of Z is the zero ideal 0.

Proposition 6.3.3. x ∈ R ⇔ 1− xy is a unit in A for all y ∈ A.

Proof. If 1 − xy is not a unit, then it belongs to a maximal ideal m. Sincex ∈ M , xy ∈ m, hence it implies 1 ∈ m, which violates the definition ofmaximal ideal.

If x ∈ m for some maximal ideal m, then m and x generate 1, i.e. xy+m =1. Hence 1− xy is not a unit.

Let A = F2[x, y]/(x2−y) where x, y are indeterminant. Find the nilradicalof A and Jacobson radical of A.

Exercise 6.3.1. In the ring A[x], the nilradical is equal to the Jacobson radical.

6.4 Rings and modules of fractions

A multiplicatively closed set of a ring A is a subset S of A such that 1 ∈ S andS is closed under multiplication. Then for the order pairs (a, s), a ∈ A, s ∈ S,the following relation is an equivalent relation

(a, s) ∼ (b, t) ⇔ u(at− bs) = 0, for some u ∈ S.


Let S−1A denote the equivalent classes of (a, s). It is called the ring offractions respect to S. We can think of the class (a, s) as a

s.

There is a ring homomorphism from f : A → S−1A by sending a 7→ (a, 1).

Example 6.4.1. The following sets are multiplicative

• S = {ak}∞k=0, where a ∈ A.

• S = A− p where p is a prime ideal of A.

• S = A×

Example 6.4.2. A = Z, S = {2k}∞k=0, then S−1Z = {m2n , m, n ∈ Z, n ≥ 0}.

Denote this new ring as Z[1/2] (means 2 is invertible here.) For any primep ∈ Z different from 2, (p) is a prime ideal of Z[1/2].

Example 6.4.3. A = Z. If S = A − (2), then S−1Z = {mn, m, n ∈ Z, n = 1

mod 2}. This ring has only one maximal ideal (2) and hence is a local ring.Its residue field is F2.

Proposition 6.4.1. Let p be a prime ideal of A, let S = A − p. ThenAp = S−1A is a local ring with the maximal ideal M = {a

s, a ∈ p, b ∈ S}.

The process of passing from A to Ap is called the localization of A at p.

6.5 Nakayama’s lemma

Proposition 6.5.1. Let M be a finite generated A-module, and let a bean ideal of A, and let φ be an A-module endomorphism of M such thatφ(M) ⊆ aM . The φ satisfies an equation of the form

πn + a1φn−1 + · · ·+ an = 0, ai ∈ a.

Corollary 6.5.1. Let M be a finitely generated A-module and let a be anideal of A such that aM = M . Then there exist x ≡ 1( mod a) such thatxM = 0.

Theorem 6.5.1 (Nakayama’s lemma). Let M be a finitely generated A-module and a an ideal of A contained in the Jacobson radical R of A. ThenaM = M implies that M = 0.

Proof. By the previous corollary we have xM = 0 for some x ≡ 1( mod R.By Proposition 6.3.3 x is a unit in A, hence M = 0.

6.6. INTEGRAL DEPENDENCY 59

6.6 Integral dependency

Let B be a ring and A be a subring of B. An element of B is said to beintegral over A if x is a root of a monic polynomial with coefficients in A,that is x satisfies an equation of the form

xn + an−1xn−1 + · · ·+ a0 = 0, ai ∈ A. (6.1)

Example 6.6.1. When A = Z, B = Q, any a ∈ Q which is integral over Zsatisfies a minimal polynomial x− a ∈ Z[x]. Hence a ∈ Z.

Example 6.6.2. When A = Z and B = Z[i], then every a+bi ∈ Z[i] is integralover A as it satisfies x2 − 2ax + (a2 + b2) = 0.

Proposition 6.6.1. The following are equivalent:

1. x ∈ B is integral over A;

2. A[x] is a finitely generated A-module;

3. A[x] is contained in a subring C of B such that C is a finitely generatedA-module;

4. There exists a faithful A[x]-module M which is finitely generated as anA-module.

Proof. 1 ⇒ 2 : A[x] is generated by 1, x, ..., xn−1.2 ⇒ 3: take C = A[x].3 ⇒ 4 : Take M = C, which is a faithful A[x]-module as yC = 0 ⇒ y ·1 =

0.4 ⇒ 1: Since M is an A[x]-module, we have xM ⊂ M . Since M is

faithful, then we have xn + a1xn−1 + ...an = 0 for some ai ∈ A.

Corollary 6.6.1. Let xi, 1 ≤ i ≤ n be elements of B, each integral over A.Then the ring A[x1, · · · , xn] is a finitely generated A-module.

Proof. Ny induction on n. When n > 1, let Ar = A[x1, ..., xr]. Then by hy-pothesis, An−1 is a finitely generated A-module, say generated by b1, ..., bm.An = An−1[xn] is a finitely generated An−1-module and an A-module gener-ated by b1, ..., bm, b1xn, ..., bmxn.


The elements C of B which are integral over A form a subring of Bcontaining A. The set C is called the integral closure of A in B. If C = A,then A is said to be integrally closed in B. If C = B, then the ring B is saidto be integral over A.

For example, Z is integrally closed in Q, and Z[i] is integral over Z.

Proposition 6.6.2. Let A ⊆ B be rings, B integral over A.

1. If b is an ideal of B and a = bc = A∩ b, then B/b is integral over A/a.

2. If S is a multiplicatively closed subset of A, then S−1B is integral overS−1B.

Proof. If x ∈ B we have xn + an−1xn−1 + · · ·+ a0 = 0, ai ∈ A. Reduce by b

we have we have x modb satisfied a monic equation with coefficients in A/a.Hence x mod b is integral over A/a.

Let x/s ∈ S−1B, x ∈ B, s ∈ S. Then the equation above gives

(x/s)n + (an−1/s)(x/s)n−1 + · · ·+ a0/sn = 0

which shows that x/s is integral over S−1A.

Example 6.6.3. Since Z[i] is integral over Z, pick S = Z−(2) a multiplicativeset in Z, then S−1Z[i] = { b

s| b ∈ Z[i], s ∈ Z− (2)} = S−1Z + iS−1Z. This

ring has only one maximal ideal (1+i) hence is local By the above proposition,S−1 + iS−1Z is integral over S−1Z.

Proposition 6.6.3. Let A ⊆ B be integral domains and suppose B is integralover A. Then B is a field if and only if A is a field.

Proof. If B is a field, then for any x ∈ A, x−1 ∈ B, and hence x−1 is integralover A. So it satisfies

x−n + an−1x1−n + · · ·+ a0 = 0, ai ∈ A.

So x−1 = −an−1 − an−2x− · · · − a0xn−1 ∈ A. So A is a field.

If A is a field and x ∈ B is integral over A and satisfies

xn + an−1xn−1 + · · ·+ a0 = 0, ai ∈ A, a0 6= 0

then

x−n +a1

a0

x1−n + · · ·+ 1

a0

= 0.

Since A is a field, we have x−1 is integral over A and hence x−1 ∈ B since Bis integral over A. So B is also a field.

6.7. INTEGRALLY CLOSED DOMAINS 61

Corollary 6.6.2. Let A ⊆ B be rings, and B be integral over A. Let q be aprime ideal of B and let p = qc = a ∩A. Then q is maximal if and only if p

is maximal.

Proof. B/q is integral over A/p, and both rings are integral domains. Thenit followings from Proposition 6.6.3.

Proposition 6.6.4. Let A ⊆ B be rings, and B be integral over A. Let qand q′ be prime ideals of B such that a ⊂ q′ and qc = q′c = p. Then q = q′.

Proof. By Proposition 6.6.2 we know Bp is integral over Ap. Let M be theextension of p in Ap and let n, n′ be the extension of q, q′ respectively in Bp.Then M is maximal idea; of Ap; n ⊆ n′ and nc = n′c = M. By Proposition6.6.3 we know both n and n′ are maximal, hence n = n′. So q = q′.

Theorem 6.6.1. Let A ⊂ B be rings and suppose B is integral over A. Letp be a prime ideal of A. Then there exists a prime ideal q of B such thatq ∩ A = p.

Proof. By Proposition 6.6.2, Bp is integral over Ap and the diagram

A → Bα ↓ ↓ βAp → Bp

is commutative. Let n be the maximal ideal of Bp; then M = n ∩ Ap ismaximal by Proposition 6.6.3, hence is the unique maximal ideal of the localring Ap. If q = β−1(n), then q is prime and we have q∩A = α−1(M) = p.

Definition 6.6.1. An integral domain is said to be integrally closed if it isintegrally closed in its field of fractions. For example, Z is integrally closed.

6.7 Integrally closed domains

Proposition 6.7.1. Let A ⊆ B be rings, C the integral closure of A in B.Let S be a multiplicatively closed subset of A. Then S−1C is the integralclosure of S−1A in S−1B.


Proof. By Proposition 6.6.2, S−1C is integral over S−1A. Then elementb/s ∈ S−1B which is integral over S−1A satisfies

(b/s)n + (an−1/sn−1)(b/s)n−1 + · · ·+ a0/s0 = 0,

for some ai ∈ A, si ∈ S. Let t = sn−1sn−2 · · · s0 and multiply the aboveequation by (st)n throughout. Then the above equation becomes an equationof integral dependence over A. Hence bt ∈ C and b/s = bt/st ∈ S−1C.

Proposition 6.7.2. Let A be an integral domain. Then following are equiv-alent.

1. A is integrally closed;

2. Ap is integrally closed for each prime ideal p;

3. AM is integrally closed for each maximal ideal M.

Proof. Let K be the field of fractions of A, let C be the integral closureof A in K. Let f : A → C be the identity mapping of A onto C. Then Abeing integrally closed is equivalent to f being surjective, and by the previousProposition Ap (resp. AM) being integrally closed, is equivalent to fp (resp.fM is surjective.

Chapter 7

Primary decomposition

7.1 Primary ideals

Example 7.1.1. In Z, the ideal (12) = (22) ∩ (3). The ideal (3) is prime and(22) is not.

Recall, given an ideal a of A, the radical of a, denoted by r(a) =√

a ={x ∈ A|xn ∈ α, for some n > 0}.

Proposition 7.1.1. r(a) is the intersection of all prime ideals of A whichcontains a.

Proof. f : A → A/a sending r(a) to the nilradical of A/a which consistsof nilpotent elements of A/a. The nilradical R is the intersection of primeideals of A/a. Hence r(a) = f−1(R) is the intersection of prime ideals of Acontaining a.

An ideal a in a ring A is primary if a 6= A and if

xy ∈ a ⇒ either x ∈ a or yn ∈ a for some n > 0.

In other words, if ab ∈ α, then either a ∈ a or b ∈ r(a). Equivalently

α is primary ⇔ A/a 6= 0 and every zero divisor in A/a is nilpotent.

Proposition 7.1.2. If a is primary, then r(a) is the smallest prime contain-ing a.

63

64 CHAPTER 7. PRIMARY DECOMPOSITION

Proof. The smallest part is clear as r(a) is the intersection of all prime idealscontaining a. We only need to show it is a prime ideal. If xy ∈ r(a), then(xy)n ∈ a for some n > 0. If xn /∈ a and because a is primary, hence yn ∈ r(a)and hence y ∈ r(a).

Example 7.1.2. If k is a field and x and y are two variables, then (x, y2)is primary in A = k[x, y] as the zero divisors of A/(x, y2) = A[y]/(y2) arenilpotent. Moreover (x, y) is a maximal ideal as A/(x, y) = k is a field. wehave (x, y)2 = (x2, xy, y2) ( (x, y2) ( (x, y).

Example 7.1.3. Let A = k[x, y, z]/(xy−z2), let x, y, z be the images of x, y, zin A. Then p = (x, z) is prime as A/a = A[y] is an integral domain. Butp2 = (xz, x2, z2) is not primary as xy = z2 ∈ p2, but x /∈ p2 andy /∈ r(p2) = p.Hence p2 is not primary. So not all powers of prime ideals are primary.

Proposition 7.1.3. Let a be an ideal in A, if r(a) = M is a maximal ideal,then a is primary. Consequently, every power of a maximal ideal is primary.

Proof. Since r(a) is the intersection of all prime ideals containing a andit ends up with a maximal ideal, hence only one prime ideal, which is M

containing a. Now f : A → A/a maps r(a) to the nilradical of A/a. Itimplies A/a is has also only one prime ideal nil(A/a). So every element inA/a is either a unit or a zero divisor which is nilpotent. So a is primary.

7.2 Primary decomposition

An ideal a is said to have a primary decomposition if a is a finite intersectionof primary ideals.

Example 7.2.1. (x2, xy) = (x)∩(x, y)2 is a primary decomposition of (x2, xy).It is because (x) is prime and (x, y) is maximal, hence (x, y)2 is primary.

Example 7.2.2. Every ideal in Z has a primary decomposition. It is becauseevery ideal is principal, so can be written as (n) =

∏ni pdi

i where pi are distinctprimes. Then (n) =

⋂ni=1(p

bii ) and each (pdi

i ) is primary.

7.3 Minimal primary decomposition

Lemma 7.3.1. Let ai(1 ≤ i ≤ n) be p-primary, then a = ∩ni=1ai is wp-

primary.

7.3. MINIMAL PRIMARY DECOMPOSITION 65

Proof. r(∩ni=1ai) = ∩r(ai) = p. Now we need to show a is primary. If

xy ∈ a, y /∈ a, then for some i we have xy ∈ ai and y /∈ ai, hence x ∈ p, sinceai is primary.

Definition 7.3.1. Let a and b be two ideals of a ring A, then we use (a : b)

to denote the following set {x ∈ A∣∣∣ xb ∈ a}. When b = (x) is a principal

ideal then (a : b) is denoted by (a : x).

Example 7.3.1. In A = Z, let a = (2) and b = (3), then ((2) : 3) = (2) and((4) : 2) = (2).

Example 7.3.2. (0 : x) = Ann(x).

Lemma 7.3.2. Let a be a p-primary ideal, x ∈ A. Then

1. if x ∈ a, then (a : x) = (1);

2. if x /∈ a then (a : x) is p-primary, and therefore r(a : x) = p.

3. if x /∈ p then (a : x) = a

Proof. 1) is easy.2) (a : x) = {y ∈ A|xy ∈ a}, since a is primary, y ∈ r(a) = p. So

a ⊆ (a : x) ⊆ p, so by the previous lemma, (a : x) is p-primary.3) Follows from the definition.

Given a primary decomposition of an ideal a = ∩ni=1ai where ai are pri-

mary, it is said to be minimal if the following two conditions are satisfied.

1. The r(ai) are all distinct.

2. ai + ∩j 6=iaj(1 ≤ i ≤ n).

Proposition 7.3.1. Any primary decomposition can be reduced to a minimalone.

66 CHAPTER 7. PRIMARY DECOMPOSITION

Chapter 8

Chain condition

8.1 Chain condition

Definition 8.1.1. A module A is said to satisfied the ascending chain condi-tion (a.c.c) on submodules (or to be Noetherian) if every chain A1 ⊂ A2 ⊂ · · ·of submodules of A, there is an integer n such that Ai = An for all i ≥ n.

A module B is said to satisfied the descending chain condition (d.c.c) onsubmodules (or to be Artinian) if every chain A1 ⊃ A2 ⊃ · · · of submodulesof A, there is an integer n such that Ai = An for all i ≥ n.

For example, the Z-module Z satisfied the ascending chain condition butnot the descending chain condition on the submodules.

If a ring R is considered as a left [resp. right] module over itself, then itis easy to see that the submodules of R are the left [resp. right] ideals of R.So we can speak of chain conditions on left or tight ideals.

Definition 8.1.2. A ring R is left [resp. right] Noetherian if R satisfiesthe ascending chain condition on left [resp. right] ideals. R is said to beNoetherian if R is both left and right Noetherian.

A ring R is left [resp. right] Artinian if R satisfies the descending chaincondition on left [resp. right] ideals. R is said to be Artinian if R is both leftand right Artinian.

Example 8.1.1. A division ring D is both Noetherian and Artinian since theonly left and right ideals are D and 0.

Example 8.1.2. Every commutative principal ideal ring is Noetherian. Specialcases including Z, Zn, and F [x] with F a field.

67

68 CHAPTER 8. CHAIN CONDITION

Example 8.1.3. The ring MatnD of all n× n matrices over a division ring isboth Noetherian and Artinian.

Example 8.1.4. The ring of all 2× 2 matrices

(a b0 c

)such that a ∈ Z and

b, c ∈ Q is right Noetherian but not left Noetherian.

Example 8.1.5. The ring of all 2× 2 matrices

(d r0 s

)such that d ∈ R and

s, r ∈ R is right Artinian but not left Artinian.

Example 8.1.6. If I ia a nonzero ideal in a principal ideal domain R, thenthe ring R/I ia both Noetherian and Artinian.

Example 8.1.7. The ring k[x1, x2, ...] is not Noetherian nor Artinian. Butits field of fraction is both Noetherian and Artinian. Hence a subring of aNoetherian ring need not be Noetherian.

Let Σ be a set of partially ordered by a relation ≤

Proposition 8.1.1. The following conditions are equivalent:

1. Every increasing sequence x1 ≤ x2 ≤ · · · in σ is stationary;

2. Every nonempty subset of σ has a maximal element;

Proof. 1) ⇒ 2) If 2) fails then there is a nonempty subset T of σ withno maximal element, and we can construct inductively a non-terminatingstrictly increasing sequence in T .

2) ⇒ 1) The set (xm) has a maximal element, say xn. Hence (xm) isstationary.

8.2 Basic properties of Noetherian and Ar-

tinian rings

Let A be a commutative ring with unity and let M be an abelian groupwhose binary operation is written additively. Suppose M is an A-module.

Proposition 8.2.1. The group M is a Noetherian A-module if and only ifevery submodule of M is finitely generated.

8.2. BASIC PROPERTIES OF NOETHERIAN AND ARTINIAN RINGS69

Proof. Let N be a submodule of M , and let Σ be the set of all finitelygenerated submodules of N . Then Σ is nonempty as 0 ∈ Σ, and thereforehas a maximal element, say N0. IfN0 6= N , consider the submodule N0 +Axwhere x ∈ N −N0; this is finitely generated and strictly contains N0. HenceN0 = N .

Let M1 ⊂ M2 ⊂ · · · be an ascending chain of submodules of M . ThenN = ∪∞n=1Mn is a submodule of M , hence is finitely generated by x1, ..., xr.Say xi ∈ Mni

, and let n = maxri=1ni; ten each xi ∈ Mn, hence the chain is

stationary.

Because of this result, Noetherian modules are more important than Ar-tinian modules.

Proposition 8.2.2. Let 0 → M ′ α→Mβ→ M ′′ → 0 be an exact sequence of

A-modules. Then

1. M is Noetherian if and only if M ′ and M ′′ are Noetherian;

2. M is Artinian if and only if M ′ and M ′′ are Artinian.

Proof. We shall prove 1) and the proof of 2) is similar.⇒ An ascending chain of submodules of M ′ (or M ′′) gives rise to a chain

in M , hence is stationary.⇐ Let (Ln)n≥1 be an ascending chain of submodules; then (α−1(Ln)) is

a chain in M ′ and (β(Ln)) is a chain in M ′′. For large enough n both thesechains are stationary, and it follows that the chain (Ln) is stationary.

Corollary 8.2.1. If Mi(1 ≤ i ≤ n) are Noetherian (resp. Artinian) A-modules, so is ⊕n

i=1Mi.

Proof. Applies the above Proposition to

0 → Mnα→⊕n

i=1 Miβ→ ⊕n−1

i=1 Mi → 0.

The result follows from induction.

Proposition 8.2.3. Let A be a Noetherian (resp. Artinian ring). M afinitely generated A-module. Then M is Noetherian (resp. Artinian).

Proof. M is a quotient of An for some n. The results from Corollary 8.2.1and Proposition 8.2.2.


Proposition 8.2.4. Let A be a Noetherian (resp. Artinian), a an ideal ofA. Then A/a is a Noetherian (Artinian) ring.

Proof. By Proposition 8.2.2, A/a is Noetherian (resp. Artinian) as an A-module, hence also as an A/a-module.

A chain of submodules of a module M is a sequence (Mi)(0 ≤ i ≤ n) ofsubmodules of M such that

M = M0 ⊃ M1 ⊃ · · · ⊃ Mn = 0 strict inclusions.

The length of the chain is n. A composition series of M is a maximalchain, that is one in which no extra submodules can be inserted: which isequivalent to say each Mi−1/Mi(1 ≤ i ≤ n) is simple.

Proposition 8.2.5. Suppose that M has a composition series of length n.Then every composition series of M has length n, and every chain in M canbe extended to a composition series.

Proof. Let l(M) be the least length of a composition series of a module M .(l(M) = ∞ if M has no composition series.)

i) N ⊂ M ⇒→ l(N) < l(M). Let (Mi) be a composition series of M ofminimum length and consider Ni = NMi of N . Since Ni−1/Ni ⊂ Mi−1/Mi

and the latter is a simple module, we have either Ni−1/Ni = Mi−1/Mi orNi−1 = Ni; hence removing repeated terms, we have a composition series ofN , so that l(N) ≤ l(M). If l(N) = l(M) = n, then Ni−1/Ni = Mi−1/Mi foreach i; hence Mi = Ni and N = M .

ii) Any chain in M has length ≤ l(M). Let M = M0 ⊃ M1 ⊃ · · · bea chain of length k. By i) we have l(M) > l(M1) > · · · l(Mk) = 0 hencel(M) ≥ k.

iii) Consider any composition series of M . If it has length k, then k ≥l(M). Hence k = l(M). For any chain, if its length is l(M) then it must be acomposition series by ii); if its length is < l(M), it is not a composition series,hence not maximal. So new terms can be inserted until length is l(M).

Proposition 8.2.6. M has a composition series if and only if M satisfiesboth chain conditions.

Proof. ⇒ All chains in M are of bounded length, hence both a.c.c and d.c.c.⇐ Construct a composition series of M as follows. Since M = M0 satisfies

the maximum condition by Proposition 8.1.1, it has a maximal submodule

8.2. BASIC PROPERTIES OF NOETHERIAN AND ARTINIAN RINGS71

M1 ⊂ M0. Similarly M1 has a maximal submodule M2 ⊂ M1, and so on.Thus we have a strictly descending chain M0 ⊃ M1 ⊃ · · · which is d.c.c.must be finite, and hence is a composition series of M .

A module satisfying both a.c.c and d.c.c is called a modules of finitelength. l(M) is called the length of M . The Jordan-Holder theorem appliesto modules of finite length: if (Mi)

ni=0 and (M ′)n

i=0 are any two compositionseries of M , there is a one-to-one correspondence between the set of quotients(Mi−1/Mi)1≤i≤n and (Mi−1/Mi)1≤i≤n such that the corresponding quotient isisomorphic.

Proposition 8.2.7. The length l(M) is an additive function on the class ofall A-modules of finite length.

Proof. If 0 → M ′ a→Mβ→ M ′′ → 0 is an exact sequence, then l(M) =

l(M ′) + l(M ′′). Take the image under a of any composition series of M ′ andthe inverse image under β of any composition series of M ′′; these fits togetherto give a composition series of M .

Example 8.2.1. Let M be a Noetherian A-module and let a be the annihilatorof M in A. Prove that A/a is a Noetherian ring.

Chapter 9

Noetherian rings

9.1 Properties of Noetherian rings

A ring is said to be Noetherian if it satisfies the following three equivalentconditions:

1. Every nonempty set of ideals has a maximal element.

2. Every ascending chain of ideals in A is stationary.

3. Every ideal in A is finitely generated.

Proposition 9.1.1. If A is Neotherian and φ is a homomorphism of A ontoa ring B, then B is Noetherian.

Proof. This follows from Proposition 8.2.4 since B ∼= A/Ker(φ).

Proposition 9.1.2. Let A be a subring of B; suppose that A is Noetherianand that B is finitely generated as an A-modules. Then B is Noetherian.

Proof. By Proposition 8.2.3, B is a Noetherian A-module. Hence it is aNoetherian B-module.

Theorem 9.1.1 (Hilbert Basis Theorem). If A is Noetherian, then thepolynomial ring A[x] is Noetherian.

Proof. Let a be an ideal in A[x]. The leading coefficients of the polynomialsin a form an ideal ı in A. Since A is Noetherian, ı is finitely generated, sayby a1, ..., an. For i = 1, ..., n there is a polynomial fi ∈ A[x] of the form

73

74 CHAPTER 9. NOETHERIAN RINGS

fi = aixri+ (lower terms). Let r = maxr

i=1ri. The fi generate an ideal a′ ⊂ a

in A[x]. Let f = axm+ (lower terms) be nay element of a; we have a ∈ ı.If m ≥ r, write a =

∑ni=1 uiai where ui ∈ A; then f −

∑uifix

m−ri is ina and has degree < m. Proceeding in this way, we can go on subtractingelements of a′ from f until we get a polynomial g of degree < r. Hencef = g + h, h ∈ a′. Let M be the A module generated by 1, x, ..., xr−1; thena = (aM) + a′. Now M is finitely generated A-module. hence is Noetherian.So a is finitely generated and is Noetherian.

Corollary 9.1.1. If A is Noetherian so if A[x1, ..., xn].

Recall that an A-algebra B is a ring B together with a ring homomor-phism f : A → B. A ring homomorphism f : A → B is finite and B is a finiteA-algebra, if B is a finitely generated as an A-module. The homomorphismf is of finite type and B is a finitely generated A-algebra, if there exists afinite set of elements x1, ..., xn in B such that every elements of B can bewritten as a polynomial in x1, ..., xn with coefficients in f(A); or equivalentlythere is an A-algebra homomorphism from a polynomial ring A[t1, ..., tn] toB.

Corollary 9.1.2. Let B be a finitely generated A-algebra. If A is Noetherian,then so is B. In particular, every finitely generated ring, and every finitelygenerated algebra over a field is Noetherian.

Proof. The ring B is a finitely generated A-algebra, then B is a homomorphicimage of a polynomial ring A[x1, ..., xn], which is Noetherian by Corollary9.1.1.

We know let xi, 1 ≤ i ≤ n be elements of B each integral over A. Thering A[x1, ..., xn] is a finitely generated A-module. Consequently, the set Cof elements of B which are integral over A is a subring of B containing A.This ring C is called the integral closure of A in B. If C=A then we said Ais algebraically closed in B. If C = B then the ring B is said to be integralover A.

Proposition 9.1.3. Let A ⊆ B ⊆ C be rings. Suppose that A is Noetherian,that C is finitely generated as an A-algebra and that C is finitely generatedas a B-module. Then B is finitely generated as an A-algebra.

9.1. PROPERTIES OF NOETHERIAN RINGS 75

Proof. Let x1, ..., xm generated C as an A-algebra, and let y1, ..., yn generatedC as a B-module. The there exist expressions of the form

xi =∑

j

bijyj (bij ∈ B (9.1)

yiyj =∑

k

bijkyk (bijk ∈ B). (9.2)

Let B be the algebra generated over A by the bij and bijk. Since A is Noethe-rian, so is B0, and A ⊆ B0 ⊆ B.

An element of C is a polynomial in the Xi with coefficients in A. Substi-tuting (9.2) and making repeated use of (9.2) show that each element of C isa linear combinations of the yi with coefficients in B0, and hence C is finitelygenerated as a B0-module. Since B0 is Noetherian, and B is a submoduleof C, it follows that B is finitely generated as a B0-module. Since B0 isfinitely generated as an A-algebra, it follows that B is finitely generated asA-module.

Proposition 9.1.4. Let K be a field and E is a finitely generated Kalgebra.If E is a field, then it is a finite algebraic extension of K.

Proof. Let E = K[x1, ..., xn]. If E is not algebraic over E then we canrenumber xi so that x1, ..., xr are algebraically independent over K and eachxr+1, ..., xn is algebraically over the field F = K(x1, ..., xr). Hence E is analgebraic extension over F and therefor finitely generated as an F -module.So apply the previous Proposition to F ⊆ F ⊆ E, it follows that F is finitelygenerated as a K-algebra, say F = K[y1, ..., ys] where yi = fi/gi, fi, gi ∈K[x1, ..., xr].

There are infinitely many polynomials in K[x1, ..., xr], hence there is anirreducible polynomial h which is coprime to all gi and the element h−1 of Fcould not be a polynomial in yi. This is a contradiction. Hence E is algebraicover K, therefore, finite algebraic.

Corollary 9.1.3. Let k be a field, A a finitely generated k-algebra. let M bea maximal ideal of A. The the field A/M is a finite algebraic extension ofK. In particular, if K is algebraically closed then A/M = K.

Proof. Take E = A/M in the previous Proposition.


9.2 Primary decomposition for Noetherian rings

We are going to show every ideal in a Noetherian ring has a primary decom-position.

An ideal a is said to be irreducible if a is the intersection of two ideals,i.e a = b ∩ c then either a = b or a = c.

Example 9.2.1. In the ring Z, every ideal is principal. Hence (n) ∩ (m) =(lcm(n, m)). Hence every ideal (p) is irreducible if an only if p is a prime.

Similarly, for the any principal ideal domain, and ideal a = (f) is irre-ducible if and only if the element f is not a product of two non-units.

Proposition 9.2.1. In a Neotherian ring, every ideal is a finite intersectionof irreducible ideals.

Proof. If not, let Σ be the set of all ideals in A which are not finite intersectionof ideals. By assumption, Σ is nonempty. Since A is a Noetherian ring(hence every nonempty set of ideals contains a maximal one), we assume a

is a maximal ideal in Σ. a itself won’t be irreducible by the assumption, soit can be written as intersection of two ideals b and c such that both b and c

contain a as a proper subset. It follows b and c are not in Σ so they can bewritten as finite intersection of irreducible ideals. So is a, which contradictsthe choice of a.

Exercise 9.2.1. Every irreducible ideal in a Noetherian ring is primary.

Proof. We look at the quotient ring A/a and suppose now 0 is irreducibleand we want to show that it is primary. Let xy = 0 and suppose y 6= 0. Thechain of annihilators

Ann(x) ⊆ Ann(x2) ⊆ · · ·is stationary, so we have Ann(xi) = Ann(xn) when i ≥ n. Now consider theintersection (xn) ∩ (y). If a ∈ (xn) ∩ (y) then a = cxn = by ∈ Ann(x), i.e.byx = 0 = cxnx = cxn+1 So c ∈ Ann(xn+1) = Ann(xn), hence cxn = a = 0.So (xn) ∩ (y) = 0. As 0 is irreducible and y 6= 0, hence xn = 0 so 0 isprimary.

Theorem 9.2.1. In a Noetherian ring A every ideal has a primary decom-position.

Proposition 9.2.2. In a Noetherian ring A, every ideal a contains a powerof its radical.

9.2. PRIMARY DECOMPOSITION FOR NOETHERIAN RINGS 77

Proof. Let x1, ..., xk generate r(a): say xnii ∈ a(1 ≤ i ≤ k). Let m =∑k

i=1(ni−1)+1. Then r(a)m is generated by products xr11 · · ·x

rkk with ri = m;

so we have some ri ≥ ni for at least one index i. Hence r(a)m ⊆ a.

Corollary 9.2.1. In a Noetherian ring the nilradical is nilpotent.

Proof. Take a = 0 in the previous proposition.

Corollary 9.2.2. Let A be Noetherian, M a maximal ideal of A, a any idealof A, then the following are equivalent.

1. a is M-primary.

2. r(q) = M.

3. Mn ⊆ a ⊆ M for some n > 0.

Proof. 1) ⇔ 2) ⇒ 3) are easy, 3) ⇒ 2) by taking radicals.

Example 9.2.2. In A = k[x, y], the ideal (x2, xy) = (x)∩(x, y)2 = (x)∩(x2, y)has two different primary decompositions. Both of them are minimal.

Theorem 9.2.2. Let a 6= (1) be an ideal in a Noetherian ring. Then theprime ideals which belong to a are precisely the prime ideals which occur inthe set (a : x)(x ∈ A).

Proof. Passing to A/a we may assume a = 0. Let ∩ni=1ai = 0 be a minimal

primary decomposition of the zero ideals, and pi be the radical of ai. Letbi = ∩j 6=iai 6= 0. Then from Lemma 7.3.2 we have r(Ann(x)) = pi for anyx 6= 0 in bi, so Ann(x) ⊆ pi. Since ai is pi-primary, there exists an integerm such that pm

i ⊂ ai, and therefore, bipmi ⊆ bi ∩ pm

i ⊆ ai ∩ bi = 0. Letm ≥ 1 be the smallest integer such that bip

mi = 0, and let x be a non-

zero element in bipm−1i . Then pix = 0, therefore for such an x we have

Ann(x) ⊇ pi and hence Ann(x) = pi. Conversely, if px = r(Ann(x)) isprime, then px = r(Ann(x)) = ∩(ai : x) = ∩x/∈ai

pi and hence px = pi forsome i, so px is a prime ideal belongs to 0.

Chapter 10

Artinian Rings

10.1 Further properties of Artinian rings

An Artinian ring is one satisfying the d.c.c on ideals.

Proposition 10.1.1. In an Artinian ring A every prime ideal is maximal.

Proof. Let p be a prime ideal Consider A/p = B. If x ∈ B be nonzero, then(x) ⊆ (x2) ⊆ · · · which is stationary so xn = yn+1y for some y ∈ B. Since Bis an integral domain, hence xy = 1 so x has an inverse in B, so B is a fieldand hence p is maximal.

Corollary 10.1.1. In an Artinian ring, the niradical N is equal to the Ja-cobson radical R.

Proposition 10.1.2. An Artinian ring has only a finite number of maximalideals.

Proof. Consider the set of finite intersection of maximal ideals which are alsomaximal ideals. This set has a minimal element M1 ∩M2 ∩ · · ·Mn, hencefor any maximal ideal M it contains M1∩M2∩· · ·Mn, so M = Mi for somei.

Proposition 10.1.3. The Jacobson radical R of an Artinian ring is nilpo-tent.

Proof. Consider the descending chain R ⊇ R2 ⊇ R3 ⊇ · · ·Rk = Rk+1 =a ⊇ · · · which is stationary.If a 6= 0 and let Σ to be the set of all ideals in

79

80 CHAPTER 10. ARTINIAN RINGS

A such that a, b 6= 0. a ∈ Σ. So σ contains a minimal element, denoted byc, then pick a nonzero element in c. By the minimality of c we have (x) = c.Similarly we have a(x) = (x). Hence yx = x for some y ∈ a ∈ R. Byinduction we know x = yx = ykx for any k. Since y ∈ R so x = ykx = 0.This contradicts the choice of x, so a = 0.

In a ring A, a chain of prime ideals is a finite strictly increasing primeideals p0 ⊂ p1 ⊂ · · · ⊂ pn. The dimension of a ring A, denoted by dim A, isthe supremum of the lengths of all chains of primes ideals in A

Example 10.1.1. When A = Z, dim Z = 1 as 0 ⊂ (p).

Example 10.1.2. When A = k is a field, then dim A = 0.

Example 10.1.3. When A = k[x1, ..., xn] where xi are algebraically indepen-dent over k. Then dim A = n as (0) ⊂ (x1) ⊂ (x1, x2) ⊂ · · · ⊂ (x1, ..., xn).

Theorem 10.1.1. The ring is Artinian if and only if A is Noetherian andhas dimension 0.

Proof. If A is Artinian, then dim A = 0 and since R is nilpotent, we have(∏n

i=1 Mi)k = 0 for some big enough k and where Mi are all the distinct

maximal ideals. Hence being Artinian implies being Noetherian.If A has dimension one, then every prime ideal is maximal. Since A is

Noetherian, 0 has a primary decomposition, say 0 = ∩ni=1qi. Since r(qi) = Mi

is maximal, so we have Mnii ⊂ qi. Hence 0 = ∩n

i=1Mnii ⊇

∏ni=1 Mni

i . Hencebeing Noetherian implies being Artinian.

10.2 Local Artinian rings

In a local Artinian ring A, since the maximal ideal is nilpotent, hence everyelement in A is either a unit or being nilpotent.

Proposition 10.2.1. Let A be a Noetherian local ring with maximal idealM then one of the following two properties hold.

1. Mn 6= Mn+1 for all n;

2. Mn = 0 for some n and A is an Artinian local ring.

Proof. If Mn = Mn+1 then by the Nakayama’s lemma 6.5.1, we have Mn = 0.In this case, Mn contains in any prime ideal p. By taking radical we knowM is identical to p. Hence dim A = 0 and A is Artinian.

10.2. LOCAL ARTINIAN RINGS 81

Proposition 10.2.2. Any Artinian ring A is isomorphic to a direct productof a finite number of local Artinian ring.

Proof. Let Mi, 1 ≤ i ≤ n be the list of all distinct maximal ideals of A,then for some big enough k

∏Mk

i = 0. Hence the map A →∏

A/Mki is

surjective as Mki are pairwise coprime and is injective as ∩Mk

i = 0. So wehave an isomorphism.

Example 10.2.1. Let A = Z and S = Z− (p) for some prime p, then S−1Z isa local Noetherian ring, but it is not Artinian as its dimension is 1.

Example 10.2.2. Let k be a field and let A = k[x1, x2, ...]/(x1, x22, ..., x

nn...),

then this ring has only one prime ideal which is the image of (x1, x2, ...). Sothis ring has dimension 0. But it is neither Noetherian or Artinian.

Proposition 10.2.3. Let A be a local Artinian ring and let M be its max-imal ideal. Let A/M be its residue field. Then the following statements areequivalent.

1. A is a principal ideal;

2. The maximal ideal of A is principal;

3. dimk M/M2 ≤ 1.

Proof. 1) ⇒ 2) ⇒ 3) are easy.3) ⇒ 1). If dimk M/M2 = 0, then by Nakayama’s lemma 6.5.1 M = 0

so A is a field. When dimk M/M2 = 1, then we have M is principal sayM = (x). Now at a be any ideal of A, we have a ⊆ Mr, a ( Mr+1, so wehave y ∈ a such that y = axr where a /∈ a. So a is a unit in A, hence xr ∈ a

and hence a = (xr) and A is principal.

82 CHAPTER 10. ARTINIAN RINGS

Chapter 11

Discrete valuation ring andDedekind domains

We are going to consider Noetherian integral domains with dimension 1.Namely, in which rings, the nonzero prime ideals are maximal.

Proposition 11.0.4. Let A be a Noetherian domain of dimension 1, thenevery non-zero ideal a in A can be uniquely expressed as a product of primaryideals whose radicals are all distinct.

Proof. Since A is Noetherian, then a has a minimal primary decompositiona = ∩n

i=1qi, where qi is pi-primary. When a 6= 0, the pi’s are maximaland hence pairwise coprime. So

∏ni=1 qi = ∩n

i=1qi and hence a =∏n

i=1 qi.Conversely, if a =

∏qi, then the same arguments shows that a = ∩n

i=1qi;this is a minimal primary decomposition of a in which each qi is an isolatedprimary components, and is therefore unique.

11.1 Discrete valuation rings

Let K be a field. A discrete valuation on K is a mapping v of K× onto Zsuch that

1. v(xy) = v(x) + v(y), i.e. v is a homomorphism.

2. v(x + y) ≥ min(v(x), v(y)).

The set consisting of 0 and all x ∈ K× such that v(x) ≥ 0 is a ring, calledthe valuation ring of v. We denote v(0) = +∞.

83

84CHAPTER 11. DISCRETE VALUATION RING AND DEDEKIND DOMAINS

Example 11.1.1. K = Q. Let p be a fixed prime, for any x ∈ Q write x = pnywhere n ∈ Z and both denominator and numerator of n coprime to p anddefine vp(x) = n. Then vp is a discrete valuation and the valuation ring ofvp is the local ring Z(p).

Example 11.1.2. K = k(x), where k is a field and x an indeterminant. Takea fixed polynomial f ∈ k[x] and define vf similarly as above. Then vf is adiscrete valuation and the valuation ring is the local ring of k[x] with respectto the prime ideal f .

An integral domain A is a discrete valuation ring if there is a discretevaluation v of its field of fractions K such that A is the valuation ring of v.A is a local ring, and its maximal ideal M is the set of all x ∈ K such thatv(x) > 0.

Proposition 11.1.1. Let A be a Noetherian local domain of dimension 1,M its maximal ideal, k = A/M its residue field. Then the following areequivalent:

1. A is a discrete valuation ring;


3. M is a principal deal;

4. dimk(M/M2) = 1;

5. Every non-zero ideal is a power of M; i.e. there exists x ∈ A such thatevery non-zero ideal is of the form (xk), k ≥ 0.

Proof. 1) ⇒ 2) If A is a discrete valuation ring, let F be its field of fractions.Then for any x ∈ F , either x ∈ A or x−1 ∈ A. If y ∈ K be integral over A,it satisfies

yn + an−1yn−1 + · · ·+ a0 = 0, ai ∈ A.

If y ∈ A, then we are done. If not, then y = −(an−1+an−2y−1+· · ·+a0y

1−n ∈A.

2)⇒ 3) For any ideal 0 6= a = (a) ⊂ M, there exist an integer n such thatMn ⊆ a and Mn−1 * a. Choose b ∈ Mn−1 and let x = a/b ∈ K ∈ F . Thenwe have x−1 /∈ A since b /∈ (a), hence x−1 is not integral over A. So x−1M *M otherwise M would be a faithful A[x−1]-module, finitely generated as an

11.2. DEDEKIND DOMAINS 85

A-module. But x−1M ⊆ A by construction. Hence x−1M = A and thereforeM = Ax = (x).

3) ⇒ 4) Since M is principal, then dimk M/M2 ≤ 1. If M/M2 = 0, thenA has dimension 0.

4) ⇒ 5) Pick x such that x ∈ M and x /∈ M2. Since M is principal,M = (x). For any ideal a of A, a ⊇ (xk) for some k. Then a/Mk = a/Mk isan ideal in the Artinian local ring A/Mk. Hence Proposition 10.2.3 impliesthat a = Mn = (xn) for some n.

5) ⇒ 1) For any a ∈ A, a ∈ (xk) and x /∈ (xk+1). Define v(a) = k. Wecan extend v to the field of fractions F of A by v a

b= v(a) − v(b). We can

check that v is a discrete valuation on F .

11.2 Dedekind domains

Theorem 11.2.1. Let A be a Noetherian domain of dimension one. Thenthe following are equivalent:


2. Every primary ideals in A is a prime power;

3. Every local ring Ap, (p 6= 0) is a discrete valuation ring.

Definition 11.2.1. A ring satisfying the conditions of above is called aDedekind domain.

Corollary 11.2.1. In a Dedekind domain, every non-zero ideal has a uniquefactorization as a product of prime ideals.

We will give two types of Dedekind domains.

86CHAPTER 11. DISCRETE VALUATION RING AND DEDEKIND DOMAINS

Chapter 12

Algebraic integers

12.1 Algebraic integers form a ring and a Z-

module

Definition 12.1.1. A subset W ⊂ C is called a Z-module if

1. γ1, γ2 ∈ W implies that γ1 ± γ2 ∈ W ;

2. There exist elements γ1, ..., γn ∈ W such that every γ ∈ W is of theform

∑ni=1 biγi with bi ∈ Z.

Proposition 12.1.1. Let W be a Z-module and suppose that ω ∈ C is suchthat ωγ ∈ W for all γ ∈ W . Then ω is an algebraic integer.

Proof. ωγi ∈ W hence ωγi =∑n

j=1 aijγj, aij ∈ Z. Hence det(aij − δijω) = 0.ω is a root of the characteristic polynomial of (aij) which is monic and haveZ-coefficients.

A complex number is said to be an algebraic number is it is algebraic overQ and an algebraic integer if it is the root of a monic polynomial in Z[x].

Exercise 12.1.1. 1. If u is an algebraic number, there exists an integer nsuch that nu is an algebraic integer.

2. If r ∈ Q is an algebraic integer, then r ∈ Z.

3. If u is an algebraic integer and n ∈ Z, then u + n and nu are algebraicintegers.

87

88 CHAPTER 12. ALGEBRAIC INTEGERS

4. The sum and product of two algebraic integers are algebraic integers.

Proof. 1. If u is a root of anxn + an−1x

n−1 + ...a1x + a0 = 0. We mayassume ai ∈ Z, an 6= 0. Then anu is an algebraic integer.

2. r ∈ Q, the irreducible polynomial is x− r ∈ Z[x], then x ∈ Z.

3. It is easy to check by the definition of algebraic integers.

4. Let u and v are two algebraic integers. Let F = Q(u, v). Pick ei, i =1, ..., n a basis of F over Q, then write uivjek =

∑l ai,j,k,lel, 0 ≤ i, j ≤

n, 1 ≤ k, l ≤ n, where ai,j,k,l ∈ Q. We write every nonzero rationalnumber r = a

bwhere (a, b) = 1 and b ≥ 0. We denote dem(r) = b and

define r(0) = 1. The let

cl = lcm(dem(ai,j,k,l)0≤i,j≤n,1≤k≤n).

Let W = ⊕lZ 1clel. Then we have (u + v)W ⊂ W and uvW ⊂ W . By

Proposition 12.1.1, we know both u + v and uv are algebraic integers.

Consequently, we have

Theorem 12.1.1. Let F be an algebraic extension over Q. Then the alge-braic numbers of F form a ring.

We denote the ring of integers of an algebraic number field F by OF .

Theorem 12.1.2. If F is an algebraic number field, then ring of algebraicintegers of F form a finitely Z-module with finite rank.

Example 12.1.1. The ring of algebraic integers of the field Q(i) is a rankstwo lattice. Explicitly, OQ(i) = Z ⊕ Zi. I.e for any z ∈ OQ(i), then z =a + bi, a, b ∈ Z.

Example 12.1.2. The ring of algebraic integers of the field Q(√−7) is also a

ranks two lattice. Explicitly, OQ(√−7) = Z⊕ Z

√−7+12

.

More generally we have the following.Let d > 0 be an integer. Let K = Q(

√−d). Then

OK =

{Z⊕

√−d · Z if d = 1(4)

Z⊕√−d+12

· Z if d = 3(4).

12.2. RING OF INTEGERS OF A NUMBER FIELD IS A DEDEKIND DOMAIN89

12.2 Ring of integers of a number field is a

Dedekind domain

By theorem 12.1.1 we know that given an algebraic number field, which is afinite extension of Q, the algebraic integers of F form a ring and a Z-modulewith finite rank.

Theorem 12.2.1. The ring of integers in an algebraic number field K is aDedekind domain.

Proof. Let K be an algebraic number field. Let A be the set of all algebraicintegers. The field K is a separable extension of Q (because the characteristicis zero). By Theorem 12.1.2, A is a Z-module with finite rank and hence aNoetherian ring over Z. Also A is integrally closed. Since every nonzero idealis maximal, hence A is a Dedekind domain.

12.3 Primary decomposition for imaginary quadratic

number fields

Example 12.3.1. Let K = Z(i). Its ring of integers is Z[i] which is a p.i.d.Let p be a prime ideal of Z[i]. Let f : Z → Z[i] be the embedding map. Thenpc = p ∩ Z = (p) is a prime ideal in Z. We have the following examples ofprime factorization over Z[i].

(1 + i)2 (1 + 2i)(1− 2i) (3) over Z[i]| \/ |

(2) (5) (3) over Z

In algebraic number theory, we say over Z[i] the ideal (2) ramifies with ram-ification degree 2; the ideal (5) splits completely; and the ideal (3) is inert.In general, for any prime p ∈ Q,

p2 p p (p) over Z[i]| \/ |

(2) (p) p = 1(4) (p) p = 3(4) over Z

90 CHAPTER 12. ALGEBRAIC INTEGERS

More generally, we have the following. For any prime q > 0, let K =Q(√−q). For any a ∈ Z, define the Legendre symbol

(a

q

)= a

p−12 mod q =

0 if x2 − a = 0 mod q has only one solution;1 if x2 − a = 0 mod q has two different solutions;−1 if x2 − a = 0 mod q has no solution;

The factorization of a prime ideal (p) in Z over Z[i] is indicated by thefollowing:

p2 p p (p) over OK

| \/ |(p) if

(aq

)= 0; (p) if

(aq

)= 1; (p) if

(aq

)= −1 over Z

Moreover, we can generalize the above results to any imaginary quadraticnumber field Q(

√−d), d > 0. Accordingly, we can generalize the Legendre

symbol to Jacobi symbol. For more details, please refer to [IR90].

Chapter 13

Algebraic curves

13.1 Affine varieties and affine curves

Definition 13.1.1. Given a commutative ring A with unit, let V = Spec(A) ={maximal ideals of A} be the spectrum of the ring A. In algebraic geome-try, we can endow with the spectrum of a ring a special topology, called theZarisk topology, where the closed sets correspond to the spectra of ideals ofA. Under such a topology, Spec(A), as a topological space, is called an affinevariety.

Example 13.1.1. Let k be a field. Let xi be indeterminants. Let

Ank = Spec(k[x1, · · · , xn]) = {maximal ideals of k[x]}

= {(x1 − a1, · · · , xn − an)∣∣ ai ∈ k}.

It is called the n-dimensional affine space over the field k.

Definition 13.1.2. On the other hand, we call A to be the ring of regularfunctions of the affine variety V . More formally, given an affine variety V , itcan be embedded into the an affine space φ : V → An

k . Let

I(V ) = {f ∈ k[x1, · · · , xn]∣∣∣ f(P ) = 0 for all P ∈ V.

It is easy to check I(V ) is an ideal in k[x1, · · · , xn]. Then the ring

k[V ] = k[x1, · · · , xn]/I(V )

is called the ring of regular functions of V over the field k. Its field offractions, denoted by k(V ) called the field of regular functions of V over thefield k.

91

92 CHAPTER 13. ALGEBRAIC CURVES

Definition 13.1.3. Let V1, V2 be affine varieties. A rational map from V −1to V2 is map of the form

φ : V1 → V2

φ = [f1, · · · , fn]

where fi ∈ k(V1) such that for any P ∈ V1,

φ(P ) = [f1(P ), · · · , fn(P )] ∈ V2.

If all fi ∈ k[x1, · · ·xm], then φ is said to be defined over K. Such a rationalmap defined over k induces a function

φ∗ : k(V2) → k(V1)

φ∗(f) = f ◦ φ

Definition 13.1.4. An affine C satisfying dimk k(C) = 1 is called a curveover the field k.

Definition 13.1.5. Let φ : C1 → C2 be a map of curves defined over k. If φis a constant, then we define the degree of φ to be 0; otherwise we say φ isfinite, and define its degree by

deg φ = [k(C1) : φ∗k(C2)].

We say that φ is separable (inseparable, purely inseparable) if the extensionk(C1)/φ

∗k(C2) has the corresponding property and we denote the separableand inseparable degrees of the extension by degs φ and degi φ.

For more details, please refer to chapter 1 of [Har77] or chapter 1 and 2of [Sil86].

13.2 An example -Elliptic curve

Example 13.2.1. Let k be an algebraically closed field. We pick k = C here.A = C[x, y]/(y2 = x3 + x). It is a domain with dimension 1 as any nonzeroprime ideal of A is of the form (x− x0, y− y0) where P0 = (x0, y0) is a pointon the locus of the equation C : y2 = x3+x, which is a complex curve (due to

13.2. AN EXAMPLE -ELLIPTIC CURVE 93

dim A = 1). Let f(x, y) = y2−x3−x and MP = {f ∈ A, f(P0) = f(x0, y0) =0}. It is easy to see

Φ : A → Cf 7→ f(P )

is a surjective ring homomorphism. Since C is a field, hence MP , which isthe kernel of Φ, is maximal. Moreover, A/MP = C. Now dimC MP /M2

P ≤ 2and MP = (x−x0, y−y0), M

2P = ((x−x0)

2, (y−y0)2, (x−x0)(y−y0)). Then

x− x0 = (y − y0)2 − (x− x0)

3 ∈ M2P .

HencedimC MP /M2

P = 1 = dim A (13.1)

On the other hand, condition (13.1) implies that when localized at a nonzeroprime ideal we obtained a discrete valuation ring. So A itself is a Dedekinddomain.

Remark 13.2.1. When condition (13.1) is satisfied at a point P of a curve C,then the curve is said to be smooth at P . Otherwise, the point P is said tobe a singular point. If C is smooth at all P , then C is said to be a smoothcurve.

Consider the homomorphism f : C[x] ↪→ A. The contraction of any primep ideal in A is still a prime ideal in C[x]. For example

(x− 0, y − 0)c = (x− 0, y − 0) ∩ C[x] = (x− 0).

(x− 1, y −√

2)c = (x− 1) = (x− 1, y +√

2).

Similar to the factorization for quadratic imaginary number field, we havethe factorization for any ideal (x− a), a ∈ C over A as follows:

(x− a, 0)2 (x− a, y −√

b) (x− a, y +√

b) over A| \/

(x− a) (x− a) over Cif b = a3 + a = 0; if b = a3 + a 6= 0;

Now let k be a field with characteristic char(k) = 2. Let A = k[x, y]/(y2 =x3+x). In terms of field extension, we can say that the field of fractions k(C)

94 CHAPTER 13. ALGEBRAIC CURVES

of A is a quadratic extension of the field of fractions k(x) of k[x]. In otherwords, k(C) = k(x)(u), where u is a solution of the equation y2−x3−x = 0.

The mapf ′ : k(x) ↪→ k(C)

induced a mapC = Spec(A) → A1

k = Spec(k[x])

( a rational map between the affine variety C and the affine variety A1k.)

Then deg f ′∗ = 2. When char(k) = 2, degi f′∗ = 2. When char(k) 6= 2,

degs f ′∗ = 2.

13.3 An example of singular algebraic curve

Example 13.3.1. Similarly, when we localize A = C[x, y]/(y2 = x3+x2) at theprime ideal (x, y), we obtained that dimC MP /M2

P = 2. Hence y2 = x3 + x2

has a singular point at (0, 0) while it is smooth everywhere else. In this caseA is not a Dedekind domain.

Bibliography

[AM69] M. F. Atiyah and I. G. Macdonald, Introduction to commutativealgebra, Addison-Wesley Publishing Co., Reading, Mass.-London-Don Mills, Ont., 1969. MR MR0242802 (39 #4129)

[Har77] Robin Hartshorne, Algebraic geometry, Springer-Verlag, New York,1977, Graduate Texts in Mathematics, No. 52. MR MR0463157 (57#3116)

[Hun80] Thomas W. Hungerford, Algebra, Graduate Texts in Mathematics,vol. 73, Springer-Verlag, New York, 1980, Reprint of the 1974 orig-inal. MR MR600654 (82a:00006)

[IR90] Kenneth Ireland and Michael Rosen, A classical introduction tomodern number theory, second ed., Graduate Texts in Mathemat-ics, vol. 84, Springer-Verlag, New York, 1990. MR MR1070716(92e:11001)

[Jac85] Nathan Jacobson, Basic algebra. I, second ed., W. H. Freeman andCompany, New York, 1985. MR MR780184 (86d:00001)

[Sil86] J. H. Silverman, The arithmetic of elliptic curves, Springer-Verlag,New York, 1986.

95

Index

Z-module, 87

affine space, 91affine variety, 91algebra, 74algebra, finitely generated, 74algebraic closure of a field, 29algebraic curve, 91algebraic element, 18algebraic number, 87algebraic number field, 21, 89algebraically closed field, 28Artinian ring, 67Artinian ring, left, 67Artinian ring, local, 80Artinian ring, right, 67

chain condition, acceding, 67chain condition, descending, 67characteristic, 15composite of fields, 18composition series, 70contraction, ideal, 55

Dedekind domain, 85degree of map between curves, 92degree, inseparable, 41degree, of an extension, 16degree, separable, 41dimension of a ring, 80discrete valuation, 83

discrete valuation ring, 84

extension, 16extension finite, 16extension, abelian, 47extension, cyclic, 47extension, ideal, 55extension, simple, 20

field, 15field of invariants, 30field of rational functions, 16field of regular functions of an affine

variety, 91field, residue, 56field, splitting, 24finite field, 49fraction, ring or module, 57

Galois extension, 33Galois group, 29

Hilbert Basis Theorem, 73

ideal, irreducible , 76ideal, primary, 63inseparable, purely, 37integral closure, 60integral element, 59integrally closed domain, 61isomorphism of fields, 23isomorphism, fields, 23

96

INDEX 97

Jacobson radical, 57

Legendre symbol, 90length of the chain of modules, 70linearly independent, automorphisms,

46localization, 58

minimal polynomial, 18module, Artinian, 67module, Noetherian, 67multiplicatively closed set, 57multiplicity of a root, 26

nilradical, 56Noetherian ring, 67Noetherian ring, left, 67Noetherian ring, right, 67norm, 45normal extension, 31

perfect field, 27primary decomposition, 64, 76primary decomposition, minimal, 65prime finite field, 15principal ideal domain, 16purely inseparable closure, 40purely inseparable extension, 37

radical, 63rational map between affine vari-

eties, 92ring of regular functions of an affine

variety, 91ring, commutative, 55ring, local, 56root, multiple, 26

separable closure, 40separable polynomial, 27

separable, field, 27simple extension, 17simple root, 26singular point, 93smooth curve, 93spectrum of a ring, 91

transcendental element, 18

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