INTRODUCTION TO COMMUTATIVE ALGEBRA AND ALGEBRAIC . · PDF fileINTRODUCTION TO COMMUTATIVE...

INTRODUCTION TO COMMUTATIVE

ALGEBRA AND ALGEBRAIC GEOMETRY.

The purpose of these notes is to give a quick and relatively painless introduc-tion to the language of modern commutative algebra and algebraic geometry. Thepoint of view which we will emphasize is that commutative algebra and algebraicgeometry (at least the local theory) are two different languages for talking aboutthe same thing. The algebraic and the geometric aspects of every phenomenon willbe introduced and discussed concurrently. The first part of the course will consistof standard material; towards the end we will treat some topics of current inter-est such as valuation theory and singularities. Due to spacetime limitations, wewill omit some important topics, notably cohomology, traditionally present in anintroductory course.

Unless otherwise stated, all rings in these notes will be commutative with 1. Nwill denote the set of natural numbers, N0 the set of non-negative integers.

§1. Varieties and ideals.

Let k be a field and x1, . . . , xn independent variables. The following are somebasic examples of rings appearing in algebraic, analytic and formal geometry. Thereader should keep them in mind in order to test and illustrate the general theory.

(1) Polynomial rings k[x1, . . . , xn] and D[x1, . . . , xn], where D = Z or, more

generally, D is the ring of integers in some number field, such as Z[√

2] or

Z[√

5i].(2) Formal power series rings

k[[x1, . . . , xn]] :=

∑

α∈Nn0

cαxα11 xα2

2 . . . xαnn

∣∣∣∣∣∣α = (α1, . . . , αn) and cα ∈ k

and Zp[[x1, . . . , xn]], where p is a prime number and Zp is the ring of p-adicintegers.

(3) Convergent power series rings C{x1, . . . , xn} and Zp{x1, . . . , xn}.The beauty and power of commutative algebra lie in the fact that, it provides a

universal language and methods for studying number theory, local algebraic geom-etry and local questions in complex analysis, as illustrated by the above examples.

Given rings R and S, a homomorphism from R to S is a map R → S whichpreserves the ring operations + and ·.

Typeset by AMS-TEX

1

2 INTRODUCTION TO COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY.

Definition 1.1. Let R be a ring and I a non-empty subset of R. I is an ideal ofR if I is closed under addition and for any a ∈ R, x ∈ I, we have ax ∈ I.

For example, {0} is an ideal (usually denoted by (0)); the entire ring R is alsoan ideal, sometimes called the unit ideal and denoted by (1).

Let B = {eλ}λ∈Λ be a subset of R. Let BR denote the set of all linear combina-tions of the form

BR =

{n∑

i=1

eλibλi | bλi ∈ R, eλi ∈ B, n ∈ N

}.

Exercise 1. If I = BR, prove that I is the smallest ideal of R containing B.

If I = BR, we say that I is the ideal generated by B, or that B is a set ofgenerators or a base of I. Of course, a given ideal I may have many different setsof generators. An ideal I is said to be finitely generated if it has a finite set ofgenerators.Definition 1.2. A ring R is Noetherian if every ascending chain I1 ⊂ I2 ⊂ I3 ⊂ . . .of ideals of R stabilizes, that is, there exists n0 ∈ N such that In = In0

for all n > n0.

Exercise 2. Prove that a ring R is Noetherian if and only if every ideal of R isfinitely generated.

Commutative algebra grew out of number theory and geometric invariant theoryat the turn of the century. One of the starting points was Hilbert’s basis theorem:

Theorem 1.3. Let R be a Noetherian ring and x1, . . . , xn independent variables.Then R[x1, . . . , xn] is also Noetherian for any n.

Proof. It is sufficient to consider the case n = 1. We want to establish that R[x] isNoetherian, provided R is Noetherian. Let I be an ideal of R[x]. Let J denote theideal of R, consisting of all the leading coefficients of elements of I (the reader shouldcheck that J is, indeed, an ideal). Since R is Noetherian, J is finitely generated.Hence there exists a finite collection of polynomials f1, . . . ,fm ∈ R[x], whose leadingcoefficients generate J . Let d = max

1≤i≤mdeg fi. Then for every f ∈ I such that

deg f ≥ d, there exist h1, . . . ,hm ∈ R[x] such that∑mi=1 hifi has the same degree

and leading coefficient as f , in other words,

(1.1) deg

(f −

m∑

i=1

hifi

)< deg f.

Applying (1.1) repeatedly and using induction on deg f , we can find g1, . . . ,gs ∈R[x] such that

(1.2) deg

(f −

m∑

i=1

gifi

)< d.

INTRODUCTION TO COMMUTATIVE ALGEBRA AND ALGEBRAIC GEOMETRY. 3

By (1.2), there exists the smallest l ∈ N0 having the following property. There existr ∈ N, f1, . . . ,fr ∈ I such that for any f ∈ I there are g1, . . . ,gr ∈ R[x] satisfying

(1.3) deg

(f −

r∑

i=1

gifi

)< l

(where, by convention, we take deg 0 to be −1).Now, suppose I is not finitely generated. Then l > 0. Let Il denote the set

of those elements of I which have degree exactly l − 1. Let Jl denote the idealof R consisting of 0 and all the leading coefficients of elements of Il. Since Ris Noetherian, Jl is finitely generated. Hence there exist finitely many elementsfr+1, . . . ,fs of Il, such that for every f ∈ Il there exist gr+1, . . . ,gs ∈ R satisfying

deg

(f −

s∑i=r+1

gifi

)< l − 1. In view of (1.3), this contradicts the minimality of

l. �

Corollary 1.4. Let k be a field and x1, . . . , xn independent variables. Then thepolynomial ring k[x1, . . . , xn] is Noetherian. Similarly for Z[x1, . . . , xn].

Let R be a ring and I an ideal of R. The quotient of R by I is the set ofcosets R

I (viewed as abelian groups with respect to +), endowed with the obvious

ring operations. We have the natural surjective homomorphism R → RI, whose

kernel is I. We also say that RI is a homomorphic image of R. Note also that

every surjective ring homomorphism π : R → S is the quotient of R by some ideal(namely, by Ker π).

We are now ready to introduce our basic dictionary between algebra and geome-try, at first in the classical setting with algebraic varieties on the geometric side andfinitely generated algebras over a field on the algebraic side. Soon we will switch tothe more general modern setting with arbitrary schemes on the geometric side andarbitrary rings on the algebraic side.

Classically, algebraic geometry studied algebraic subvarieties of kn (i.e. subsetsof kn defined by finitely many polynomial equations), where k is a field and n ∈ N.In fact, k was usually taken to be C.

Definition 1.5. Let k be a field and let I be an ideal of k[x1, . . . , xn]. The alge-braic variety defined by I in kn is

V (I) = {a ∈ kn | f(a) = 0 for all f ∈ I}.

Since k[x1, . . . , xn] is Noetherian, we can choose a finite base (f1, . . . ,fm) for I,so that V (I) = {a ∈ kn | f1(a) = · · · = fm(a) = 0}.

Associated to every ideal I of k[x1, . . . , xn] we have the algebraic variety V (I).Conversely, given any set V ⊂ kn, we may consider the ideal I(V ) defined byI(V ) = {f ∈ k[x1, . . . , xn] | f(a) = 0 for all a ∈ V }. A natural quesiton arises:


what is the relation between I and I(V (I))? To answer this question, we needanother definition.Definition 1.6. Let R be a ring and I an ideal of R. The radical of I , denotedby√I, is defined to be

√I = {x ∈ R | there exists n ∈ N0 such that xn ∈ I}.

We say that I is radical if I =√I.

Exercise 3. Prove that√I is an ideal and that I ⊂

√I.

Let k be a field and I an ideal of k[x1, . . . , xn]. Clearly, V (I) = V (√I). It is

also clear that I ⊂ I(V (I)) for any ideal I ⊂ k[x1, . . . , xn]. Hence for any ideal I,√I ⊂ I(V (

√I)) = I(V (I)). It turns out that if k is an algebraically closed field (this

means that every non-constant polynomial in k[x] has a root in k), we always have√I = I(V (I)). This is a non-trivial theorem, called Hilbert’s Nullstellensatz.

We will state and prove several versions of the Nullstellensatz, all of which aresometimes referred to by this name.

Theorem 1.7 (The strong form of Hilbert’s Nullstellensatz). Let k be an

algebraically closed field and I an ideal of R = k[x1, . . . , xn]. Then I(V (I)) =√I.

In other words,√I = {f ∈ R | f(a) = 0 ∀ a ∈ V (I)}.

In other words, there is a one-to-one correspondence between algebraic subvari-eties of kn and radical ideals of k[x1, . . . , xn].

A proof of Hilbert’s Nullstellensatz will be given shortly. First, we would like toadvance a little more in our task of constructing a dictionary between algebra andgeometry. To do this, we need to define the notions of maximal and prime ideals.

Definition 1.8. Let R be a ring and I $ R an ideal. We say that I is prime iffor any x, y ∈ R such that xy ∈ I, we have either x ∈ I or y ∈ I. We say that I ismaximal if I 6= R and the only ideal of R properly containing I is R itself. Theset of all the maximal ideals of R will be denoted by Max(R), the set of all theprime ideals of R by Spec R.

Associated to the properties of ideals we have encountered so far —radical, primeand maximal — we have the corresponding properties of rings (the precise connec-tion will be explained in a moment).

Definition 1.9. Let R be a ring and x an element of R. We say that x is nilpotentif there exists n ∈ N such that xn = 0. We say that x is a zero divisor if thereexists y ∈ R \ {0} such that xy = 0. The set of all nilpotent elements of R is called

the nilradical of R, denoted by√

(0) or nil(R).

Remark 1.10. Since√

(0) is the radical of the zero ideal (0), it is itself an ideal.Describing the set of zero divisors of R is more tricky; we will do it somewhat laterin the course.


Definition 1.11. Let R be a ring. R is said to be reduced if it has no non-zero nilpotent elements. R is an integral domain (sometimes abbreviated to“domain”) if R has no non-zero zero divisors.

Every field is an integral domain. If R is a ring, we will denote by Rred the ringR√(0)

; of course, Rred is reduced.

Exercise 4. Prove that√

(0) equals the intersection of all the prime ideals of R(Hint: to prove one of the two inclusions, you will need Zorn’s lemma).

The intersection of all the maximal ideals of R is called the Jacobson radicalof R, or simply the radical of R), sometimes denoted by J(R).Exercise 5. Prove the following implications: I is maximal ⇐⇒ R

I is a field

=⇒ I is prime ⇐⇒ RI is an integral domain =⇒ I is radical ⇐⇒ R

I isreduced. As a special case of this, we have that R is a field if and only if the zeroideal is maximal, R is a domain if and only (0) is prime and R is reduced if andonly if (0) is radical.

By Zorn’s lemma, every ring R (whether Noetherian or not) has at least onemaximal ideal, hence, a fortiori, at least one prime ideal. For Noetherian rings theexistence of maximal ideals does not require Zorn’s lemma.

Example 1.12. Let k be a field, n ∈ N and (a1, . . . ,an) ∈ kn. Then (x1 −a1, . . . ,xn − an) is a maximal ideal of k[x1, . . . , xn].

Definition 1.13. The Zariski topology on kn is the topology whose open setsare of the form kn \ V (I), where I is an ideal of k[x1, . . . , xn].

Operations on ideals. Next, we define the operations on ideals which correspondto the topological operations of union and intersection.

Definition 1.14. Let R be a ring, {Iλ}λ∈Λ a collection of ideals of R. The sum∑λ∈Λ Iλ is defined by

∑

λ∈Λ

Iλ =

{r∑

i=1

xλi

∣∣∣∣∣ r ∈ N, λi ∈ Λ, xλi ∈ Iλi for 1 ≤ i ≤ r}.

In particular, if Λ = ∅, then∑λ∈Λ Iλ = 0.

The product IJ is defined to be the ideal of R generated by the set {xy | x ∈I, y ∈ J}. Of course, IJ can also be thought of as

IJ =

{n∑

i=1

xiyi

∣∣∣∣∣ n ∈ N, xi ∈ I, yi ∈ J for 1 ≤ i ≤ r}.

Remark 1.15. Historically, the word “ideal” first appeared in number theory. Kum-mer’s famous mistake in proving Fermat’s Last Theorem consisted in assuming


that Unique Factorization into prime numbers holds in the ring of integers in anynumber field. The mistake was soon discovered. For example, in Z[

√5i], we have

6 = 2 · 3 = (1 +√

5i)(1 −√

5i), and yet 2 is an irreducible element which divides

neither (1 +√

5i) nor (1−√

5i). It was then realized that a sensible generalizationof the Unique Factorization Theorem to rings of integers in number fields is notunique factorization for elements but unique factorization for ideals. This is howthe notion of “ideal numbers” or ideals was introduced. For instance, in the aboveexample the ideal (6), generated by 6, has a unique factorization into prime ideals:

(6) = (2, 1 +√

5i)2(3, 1 +√

5i)(3, 1−√

5i).The following properties are obvious from definitions and the Nullstellensatz:

(1) kn = V ((0))(2) ∅ = V ((1))(3) If {Iλ}λ∈Λ is a collection of ideals of k[x1, . . . , xn],then

⋂

λ∈Λ

V (Iλ) = V

(∑

λ∈Λ

Iλ

).

(4) V (I) ∪ V (J) = V (IJ) = V (I ∩ J)

(5)√J ⊂

√I =⇒ V (I) ⊂ V (J) and the converse holds if k is algebraically

closed.

In particular, this shows that Zariski topology is, indeed, a topology.The starting point for the theory of schemes is the above Example 1.12, and

the partial converse to it, known as the weak form of Hilbert’s Nullstellensatz (theconnection between the weak and the strong form will become clear in the courseof the proof).

Theorem 1.16 (A weak form of Nullstellensatz). Let k be an algebraicallyclosed field. Then (a1, . . . ,an)←→ (x1−a1, . . . ,xn−an) is a one-to-one correspon-dence between kn and Max(k[x1, . . . ,xn]). In other words, every maximal ideal ofk[x1, . . . ,xn] is of the form described in Example 1.12.

Let k be an algebraically closed field. By Theorem 1.16, we may translate allthe geometric definitions and statements about points of kn into purely algebraicstatements about ideals in the ring k[x1, . . . , xn]. For example, we can define theZariski topology on Max(k[x1, . . . , xn]) as follows: for an ideal I ⊂ k[x1, . . . , xn],V (I) = {m ∈ Max(k[x1, . . . , xn]) | I ⊂ m}. Thus maximal ideals of k[x1, . . . , xn]correspond to points of kn and radical ideals correspond to algebraic varieties.

Let σ : A→ B be a homormorphism of rings.Definition 1.17. Let I be an ideal of A. The extension IB is the ideal of Bgenerated by σ(I). If J ⊂ B is an ideal of B, the contraction J ∩ A is the ideal

σ−1(J).

Remark 1.18. The following are immediate from the definition:

(1) If J ⊆ B, is prime then J ∩ A is prime.(2) If J ⊆ B is maximal and σ is surjective, then J ∩ A is maximal.


Let x = (x1, . . . , xn). Let I be an ideal of k[x] and consider the natural homo-

morphism k[x] → k[x]I

. Then the map m → m ∩ k[x] induces a bijection between

Max(k[x]I

) and the set of maximal ideals of k[x], containing I. Hence the abovecorrespondence between kn and Max(k[x]) induces a one-to-one correspondence

between V (I) and Max(k[x]I

). k[x]I

is nothing but the ring of polynomial functions

defined on the variety V (I). We say that k[x]I

is the coordinate ring of V (I). Forexample, k[x] is the coordinate ring of kn. Given two algebraic varieties X ⊂ kn

and Y ⊂ kl, defined by ideals I ⊂ k[x1, . . . , xn], J ⊂ k[y1, . . . ,yl], it makes sense totalk about polynomial maps from X to Y . By definition, they are induced by themaps f : kn → kl such that f∗J ⊂ I. In that case, f∗ induces a homomorphism of

coordinate rings k[y]J→ k[x]

I.

Next, we show that prime ideals of k[x1, . . . , xn] correspond to irreducible subva-rieties of kn:

Definition 1.19. A closed set X in a topological space is said to be irreducibleif whenever X = X1 ∪X2, we have either X = X1 or X = X2.

Let P be a radical ideal of k[x1, . . . , xn]. We want to show that P is prime if andonly if V (P ) is irreducible. We state the result in a form which works for arbitraryrings (so that we can obtain the same fact for schemes in the next section).

Proposition 1.20. A radical ideal P in a ring R is prime if and only if wheneverP = I ∩ J with I and J ideals of R, either P = I or P = J .

Proof. “If”. First, suppose that P cannot be expressed as I∩J in a non-trivial way.If P were not prime, there would exist x, y ∈ k[x1, . . . , xn] \ P such that xy ∈ P .Let I = P + (x), J = P + (y). Then P $ I, P $ J , P ⊂ I ∩ J . On the other hand,

IJ ⊂ P , hence√IJ ⊂ P . Then I ∩ J ⊂

√I ∩ J =

√IJ ⊂ P , so that I ∩ J = P , a

contradiction.“Only if” is an immediate consequence of the following lemma.

Lemma 1.21. Let P be a prime ideal in a ring R and I1, . . . ,Ir arbitrary ideals of

R. Ifr⋂i=1

Ii ⊂ P , then Ii ⊂ P for some i, 1 ≤ i ≤ r.

Proof. We give a proof by contradiction. Suppose Ii 6⊂ P for any I. For each i

pick an element xi ∈ Ii \ P . Thenr∏i=1

xi ∈r⋂i=1

Ii ⊂ P , but xi /∈ P for any i. This

contradicts the fact that P is prime. �

§2. Schemes.

In §1 we saw that if k is algebraically closed, points of kn can be identified withmaximal ideals of k[x1, . . . , xn]. This suggests generalizing algebraic geometry fromkn to Max(R), where R is an arbitrary ring, and where we think of maximal idealsas points in our space. The idea of a scheme, developed in the twentieth century,


consists, in addition, in enlarging our notion of a point to include all the primeideals of R.

Let R be a ring. Motivated by the above considerations, we define the Zariskitopology on Spec R to be the topology whose closed sets are of the form V (I) ={P ∈ Spec R | I ⊂ P}, where I is an ideal of R.

Definition 2.1. An affine scheme is a pair (R, SpecR) of a ring and a topologicalspace, where Spec R is endowed with its Zariski topology.

Remark 2.2. Note that single point sets need not be closed in Spec R. Moreprecisely, we can describe the closure of the set {P} for P ∈ Spec R. Let Q ∈ V (P ),so that Q is a prime ideal,containing P . For every ideal I such that P ∈ V (I), we

have I ⊂ P ⊂ Q, so that Q ∈ V (I). Hence V (P ) ⊂ {P}. Conversely, V (P ) is a

closed set containing P , hence {P} ⊂ V (P ). This proves that {P} = V (P ). Thepoint P is characterized by being the unique point whose closure is V (P ). P iscalled the generic point of the closed set V (P ). For example, if R is an integral

domain, then (0) is a prime ideal. In this case, {(0)} = Spec R, so that (0) is thegeneric point of the whole space. More generally, the same conclusion holds for√

(0) in a ring in which√

(0) is prime.

Remark 2.2 proves that, given P ∈ Spec R, the one-point set {P} is closed if andonly if P is maximal. In particular, Spec R is not a Hausdorff space, unless everyprime ideal of R is maximal.

In the category of algebraic varieties, one had the notion of a morphism betweentwo varietiesX and Y , which is, by definition, a polynomial map in several variables.We saw that each such morphism f induces a dual homomorphism f∗ from thecoordinate ring of Y to the coordinate ring of X . In scheme theory, the situation isanalogous, except, since our basic objects are rings, we start with a homomorphismof rings A→ B and define the induced morphism Spec B → Spec A.

Let σ : A → B be a homomorphism of rings. Define the map σ∗ : Spec B →Spec A by σ∗(P ) = P ∩ A ≡ σ−1(P ) for P ∈ Spec B. Maps Spec B → Spec Awhich arise in this way are called morphisms of affine schemes. It is immediatefrom definitions that morphisms are continuous with respect to Zariski topology.

Two kinds of morphisms are of particular importance in geometry: open andclosed embeddings. Open embeddings correspond to localization of rings, whichwill be the subject of the next section.

Ring homomorphisms corresponding to closed embeddings are the surjective ho-momorphisms, which we will discuss right now. Let I be an ideal of A. Considerthe natural homomorphism σ : A → A

I. Then σ∗ : Spec A

I→ Spec A is injective

and Im(σ∗) = V (I). Also, σ∗ is a closed map, for if J is an ideal of AI, then

σ∗(V (J)) = V (J ∩ A). Thus, σ∗ is a closed embedding from Spec AI into Spec A,

i.e., σ∗ is a homeomorphism; σ∗ : Spec AI → V (I) (with the topology on V (I)

induced from Spec A). This shows that the closed subset V (I) of Spec A itselfhas the structure of an affine scheme. We say that V (I) is a closed subscheme


of Spec A. AI is called the coordinate ring of V (I). As we switch from naive

algebraic geometry to this new and more sophisticated view of life, kn is replacedin our minds by Spec k[x1, . . . , xn] and the algebraic subvariety V (I) by the closed

subscheme Spec k[x1,...,xn]I ⊂ Spec k[x1, . . . , xn].

Geometrically, prime ideals correspond to irreducible varieties. The point ofscheme theory is that we agree to regard irreducible subvarieties themselves aspoints. One of the advantages of doing so is that it allows to make rigorous all theinformal classical arguments of the Italian geometers of the early twentieth centurywhich began with the words “Take a sufficiently general point on the algebraicvariety...”. In scheme theory, the canonically defined generic point P takes theplace of the intuitive notion of choosing “a generic point” on the irreducible varietyV (P ).

Apart from rigor, there are other important advantages that the language ofschemes has over the classical language of algebraic varieties. The fact that we cando geometry over arbitrary rings, rather than just quotients of k[x1, . . . , xn] meansthat we can work over fields of positive characteristic and even over Z itself, whichis very important for number theory. In fact, most of the recent advances in numbertheory would be inconceivable without the use of the geometric language of schemes.For instance, a large part of Faltings’ famous proof of Mordell’s conjecture consistsof generalizing the classical theory of algebraic surfaces to arithmetic surfaces, i.e.

surfaces of the form Spec Z[x,y](f(x,y)) . Along with rings over Z, we can consider complex

analytic local rings, formal power series rings (which appear naturally in studyinglocal geometry and singularities), rings over the p-adic numbers, etc. Scheme the-ory provides us with a uniform language for doing geometry in all these differentcontexts, local and global.

Another great advantage of schemes over varieties is the presence of nilpotentstructure, which makes the theory much richer. Namely, in classical algebraic ge-ometry, given a non-radical ideal I ⊂ k[x1, . . . , xn], one did not distinguish between

V (I) with V (√I); indeed, they are identical as topological spaces. In scheme the-

ory, however, they are regarded as different closed subschemes of Spec k[x1, . . . , xn]by definition. More generally, if R is a ring with nilpotent elements, Spec R andSpec Rred are identical as topological spaces, but are different as schemes. This

opens interesting geometric possibilities. For example, Spec k[x,y](x2)

is a double line:

it is the y-axis counted twice. We also have a completely new notion of embeddedpoints (which will be discussed in more detail later on, in the section on primary

decomposition). For instance, the scheme Spec k[x,y](x2,xy)

is usually thought of as the

union of the y-axis with the origin (a meaningless notion in classical algebraic ge-ometry), because (x2, xy) = (x) ∩ (x, y)2. In this case, we say that (x, y) is an

embedded point of the scheme Spec k[x,y](x2,xy) . Even if one is only interested in the

usual, reduced algebraic varieties, such non-reduced objects arise in a natural wayas limits of families of varieties under deformations.

To summarize the nature of the generalization from varieties to schemes, we give


a definition of finitely generated algebras and morphisms of finite type.

Let A be a ring. An A-algebra is a ring B together with a fixed homomor-phism σ : A → B. The homomorphism σ is sometimes called the structurehomomorphism for the A-algebra B. An A-algebra homomorphism is a ringhomomorphism B → C between two A-algebras, which respects the structure ho-momorphisms.

Definition 2.3. Let B be an A-algebra and x = {xλ}λ∈Λ a subset of B. Let A ⊂ Bdenote the image of A under the structure homomorphism. The A-subalgebra A[x]of B, generated by x, is defined to be the smallest A-subalgebra of B, containingx. In other words, A[x] is the set of all elements of B which can be written asmultivariable polynomials in the xλ with coefficients in A. We say that B is afinitely generated A-algebra, or that B is of finite type over A, if there is afinite subset x ⊂ B such that B = A[x]. A morphism Spec B → Spec A is said tobe of finite type if B is finitely generated over A.

Remark 2.4. B is finitely generated over A if and only if B is isomorphic to

an A-algebra of the form A[x1,...,xn]I

, where x1, . . . , xn are independent variables,A[x1, . . . , xn] denotes the ring of polynomials in n variables with coefficients in Aand I is an ideal of A[x1, . . . , xn].

Coming back to varieties and schemes, we can summarize the situation as follows.Initially, we generalized the naive theory of algebraic varieties in two ways. First,we enlarged our notion of points by agreeing to consider irreducible subvarietiesthemselves as points. Secondly, we agreed to distinguish between two varieties hav-ing different nilpotent structures, even though they may be identical as topologicalspaces. The effect of these two generalizations is that of replacing the categoryof algebraic varieties over a field k by the category of affine schemes which are offinite type over k. Rings which are of finite type over some field still form a ratherrestricted class (for instance, none of the rings listed in the beginning of the firstlecture are of this type, except k[x1, . . . , xn]). Our last generalization consisted inallowing arbitrary rings instead of limiting ourselves to finitely generated algebrasover a field.

Definition 2.5. An affine scheme Spec A is said to be reduced if A is reduced.

Remark 2.6. Just as in the case of algebraic varieties, prime ideals correspond toirreducible closed subschemes. More precisely, let A be a ring and I a radical ideal.Then by Proposition 1.20, Spec A

Iis irreducible if and only if I is prime. Another

way of stating the same fact is that for a reduced ring A, Spec A is an irreducibletopological space if and only if (0) is a prime ideal of A if and only if A is an integraldomain. In this case, we say that Spec A is an integral scheme. An importantobject associated with an integral scheme is its field of rational functions (which isthe analogue of the field of meromorphic functions on a complex manifold):

Definition 2.7 Let A be an integral domain. The field of fractions K of A is


the set of equivalence classes of fractions

K ={ ab

∣∣∣ a, b ∈ A, b 6= 0}/ ∼

where ∼ is the obvious equivalence relation ab ∼ c

d ⇐⇒ ad = bc. K is also calledthe field of rational functions of the affine scheme Spec A. If P ∈ Spec A, thefield of fractions of AP is called the residue field of P , sometimes denoted by κ(P ).κ(P ) is nothing but the field of fractions of the integral scheme V (P ).

We end this section by mentioning the semi-classical precursor of the modernnotion of the generic point of an integral subscheme, in order to put things in ahistorical perspective.

Let X be an affine subvariety of kn, where k is a field. Assume X is irreducible,

so that X = V (P ), where P is a prime ideal of k[x1, . . . , xn]. Let A = k[x1,...,xn]P

and let K be the field of fractions of A. We have k ⊂ K, so that k[x1, . . . , xn] ⊂K[x1, . . . , xn]. Let P ′ = PK[x1, . . . , xn]. Let X ′ = V (P ′) ⊂ Kn. There exists apoint ξ ∈ X ′ such that for any f ∈ k[x1, . . . , xn], f(ξ) = 0 implies that f ∈ P .Such a point ξ is called a k-generic point of X and used to play the role which themodern notion of generic point plays today. One advantage of the modern approachis that the generic point of an irreducible subscheme is canonically defined, whilethe k-generic point ξ above is not, in general, unique.

§3. Localization.

By definition, we have a one-to-one correspondence between morphisms of affineschemes and homomorphisms of rings. We saw in §2 that closed embeddings ofaffine schemes correspond to surjective homomorphisms of rings. In this section,we consider the case of open embeddings, which correspond, appropriately, to theoperation of localization in rings.Definition 3.1. Let A be a ring. A subset S ⊂ A \ {0} is said to be multiplicativeif 1 ∈ S and S is closed under multiplication.

Examples:

(1) Let f ∈ A \√

(0). Then {1, f, f2, f3, . . .} is multiplicative.(2) If P ∈ Spec A, then A \ P is a multiplicative set.

Definition 3.2 Let A be a ring and S ⊆ A a multiplicative subset. The local-ization of A at S is defined to be AS = {as | a ∈ A, s ∈ S}/ ∼, where ∼ is

the following equivalence relation: as ∼ b

t whenever there exists u ∈ S such thatu(at − bs) = 0. It is clear that AS is a ring with the operations of addition andmultiplication, well known from elementary school. 0

1 is its zero element and 11 the

unit element.Let A be a ring, S ⊂ A a multiplicative subset. We have the natural homomor-

phism σ : A → AS , given by a 7→ a1. σ is injective if A is an integral domain, but

not in general. We have Ker σ = {x ∈ A | sx = 0 for some s ∈ S}.


An element x in a ring A is said to be invertible, or a unit, if there exists y ∈ Asuch that xy = 1. Then every element of σ(S) is invertible in AS .Examples:

(1) Let A be a ring, f ∈ A. Let S = {1, f, f2, f3, . . .}. We write Af for AS . Afis the ring consisiting of all the fractions whose denominator is a power off .

(2) Let A be a ring and P ∈ Spec A. Let S = A \ P . By abuse of notation,which has become completely standard, we will write AP for AA\P .

Definition 3.3. A ring A is local if A has only one maximal ideal.

If A is a local ring with maximal ideal m, the field k = Am is called the residue

field of A. We will sometimes denote such an A by (A,m) or (A,m, k) to indicatethat m is the maximal ideal and k the residue field.

Remark 3.4. Let A be a ring and m ⊂ A and ideal. A is local with the maximalideal m if and only if every element of A \m is invertible. Indeed, suppose (A,m)is local and let a ∈ A \m. Suppose a is not invertible. Then (a) is a proper idealof A. By Zorn’s lemma, (a) is contained in a maximal ideal, which must be msince m is the only maximal ideal of A. This means that a ∈ m, a contradiction.Conversely, suppose every element of A \ m is invertible, but m is not maximal.Then there exists m′ ∈ Max(A) such that m $ m′ $ A. There exists a ∈ m′ \m.By assumption a is invertible. Then (a) = A and so m′ ⊃ (a) = A, a contradiction.

Remark 3.5. Let A be a ring and P ∈ Spec A. Then AP is a local ring with themaximal ideal PAP .

Remark 3.6. Let A be a ring, S ⊂ A a multiplicative set, and consider the naturalhomomorphism

σ : A→ AS.

Then

(3.1) σ∗(Spec AS) = {P ∈ Spec A | P ∩ S = ∅}.

Indeed, let P ∈ Spec AS. We must show that σ∗(P ) ∩ S = σ−1(P ) ∩ S = ∅. Ifa ∈ σ−1(P )∩S, then a

1 ∈ P and 1a ∈ AS, hence 1 ∈ P and P = AS , a contradiction

because P is a prime ideal of AS .

Exercise 3.7. Show the opposite inclusion σ∗(Spec AS) ⊃ {P ∈ Spec A | P ∩S =∅} in (3.1). Also, show that σ∗ is injective and hence induces a bijection betweenSpec AS and {P ∈ Spec A | P ∩ S = ∅}.σ∗ is called the restriction map from Spec A to the subset {P ∈ Spec A | P ∩

S = ∅}. In the case of algebraic varieties, σ∗ corresponds to restricting a polynomial(i.e. an element of A) to the subset where no element of S vanishes.

For example, let f ∈ A \√

(0), S = {1, f, f2, . . .}. Consider σ∗ : Spec Af −→Spec A. Then σ∗ induces a bijection between Spec Af and Spec A \ V (f).


Let I ⊆ A be an ideal. Let {fλ}λ∈Λ be a set of generators of I. Let P ∈ Spec A.Then I ⊂ P if and only if fλ ∈ P for all λ ∈ Λ.

Let Uλ = Spec A \ V (fλ) ∼= Spec Afλ . Then Spec A \V (I) = {P ∈ Spec A | I 6⊂P} = {P ∈ Spec A | fλ /∈ P for some λ ∈ Λ} =

⋃λ∈Λ

(Spec A \ V (fλ)) =⋃λ∈Λ

Uλ.

Thus the open set Spec A \ V (I) is a union of the open sets Uλ = Spec A \V (fλ) ∼= Spec Afλ . Since the ideal I was arbitrary, we see that the collection{Spec A \ V (f) | f ∈ A} form a basis for the Zariski topology in Spec A. We willsometimes refer to the sets Spec Af as the basic open sets of Spec A.

Remark 3.8. If A is Noetherian, we may take Λ above to be finite. Hence in thatcase any open set is a finite union of basic open sets.

We end this section by giving another point of view on localizations in terms ofinductive limits.

Definition 3.9. Let Λ be a partially ordered set. We say that Λ is a directed setif for any i, j ∈ Λ, there exists l ∈ Λ such that i ≤ l and j ≤ l.Definition 3.10. Let {Aλ}λ∈Λ be a collection of sets, indexed by a directed setΛ. Suppose that for any i, j ∈ Λ with i ≤ j we are given a map µij : Ai → Aj suchthat µii = id for all i ∈ Λ and µjl ◦ µij = µil for all triples i, j, l ∈ Λ such thatl ≥ j ≥ i. {Aλ}λ∈Λ is called a direct system or an inductive system.

Let {Bλ}λ∈Λ be another collection of sets indexed by Λ. Suppose that for anyi, j ∈ Λ with i ≥ j we are given a map φij : Bi → Bj such that φii = id for all i ∈ Λand φjl ◦ φij = φil for all triples i, j, l ∈ Λ such that l ≤ j ≤ i. {Bλ}λ∈Λ is calledan inverse system or a projective system.

Definition 3.11. Let {Aλ}λ∈Λ be a direct system. The direct limit lim→λ

Aλ is

defined to be lim→λ

Aλ =∐λ∈Λ

Aλ/ ∼, where ∼ is the following equivalence relation.

Given xi ∈ Ai, xj ∈ Aj , we say that xi ∼ xj if there exists l, l ≥ i, j such thatµil(xi) = µjl(xj).

The inverse, or projective limit lim←Bλ is defined to be

lim←Bλ =

{{xλ}λ∈Λ ∈

∏

λ∈Λ

Bλ

∣∣∣∣∣ φij(xi) = xj for all i, j such that i ≥ j}.

Exercise 3.12. Show that lim→λ

Aλ and lim←Bλ are characterized by the following

universal mapping properties. Let A = lim→λ

Aλ. Then there are natural maps πi :

Ai → A, i ∈ Λ, compatible with all the µij. Moreover, given any other set A′ and acollection of maps π′i : Ai → A′ compatible with the µij, there exists a unique mapA→ A′, compatible with all the πi and π′i.

Similarly, let B = lim←λ

Bλ. Then there are natural maps πi : B → Bi, i ∈ Λ,

compatible with all the φij . Moreover, given any other set B′ and a collection of


maps π′i : B′ → Bi compatible with the φij, there exists a unique map B′ → B,compatible with all the πi and π′i.

If the sets Aλ and Bλ are endowed with an additional structure (such as rings,groups, etc.), this structure usually extends in an obvious way to projective andinductive limits. We will take this fact for granted without mentioning it explicitly.

Exercise 3.13. Let A be a ring, S ⊂ A a multiplicative subset. Define a partialorder in S by saying that f ≤ g if there exists h ∈ S such that g = fh. Then S isa directed set. Show that the localizations Af form a directed system and that thebasic open sets Uf = Spec Af form a projective system. Show that

(3.2) lim→

f∈S

Af = AS .

In particular, if P ∈ Spec A, AP = lim→

f /∈P

Af .

By passing to Specs and reversing all the arrows in (3.2), obtain

lim←

f∈S

Spec Af = Spec AS

(as sets).

§4. A proof of Hilbert’s Nullstellensatz.

In this section, we give a proof of the various versions of the Nullstellensatz.An injective ring homomorphism is called a ring extension. Let σ : A → B be

an extension of rings. Let x ∈ B.

Definition 4.1. We say that x is algebraic over A, if it satisfies a polynomialequation of the form

(4.1) anxn + an−1x

n−1 + · · ·+ a1x+ a0 = 0,

with ai ∈ A. x is integral, or finite over A if, in addition, we may take an = 1in (4.1). If x is not algebraic, it is said to be transcendental. We say that σ isalgebraic, or that B is algebraic over A, if every x ∈ B is algebraic. σ is integralif every x ∈ B is integral over A. Extensions which are not algebraic are calledtranscendental.

Note that if A is a field then “integral over A” and “algebraic over A” are thesame thing.

Proposition 4.2 (an intermediate version of the Nullstellensatz). Let k bea field and A a finitely generated k-algebra. Then A is a domain, algebraic over k,if and only if A is a field.


First, let us assume Proposition 4.2 and show that it implies the two otherversions of Nullstellensatz which were stated previously.

Proof of Theorem 1.16. It is obvious that for every a = (a1, . . . ,an) ∈ kn, (x1 −a1, . . . ,xn−an) is a maximal ideal of k[x1, . . . , xn], and that distinct points a, a′ ∈ kngive rise to distinct maximal ideals. It remains to show that every maximal ideal isof the form (x1−a1, . . . ,xn−an), a ∈ kn. Take any m ∈Max(k[x1, . . . , xn]). Thenk[x1,...,xn]

m is a field, which is a finitely generated k-algebra (since it is generated

by the images of the xi). By Proposition 4.2, k[x1,...,xn]m is algebraic over k, hence

equals k (k being algebraically closed). Then for each i, 1 ≤ i ≤ n, there existsai ∈ k such that xi ≡ ai mod m. Then m ⊃ (x1 − a1, . . . ,xn − an), hence m =(x1 − a1, . . . ,xn − an), as desired. �

To prove Theorem 1.7, we state and prove the following more general result,which holds without the assumption that k is algebraically closed.

Proposition 4.3. Let k be any field and R a finitely generated k-algebra. For anideal I ⊂ R, let V (I) = {m ∈ Max(R) | I ⊂ m}. For a subset V ⊂ Max(R), let

I(V ) = {f ∈ R | f ∈ m for all m ∈ V }. Then for any ideal I, I(V (I)) =√I.

By Theorem 1.16, Proposition 4.3 implies Theorem 1.7.

Proof of Proposition 4.3 (assuming Proposition 4.2). Clearly,√I ⊂ I(V (I)). We

must prove the opposite inclusion.Take f /∈

√I. We may consider B = (RI )f = R

I [ 1f ] (where by abuse of notation

we identify f with its natural image in RI ). We have a natural map σ : R → B.

Take any m ∈ Max(B). Then I ⊂ m ∩ R and f /∈ m ∩ R. If we can show that

m ∩R is maximal, we will have m ∩R ∈ V (I) and hence f /∈ I(V (I)), as desired.To see that m ∩ R is maximal, apply Proposition 4.2 to B

m and Rm∩R . B is a

finitely generated k-algebra (it is generated by the generators of R and 1f ). By

Proposition 4.2, we have m maximal =⇒ Bm is a field =⇒ B

m is algebraic over

k =⇒ RR∩m isa domain, algebraic over k (since R ∩ m is prime and R

R∩m ⊂ Bm)

=⇒ RR∩m is a field =⇒ R ∩m is a maximal ideal of R. �

To complete the proof of Proposition 4.3, and hence of Theorem 1.7, it remainsto prove Proposition 4.2.

Proof of Proposition 4.2. First, suppose A is a domain, algebraic over k. Take anyx ∈ A. We must show that x is invertible in A. Since A is algebraic, x satisfies anequation of the form anx

n + · · · + a0 = 0, with ai ∈ k. Since A is a domain, wemay assume that a0 6= 0. Then x(− 1

a0)∑n−1i=0 ai+1x

i = 1, which proves that x isinvertible.

Conversely, suppose A is a field, which is not algebraic over k. Since A is finitelygenerated, we may write A = k[x1, . . . , xn] for some x1, . . . , xn ∈ A. We canrenumber the xi such that x1, . . . ,xr are algebraically independent over k and A isalgebraic over k[x1, . . . ,xr]. Replacing k by the field k(x1, . . . ,xr−1) does not change


the problem, so that we may assume r = 1. For 2 ≤ i ≤ n, let bi ∈ k[x1] be theleading coefficient of a polynomial equation over k[x1], satisfied by xi. Then all thexi are integral over k[x1]b2...bn . At this point, we need the following basic fact aboutintegral extensions:

Lemma 4.4. Let A ⊂ B be a finitely generated ring extension. If each of thegenerators is integral over A, then B is integral over A.

First, we finish the proof, assuming Lemma 4.4. Since each of the xi is integralover k[x1]b2...bn , by Lemma 4.4 A is integral over k[x1]b2...bn . Since A is a field, itcontains k(x1), so that k(x1) is integral over k[x1]b2...bn .

Definition 4.5. Let A→ B be a ring extension. The integral closure of A in Bis the A-subalgebra A of B consisting of all the elements of B, integral over A (thefact that A is a subalgebra follows from Lemma 4.4). A is said to be integrallyclosed in B, if A = A. If A is a domain, we will say that it is integrally closedor normal if A is integrally closed in its field of fractions.

Definition 4.6. A domain A is said to be a unique factorization domain, orUFD, if every element of A can be written uniquely (up to multiplication by unitsof A) as a product of irreducible elements.

Let b = x1b2 . . . bn + 1, so that deg b ≥ 1 and b is relatively prime to b2 . . . bn.We have k[x1]b2...bn $ k(x1), because 1

b ∈ k(x1) \ k[x1]b2...bn . At the same time,k(x1), which is the field of fractions of k[x1]b2...bn , is integral over k[x1]b2...bn . Toget a contradiction, it is sufficient to show that k[x1]b2...bn is normal. Now, it is wellknown that k[x1] is a UFD (this is proved by using Euclidean algorithm, in otherwords, division with remainder). Thus the normality of k[x1]b2...bn follows from twofacts, which we leave to the reader as easy exercises:

Exercise 4.7. If B is a UFD and S a multiplicative subset of B, then BS is aUFD.

Exercise 4.8. Any UFD is normal.

To complete the proofs of all the Nullstellensatzes, it remains to prove Lemma4.4. To do this, we introduce the concept of a module over a ring.

Definition 4.9. Let A be a ring. An A-module is an abelian group M togetherwith an operation A×M →M (multiplication by elements of A), such that for alla, b ∈ A, x, y ∈M , we have:

(1) a(bx) = (ab)x(2) (a+ b)x = ax+ bx(3) a(x+ y) = ax+ ay(4) 1x = x.

Any ideal of A is an A-module. Any A-algebra is an A-module.


Definition 4.10. Let A be a ring and M an A-module. An A-submodule of Mis an abelian subgroup of M , which is an A-module under the operations inheritedfrom M .

For example, the A-submodules of A are precisely the ideals of A.

Definition 4.11. Let M be an A-module and x = {xλ}λ∈Λ a subset of M . TheA-submodule B, generated by x (sometimes denoted by

∑λ∈ΛAxλ), is defined

to be the smallest A-submodule of M , containing x. In other words,∑λ∈ΛAxλ is

the set of all elements of M which can be written as A-linear combinations of thexλ. We say that M is a finitely generated or a finite A-module if there is a finitesubset {x1, . . . ,xn} ⊂ M such that M =

∑ni=1Axi. An A-algebra B is said to be

finite over A if B is a finite A-module (of course, this is stronger than saying thatB is a finitely generated A-algebra).

Exercise 4.12. Let A ⊂ B ⊂ C be ring extensions. If B is finite over A and Cfinite over B, then C is finite over A.

Now we are in the position to prove Lemma 4.4. We prove Lemma 4.4 in thefollowing stronger form.

Lemma 4.13. Let A → B be an extension of rings. The following conditions areequivalent.

(1) B is finite over A.(2) B is integral and finitely generated over A (as an algebra).(3) B is finitely generated over A as an algebra, and each of the generators is

integral over A.

Proof. (2) =⇒ (3) is trivial.(1) =⇒ (2). Assume that B is finite over A. Clearly, B is finitely generated as

an A-algebra. Let x1, . . . ,xn be a set of generators of B as an A-module. Take anyy ∈ B; we want to show that y is integral over A. There exist aij ∈ A, 1 ≤ i, j ≤ n,such that yxi =

∑nj=1 aijxj . Let T denote the n× n matrix with entries aij , I the

n× n identity matrix. Consider the n× n matrix yI − T with entries in B. Let xdenote the column n-vector with entries xi. By construction, (yI−T )x = 0. Hence,by linear algebra, det(yI − T )xi = 0 for 1 ≤ i ≤ n. Since the xi generate B as anA-module, det(yI − T )x = 0 for every x ∈ B, so that det(yI − T ) = 0. We haveconstructed a monic polynomial equation satisfied by y over A, which proves thaty is integral over A.

(3) =⇒ (1) Suppose B is finitely generated, with generators integral over A. Sincea composition of finite extensions is finite (Exercise 4.12), we may assume that, asan algebra, B is generated over A by a single element x, integral over A. Thismeans that as an A-module, B is generated by 1, x, x2, . . . ,xn, . . . . By assumption,x satisfies a monic equation xn+an−1x

n−1 + · · ·+a0 = 0 over A. Hence xn belongsto the A-submodule of B generated by 1, x2, . . . ,xn−1. Then by induction on l, anyxl with l ≥ n belongs to the A-submodule of B generated by 1, x2, . . . ,xn−1. Hence


B is generated by 1, x2, . . . ,xn−1 as an A-module, which proves that B is finite overA.

This completes the proof of Lemma 4.13, Propositions 4.2 and 4.3 and Theorem1.7. �

§5. Primary decomposition.

Definition 5.1. A topological space X is said to be Noetherian if any descendingchain of closed sets in X stabilizes.

Exercise 5.2. Show that in a Noetherian topological space every closed set can bewritten uniquely as a finite union of irreducible closed sets.

Clearly, if R is a Noetherian ring, then the affine scheme Spec R is Noetherian.Let I be a radical ideal of R. We saw earlier (Proposition 1.20) that the reducedclosed subscheme V (I) ⊂ Spec R is irreducible if and only if I is prime. Hence,Exercise 5.2 applied to Spec R is equivalent to saying that in a Noetherian ring everyradical ideal can be written uniquely as an intersection of primes (the uniquenessfollows easily from Lemma 1.21).

Since in scheme theory we are interested in including the case of non-reducedschemes, we would like to extend this result to ideals of R which are not neces-sarily radical. The corresponding result in the non-radical case is called primarydecomposition.

First, we note that the elementary argument of Exercise 5.2 applies verbatim toshow that every ideal can be written as an intersection (not necessarily unique) ofirreducible ideals:

Definition 5.3. Let I be an ideal of A. I is irreducible if it cannot be writtenin the form I = I1 ∩ I2, where I1 and I2 properly contain I.

Lemma 5.4. Let A be a Noetherian ring. Then every ideal of A can be written asa finite intersection of irreducible ideals in A.

Proof. Let∑

denote the collection of all the proper ideals of A which cannot beexpressed as an intersection of finitely many irreducible ideals of A. We want toshow that Σ = ∅. Suppose not. Since A is Noetherian, Σ contains a maximalelement I. Since I ∈ Σ, I is not irreducible. Then I can be written as I = I1 ∩ I2for some ideals I1, I2 of A such that I $ I1 and I $ I2. By the maximality of I, wehave I1, I2 /∈ ∑. Hence both I1 and I2 can be expressed as finite intersections ofirreducible ideals of A and hence so can I. Then I /∈ Σ, a contradiction. Therefore∑

= ∅. �

The point of primary decomposition is to give a more explicit description ofirreducible ideals and to relate the irreducible ideals appearing in a decompositionof I to the set of zero divisors in A

I .Definition 5.5. Let A be a ring and Q $ A an ideal. Q is primary if for anyx, y ∈ A such that xy ∈ Q, either x ∈ Q or y ∈ √Q.


Note that in the above definition we think of x, y as an ordered pair. The defini-tion is not symmetric in x and y.Remark 5.6. Let Q be a primary ideal of A. Then P =

√Q is prime. Indeed,

xy ∈ P =⇒ (xy)n ∈ Q for some n ∈ N =⇒ xn ∈ Q or (yn)m ∈ Q for somem ∈ N. Thus xn ∈ Q or ynm ∈ Q, so that either x ∈ P or y ∈ P . This proves thatP is prime. We say that Q is a P -primary ideal or that it is primary to P .

Let Q be an ideal such that√Q = P is prime. Then Q is primary if and only if

Q = QAP ∩A. In particular, the intersection of two P -primary ideals is P -primary.

Next, we show that every irreducible ideal is primary, so that every ideal in aNoetherian ring is a finite intersection of primary ideals. In the rest of this section,we will discuss to what extent this primary decomposition is unique.

Exercise 5.7. Let m ∈ Max(A). Show that an ideal Q is m-primary if and onlyif√Q = m.

Lemma 5.8. Let A be a Noetherian ring, I ⊂ A an irreducilbe ideal. Then I isprimary.

Proof. Suppose ab ∈ I and b /∈ I. Let I : a = {x ∈ A | ax ∈ I}. I : a is anideal. I : a ⊂ I : a2 ⊂ (I : a3) ⊂ . . . is an ascending sequence of ideals in A. SinceA is Noetherian, there exists n ∈ N such that I : an = I : an+i for all i ≥ 1.

Now, I ⊂ (I + (an)) ∩ (I + (b)). Conversely, let r ∈ (I + (an)) ∩ (I + (b)). Thenr = g + can = h+ db where g, h ∈ I and c, d ∈ A. So ra = ga+ can+1 = ha + dab,so that g, h, ab ∈ I. Then c ∈ (I : an+1) = (I : an), so that r = g + can ∈ I. Weobtain I = (I + (an)) ∩ (I + (b)).

Now, I is irreducible and I $ I + (b) because b /∈ I. Hence I = I + (an), so

an ∈ I. We have proved that a ∈√I. Therefore I is primary, as desired.

Corollary 5.9. Let A be a Noetherian ring, I ⊂A an ideal. There exists a decom-position

(5.1) I =

n⋂

i=1

Qi,

where each Qi is primary.Example 5.10. Primary decomposition is not unique. Let A = k[x, y]. We have(x2, xy) = (x)∩(y, x2) = (x)∩(y, x)2, which are two different primary decompositionof the same ideal.

We will now further analyze primary decomposition of a given ideal.Definition 5.11. Let A be a ring, I ⊂A an ideal. Consider a decomposition

I =n⋂i=1

Qi where each Qi is primary. Let Pi =√Qi; by Remark 5.6, the Pi are

prime. The Pi are called associated primes of I. The minimal ones among the Piare called the minimal primes of I. Associated primes which are not minimal arecalled the embedded components or embedded points of I. Sometimes the


Pi are also referred to as the associated (resp. minimal) primes of the A-module AI .

In particular, the associated (resp. minimal) primes of (0) are called the associated(resp. minimal) primes of A. This terminology seems ambiguous, but usually itcauses no confusion.

Note 5.12. By Lemma 1.21, the minimal primes of I are precisely the minimalelements of V (I) under inclusion, so that the name is appropriate.

For instance, in Example 5.10, (x) and (x, y) are the associated primes, (x) isthe only minimal prime and (x, y) is an embedded point. In this example, we seethat the collection of associated primes depends only on the ideal I and not on thechoice of the primary decomposition. We will now show that this is true in generalby describing the Pi in terms of zero divisors mod I.

Lemma 5.13. Let Q ⊆ A be a primary ideal, P =√Q, a an element of A.

(1) If a ∈ Q, then (Q : a) = A.(2) If a ∈ P \Q, then Q $ (Q : a) and Q : a is P -primary.(3) If a /∈ P , then (Q : a) = Q.

Proof. (1) is obvious.We always have Q ⊂ (Q : a) because Q is an ideal.(3) Let b ∈ (Q : a); then ab ∈ Q. Since a /∈ P =

√Q, b ∈ Q. This proves that

(Q : a) = Q.(2) First, we show that (Q : a) ⊂ P . Indeed, take b ∈ (Q : a); then ab ∈ Q. Since

a /∈ Q and Q is primary, we have b ∈ P . This proves that (Q : a) ⊂ P .

Now, P =√Q ⊆

√(Q : a) ⊆

√P = P (the last equality holds since P is prime),

therefore√

(Q : a) = P .

Now, take b, c ∈ A such that bc ∈ (Q : a) and b /∈√

(Q : a) = P . Then abc ∈ Q.Since b /∈ P =

√Q and Q is primary, ac ∈ Q. Hence c ∈ (Q : a). This proves that

(Q : a) is primary.It remains to show that Q : a 6= Q. Indeed, since a ∈ P \Q, there exists a unique

n ∈ N such that an /∈ Q but an+1 ∈ Q. Then an ∈ (Q : a) \Q. This completes theproof of Lemma 5.13. �

We will now use Lemma 5.13 to interpret primary decomposition in terms of theset of zero divisors mod I.

Theorem 5.14. Let I be an ideal in a Noetherian ring A. Consider a primarydecomposition (5.1). Let Pi =

√Qi, 1 ≤ i ≤ n. Then:

(1) P1, . . . ,Pn are precisely all the prime ideals which are of the form I : a forsome a ∈ A.

(2) Let Pi = PiAI , 1 ≤ i ≤ n. Then the set of zero divisors in the ring A

I is

∪ni=1Pi.

In particular, the set of zero divisors in A is precisely the union of the associatedprimes of A (recall that nil(A) is the intersection of the associated primes of A).


Corollary 5.15. The collection {Pi}1≤i≤n depends only on I and not on the choiceof primary decomposition.

If A is a ring and a ∈ A, the annihilator of a is the ideal Ann a = (0) : a. Thusanother way of stating Theorem 5.14 is to say that the associated primes of A areprecisely all the prime ideals of the form Ann a for a ∈ A.Proof of Theorem 5.14. (1) We may assume that for all i = 1, 2, . . . ,n, Qi 6⊃n⋂j=1j 6=i

Qj . Take any i, 1 ≤ i ≤ n and any a ∈n⋂j=1j 6=i

Qj \Qi. We have

(5.2) (I : a) =

n⋂

j=1

Qj

: a =

n⋂

j=1

(Qj : a) = Qi : a ⊂ Pi,

where the last two statements hold by Lemma 5.13. Let Pi denote a maximal

element among all the ideals of the form I : a, where a ∈n⋂j=1j 6=i

Qj \Qi. By (5.2) and

Lemma 5.13 (2),(3),

(5.3) Pi =

√Pi.

We want to show that Pi = Pi, that is, all the Pi are of the form (I : a) for some

a ∈ A. Suppose not. Let a ∈n⋂j=1j 6=i

Qj \Qi be such that Pi = I : a and take b ∈ Pi\Pi.

By (5.3), there exists l ∈ N such that bl /∈ Pi but bl+1 ∈ Pi. Since bl /∈ Pi = I : a,

abl /∈ I. Since a ∈n⋂j=1j 6=i

Qj,

(5.4) abl ∈n⋂

j=1j 6=i

Qj.

Combined with abl /∈ I, (5.4) gives abl ∈n⋂j=1j 6=i

Qj \Qi. On the other hand, bl+1a ∈ I

by the choice of l. Then Pi = I : a $ I : (abl), where the inclusion is strict because

b ∈ I : (abl) \ Pi. This contradicts the maximality of Pi. We have proved that allthe Pi, 1 ≤ i ≤ n are of the form I : a for some a ∈ A.

Conversely, let P ∈ Spec A be of the form P = I : a for some a ∈ A. Then

P =n⋂j=1

(Qj : a). Since prime ideals are irreducible (Proposition 1.20), we must

have P = Qi : a for some i, 1 ≤ i ≤ n. Hence Qi : a 6= A, so that, by Lemma 5.13,


√(Qi : a) = Pi. We obtain P =

√P =

√(Qi : a) = Pi. This completes the proof

of (1).

(2) We may replace A by AI, I by (0) and Pi by Pi. In particular, we take the

Qi to be primary components of the zero ideal: (0) = ∩ni=1Qi. By (1), each of thePi is the annihilator ideal of some a ∈ A, hence every element in ∪ni=1Pi is a zerodivisor.

Conversely, take any a ∈ A \ ∪ni=1Pi. Then by Lemma 5.13 (3), Qi : a = Qifor 1 ≤ i ≤ n. We have Ann a ≡ (0) : a = ∩ni=1(Qi : a) = ∩ni=1Qi = (0). Thusa ∈ A \ ∪ni=1Pi implies that a is not a zero divisor, as desired. This completes theproof of Theorem 5.14. �

§6. Dimension Theory.

Continuing with our dictionary between algebra and geometry, we come to theconcept of dimension of an algebraic variety or scheme. Intuitively, we all know whatis meant by dimension in geometry; for example, everybody knows that the n-spacekn has dimension n. Therefore, it is natural to look for the corresponding definitionon the algebraic side: that of the dimension of a ring. There are several alternativeways of thinking about dimension; the purpose of this section is to describe theiralgebraic interpretation and to prove that all give the same answer.

The first one is the maximal length of a descending chain of irreducible sub-varieties. For example, a maximal chain of irreducible subvarieties in kn is kn ⊃kn−1 ⊃ · · · ⊃ k2 ⊃ k ⊃ {0}; it has length n (not counting kn itself). We model ourdefinition of the dimension of a ring on this geometric idea.

The second approach is to consider the number of elements in a local coordinatesystem on X . More precisely, since we want to include singular varieties, we shouldconsider the smallest number of equations, whose set of common zeroes inX consistsof isolated points. This is the idea behind the integer δ(A), associated to a localring A later in this section.

A third way of defining dimension, of great importance both in theory and incomputations, is the degree of the Hilbert-Samuel polynomial. Finally, in the case ofan algebraic variety X over a field k, the dimension equals the largest transcendencedegree of the field of rational functions on an irreducible component of X . We nowgive precise definitions and prove the equality of all four numbers.

Definition 6.1. Let A be a ring. The Krull dimension of A, denoted dimA, isthe maximal length of an ascending chain of prime ideals in A:

(6.1) P0 $ P1 $ · · · $ Pd

not counting the minimal prime P0. In other words, if (6.1) is the longest, we saythat dim A = d. If there are arbitrarily long chains of primes in A, we say thatdim A = ∞. For P ∈ Spec A, the height of P is defined by ht P = dim AP(which is the maximal length of a chain of prime ideals contained in P ).


Exercise 6.2. Let A be a ring. Prove that dim A = max{dim Am | m ∈Max(A)}.(Hint: first, show that for any P ∈ Spec A, the natural map Spec AP → Spec Ainduces a bijection between Spec AP and {Q ∈ Spec A | Q ⊂ P}.)

Soon we will show that dim A <∞ if A is a Noetherian local ring.

Problem 6.3. Give an example of an infinite dimensional Noetherian ring.

Exercise 6.2 says that dimension can be computed locally. In other words, inorder to study dimension in arbitrary rings, it is sufficient to understand the caseof local rings.Definition 6.4. Let (A,m, k) be a Noetherien local ring. Let δ(A) denote thesmallest number of generators of an m-primary ideal in A.

Definition 6.5. Let A be a ring and M an A-module. The length of M isdefined to be the maximal length of a chain of submodules (0) $ M1 $ M2 $ · · · $Mn = M , not counting (0). If there exist arbitrarily long such chains, we say thatlength(M) =∞.

For example, if V is a vector space over a field k, length(V ) as a k-module isnothing but dimk V .

A homomorphism of A-modules is a homomorphism of abelian groups, whichrespects multiplication by elements of A.

Given A-modules N ⊂ M , the quotient MN

of M by N is defined to be thequotient of abelian groups, endowed in the obvious way with the operation of mul-tiplication by elements of A. In particular, we may talk about the quotient I

Jof

two ideals of A, J ⊂ I. We may view IJ as an ideal in the ring A

J .Exercise 6.6. Show that length is an additive function, that is, if 0 → M ′ →M →M ′′ → 0 is an exact sequence, then length(M) = length(M ′) + length(M ′′).

Definition 6.7. Let (A,m, k) be a Noetherian local ring. The Hilbert Samuelfunction of A is the function HA,m : N→ N, defined by HA,m(n) = length

(A

mn+1

)

(considered as an A-module). By additivity of length,

(6.2) length

(A

mn+1

)=

n∑

i=0

dimkmi

mi+1,

where the mi

mi+1 are k-modules, that is, k-vector spaces.

Note that since A is Noetherian, each of mi is finitely generated, so that all thequantities in (6.2) are finite.

Theorem 6.8. HA,m(n) is a polynomial for n≫ 0. In other words, there exists apolynomial P (n) with rational coefficients, such that P (n) = HA,m(n) for n ≫ 0.P (n) is sometimes called the Hilbert polynomial of A.

A proof of this Theorem 6.8 will be given shortly. First, we finish our list of thedifferent characterizations of dim A.


Notation 6.9. Let d(A) denote the degree of the Hilbert polynomial of A.

Finally, let A be a k-algebra. Let x1, . . . , xn be elements of A.

Notation. Given x = (x1, . . . , xn) and α = (α1, . . . , αn) ∈ Nn0 , xα will denote themonomial xα = xα1

1 xα22 . . . xαnn .

x1, . . . , xn are said to be algebraically independent over k if they satisfy nonon-trivial polynomial relations of the form

∑aαx

α = 0, aα ∈ k. The transcen-dence degree of A over k, denoted tr. deg(A/k), is defined to be the maximalnumber of elements of A, algebraically independent over k. If A is a finitely gener-ated k-algebra, then tr. deg(A/k) <∞.

Example 6.10. Let k be a field, A = k[x1, . . . ,xd] the polynomial ring in d vari-ables. (0) $ (x1) $ (x1, x2) $ · · · $ (x1, . . . ,xd) is a chain of prime ideals of lengthd.

Now let m = (x1, . . . ,xd) be the maximal ideal corresponding to the origin in kd.Consider the localization Am. m is generated by d elements. The Hilbert-Samuelfunction of Am is HAm,m(n) = length( A

mn+1 ) =(n+dd

), which is a polynomial in

n of degree d. In this case, HAm,m(d) is a polynomial for all n, not merely for nsufficiently large.

Thus, in this example, we have

(6.3) tr. deg(A/k) = d(Am) = d ≥ dim A ≥ dim Am ≥ δ(A).

We will soon show that all the inequalities in (6.3) are, in fact, equalities.

We now turn to the main purpose of this section, which is to prove that for anyNoetherian local ring A, dim A = d(A) = δ(A). If A is a finitely generated algebraover a field k, we will prove that dim A = tr. deg(A/k).

The first step is to prove that d(A) is well defined, that is, HA,m(n) is, indeed, apolynomial for n≫ 0. To this end, we introduce the notion of graded algebras andgraded modules.

Definition 6.11. Let G0 be a ring. A graded G0-algebra is a G0-algebra of the

form G =∞⊕n=0

Gn, such that for all n, l ∈ N0, we have GnGl ⊂ Gn+l (in particular,

each Gn is a G0-module under the ring operations in G).If G is a graded G0-algebra, a graded G-module is a G-module of the form

M =∞⊕n=0

Mn where each Mn is a G0-module and for all n, l ∈ N0, we have GnMl ⊂Mn+l. For n ∈ N0, elements of Mn are called homogeneous of degree n.

Example 6.12. Let A be a ring and I ⊂ A an ideal. Associated to A and I is

the graded AI -algebra grIA =

∞⊕n=0

In

In+1 . If I is a finitely generated ideal, then grIA

is a finitely generated AI -algebra (its generators as an A

I -algebra are precisely the


generators of I as an ideal, viewed as elements of grIA of degree 1). By the Hilbertbasis theorem, if A is Noetherian then grIA is Noetherian (cf. Exercise 6.13 below).In particular, if (A,m, k) is a local Noetherian ring, grmA is a finitely generated

graded k-algebra. Moreover, for any n ∈ N0, dimkmn

mn+1 < ∞. In other words, thehomogeneous elements of degree n form a finite dimensional vector space. For themoment, this last example of a graded algebra is our main interest for applications.Later in the course, other important examples of graded algebras will appear.

Exercise 6.13. Let G be a graded G0-algebra. Prove that G is Noetherian if andonly if G0 is Noetherian and G is finitely generated over G0. Now, suppose thatthis is the case and let M be a finitely generated graded G-module. Show that eachMn is a finite G0-module.

The Hilbert-Samuel function of a Noetherian local ring (A,m, k) is

HA,m(n) =

n∑

i=0

lengthmi

mi+1.

We start our proof of Theorem 6.8 by the following simple observation. Let P (n)be a real valued function of n. Let P (1)(n) =

∑ni=0 P (n) and let d ∈ N0. Then P

is a polynomial in n of degree d if and only if P (1) is a polynomial of degree d+ 1.Hence to prove Theorem 6.8 it is sufficient to prove, for a local Noetherian ring

(A,m, k), that for n≫ 0 length mn

mn+1 is a polynomial in n. We can generalize thisstatement as follows.

Let G be a graded algebra, M =∞⊕n=0

Mn a finite graded G-module, as in the

Example 6.12 (in particular, each Mn is a finite G0-module). Let λ be an additivefunction from the category of finite G0-modules to N0. For n ∈ N0, we may considerλ(Mn) as a function of n. To prove Theorem 6.8, it is enough to show that λ(Mn)is a polynomial for n ≫ 0. To this end, we consider the generating function of

M : P (M, t) =∞∑n=0

λ(Mn)tn, where t is an independent variable.

Theorem 6.14 (Hilbert, Serre). Let x1, . . . ,xd be a set of homogeneous gener-

ators of M , ki = deg xi. Then P (M, t) = f(t)d∏i=1

(1−tki )

, where f is a polynomial in t

with integer coefficients.Proof. Induction on d. First, let d = 0. Then A = A0, hence Mn = 0 for n≫ 0,

so that λ(Mn) = 0 for n≫ 0 and the result is true in this case.Next, suppose the result is known for d− 1. For each n ∈ N0, multiplication by

xd is a G0-module homomorphism from Mn to Mn+kd . Let Kn and Ln+kd denote,respectively, the kernel and the cokernel of this homomorphism. For each n ∈ N0,we obtain an exact sequence

(6.4) 0→ Kn →Mn →Mn+kd → Ln+kd → 0.


By additivity of λ, (6.4) implies that

(6.5) λ(Kn)− λ(Mn) + λ(Mn+kd)− λ(Ln+kd) = 0.

Let K =∞⊕n=0

Kn, L =∞∑

n=kd

Ln. Multiply (6.5) by tn+kd and sum over n. We obtain

(6.6) tkdP (K, t)− tkdP (M, t) + P (M, t)− P (L, t) + g(t) = 0,

where g(t) is a polynomial in t. Since both K and L are annihilated by xd, both Kand L are graded G0[x1, . . . , xd−1]-modules. Hence by the induction assumption,d−1∏i=1

(1− tki)P (K, t) andd−1∏i=1

(1− tki)P (L, t) are polynomials in t. By (6.6), we obtain

(1 − tkd)P (M, t) = −tkdP (K, t) + P (L, t) − g(t). Hence P (M, t) = f(t)d−1∏i=1

(1−tki )

for

some polynomial f(t) and the Theorem is proved. �

We have the following Corollary, which obviously implies Theorem 6.8.Corollary 6.15. Let G =

⊕Gn, M =

⊕Mn, λ, xi be as in Theorem 6.14.

Suppose that ki ≡ deg xi = 1, 1 ≤ i ≤ d. Then, for n≫ 0, λ(Mn) is a polynomialin n of degree at most d− 1.

Proof. By Theorem 6.14, P (M, t) = f(t)(1−t)d , where f is a polynomial in t. Write

f(t) =N∑k=0

aktk. Since 1

(1−t)d=

∞∑n=0

(n+d−1d−1

)tn, we have, for n ≫ 0, λ(Mn) =

N∑k=0

ak(n−k+d−1

d−1

), which is a polynomial in n of degree at most d− 1. �

Remark 6.16. Notice in the above proof that the degree of λ(Mn) for large n is

precisely the order of the pole of f(t)(1−t)d

at t = 1 minus 1.

We now state the main result of this section.

Theorem 6.17. Let (A,m, k) be a Noetherian local ring. Then dim A = d(A) =δ(A).

First, we study the case dimA = 0.

Definition 6.18. A ring A is said to be Artinian if every descending chain ofideals in A stabilizes.

Despite the apparent similarity of this definition with the Noetherian property,we will now show that the Artinian property is much stronger. Namely, we willshow that A is Artinian if and only if A is Noetherian and dim A = 0.Proposition 6.19. Let A be an Artinian ring. Then every prime ideal of A ismaximal (in other words, dim A = 0).

Proof. Let P ⊂ A be a prime ideal. Then AP is an Artinian domain. We want

to show it is a field. Take any x ∈ AP . It suffices to show that x is invertible.


Consider (x) ⊃ (x2) ⊃ (x3) ⊃ . . . . This chain of ideals stabilizes, so that(xn) = (xn+1) for some n. Hence there exists b ∈ A

P such that bxn+1 = xn. SinceAP is a domain, bx = 1, so that x is invertible. This proves that A

P is a field, i.e. Pis maximal. �

Corollary 6.20. Let A be an Artinian ring. Then nil(A) = J(A) (recall that J(A)stands for the Jacobson radical of A).

Proof. This follows from Exercise 4 of §1. �

Exercise 6.21. Let A be any ring, m1, m2 two distinct maximal ideals of A. Provethat m1m2 = m1 ∩m2.

Theorem 6.22. Let A be an Artinian ring. Then A has finitely many maximalideals.

Proof. Suppose not. Then there is an infinite set m1, m2, . . . ,mn, . . . of distinctmaximal ideals of A. Consider the descending chain of ideals m1 ⊃ m1 ∩ m2 ⊃m1 ∩ m2 ∩ m3 ⊃ . . . . Since A is Artinian, we must have m1 ∩ m2 ∩ · · · ∩ mn =m1 ∩ · · · ∩mn ∩mn+1 for some n. By Lemma 1.21, mn+1 = mi for some i ≤ n.This is a contradiction and the Theorem is proved. �

Exercise 6.23. Let A be a ring and I a finitely generated ideal, contained in nil(A).Show that there exists n ∈ N such that In = (0). More generally, show that if I is

any finitely generated ideal, there exists n ∈ N such that (√I)n ⊂ I.

Theorem 6.24. Let A be an Artinian ring. Let Max(A) = {m1, . . . ,ms}. Then(m1m2 . . .ms)

n = 0 for some n ∈ N.Proof. Let I = m1m2 . . .ms. By Exercise 6.21, I = J(A). By Corollary 6.20,

I =√

(0). Now, we cannot apply Exercise 6.23, because we have not yet provedthat I is finitely generated.

Consider the descending chain of ideals I ⊃ I2 ⊃ I3 ⊃ . . . ⊃ In ⊃ . . . . ThenIn = In+1 for some n. We claim that In = (0). Suppose not. Let Σ denote the

collection of all the ideals H ⊂ A, for which HIn 6= 0. Σ 6= ∅ since I ∈ Σ. SinceA is Artinian, Σ contains a minimal element H (in the sense of inclusion). SinceHIn 6= 0, there exists x ∈ H such that xIn 6= 0. We have (x) ⊂ H, hence (x) = Hby the minimality of H. Now, (xIn)In = xI2n = xIn 6= (0) and xIn ⊂ (x), hencexIn = (x) by the minimality of H. Then there exists y ∈ In such that x = xy.Then

(6.7) x = xy = xy2 = · · · = xyl = . . . .

Since y ∈ In ⊂ I =√

(0), (6.7) implies that x = 0, which is a contradiction. Thisproves Theorem 6.24. �

Theorem 6.25. Let A be a ring. Then A is Artinian if and only if length(A) <∞(where we view A as an A-module).

Proof. ⇐ is obvious.


⇒ Let Max(A) = {m1, . . . ,ms} and let I = m1 . . .ms = J(A). By the previoustheorem, In = 0 for some n ∈ N. Then there exists r ≥ s and maximal idealsm1, . . . ,mr, not necessarily all distinct, such that m1m2 . . .mr = 0. Consider thechain A ⊃ m1 ⊃ m1m2 ⊃ . . . ⊃ m1m2 . . .mr−1 ⊃ (0) of submodules of A. By the

additivity of length, it is sufficient to show that length(

m1...mim1m2...mimi+1

)<∞. Now,


is annihilated by mi+1, hence it is an Ami+1

-module, that is, an Ami+1

-

vector space. In a vector space, a submodule is the same thing as a vector subspace,so that a vector space has finite length if and only if it has finite dimension if andonly if every descending chain of submodules stabilizes. This last condition holdsby the Artinian property of A, hence dim A

mi+1


<∞. This proves that

length(A) is finite, as desired. �

Theorem 6.26. A ring A is Artinian if and only if A is Noetherian and dim A = 0.

Proof. Suppose A is Artinian. By the previous theorem, length(A) <∞, hence Ais Noetherian. Since every prime ideal in A is maximal, dimA = 0.

Conversely, suppose A is Noetherian and dim A = 0. Let (0) =s⋂i=1

Qi be a

primary decomposition of (0). Let Pi =√Qi be the associated primes. Since

dim A = 0, each of the Pi is maximal. Moreover,√

(0) = ∩si=1Pi, hence byExercise 6.23 there exists n such that (P1 . . . Ps)

n = 0.Now we can use the same reasoning as in the last theorem. We have A ⊃

P1 ⊃ P1P2 ⊃ . . . ⊃ P1P2 . . . Pr−1 ⊃ (0), where r > s and P1, . . . ,Pr is a sequenceconsisting of P1, . . . ,Ps with some repetitions. Since A is Noetherian, P1 . . . Pi isa finitely generated ideal, hence P1...Pi

P1...Pi+1is a finite A-module, hence a finite A

Pi+1-

module, i.e. a finite dimensional APi+1

-vector space. Now the reasoning in the proof

of the last theorem applies, and we conclude that length(A) < ∞, so that A isArtinian. �

Now, we shall prove a special case of Theorem 6.17.Theorem 6.27. Let (A,m, k) be a Noetherian local ring. Then dim A = 0 if andonly if δ(A) = 0 if and only if d(A) = 0.

Proof. dim A = 0 ⇐⇒ m is the only prime ideal ⇐⇒ m =√

(0) ⇐⇒ (0)

is m-primary ⇐⇒ δ(A) = 0 ⇐⇒ ∃n0 ∈ N such that A = Amn ∀n ≥ n0 ⇐⇒

HA,m(n) = const. for n≫ 0 ⇐⇒ d(A) = 0. �

This takes care of the zero dimensional local rings.Let (A,m, k) be any Noetherian local ring and Q ⊂ A an m-primary ideal. Next,

we show that d(A) ≤ δ(A).

Proposition 6.28.

(1) length AQn

<∞.

(2) length AQn is a polynomial HQ(n) for n ≫ 0 of degree at most s where s is

the minimal number of generators of Q.


(3) deg HQ is independent of Q (hence deg HQ = d(A) ≤ δ(A)).

Proof. 1) is clear, since AQn is Artinian.

2) Apply Theorem 6.14 to the AQ

-algebra grQA =∞⊕n=0

Qn

Qn+1 , generated over AQ

by

s elements. By Theorem 6.14, length Qn

Qn+1 is a polynomial in n of degree at most

s − 1. Since HQ(n) = length(

AQn+1

)=

n∑i=0

length(

Qi

Qi+1

), we have deg H ≤ s, as

desired.

3) It is enough to prove that deg HQ = deg Hm. Since Q is m-primary, thereexists r ∈ N such that mr ⊂ Q ⊂ m. Then mrn ⊂ Qn ⊂ mn for any n ∈ N, so that

length(Amrn

)≥ length

(AQn

)≥ length

(Amn

).

Since Hm(rn) ≥ HQ(n) ≥ Hm(n) for all n ∈ N and since both Hm and HQ arepolynomials in n, we have deg HQ = degHm. QED

To complete the proof of Theorem 6.17, it remains to show that d(A) ≥ dim A ≥δ(A). We now state a crucial lemma, whose proof will be the subject of the nexttwo sections.

Lemma 6.29. Let x ∈ A. Assume that x is not a zero divisor. Then d(A(x)

)≤

d(A)− 1.

The rest of this section is devoted to proving Theorem 6.17, assuming Lemma6.29.

Proposition 6.30. d(A) ≥ dim A.

Proof. Induction on d(A). If d(A) = 0, the Proposition follows from Theorem6.27.

Suppose d(A) > 0.

Let P0 $ P1 $ · · · $ Pr be a chain of prime ideals. We want to show r ≤ d(A).

Let A = AP0

. Take a non-zero element x ∈ P1

P0. Since A is a domain, x is not

a zero divisor in A. By Lemma 6.29, d(A(x)

)≤ d(A) − 1 ≤ d(A) − 1. On the

other hand, P1A(x) $ P2A

(x) · · · $ PrA(x) is a chain of distinct prime ideals of A

(x) . Hence

dim(A(x)

)≥ r− 1. Using the induction hypothesis, we obtain r− 1 ≤ dim

(A(x)

)≤

d(A(x)

)≤ d(A)− 1 therefore r ≤ d(A), as desired. Q.E.D.

Corollary 6.31. If A is a Noetherian local ring, then dimA <∞. In particular, ifP ∈ Spec A, then ht P <∞.

It remains to show that dim A ≥ δ(A).

Given any ideal I ⊂ A, define ht I = min{ht P | I ⊂ P, P ∈ Spec A}. ByCorollary 6.31, ht I <∞ for any ideal I in a Noetherian ring A.


Lemma 6.32. dim A ≥ δ(A).

Proof . Let d = dim A. We will construct, recursively in i, elements x1, . . . ,xi ∈m such that ht(x1, . . . xi) ≥ i. Assume that i ≤ d and that x1, . . . ,xi−1 withht(x1, . . . ,xi−1) ≥ i−1 are already constructed. If ht(x1, . . . xi−1) ≥ d, then m is theonly minimal prime of (x1, . . . xi−1), hence (x1, . . . xi−1) is m-primary, hence δ(A) ≤i− 1 ≤ d and the Lemma is proved. Thus we may assume that ht(x1, . . . xi−1) < d.Let P1, . . . ,Pr be the minimal primes of (x1, . . . ,xi−1). By assumption, ht Pj < d,1 ≤ j ≤ r, so that Pj 6= m, 1 ≤ j ≤ r. At this point, we need an elementary lemmaabout unions of prime ideals.

Lemma 6.33. Let I, P1, . . . ,Pr be ideals in a ring A, where all except possibly two

of the Pj are prime. If I ⊂r⋃j=1

Pj, then I ⊂ Pj for some j.

We finish the proof of Lemma 6.32 assuming Lemma 6.33. By Lemma 6.33, m 6=r⋃j=1

Pj . Take an element xi ∈ m \r⋃j=1

Pj . We want to show that ht(x1, . . . ,xi) ≥ i.

Indeed, take any minimal prime P of (x1, . . . ,xi). Since (x1, . . . ,xi−1) ⊂ P , P ⊃ Pjfor some j, 1 ≤ j ≤ r. Since xi ∈ P \ Pj , Pj $ P . Since ht Pj ≥ ht(x1, . . . ,xi−1) ≥i − 1, we have ht P ≥ ht Pj + 1 = i. We have constructed the desired elements(x1, . . . ,xi). For some i ≤ d, we obtain ht(x1, . . . ,xi) ≥ d, hence (x1, . . . ,xi) ism-primary, hence d = ht m ≥ δ(A). It remains to prove Lemma 6.33.

Proof of Lemma 6.33. We may assume that P3, . . . ,Pr are prime. The proof isby induction on r. First, suppose r = 2. Suppose that I 6⊂ P1 and I 6⊂ P2.Pick elements x1 ∈ I \ P1 and x2 ∈ I \ P2. Then x1 ∈ P2 and x2 ∈ P1, so thatx1 + x2 ∈ I \ (P1 ∪ P2), a contradiction.

Next, suppose the result is known for r − 1, with r ≥ 3. Suppose I 6⊂ Pj for1 ≤ j ≤ r. By the induction assumption, we may assume that I is not contained inthe union of any smaller subcollection of the Pj . For each j, pick xj ∈ I \ ∪ri=1

i6=jPi.

Now consider the element x = xr +r−1∏j=1

xj . Clearly, x ∈ I \r−1⋃j=1

Pi. On the other

hand, since xj /∈ Pr for j < r and Pr is prime, we have x /∈ Pr. This contradicts the

fact that I ⊂r⋃i=1

Pi. This completes the proof of Lemma 6.33 and with it Lemma

6.32. �

Let (A,m, k) be a Noetherian local ring of dimension d. Let x1, . . . ,xd ∈ m

be such that√

(x1, . . . ,xd) = m. The elements x1, . . . ,xd are called a system ofparameters of A.

To complete the proof of Theorem 6.17, it remains to prove Lemma 6.29. To doso, we need to study filtrations, ideal-adic topologies and the Artin-Rees lemma.This is the content of the next section.


§7. Filtrations, ideal-adic topologies and the Artin-Rees lemma.

Let A be a ring, I an ideal of A, M an A-module. The notion of multiplying Mby an ideal is defined in the obvious way. Namely IM is the submodule of M givenby IM = {ax ∈M | a ∈ I, x ∈ M}.Definition 7.1 An I-filtration of M is a chain of submodules M = M0 ⊃ M1 ⊃. . . ⊃Mn such that IMn ⊂Mn+1 for all n ∈ N0.

A filtration is stable if IMn = Mn+1 for n ≫ 0. The I-adic topology onM is the topology whose basic open sets are all the sets of the form x + InM ,x ∈M,n ∈ N.

Example 7.2. Let A = M = Z and let I = (p), where p is a prime number. Wehave the p-adic filtration and the p-adic topology on Z. Clearly this filtration isstable.

Remark 7.3. Let {Mn} and {M ′n} be two stable filtrations. Then {Mn} and {M ′n}have bounded difference, that is, there exists n0 ∈ N such that M ′n+n0

⊂ Mn ⊂M ′n−n0

for all n ≫ 0 (this is so because there exists n0 such that, for all n ∈ N,M ′n+n0

= InM ′n0⊂ InM ⊂Mn and similarly in the other direction).

Lemma 7.4 (the Artin-Rees Lemma). Let A be a Noetherian ring, M ′ ⊂ Mfinitely generated A-modules. Let Q be an ideal. Let {Mn} be a stable Q-filtrationof M . Let M ′n = Mn ∩M ′. Then {M ′n} is a stable Q-filtration of M ′.

Proof. Consider the graded A-algebra G =∞⊕n=0

Qn and the graded G-module

M∗ =∞⊕n=0

Mn.

Lemma 7.5. {Mn} is stable if and only if M∗ is a finite G-module.

Proof. For each n, let M∗n denote the graded G-module M∗n = M⊕M1⊕· · ·⊕Mn⊕QMn ⊕Q2Mn ⊕ . . . . We get a chain of submodules

(7.1) M∗1 ⊂M∗2 ⊂ . . . ⊂M∗n ⊂ . . .

of M∗.By Exercise 6.13, each Mn is a finite A-module. Hence since G is Noetherian, M∗

is finitely generated as a G-module ⇔ it is generated by its homogeneous elementsof degree less than some fixed number ⇔ (7.1) stabilizes⇔ there exists n0 ∈ N suchthat Mn+n0

= QnMn0for all n ∈ N ⇔ the filtration is stable. �

We return to the proof of the Artin–Rees Lemma. We need the following char-acterization of finite modules over a Noetherian ring, which we leave to the readeras an easy exercise.Exercise 7.6. Let A be a Noetherian ring, M an A-module. Then M is finiteover A if and only if it is Noetherian, that is, any ascending chain of submodulesN1 ⊂ N2 ⊂ . . . ⊂ Nk ⊂ . . . stabilizes. In particular, if A is Noetherian, M a finite


A-module and M ′ ⊂ M a submodule, then M ′ is also a finite A-module. (Hint:if M is a finite A-module then M can be written as a quotient of a direct sum offinitely many copies of A.)

It is easily seen that {Mn ∩M ′} is a Q-filtration.

Let M ′∗

=∞⊕n=0

M ′n. We have M ′∗ ⊂ M∗. Since M∗ is a finite G-module, by

Exercise 7.6 so is M ′∗. By Lemma 7.5, this implies the Artin-Rees lemma.

Corollary 7.7. Let A be a Noetherian ring, Q ⊂ A an ideal and M a finiteA-module. Let M ′ ⊂ M be a submodule. Then there exists n0 ∈ N such that(Qn+n0M) ∩M ′ = Qn(Qn0M ∩M ′) for all n ∈ N0.

Corollary 7.8. Keep the above notation. Let {M ′n} be a stable Q-filtration of M ′.There exists n0 ∈ N such that (Qn+n0M) ∩M ′ ⊂ M ′n ⊂ (Qn−n0M) ∩M ′ for alln ≥ n0.

§8. Dimension theory, continued.

We are now in a position to prove Lemma 6.29. First, we prove the followinggeneralization of Proposition 6.28.

Proposition 8.1. Let (A,m) be a Noetherian local ring, Q ⊂ A an m-primaryideal. Let M be a finite A-module, {Mn} a stable Q-filtration of M . Let s be theminimal number of generators of Q. Then:

(1) length(MMn

)<∞ for all n ∈ N0.

(2) length(MMn

)is a polynomial HM,Q(n) for n≫ 0, of degree at most s.

(3) deg HM,Q depends only on M , not on Q and {Mn}. The highest coefficientof HM,Q depends only on M and Q, not on the filtration {Mn}.

Proof . (1) Since {Mn} is a Q-filtration, Mn

Mn+1is an A

Q-module. Since Mn is

a finite A-module, Mn

Mn+1is a finite A

Q-module. Hence it is a quotient of the direct

sum of finitely many copies of AQ . Now the fact that lengthAQ < ∞ implies that

length Mn

Mn+1<∞ and hence length M

Mn+1<∞.

(2) Follows immediately from Corollary 6.15 of the Hilbert-Serre Theorem, ap-

plied to the graded AQ -algebra G =

∞⊕n=0

Qn

Qn+1 and the graded G-module∞⊕n=0

Mn

Mn+1.

(3) Take any other stable Q-filtration {M ′n}. There exists n0 ∈ N such that

(8.1) Mn−n0⊂M ′n ⊂Mn+n0

for all n ∈ N.

Let H ′M,Q denote the Hilbert polynomial associated to {M ′n}. By (8.1) HM,Q(n−n0) ≥ H ′M,Q(n) ≥ HM,Q(n−n0) for all n ∈ N, hence HM,Q and H ′M,Q have the samedegree and highest coefficient. To show that, in addition, deg HM,Q is independent


of Q, compare deg HM,Q with deg HM,m. Since the degree is independent ofthe choice of the filtration, we may take the filtrations to be QnM and mnM ,respectively. Since Q is m-primary, there exists r ∈ N such that mr ⊂ Q ⊂ m(Exercise 6.23). Then mrn ⊂ Qn ⊂ mn for any n ∈ N. Hence HM,m(rn) ≥HM,Q(n) ≥ HM,m(n) for all n ∈ N. This proves that deg HM,Q = deg HM,m, asdesired. QED

Notation. The common degree of the polynomials HM,Q will be denoted by d(M).We now prove Lemma 6.29 in the following stronger form.

Lemma 8.2. Let (A,m), M be as above. Let x ∈ m. Assume that x is not a zerodivisor for M , that is, xa 6= 0 for any a ∈M \ {0}. Then d

(MxM

)≤ d(M)− 1.

Proof . Let N = xM , M ′ = MN . Then N ∼= M since x is not a zero divisor. Let

Nn = mnM ∩N . We have an exact sequence

0→ N →M →M ′ → 0,

which induces an exact sequence

(8.2) 0→ N

Nn→ M

mnM→ M ′

mnM ′→ 0.

for each n ∈ N. By additivity of length, (8.2) gives

(8.3) HN,m(n)−HM,m(n) +HM ′,m(n) = 0.

By the Artin–Rees Lemma , {Nn} is stable. By Proposition 8.1, HN,m and HM,m

have the same degree and highest coefficient, hence, by (8.3),

deg HM ′,m < deg HM,m.

This proves that d(M ′) ≤ d(M)− 1. �

This completes the proof of the Dimension Theorem 6.17.

Exercise 8.3. Let G =∞⊕n=0

Gn be a Noetherian graded G0-algebra, with G0 Ar-

tinian. Let M =∞⊕n=0

Mn be a finite graded G-module and let d(M) denote the

degree of the Hilbert polynomial

H(n) =n∑

i=0

length(Mi)

for n ≫ 0. Let x ∈ Gs be a homogeneous element of degree s, which is not a zerodivisor in M . Show that d( M

(x)M ) ≤ d(M)− 1.

Now we can relax and derive several important corollaries from Theorem 6.17.


Corollary 8.4. Let (A,m, k) be a Noetherian local ring. Then dim A ≤ dimkmm2 .

Proof. Let x1, . . . ,xs be elements of m whose images in mm2 generate m

m2 . Then

(8.4) m = (x1, . . . ,xs) +m2.

We now invoke one of the most famous lemmas in all of commutative algebra—theLemma of Nakayama—to conclude that (8.4) implies m = (x1, . . . ,xs). This willshow that dim A = δ(A) ≤ s, as desired.

It remains to state and prove Nakayama’s Lemma.

Lemma 8.5 (Nakayama’s Lemma). Let A be a ring, I an ideal of A, M an A-module, N ⊂M a submodule. Assume either that M is finite or that I is nilpotent(that is, In = (0) for some n ∈ N). Suppose that M = IM +N . Then there existsx ∈ I such that (1 + x)M = N . If, in addition, I ⊂ J(A), then M = N .

Applied to (8.4) with I = M = m, (x1, . . . ,xs) = N , Nakayama’s Lemma impliesthat m = (x1, . . . ,xs).

Proof of Nakayama’s Lemma. If I is nilpotent, we have In = (0) for somen ∈ N. Substituting IM for M recursively in the expressions IiM +N , we obtainM = IM +N = I2M +N = · · · = InM +N = N , and the Lemma is true in thiscase.

Next, suppose M is finite over A. Let w1, . . . ,wn be the generators of M . Replac-ingM by M

N, we may assume thatN = (0). ThenM = IM , hence w1, . . . ,wn ∈ IM .

Then there exist aij ∈ I, 1 ≤ i, j ≤ n, such that

(8.5)

w1 = a11w1 + a12w2 + · · ·+ a1nwn

w2 = a21w1 + a22w2 + · · ·+ a2nwn

...

wn = an1w1 + an2w2 + · · ·+ annwn

Let

T =

a11 a12 . . . a1n

a21 a22 . . . a1n...

an1 an2 . . . ann

.

By (8.5),

(8.6)

w1 − a11w1 − . . . − a1nwnw2 − a21w1 − . . . − a2nwn

...wn − an1w1 − . . . − annwn

=

00...0

.


Let 1n denote the n×n identity matrix. (8.6) says (1n−T )

w1...wn

=

0...0

. Since

det(1n − T )1n = adj(1n − T )(1n − T ), we have det(1n − T )wi = 0, i = 1, 2, . . . ,n.Since all the aij ∈ I, det(1n − T ) = 1 + x for some x ∈ I. We have

(8.7) (1 + x)wi = 0, 1 ≤ i ≤ n.

Since M =∑ni=1Awi, (8.7) implies (1+x)M = (0). If, in addition, x ∈ J(A), then

1 + x is not in any maximal ideal of A, hence 1 + x is invertible. Then M = (0), asdesired. �

dimkmm2 is called the embedding dimension, m

m2 is called the Zariski cotan-

gent space of A. Its dual Homk

(mm2 , k

)is called the Zariski tangent space.

We continue with Corollaries of Theorem 6.17.Corollary 8.6. Let A be a Noetherian ring and let x1, . . . ,xr be elements of A,such that (x1, . . . ,xr) 6= A (in other words, there exists m ∈ Max(A), containingall the xi). Then for any minimal prime P of (x1, . . . ,xr), we have ht P ≤ r (inparticular, ht(x1, . . . ,xr) ≤ r).

Proof. Let P be any minimal prime of (x1, . . . ,xr).

Exercise 8.7. Let I be an ideal in a Noetherian ring A, P a minimal prime of I.Show that

√IAP = PAP .

By Exercise 8.7,√

(x1, . . . ,xr)AP = PAP , hence ht P ≡ dim AP = δ(AP ) ≤ r.Theorem 8.8 (Krull’s Principal Ideal Theorem). Let A be a Noetherian ring,x an element of A, which is neither a unit nor a zero divisor. Then every minimalprime ideal of (x) has height 1.

Proof. Let P be a minimal prime of (x). By Corollary 8.6, ht P ≤ 1. It remainsto show that ht P 6= 0. Indeed, suppose ht P = 0. Then P is a minimal prime ofA. By Theorem 5.14, every element of P is a zero divisor, which contradicts thefact that x ∈ P .

Exercise 8.9. Let A be a ring, S ⊂ A a multiplicative set and P a prime idealsuch that P ∩ S = ∅. Show that ht P = ht(PAS).

Exercise 8.10. For any n ∈ N, give an example of a Noetherian local ring A andx ∈ A which is not a zero divisor, such that (x) has an associated prime of heightn.

Let (A,m) be a Noetherian local ring, x an element of m which is not a zerodivisor. From Lemma 6.29 and the Dimension Theorem we know that

(8.8) dimA

(x)≤ dim A− 1.

The next Corollary says that the inequality is, in fact, an equality.


Corollary 8.11. In this situation, we have dim A(x) = dim A− 1.

Proof. Let d′ = dim A(x) = δ

(A(x)

). Let x1, . . . ,xd′ be elements of A whose

images in A(x)

generate an m(x)

-primary ideal. Then x1, . . . ,xd′ , x generate an m-

primary ideal in A. Then dim A = δ(A) ≤ d′ + 1 = dim A(x) + 1, which proves the

opposite inequality of (8.8). Hence the Corollary is proved.

§9. Dimension and transcendence degree in

finitely generated algebras over a field.

Notation. If P ∈ Spec A for a ring A, let κ(P ) = APPAP

denote the residue field ofthe local ring AP .

Let k be a field and A a finitely generated k-algebra. If x1, . . . , xn are thegenerators of A over k, A can be written as the quotient of the polynomial ring

k[x1, . . . , xn] by an ideal I ⊂ k[x1, . . . , xn]; that is, A = k[x1,...,xn]I

.Theorem 9.1.

(1) dim A = tr. deg(A/k).(2) If P ∈ Spec A and K is the field of rational functions of some irreducible

component of Spec A containing P (in other words, K = κ(P0) for some min-imal prime P0 of A, contained in P ), we have dim AP

P0AP+tr. deg(κ(P )/k) =

tr. deg(K/k).

Proof. By the following exercise, we may assume that A is a domain in (1):

Exercise 9.2. Let A be a finitely generated k-algebra and P1, . . . ,Pr the minimalprimes of A. Show that

(1) dim A = max1≤i≤r

{dim APi}

(2) tr. deg(A/k) = max1≤i≤r

{tr. deg( APi/k)}.

In (2) of Theorem 9.1, we may replace A by AP0

. From now on, we will assumethat A is a domain.

Let d = dim A and let (0) $ P1 $ · · · $ Pd be a maximal chain of primeideals. Then dim APd = d. Moreover, Pd is maximal, hence by Proposition 4.2,tr. deg( A

Pd/k) = 0. Since tr. deg(A/k) is obviously unaffected by localization (here

we are using the fact that A is a domain), we have (2) =⇒ (1) in Theorem 9.1.

It remains to prove (2). Let P ∈ Spec A. Let t = tr. deg(APPAP

/k). Renumber-

ing the xi, we may assume that the images of x1, . . . ,xt in AP form a transcen-

dence base of APPAP

over k. Let L = k(x1, . . . ,xt); we have the obvious inclusion

L → AP (this is where we are using the fact that A is a domain: the naturalmap A → AP is injective). We have a natural map φ : L[xt+1, . . . ,xn] → AP ;by definition, AP = L[xt+1, . . . ,xn]φ−1(P ). Let K be the field of fractions of A.Since transcendence degree is additive in field extensions, we have tr. deg(A/k) =


tr. deg(AP /k) = tr. deg(K/k) = t + tr. deg(K/L). Therefore we may replace k byL and A by L[xt+1, . . . ,xn]. In other words, we may assume that P is maximal andt = 0. It remains to prove that under these assumptions, dim AP = tr. deg(A/k).Lemma 9.3. Let (A,m, k) be a Noetherian local ring of dimension d. Let x1, . . . ,xdbe a system of parameters of A and let Q = (x1, . . . ,xd). Let f ∈ A[t1, . . . ,td] be ahomogeneous polynomial of degree s in d variables t1, . . . ,td. If f(x1, . . . ,xd) ∈ Qs+1,then f ∈ mA[t1, . . . ,td].

Proof . Let f denote the natural image of f in AQ

[t1, . . . ,td]. Consider the map

(9.1) α :A

Q[t1, . . . ,td]→

∞⊕

i=0

Qi

Qi+1,

defined by α(ti) = xi, 1 ≤ i ≤ d. By Proposition 6.28, d(⊕ Qi

Qi+1

)= dim A = d.

Also length (t1,...,td)i

(t1,...,td)i+1 = length(AQ

)·(d+i−1d−1

), so that d

(AQ [t1, . . . ,td]

)= d.

By assumption, f ∈ ker α. Since both algebras in (9.1) have the same degree ofHilbert polynomial, by Exercise 8.3 f is a zero divisor in A

Q[t1, . . . ,tn]. We leave it

as an Exercise to finish the proof:

Exercise 9.4. Show that (0) is a primary ideal in AQ

[t1, . . . ,td] and that

mA

Q[t1, . . . ,td] = nil

(A

Q[t1, . . . ,td]

)

is the only associated prime of AQ [t1, . . . ,td]. In particular, every zero divisor in

AQ

[t1, . . . ,td] belongs to mAQ

[t1, . . . ,td].

Q.E.D.Corollary 9.5. Let (A,m) be a Noetherian local ring of dimension d, x1, . . . ,xda system of parameters of A. Suppose A contains a field k0. Then x1, . . . ,xd arealgebraically independent over k0.

Proof . Let f(x1, . . . ,xd) = 0 be an algebraic relation among the xi over k0.Write f = fs + fs+1 + · · · + ft where each fi is homogeneous of degree i. Thenfs(x1, . . . ,xd) ∈ Qs+1, hence by Lemma 9.3 f ∈ mA[t1, . . . ,td], which implies f ≡ 0since f is a polynomial with coefficients in k0.

Corollary 9.5 proves that dim AP ≤ tr. deg(A/k). For the opposite inequality,we need the Noether Normalization Lemma.Lemma 9.6 (Noether Normalization Lemma). Let k be a field and A =k[x1,...,xn]

I a finitely generated k-algebra. Then there exist y1, y2, . . . ,yd ∈ A, alge-braically independent over k, such that A is integral over k[y1, . . . ,yd].

Proof. We may assume that x1, . . . ,xd are algebraically independent over k andxd+1, . . . ,xn are algebraic over k[x1, . . . ,xd]. We will prove, by induction on i,


that there exists a choice of generators xj such that xd+1, . . . ,xi are integral overk[x1, . . . ,xd]. Suppose this is known for i, d ≤ i < n. We must prove the samestatement for i+ 1. We will make a change of variables of the form

(9.2) xj = xj + xλji+1,

1 ≤ j ≤ d. By assumptions, there is an algebraic dependence relation

(9.3) F (xi+1) = a0xli+1 + a1x

l−1i+1 + · · ·+ al = 0,

aj ∈ k[x1, . . . ,xd]. It is enough to find λj ∈ N, 1 ≤ j ≤ d, such that after thesubstitution (9.2), (9.3) becomes an integral equation for xi+1 over k[x1, . . . ,xd].Then xi+1 will be integral over k[x1, . . . ,xd] and xd+1, . . . , xi over k[x1, . . . ,xd, xi+1],hence also over k[x1, . . . ,xd].

Let x = (x1, . . . ,xd), xα = xα1

1 . . . xαdd . Write

(9.4) F (xi+1) =∑

αj

aαxαxji+1.

Choose λ1 ≫ λ2 ≫ · · · ≫ λd ≫ 2 such that∑ds=1 αsλs + j 6= ∑d

s=1 α′sλs + j′ for

any two monomials xαxji+1 6= xα′

xj′

i+1 appearing in (9.4) with non-zero coefficients(the existence of such λj is easy to see by induction on d).

Substituting (9.2) into (9.3), the coefficient of the highest order term in xi+1 is anon-zero element of k by the choice of λj . Hence after the substitution (9.2), (9.3)becomes an integral dependence relation of xi+1 over k[x1, . . . ,xd], as desired.

By induction in i, we find a choice of generators xi such that k[x1,...,xn]I

is integralover k[x1, . . . ,xd]. QED

Remark 9.7. If k is infinite, we can achieve the same result by a transformationof the form xj = xj + cjxi+1, 1 ≤ j ≤ d, where cj are sufficiently general constantsin k.

It is also worth mentioning that the analogue of Noether normalization for formalor convergent power series ring is stronger. Namely, we have

Theorem 9.8. Let (A,m, k) be a formal (resp. convergent) power series ring overk (resp. C). Let y1, . . . ,yd be a system of parameters of A. Then A is finitelygenerated as a module over k[[y1, . . . ,yd]] (resp. C{y1, . . . ,yd}) (it is easy to seethat there are no formal or convergent relations among the yi, that is, no formal orconvergent power series in the yi is equal to 0 in A).

Thus in the formal and convergent cases it is much easier to choose y1, . . . ,yd:any system of parameters will do.

Theorem 9.8 can be proved rather easily using the Weierstrass preparation the-orem, but we will not do it here (see [4, Chapter V, (31.6) and Chapter VII, (45.3)and (45.5)]).


Let A, y1, . . . ,yd be as in Noether Normalization Lemma. Clearly,

d = tr. deg(A/k).

To prove Theorem 9.1, it remains to prove that dim Am = d for everym ∈Max(A).It is a little easier to prove that dim Am = d for some m ∈Max(A); this is alreadyenough for Theorem 9.1 (1). We prove this weaker result first in order to warm up.

The plan is to show that dim k[y1, . . . ,yd] = d and then analyze the behaviourof dimension under integral extensions and show that it is preserved by such exten-sions.

Lemma 9.9. Let m ∈Max(k[y1, . . . ,yd]). Then ht m = d. In particular,

dim k[y1, . . . ,yd] = d.

Proof. Take any m ∈Max(k[y1, . . . ,yd]) and let k′ = κ(m) = k[y1,...,yd]m

. By Propo-sition 4.2, k′ is a finite algebraic extension of k. Let yi = yi mod m be the naturalimage of yi in k′; then k′ is generated by y1, . . . ,yd as a k-algebra. For 1 ≤ i ≤ d,let fi denote the minimal polynomial of yi over the field k[y1, . . . ,yi−1]. For each i,pick and fix a polynomial fi ∈ k[y1, . . . ,yi] whose natural image in k[y1, . . . ,yi−1, yi]

is fi. Then, by induction on i, k′ = k[y1,...,yd](f1,...,fd)

. Hence m = (f1, . . . ,fd); in par-

ticular, dim Am = δ(m) ≤ d. On the other hand, let Pi ⊂ k[y1, . . . ,yi] denotethe kernel of the natural map k[y1, . . . ,yi] → k[y1, . . . ,yi]. It is easy to see that(0) $ P1k[y1, . . . ,yd] $ · · · $ Pd = m is a chain of d distinct prime ideals, so thatdim Am = d. �

To complete the proof of Theorem 9.1, it remains to analyze the behaviour ofdimension in integral ring extensions.Proposition 9.10. Let A ⊂ B be an integral ring extension. Then dim A =dim B.Lemma 9.11. Let A ⊂ B be an integral ring extension and P ∈ Spec A, Q ∈Spec B such that Q ∩ A = P . Then P is maximal if and only if Q is maximal.

Proof. AP→ B

Qis an integral extension of integral domains. Thus we can further

reduce Lemma 9.11 toLemma 9.12. Let A ⊂ B be an integral extension of integral domains. Then A isa field if and only if B is a field.

Proof. =⇒ is proved by the same argument as =⇒ in Proposition 4.2.(⇐=) Assume that B is a field. Let x ∈ A. Then 1

x ∈ B. Since B is a domain,

integral over A, 1x satisfies an integral dependence relation x−m + b1x

−m+1 + · · ·+bm = 0, where bi ∈ A. Then 1

x = −b1 − b2x− · · ·− bmxm−1 ∈ A, which proves thatA is a field. QEDCorollary 9.13. Let A ⊂ B be an integral ring extension, P ∈ Spec A. Then

(1) There exists Q ∈ Spec B such that Q ∩A = P(2) There are no inclusion relations between primes Q ⊂ B, satisfying (1).


Proof.(1) Localize both A and B by the multiplicative set S = A \ P . ThenAP → BS , is an integral ring extension. Let Q be any maximal ideal of BS. ThenQ ∩AP must be maximal by Lemma 9.11, hence Q ∩AP = PAP .

Then Q∩B lies over P : (Q∩B)∩A = Q∩A = (Q∩AP )∩A = PAP ∩A = P ./par

2) Suppose Q1 ⊂ Q2 are two primes of B such that Q1 ∩A = Q2 ∩A = P . ThenQ1BS and Q2BS are both maximal in BS which contradicts the fact that Q1 $ Q2.

Corollary 9.14 (The going up theorem). Let A ⊂ B be an integral extensionof integral domains. Let P ⊂ P ′ ⊂ A be prime ideals. Let Q ⊂ B be a prime idealsuch that Q ∩ A = P . Then there exists a prime Q′ of B, such that Q ⊆ Q′ andP ′ = Q′ ∩ A.

Proof. Apply Corollary 9.13 (1) to the extension AP → B

Q and the ideal P ′

P .

There exists Q′′ ∈ SpecBQ

such that Q′′ ∩ AP

= P ′

P. Put Q′ = Q′′ ∩B.

Proof of Proposition 9.10. Let A ⊂ B be an integral ring extension. LetQ0 $ · · · $ Qn be a chain of prime ideals of B. By Corollary 9.13 (2), Q0∩A $ · · · $Qn ∩ A is a chain of distinct prime ideals of A. This proves that dim B ≤ dim A.Conversely, let P0 $ · · · $ Pn be a chain of prime ideals of A. By Corollary 9.13(1), there exists Q0 ∈ Spec B lying over P0. By the going up theorem and inductionon i, we can successively construct a chain of prime ideals Q0 $ · · · $ Qn of B suchthat Qi ∩A = Pi. This proves that dim A ≤ dim B, so that dim B = dim A. �

We come back to the proof of Theorem 9.1. We have an integral domain A of finitetype over k. By Noether Normalization Lemma, there exist y1, . . . ,yd, algebraicallyindependent over k, such that A is finite over k[y1, . . . , yd]. By Lemma 9.9 andProposition 9.10, dim A = dim k[y1, . . . ,yd] = d. In other words, the largestheight of a maximal ideal in A is d. This proves (1) of Theorem 9.1. To prove(2) of Theorem 9.1, we need a stronger result: every maximal ideal in A has heightexactly d. The existence of a maximal ideal of height d was proved by starting with amaximal chain of primes in k[y1, . . . ,yd] and constructing a chain of primes inA lyingabove it recursively, using the going up theorem at each step. Now, we would like tostart with a fixed maximal idealm of A. Since m∩k[y1, . . . ,yd] is maximal by Lemma9.11, we can consider a chain of d primes (0) $ P1 $ · · · $ Pd = m∩ k[y1, . . . ,yd] ink[y1, . . . ,yd]. To imitate the procedure in the proof of Proposition 9.10 and constructa chain of primes in A lying over the Pi, we now need a going down theorem. Thefact that the going down theorem holds in our context is a somewhat more delicateresult; it depends not only on the fact that A is integral over k[y1, . . . ,yd] but alsoon the fact that k[y1, . . . ,yd] is normal. Thus to complete the proof of Theorem 9.1,it remains to prove two things:

Proposition 9.15 (The going down theorem). Let φ : A → B be an integralextension of integral domains, such that A is normal. Let P, P ′ ∈ Spec A, Q′ ∈Spec B be such that P ⊂ P ′ and P ′ = Q′ ∩A. Then there exists Q ∈ Spec B suchthat Q ⊂ Q′ and P = Q ∩A.


Lemma 9.16. k[y1, . . . ,yd] is normal.

Once the going down theorem and the normality of k[y1, . . . ,yd] are established,Theorem 9.2 (2) will follow from the following result, proved by imitating the proofof Proposition 9.10.

Exercise 9.17. Let A ⊂ B be a ring extension. Let Q ∈ Spec A, P = Q∩A. Showthat:

(1) If A ⊂ B has the going down property, then ht Q ≥ ht P .(2) If for any prime P ′ ⊂ P , there are no inclusion relations among the primes

of B contracting to P ′, then ht Q ≤ ht P .

Proof of the going down theorem. Let K and L denote, respectively, the fields offractions of A and B. Replacing B by its integral closure in L and Q′ by a primeideal of the integral closure lying overQ′ does not change the problem. Furthermore,we may replace L by a normal extension of K and B by the integral closure of Ain L. Thus we will assume that L is a normal extension of K and B is integrallyclosed in L (in particular, any automorphism of L over K maps B to itself).

By Corollary 9.13 (1), there exists Q ∈ Spec B such that Q ∩ A = P . By the

going up theorem there exists a prime Q′ ⊃ Q in B such that Q′ ∩ A = P ′. Henceto prove the going down theorem, it is sufficient to prove the following lemma.

Lemma 9.18. Let A, B, K and L be as in Proposition 9.15. Assume that L isa normal extension of K and that B is integrally closed in L. Let P ∈ Spec A,Q1, Q2 ∈ Spec B be such that Q1 ∩ A = Q1 ∩ A = P . Then there exists anautomorphism σ of L over K such that σ(Q1) = Q2.

By Lemma 9.18 there exists an automorphism σ such that σ(Q′) = Q′. Then

Q = σ(Q) is the desired prime ideal. It remains to prove Lemma 9.18 (note thatfor Theorem 9.1 we only need it in the case when L is finite over K).

Proof of Lemma 9.18. First, assume that L is finite over K, so that the group G ofautomorphisms is finite: G = {σ1, . . . ,σs}. Suppose Q2 6= σi(Q1) for any i. Since

Q1 6= σi(Q2) for any i, Q1 6⊂s⋃i=1

σi(Q2) by Lemma 6.33. Take x ∈ Q1 \s⋃i=1

σi(Q2).

Then σi(x) /∈ Q2 for any i, hence

(9.5) y =

s∏

i=1

σi(x) /∈ Q2.

But y is invariant under all the automorphisms of L overK, hence there is r ∈ N suchthat yr ∈ K. Since y is integral over A, we have yr ∈ A. Then y ∈ A∩Q1 = P ⊂ Q2,which contradicts (9.5). This proves Lemma 9.18 in case L is finite over K.

Next, suppose L is not finite over K. Let K ′ denote the maximal purely insep-arable extension of K contained in L (K ′ is nothing but the invariant field of thegroup G of automorphisms). Then L is Galois over K ′. Let A′ denote the integralclosure of A in K ′.


Exercise 9.19. Show that the natural map Spec A′ → Spec A is a bijection.

By Exercise 9.19, we may replace A by A′ and K by K ′; that is, we may assumethat L is Galois over K. Let L′ be a finite Galois extension of K contained in L,and let

F (L′) = {σ ∈ G | σ(Q1 ∩ L′) = Q2 ∩ L′}.

Since K ⊂ L′ is a finite field extension, Lemma 9.18 holds with B replaced by theintegral closure of A in L′. Hence F (L′) is not empty.

The Krull topology on G is defined to be the topology whose basis consists ofall the cosets of the form σH, where σ ∈ G and H is a normal subgroup of G offinite index.

Exercise 9.20. Show that F (L′) is closed in the Krull topology.

Now, the set of finite groups which are homomorphic images of G form a pro-jective system, whose maps are the natural surjective homomorphisms: wheneverHi ⊂ Hj are two normal subgroups of finite index, we have a natural surjective mapGHi→ G

Hj. Since an automorphism σ ∈ G is completely determined by the collection

of all of its restrictions to finite subextensions K ⊂ L′ of K contained in L, we haveG = lim

←

GH

. Moreover, the Krull topology on G is, by definition, nothing but the

inverse limit of the discrete topologies on GH (that is, it is the topology whose basis

is formed by all the inverse limits of all the sets in GH or, equivalently, the coarsest

topology which makes all the natural maps G → GH of Exercise 3.12 continuous).

Since we are considering only normal subgroups of finite index, GH

is finite, so that

the discrete topology on GH is compact and Hausdorff.

Now, the inverse limit of compact Hausdorff spaces Xi is compact. Indeed, bydefinition lim

←Xi ⊂

∏iXi.

∏iXi is compact by Tychonoff’s theorem and one checks

easily that lim←Xi is closed in

∏i

Xi using the Hausdorff property of the Xi.

Hence G is compact in the Krull topology. Now, any finite collection of finiteGalois subextensions of L is contained in another finite Galois subextension. Henceany finite collection the closed sets F (L′) has a non-empty intersection. Since Gis compact, this implies that

⋂L′F (L′) 6= ∅. Any element σ ∈ ⋂

L′F (L′) satisfies the

conclusion of Lemma 9.18. �

�

This completes the proof of the going down theorem. To finish proving Theorem9.1, it remains to prove the normality of k[y1, . . . ,yd] (Lemma 9.15). By Exercise4.8, it is sufficient to prove that k[y1, . . . ,yd] is a UFD. This follows by induction ond from the Gauss lemma:

Exercise 9.21 (the Gauss lemma). Let A be a UFD and y an independentvariable. Then A[y] is a UFD. (Hint: let K be the field of fractions of A. Since K[y]


is a UFD, the question is reduced to the following: given polynomials f, g, h ∈ A[y]and an irreducible element a ∈ A such that a | fg, either a | f or a | g.)

This completes the proof of Theorem 9.1. �

We end this section by giving an alternative proof of Lemma 9.9 using 9.15–9.17.

A second proof of Lemma 9.9. . Take any m ∈ Max(k[y1, . . . ,yd]). We want to

prove that ht m = d. Let k′ = k[y1,...,yd]m . Then k′ is a finite extension of k, so

that k[y1, . . . ,yd] ⊂ k′[y1, . . . ,yd] is a finite ring extension. Let a1, . . . ,ad denote thenatural images of y1, . . . ,yd in k′. Then (y1 − a1, . . . ,yd − ad) is a maximal ideal ofk′[y1, . . . ,yd] of height d by Example 6.10. Since (y1−a1, . . . ,yd−ad)∩k[y1, . . . ,yd] =m, we have ht m = d by Exercise 9.17. �

§10. Regular Rings.

Regular rings are those rings which correspond to complex manifolds in classicalalgebraic geometry. In other words, they correspond to algebraic varieties whichare locally diffeomorphic to Cn at every point. Yet another way of saying this isthat regular varieties do not have self-intersections, corners or cusps (these latterkinds of points are called singularities). Algebraically, we express this intuitiveidea by saying that locally at every point there is a local coordinate system with dcoordinates, where d is equal to the dimension. In other words, a point ξ on a d-dimensional variety or scheme is regular if there exist d equations locally at ξ whosecommon set of zeroes is ξ with its reduced structure. Now for precise definitions.

Let (A,m, k) be a Noetherian local ring of dimension d.Definition 10.1. A is regular if m can be generated by exactly d elements

(⇐⇒

dimkmm2 = d⇐⇒

∞⊕n=0

mn

mn+1∼= k[t1, . . . ,td]

).

Definition 10.2. If A is regular of dimension d, a set {x1, . . . ,xd} of d generatorsof m is called a regular system of parameters for A (this is equivalent to sayingthat x1, . . . ,xd are linearly independent mod m2).Definition 10.3. Let A be any Noetherian ring. We say that A is regular if Amis regular for all m ∈Max(A).

Properties of regular local rings:

1) A regular local ring is a domain. This follows from the graded algebra de-scription appearing in Definition 10.1 and the following two exercises:

Exercise 10.4. Let A be a ring, I ⊂ A an ideal such that

(10.1)

∞⋂

n=1

In = 0.

If grIA is a domain then A is a domain.


Exercise 10.5. Let A be a Noetherian ring, m $ A an ideal. Assume either that

m ⊂ J(A) or that A is a domain. Then∞⋂n=1

mn = (0).

Problem 10.6. Prove that a regular local ring is normal. (Hint: prove that ifgrIA is normal, then A is normal, provided thatA is Noetherian and I ⊂ J(A) (inparticular, I satisfies (10.1)).)

More generally, it is known that a regular local ring is a UFD (this is a non-trivialtheorem of Auslander and Buchsbaum).

Example 10.7. Let k be a field. Then the polynomial ring k[x1, . . . ,xd] is regular.

Indeed, let m ⊂ k[x1, . . . ,xd] be a maximal ideal. In the first proof of Lemma 9.9,we explicitly constructed a set of d generators for m. More generally, it is easy tosee that if A is a regular ring, then so are A[x1, . . . , xn] and A

[[x1, . . . , xn]

], where

x1, . . . , xn are independent variables.

Proposition 10.8. Let (A,m, k) be a regular local ring of dimension d. Let l ≤ d.Let x1, . . . ,xl be elements of m, none of which belongs to the ideal generated by theothers. The following are equivalent:

(1) x1, . . . ,xl are linearly independent mod m2.(2) x1, . . . ,xl can be extended to a regular system of parameters of A.(3) A

(x1,...,xl)is a regular local ring.

If (1)–(3) hold, we have ht(x1, . . . ,xl) = l.

Proof. 1) ⇐⇒ 2). By Nakayama’s Lemma, x1, . . . ,xd is a regular system ofparameters if and only if xi mod m2 form a basis for m

m2 . Hence 2) follows fromthe fact that a collection of linearly independent elements in a vector space can beextended to a basis.

2) =⇒ 3). Extend x1, . . . ,xl to a system of parameters x1, . . . ,xd. Let A′ =A

(x1,...,xl), m′ = m

(x1,...,xl). Then dim A′ ≥ d − l, for if an m′-primary ideal was

generated by fewer than d− l elements, together with x1, . . . ,xl we would obtain aset of fewer than d generators for an m-primary ideal in A. On the other hand, m′

is generated by xl+1, . . . ,xd. Hence dim A′ = d− l and A′ is regular.

3) =⇒ 1). Let x1, . . . ,xr be the largest subset of x1 . . . xl which are linearlyindependent mod m2. Suppose r < l. Then by (1)=⇒(3), A

(x1,...,xr)is a regular

local ring of dimension d − r, in particular, a domain. Then xr+1 is not a zerodivisor in A

(x1,...,xr), hence by Lemma 6.29 dim A

(x1,...,xl)< dim A

(x1,...,xr)= d−r. On

the other hand, let m′ be as above. Then dimkm′

m′2= d− r by definition of r. This

contradicts the regularity of A(x1,...,xr)

.

Finally, suppose (1)–(3) hold. Then ht(x1, . . . ,xl) ≤ l by Corollary 8.6. By(3), A

(x1,...,xi)is a domain for all i ≤ d; in other words, (x1, . . . ,xi) is a prime

ideal for i ≤ d. Then 0 $ (x1) $ (x1, x2) $ · · · $ (x1, . . . ,xl). This proves thatht(x1, . . . ,xl) = l. �


Definition 10.9. Let k be a field. k is perfect if either char k = 0 or char k = p > 0and kp = k.

Definition 10.10. Let σ : k → k′ be a field extension. We say that k′ is sep-arably generated over k if σ can be decomposed into a purely transcendentalextension and a separable algebraic extension. If k(x1, . . . ,xt) is a purely transcen-dental subextension of k′ over which k′ is separable algebraic, x1, . . . ,xt is called aseparating transcendence base.

Exercise 10.11.

(1) Any finite field is perfect.(2) If k is perfect, then any algebraic extension of k is separable.(3) More generally, if k is perfect, any finitely generated field extension of k is

separably generated.

Exercise 10.12. Let k → k′ be a finitely generated, separably generated exten-sion of transcendence degree t. Let x1, . . . , xn be a set of generators of k′ over k(as a field). Show that there exists a subset of x1, . . . , xn which is a separatingtranscendence base.

Theorem 10.13. Let k be a field. Let A = k[x1,...,xn](f1,...,fl)

, P ∈ Spec A. Let Q be the

inverse image of P in k[x1, . . . ,xn]. Then:

(1) ht(f1, . . . ,fl)k[x1, . . . , xn]Q ≥ rk(∂fi∂xj

)mod P .

(2) If equality holds in (1), then AP is regular.(3) If κ(P ) is separably generated over k and AP is regular, then equality holds

in (1).

Note that by Exercise 10.11 κ(P ) is always separably generated over k in the casewhen k is perfect.

Proof. First, assume P is maximal. Let k′ = AP . Then k → k′ is a finite field

extension and k[x1, . . . , xn] → k′[x1, . . . , xn] a finite ring extension. Let ai be thenatural image of xi in k′ and consider the ideal Q′ = (x1 − a1, . . . ,xn − an) ⊂k′[x1, . . . , xn]. Then Q′ ∩ k[x1, . . . , xn] = Q. The homomorphism k[x1, . . . , xn] →k′[x1, . . . , xn] induces a map ϕ : Q

Q2 → Q′

Q′2between two k′-vector spaces of dimen-

sion n.

Lemma 10.14. ϕ is an isomorphism if and only if k′ is separable over k.

Proof. Let gi denote the minimal polynomial of ai over k(a1, . . . , ai−1) ⊂ k′, 1 ≤i ≤ n and let gi an arbitrary inverse image of gi in k[x1, . . . ,xi]. Then, as in the

proof of Lemma 9.9, g1(x1), . . . ,gn(x1, . . . ,xn) generate Q. Therefore φ( QQ2 ) is the

vector space generated by the linear forms of g1, . . . ,gn. Hence φ is an isomorphism

if and only if det∣∣∣ ∂gi∂xj

∣∣∣ /∈ Q. Now, the extension k ⊂ k′ is separable ⇐⇒ each of the

gi is separable ⇐⇒ ∂gi∂xi

/∈ Q for all i⇐⇒ det∣∣∣ ∂gi∂xj

∣∣∣ =n∏i=1

∂gi∂xi

/∈ Q, as desired. �


Now, let r = rk

(∂fi∂xj

∣∣∣Q

). Say, f1, . . . ,fr are such that rk

(∂fi∂xj

∣∣∣Q

)

1≤i,j≤r

= r.

This means that the natural images of f1, . . . ,fr in Q′

Q′2are linearly independent,

hence they are linearly independent in QQ2 ⇒ k[x1,...,xn]Q

(f1,...,fr)is a regular local ring by

Proposition 10.8 (in particular, it is a domain).Then r = ht(f1, . . . ,fr)k[x1, . . . , xn]Q ≤ ht(f1, . . . ,fl), which proves (1). Now,

suppose r = ht(f1, . . . ,fl)k[x1, . . . , xn]Q. Since (f1, . . . ,fr)k[x1, . . . , xn]Q is prime,this equality of heights implies that

(f1, . . . ,fr)k[x1, . . . , xn]Q = (f1, . . . ,fl)k[x1, . . . , xn]Q,

which shows that AP =k[x1,...,xn]Q

(f1,...,fl)is regular.

Next, assume that k′ is separable over k and AP is regular. Then ϕ is anisomorphism. Since AP is regular, (f1, . . . ,fl) = (f1, . . . ,fr) by Proposition 10.8.

rk

(∂fi∂xj

∣∣∣Q

)

i≤i≤r

is equal to the dimension of the k′-vector subspace of Q′

Q′2, spanned

by the images of f1, . . . ,fr. Since φ is an isomorphism and since f1, . . . ,fr are linearly

independent mod Q2, we have rk

(∂fi∂xj

∣∣∣Q

)

i≤i≤r

= r, as desired.

Finally, suppose P is not maximal. Let r = rk(∂fi∂xj

))

mod P . Say,

rk

(∂fi∂xj

∣∣∣∣Q

)

1≤i,j≤r

= r.

Let xi denote the natural image of xi in κ(P ). Renumbering the xi, we mayassume that xr+1, . . . ,xr+t are algebraically independent over k and xr+t+1, . . . ,xnare algebraic over k(xr+1, . . . ,xr+t). Let

k′ =k[xr+1, . . . ,xn]P∩k[xr+1,...,xn]

P ∩ k[xr+1, . . . ,xn]= k(xr+1, . . . ,xn),

k′′ = k(xr+1, . . . ,xr+t). By the choice of t, k′ is algebraic over k′′ and k′′ ⊂ AP .Since k′′[x1, . . . ,xr, xt+r+1, . . . ,xn] is obtained from k[x1, . . . , xn] by localizing withrespect to the multiplicative system k[xr+1, . . . ,xr+t], disjoint from P , replacingk by k′′ in the Theorem does not change the problem. In other words, we mayassume that t = 0 and k′ is algebraic over k. Then k[x1, . . . , xn]→ k′[x1, . . . , xn] isa finite ring extension. Let P0 = (xr+1 − xr+1, . . . ,xn − xn) ∩ k[x1, . . . , xn]. By thegoing up theorem, there exists an ideal Q′ ⊂ k′[x1, . . . , xn] lying over Q containing(xr+1 − xr+1, . . . ,xn − xn). Since height of ideals is preserved by finite extensions,none of the statements in Theorem 10.13 are affected by replacing k by k′, A byk′[x1,...,xn](f1,...,fl)

, Q by Q′, etc. For (3) in Theorem 10.13, note that if κ(P ) is separably


generated over k then by Exercise 10.12 we may choose k′ so that κ(P ) is separablygenerated over k′.

To complete the proof of Theorem 10.13, it is sufficient to show that, under theassumption t = 0, Q is a maximal ideal of k[x1, . . . , xn], since the theorem hasalready been proved in this case. By Proposition 4.2, this is equivalent to provingthat κ(P ) is algebraic over k. The assumption t = 0 means that xr+1, . . . ,xnare algebraic over k. Thus it remains to prove that x1, . . . ,xr are algebraic over

k(xr+1, . . . ,xn). Since k′ is a quotient of k(xr+1,...,xn)(f1,...,fl)k(xr+1,...,xn) , everything is reduced

to the following lemma.

Lemma 10.15. Consider elements f1, . . . ,fr in k[x1, . . . ,xr] and a prime ideal Q ∈V (f1, . . . ,fr), such that det

∣∣∣ ∂fi∂xj

∣∣∣ /∈ Q. Then Q ∈Max(k[x1, . . . ,xr]).

Note: When we apply the Lemma to finish the proof of Theorem 10.13, thefield k of the Lemma plays the role of k′ = k(xr+1, . . . ,xn) in the Theorem; theideal Q of the Lemma corresponds to the kernel of the natural homomorphismψ : k′[x1, . . . ,xr]→ κ(P ). Thus the Lemma will imply that κ(P ) is the quotient ofk′[x1, . . . ,xr] by a maximal ideal, hence is algebraic over k′ by Proposition 4.2, asdesired.

First proof of Lemma 10.15. By Nullstellensatz, there exists m ∈ V (Q) such that

det∣∣∣ ∂fi∂xj

∣∣∣ /∈ m. By Theorem 10.13, applied to the maximal ideal m ⊂ k[x1, . . . ,xr],

ht(f1, . . . ,fr)k[x1, . . . ,xr]m = r. Since k[x1, . . . ,xr]Q = (k[x1, . . . ,xr]m)Q and heightis preserved under localization (Exercise 8.9), ht(f1, . . . ,fr)k[x1, . . . ,xr]Q = r ≥dim k[x1, . . . ,xr]. Hence ht(f1, . . . ,fr)k[x1, . . . ,xr]Q = r, which proves that everyideal containing (f1, . . . ,fr) is maximal. �

Second proof of Lemma 10.15, not using Nullstellensatz. We use induction on r.Let k′ = κ(Q). If r = 1, f1 is a polynomial in one variable with non-zero deriva-tive, hence Q is generated by one of the irreducible factors of f1 and the result isimmediate.

Suppose r > 1 and the Lemma is known for r − 1. Let

k =k[x1, . . . ,xr−1]Q∩k[x1,...,xr−1]

(Q ∩ k[x1, . . . ,xr−1]).

We may assume that ∂fr∂xr

/∈ Q. Then ht(fr)k[xr] = 1 and

Qk[xr]Q∩k[xr] = (fr)k[xr]Q∩k[xr].

Hence

(10.2) (f1, . . . ,fr)k[x1, . . . ,xr]Q = ((fr) +Q ∩ k[x1, . . . ,xr−1])k[x1, . . . ,xr]Q.


Since the rank of the matirx(∂fi∂xj

)is clearly independent of the choice of the base of

the ideal (f1, . . . ,fr), (10.2) implies that there exist f1, . . . ,fr−1 ∈ Q∩k[x1, . . . ,xr−1]

such that det∣∣∣ ∂fi∂xj

∣∣∣1≤i,j≤r−1

/∈ Q. By the induction assumption,

ht(Q ∩ k[x1, . . . ,xr−1]) = r − 1,

hence ht Q = r = dim k[x1, . . . ,xr], which proves that Q is maximal. �

This completes the proof of Theorem 10.13. �

Corollary 10.16. Let k be a perfect field, A a regular ring, which is a localizationof a finitely generated k-algebra, P a prime ideal of A. Then AP is regular.

Remark 10.17. Corollary 10.16 is true for any regular local ring, whether or not itis a localization of a finitely generated k-algebra, but we will not prove it here. See[3, (19.B), Theorem 48, p. 142] for a proof.

Example 10.18. Let k be an imperfect field, a ∈ k \ kp. Let x be an independent

variable and put K = k[x](xp−a) . Then K is a field (Proposition 4.2 and Lemma 4.4),

that is, a regular local ring of dimension 0. At the same time, ∂(xp−a)∂x

≡ 0. We haveK = κ(0) for (0) ∈ SpecK. This shows that the hypothesis that κ(P ) be separablygenerated over k is necessary in Theorem 10.13 (3).

Of course, a zero dimensional local ring is regular if and only if it is a field.We end this section by considering the case of regular one-dimensional local rings.Problem 10.6 says that any regular local ring is normal. We will now show that theconverse is true for one-dimensional local rings.

Theorem 10.19. Let A be a one-dimensional Noetherian local ring. Then A isregular if and only if A is normal.

Proof. Although “only if” is a special case of Problem 10.6, we will prove it directly.Suppose A is regular. Then the maximal ideal m of A is principal, that is, generated

by a single element x. Let y be any non-zero element of m. Since∞⋂n=0

mn = (0) in

any Noetherian local ring (Exercise 10.5 or use Nakayama’s Lemma), there existsN ∈ N such that y /∈ mN . y can be written in the form y = xnz, where z /∈ (x) = m.Of course, this representation is unique. This implies thatA is a UFD, hence normal.

“If” is more subtle. We assume that A is normal and want to prove that m isprincipal. We define the notion of I−1 for an ideal I in a domain A.

Notation. Let A be a domain, K the field of fractions of A, I an ideal of A.We define I−1 = {x ∈ K | xa ∈ A for all a ∈ I}.I−1 is an A-submodule of K. Of course, II−1 is an ideal of A.

Lemma 10.20. Assume that (A,m) is local. Then I is principal if and only ifII−1 = A.

Proof. If I = (x) is principal, then I−1 = Ax−1 ⊂ K. Hence II−1 = A.


Conversely, suppose II−1 = A. Then

(10.2) 1 =n∑

i=1

aixi,

where ai ∈ I, xi ∈ I−1. Since aixi ∈ A for all i, (10.2) implies that aixi /∈ m forsome i. Say, x1a1 /∈ m. This means that a1x1 is invertible, so that 1

x1∈ I−1. Now

take any other b ∈ I. Since 1x1∈ I−1, b

x1∈ A, hence b ∈ (x1). This proves that

I = (x1), so that I is principal. �

Exercise 10.21. Let A be any domain (i.e. not necessarily local), I ⊂ A and ideal.Prove that II−1 = A if and only if I is locally principal, that is, IAP is principalfor any P ∈ Spec A. Locally principal ideals together with their inverses generate agroup (under the operation of multiplication). In geometry, elements of this groupare called line bundles on Spec A.

We continue with the proof of Theorem 10.19. By Lemma 10.20, it remains toprove that mm−1 = A for a local 1-dimensional normal Noetherian ring (A,m).Suppose mm−1 $ A. Since mm−1 is an ideal, we have mm−1 ⊂ m. Since 1 ∈ m−1,mm−1 = m. By induction on n, m(m−1)n = m for any n ∈ N.

Lemma 10.22. Let A be a domain with field of fractions K. Take x ∈ K. Let

A : x = {a ∈ A | ax ∈ A}. If x is integral over A, then∞⋂n=1

(A : xn) 6= (0). The

converse holds if A is Noetherian.

Proof. If x is integral over A, then xn ∈n−1∑i=0

Axi for some n ∈ N. Then xl ∈n−1∑i=0

Axi

for all l ≥ n by induction on l. Then A : xl ⊃n−1⋂i=1

(A : xi) for all l ≥ n. This proves

the first statement.Conversely, assume that A is Noetherian and take a non-zero element

y ∈∞⋂

n=1

(A : xn).

Then Ay−1 is isomorphic to A as an A-module, hence is Noetherian.n∑i=0

Axn form

an ascending chain of submodules of Ay−1. By the Noetherian property this chain

must stabilize. Hence for some n, xn ∈n−1∑i=0

Axi, which implies that x satisfies an

integral equation of degree n over A. �

Coming back to the proof of Theorem 10.19, take any x ∈ m−1. Since

m(m−1)n = m for all n ∈ N,


we have A : xn ⊃ m for all n ∈ N. By Lemma 10.22, x is integral over A, hencex ∈ A by normality of A. This proves that

(10.3) m−1 = A.

We now use the fact that A is one dimensional to derive a contradiction. Take a non-zero element x ∈ m. Since A is a domain, x is not a zero divisor, hence ht(x) = 1.

Since dim A = 1, this means that (x) is m-primary, so that m =√

(x). Then thereexists n ∈ N such that mn−1 6⊂ (x) but mn 6⊂ (x). Take any y ∈ mn−1 \ (x). Thenyx∈ m−1 \A, which contradicts (10.3). This completes the proof. �

Let A be an integral domain with field of fractions K. We may consider theintegral closure A of A in K. The inclusion A → A gives rise to the natural mor-phism of integral schemes π : Spec A → Spec A. If A is Noetherian and satisfiescertain mild additional conditions (they will be specified later, but they hold forany algebra of finite type over a field and any quotient or localization of a formalor convergent power series ring), A is also Noetherian. The canonical morphism πis called normalization of the scheme Spec A. It was defined (surprisingly late –in the nineteen forties) by Oscar Zariski. This is an example of the usefulness ofthe algebraic language in geometry: this notion, extremely important as it turnedout to be, did not occur to anyone until the algebraic language was developed.Normalization will be of interest to us when we discuss resolution of singularities.Geometrically, Theorem 10.19 says that normalization resolves the singularities ofcurves. More generally, it says that for an arbitrary Noetherian scheme normaliza-tion resolves the singularities in codimension 1. When normalization was defined,the theorem of resolution of singularities was known for almost a century, yet it wasquite a surprise that it had such a simple and elegant proof and that the procedurefor desingularization had such a simple description.

§11. Sheaves.

Until now, we only considered affine schemes. However, inasmuch as we areinterested in studying global questions in algebraic geometry (for instance, if wewant to define the analogue of compact complex manifolds), we are led to considermore general schemes, for which affine schemes are basic buidling blocks. Arbitraryschemes are glued from affine ones in the same way that manifolds are glued fromopen subsets of Rn. The notion which expresses the idea of glueing global objectsfrom local data in modern algebraic geometry is that of a sheaf. The definition ofsheaves is the subject of this section.

Before making the general definitions, we consider the case of an affine schemeSpec A, where A is a domain (as we will see, the absence of zero divisors makes lifeeasier). We saw that open affine subschemes of Spec A of the special form Spec Af ,f ∈ A, form a basis for the Zariski topology. It is natural to regard the ring Af asthe “coordinate ring” or the “ring of regular functions” on the affine open schemeSpec Af . However, these basic affine open sets do not, in general, exhaust all the


open subsets of Spec A. For the purposes of glueing affine schemes together, we areforced to consider these more general affine open subsets. In particular, we mustdefine what we mean by the ring of regular functions on an arbitrary open set U .The main ideas are illustrated by the following exercises.

Exercise 11.1. Let A be a domain. Prove that A =⋂

P∈Spec A

AP =⋂

m∈Max(A)

Am,

where we view all the AP as subsets of the field of fractions of A.

The point of this exercise is that an affine scheme can be reconstructed in astraightforward way from the collection of all its local rings.

Exercise 11.2. Let A be a domain and let U ⊂ Spec A be an open set. Let

(11.1) OU =⋂

P∈U

AP .

Prove that:

(1) For every P ∈ U , (OU )P∩OU = AP .(2) There are natural inclusions

(11.2) Uι→ Spec OU → Spec A.

(3) If A is noetherian then so is OU .

If ι of (11.2) is a set theoretic bijection then it is natural to regard U = Spec OUas an affine open subscheme of Spec A. Of course, it may happen that ι is notsurjective, in which case U is open but not affine. An example of that is providedby letting U be the plane without the origin. In other words, let A = k[x, y],U = Spec A \ {(x, y)}. Then OU = A. The conclusion we can draw from all of theabove is that it is natural to think of OU as the ring of regular functions on the setU .

It is not immediately obvious how to generalize the above definitions to the casewhen A is not a domain, because it is not clear what should replace

⋂P∈U

AP (if A is

not a domain, the different local rings AP are not naturally subrings of any singlering of fractions. To overcome this difficulty, we will develop the powerful machineof sheaf theory. Now for the general definitions.

Let X be a topological space.

Definition 11.3. A pre-sheaf F of sets on X is the data consisting of a set FUfor each open set U ⊂ X and a map ρUV : FU → FV for each pair U, V of opensets such that V ⊂ U , satisfying the following conditions:

(1) F∅ is a one-element set.(2) ρUU is the identity map for all U .(3) Given any three open sets W ⊂ V ⊂ U , we have ρUW = ρVW ◦ ρUV .


The set FU is called the set of sections of F over U ; the maps ρUV are calledrestriction maps.

If the sets FU have some additional structure, such as rings or groups, we maytalk about a presheaf of ring (resp. groups). In this case, we require, in addition,that the maps ρUV be homomorphisms of the structure in question. Also, Definition11.3 (1) has to be modified to say that F∅ is the zero ring (resp. the trivial group).

Let F be a presheaf.

Definition 11.4. We say that F is a sheaf if the following two conditions (identityand gluability) are satisfied for any open set U and any open covering U =

⋃λ∈Λ

Uλ.

For λ, µ ∈ Λ, let us write Uλµ for Uλ ∩ Uµ, ρλ for ρUUλ and ρλµ for ρUλUλµ . Werequire

(1) Identity: for f, g ∈ FU , if ρλ(f) = ρλ(g) for all λ ∈ Λ then f = g.(2) Gluability: given sections fλ ∈ FUλ , λ ∈ Λ such that ρλµ(fλ) = ρµλ(fµ) for

all λ, µ ∈ Λ, there exists f ∈ FU such that fλ = ρλ(f) for all λ.

Remark 11.5. The meaning of Definition 11.4 is that F is a sheaf if and only ifeach FU can be completely recovered from local information. Indeed, Definition11.4 amounts to saying that for any open covering U =

⋃λ∈Λ

Uλ,

FU ∼={{fλ}λ∈Λ ∈

∏

λ∈Λ

FUλ

∣∣∣∣∣ ρλµ(fλ) = ρµλ(fµ) for all λ, µ ∈ Λ

}.

Examples 11.6. I. The constant sheaf F on X . Let F be a set, put FU = F forall U and let ρUV : F → F be the identity map for all V ⊂ U .

II. Let X be a topological (resp. smooth) manifold and let FU be the ring ofcontinuous (resp. C∞) functions on U . Given two open sets V ⊂ U and f ∈ FU ,define ρUV (f) to be the restriction of f to V (this motivates the name restrictionmaps). Then FU together with the maps ρUV define a sheaf of rings.

Exercise 11.7. Let X = Spec A be an integral affine scheme (in other words, A is adomain). For each open set U ⊂ X, let OU denote the ring defined above. For a pairof open sets V ⊂ U , we have the obvious inclusion OU ≡

⋂P∈U

AP ⊂⋂P∈V

AP ≡ OV .

the natural projection∏P∈U

AP →∏P∈V

AP induces a ring homomorphism ρUV :

OU → OV . Prove that the rings OU , together with the maps ρUV , form a sheaf ofrings on X. This sheaf is called the structure sheaf of X.

Now let X = Spec A be any affine scheme (i.e. we no longer require A to be adomain). Let U be an open subset of X . We define OU to be the following subsetof∏P∈U

AP . For a Q ∈ Spec A, g /∈ Q, let ρgQ denote the natural localization map

ρgQ : Ag → AQ. We define(11.3)OU = {{fP ∈ AP }P∈U | ∀P ∈ U ∃g ∈ A \ P, f ∈ Ag s.t. ∀Q 6∋ g, fQ = ρgQ(f)}.


Expressing (11.3) in words, OU consists of those elements {fP ∈ AP }P∈U ∈∏P∈U

AP

such that every point P is contained in some basic open set Spec Ag such that allthe fQ, Q ∈ Ag, are images of a single element f ∈ Ag.Exercise 11.8. (a) Prove that if A is a domain, (11.3) agrees with the definitionof OU given earlier in (11.1).

(b) Keep the above notation. Let V be an open subset of U . Show that the naturalprojection

∏P∈U

AP →∏P∈V

AP induces a map ρUV : OU → OV .

(c) Prove that {OU} with the restriction maps ρUV form a sheaf of rings on X.(d) Prove that if A is noetherian then OU is noetherian for each open set U .(e)∗ Prove that OX = A (and hence OX\V (f) = Af for each f ∈ A).

Definition 11.9. OU is called the structure sheaf of X .

We now give an alternative definition of OU . The reader will appreciate, however,that all these definitions revolve around one basic idea: glueing global sections fromlocal data. More precisely, (11.3) defines sections over U by glueing them from theinfinitesimal data of fP , P ∈ U , while in the following definition we consider anopen cover U =

⋃λ∈Λ

and glue sections over U from sections over the Uλ.

Again, let X = Spec A be any affine scheme, and let U be an open subset of X .Consider an open covering of the form U =

⋃λ∈Λ

Uλ, where each Uλ = Spec Agλ is a

basic affine open subset. Then for each λ, µ ∈ Λ, Uλ ∩ Uµ = Spec Agλgµ . For eachλ, µ ∈ Λ, let ρλµ : Agλ → Agλgµ denote the localization by gµ (ρλµ is nothing butthe restriction map of the structure sheaf of Spec A.

Exercise 11.10. (a) Prove that the ring OU is naturally isomorphic to the follow-ing subring of

∏λ∈Λ

Agλ :

(11.4) OU ∼= {{fλ ∈ Agλ}λ∈Λ | ∀λ,mu ∈ Λ, ρλµ(fλ) = ρµλ(fµ)}.

(b) Given two open subsets of X, V ⊂ U , show the restriction map ρUV can bedescribed as follows in terms of (11.4). Choose an open cover of V by basic affineopen sets: V =

⋃ψ∈Ψ

Vψ, where Vψ = Spec Ahψ and where for each ψ ∈ Ψ there exists

λ(ψ) ∈ Λ satisfying gλ(ψ) | hψ (in particular, Vψ ⊂ Uλ(ψ)). Let ρλ(ψ),ψ : Agλ(ψ)→

Ahψ be the localization by hψ. For f ∈ OU , represent f by {fλ ∈ Agλ}λ∈Λ, as in(11.4). Show that ρUV (f) is the element of OV , described by {ρλ(ψ),ψ(fλ(ψ))}ψ∈Ψ.

Exercise 11.11. Show that there are natural maps Uι→ Spec OU → Spec A.

Show also that for any P ∈ U , (OU )P∩OU∼= AP , so that we have a natural map

OU → AP for each P ∈ U .

If ι is a set theoretic bijection, then we say that U = Spec OU is an affine opensubscheme of Spec A.


Stalks and germs. Let X be a topological space and F a presheaf of sets on X .Let P be a point of X . Open sets containing P , together with the inclusion mapsV ⊂ U , form an inverse system. Since, by definition, the correspondence U → FUis a contravariant functor from the open sets of X to sets, the sets FU form a directsystem.

Definition 11.12. The stalk of F at P is FP := lim−→

U∋P

FU . For an open set U ∋ P

and a section f ∈ FU , the natural image of f in FP , which we will sometimes denoteby fP , is called the germ of f at P .

Exercise 11.13. Describe the stalk at a point P in each of the three examples ofsheaves above.

Sheafication. The following three exercises describe how, given a presheaf F , onecan canonically associate to it a sheaf F .

Exercise 11.14. Let F be a presheaf of sets (resp. rings, resp. groups). For an

open set U ⊂ X, define FU to be the following subset of∏P∈U

FP :

FU = {{fP ∈ FP }P∈U | ∀ P ∈ U ∃ open V, P ∈ V ⊂ U and g ∈ FV such that

∀ Q ∈ V, fQ = gQ}.

(11.5)

Show that F is a sheaf if and only if for each open set U ⊂ X, FU = FU .

Exercise 11.15. Let F be a presheaf. For each open set U ⊂ X, let FU be definedas in (11.5). Given two open set V ⊂ U , the natural projection

∏P∈U

FP →∏P∈V

FPinduces a map ρUV : FU → FV . Show that the sets FU , together with the mapsρUV , form a sheaf. The resulting sheaf F is called the sheafication of F .

Exercise 11.16. Let X = Spec A be an affine scheme and X =⋃λ∈Λ

Uλ, with

Uλ = Spec Agλ a covering of X by basic affine open sets. Show that the structuresheaf of X is the sheafication of the presheaf A, defined as follows:

AU = OU , if U is contained in one of the Uλ

= the zero ring otherwise

Given open set V ⊂ U , the restriction map is defined to be the restriction map of thestructure sheaf of Uλ if U is contained in one of the Uλ and the zero map otherwise.

§12. Schemes.

Let A be a ring. The n-dimensional affine space over A is, by definition,AnA = Spec A[t1, . . . ,tn]. The natural inclusion A ⊂ A[t1, . . . ,tn] gives rise to the


natural projection AnA → Spec A. Intuitively, one should think of AnA as the directproduct of Spec A with the usual n-dimensional affine space. Of course, this isstrictly heuristic. If one tried to make this last statement precise, one would runinto trouble with the definition of “the usual affine space” (affine space over whatfield? and why over a field? remember that A might be something like Z). Theheuristic picture is precise only if A is of finite type over a field k.

The example of the n-dimensional affine space points to another feature of schemetheory and algebraic geometry as a whole: the built in capacity for talking aboutfamilies of varieities or schemes (or deformations of a variety or a scheme), whichis such a central notion in the subject. Namely, if Spec B → Spec A is a morphismof affine schemes, we may regard Spec B as a family of affine schemes over Spec A.Spec A is the base of the family, while the fibers V (PB), P ∈ Spec A are individualmembers of the family. For example, AnA can be regarded as the trivial family ofaffine spaces over Spec A. The fiber over P ∈ Spec A is the affine space Anκ(P ),

which is the scheme-theoretic version of the algebraic variety κ(P )n.Now that we have developed the theory of sheaves, we are ready for a definition

of a scheme.

Definition 12.1. A scheme is a topological space X together with a sheaf of ringsO, such that there exists an open covering X =

⋃λ∈Λ

Uλ, where each Uλ is of the

form Uλ = Spec Aλ for some ring Aλ, such that for each λ ∈ Λ, the restriction ofO to Uλ is the structure sheaf of Spec Aλ. The sheaf O is called the structuresheaf of X .

We now give a more explicit description of O in terms of the structure sheavesof the Uλ.

Let X be a scheme, and let V be an open subset of X . Consider an open coveringof V of the form V =

⋃ψ∈Ψ

Vψ, where each Vψ = Spec Bψ is an affine open subset.

Moreover, for each ψ, µ ∈ Ψ, Vψ ∩ Vµ is an open subset of X and therefore can becovered by open affine subschemes: Vψ∩Vµ =

⋃φ∈Φψµ

Vψµφ, where Vψµφ = Spec Bψµφ.

For each ψ, µ ∈ Ψ and φ ∈ Φψµ, let ρψµφ : Bψ → Bψµφ denote the restriction mapof the structure sheaf of Spec Bψ.

Exercise 12.2. (a) Show that OV is naturally isomorphic to the following subringof∏ψ∈Ψ

Bψ:

(12.1) OV = {{fψ}ψ∈Ψ | ∀ψ, µ ∈ Ψ, φ ∈ φ ∈ Φψµ, ρψµφ(fψ) = ρµψφ(fµ)}.

(b) Given two open subsets of X, W ⊂ V , we can choose an affine open coverW =

⋃δ∈∆

Wδ, such that for each δ ∈ ∆ there exists ψ(δ) ∈ Ψ satisfying Wδ ⊂ Vψ(δ).

For f ∈ OV , represent f by {fψ}ψ∈Ψ, as in (12.1). Show that ρVW (f) is theelement of OW , described by {ρUψ(δ)Vδ(fψ(δ))}δ∈∆.


Definition 12.3. Let X , Y be schemes. A morphism from X to Y is a mapof topological spaces X → Y , together with affine open coverings Y =

⋃λ∈Λ

Vλ,

X =⋃

λ∈Λ,φ∈Φλ

Uλφ such that π−1(Vλ) =⋃

φ∈Φλ

Uλφ and such that for each λ, φ the

map π|Uλφ : Uλφ → Vλ is induced by a morphism of affine schemes.

We now define the projective space PnA over Spec A. By definition, PnA is gluedfrom n+ 1 affine pieces, each of which is isomorphic to AnA.

Define PnA =n⋃i=0

Ui, where Ui = Spec A[x0

xi, . . . ,xi−1

xi, xi+1

xi, . . . ,xn

xi

]. The meaning

of this notation is that, in addition to saying that each Ui is isomorphic to AnA,it also describes the glueing of Ui with Uj along Ui ∩ Uj . Namely, if we writeUi = Spec A[y1, . . . ,yn], Uj = Spec A[z1, . . . ,zn], with i < j and the variables listedin the same order as above, our notation is a concise way of writing the change ofcoordinates on Ui ∩ Uj :

yl =zlzi+1

, 0 ≤ l ≤ i or j < l ≤ n

=zl+1

zi+1, i < l < j

=1

zi+1, l = j.

To restate the same definition in yet another way, Ui ∩Uj is a basic affine open setof both

Ui = Spec A

[x0

xi, . . . ,

xi−1

xi,xi+1

xi, . . . ,

xnxi

]and

Uj = Spec A

[x0

xj, . . . ,

xj−1

xj,xj+1

xj, . . . ,

xnxj

];

we have Ui ∩ Uj = Ui \ V (xjxi

) = Uj \ V ( xixj ), where the last equality is described

algebraically by the identification of the localizations

(A

[x0

xi, . . . ,

xi−1

xi,xi+1

xi, . . . ,

xnxi

])

xjxi

∼=(A

[x0

xj, . . . ,

xj−1

xj,xj+1

xj, . . . ,

xnxj

])

xixj

.

We have a natural projection PnA → Spec A, defined to be, in each coordinate chart,the natural projection AnA → Spec A: the maps

A→ A

[x0

xi, . . . ,

xnxi

]

Spec A← Spec AnA


patch together to give a morphism Spec A ← PnA. The ring A[x0, . . . ,xn] is some-times called the homogeneous coordinate ring of PnA.

We now give another, more canonical description of PnA, independent of ourchoice of coordinate system and the covering by charts, directly in terms of thegraded A-algebra A[x0, . . . ,xn].

Let G =∞⊕n=0

Gn be a graded algebra. An ideal I ⊂ G is said to be homogeneous

if it can be generated by homogeneous elements ( ⇐⇒ given x =n∑i=0

xi ∈ I, with

xi ∈ Gi, we have xi ∈ I for all i, 0 ≤ i ≤ n). For example,∞⊕n=1

Gn is a homogeneous

ideal of G.

Definition 12.4. Proj G is defined to be the set of all the homogeneous primeideals of G, not containing

⊕∞n=0Gn, endowed with the Zariski topology: the

closed sets are all the sets of the form V (I) = {P ∈ Proj G | I ⊂ P}, where Iranges over the homogeneous ideals of G.

Remark 12.5. To see that Proj G is a scheme, we can cover it by affine chartsas follows. Let G = G0[xλ]λ∈Λ, with xλ homogenous of degree kλ > 0. ThenProj G =

⋃λ∈Λ

Uλ where Uλ = Spec (Gxλ)0. Here Gxλ has the obvious structure of

a graded algebra and the subscript 0 denotes its degree 0 part.In other words, there is a 1-1 correspondence between homogeneous prime ideals

P ⊂ G, not containing xλ, and prime ideals of (Gxλ)0. This correspondence is givenby P → PGxλ ∩ (Gxλ)0. Conversely, given Q ∈ Spec (Gxλ)0, it is easy to show

that√QGxλ is a homogeneous prime ideal, whose contraction to G is the element

of Proj G corresponding to Q. Verifying the details is left as an exercise.Take λ, µ ∈ Λ. The coordinate transformations on Uλ ∩ Uµ is described by

((Gxλ)0

)xµkλ

xλkµ

=((Gxµ)0

)xλkµ

xµkλ

=(Gxλxµ

)

0

The construction globalizes naturally. Namely, if X =⋃i

Spec Ai, we can define

PnX = ∪iPnAi , where the PnAi are glued together in the obvious way. The canonicalmorphisms PnA → Spec Ai are also glued together to give the natural map PnX → X .

One of the main subjects of study in the rest of this course, and in algebraicgeometry at large, are birational projective morphisms, or blowings up. We endthis section by defining birational and projective. In the next section we will givea more constructive definition of blowing up, rather than characterizing it by itsabstract properties of being birational and projective.

Definition 12.6. Let π : Y → X be a morphism of schemes. We say that π isprojective, or that Y is projective over X , if there exists an open affine coverX =

⋃i

Spec Ai and n ∈ N with the following property. Let Vi = π−1(Spec Ai)


and let πi = π|Vi . Then we require that for each i, πi have a factorization of the

form Viιi→ PnAi

αi→ Spec Ai, where ιi is a closed embedding and αi is the naturalprojection. To say that π is projective is a little weaker than saying that π has afactorization of the form

(12.2) Y → PnX → X,

since the definition of projective requires that a factorization of type (12.2) existlocally on X .

Definition 12.7. Let X be a scheme, ξ ∈ X a point. Let Spec A ⊂ X be an openaffine chart of X containing ξ and let P be the prime ideal of A corresponding to ξ.The ring AP is called the local ring of ξ in X and is denoted by OX,ξ. Of course,OX,ξ is independent of the choice of the affine open chart Spec A.

Definition 12.8.

(1) A scheme X =⋃i

Spec Ai is said to be reduced if all the Ai are reduced.

(2) X is said to be integral if it is reduced and irreducible as a topologicalspace (i.e. is not a union of two non-trivial closed sets).

Remark 12.9. If X is integral, then each Spec Ai is a dense open set (otherwise itsclosure and its complement would express X as a union of two closed sets). Henceeach Ai is a domain, otherwise there would exist a non-trivial decomposition ofSpec Ai into closed sets, which could be extended to a non-trivial decomposition ofX by passing to closures. We also have that Spec Ai ∩Spec Aj 6= ∅ for any i and j,otherwise we would have X = (X \ Spec Ai)∪ (X \ Spec Aj). Then the (0) ideal iscommon to both Spec Ai and Spec Aj . Let ξ denote the common point of all the

Spec Ai which corresponds to the (0) ideal in Ai. Since {ξ} ∩ Spec Ai = Spec Ai,

we have {ξ} = X . This point ξ is called the generic point of X .Definition 12.10. Let X be an integral scheme and ξ its generic point. ThenOX,ξ = K(X) is a field, called the field of rational functions on X or thefunction field of X . K(X) is nothing but the common field of fractions of all ofthe Ai.

Let π : X → Y be a morphism of schemes. Let ξ ∈ X and η = π(ξ) ∈ Y . Thenπ induces a map π∗ : OY,η → OX,ξ. We have mY,η = mX,ξ ∩OY,η.Exercise 12.11. Let σ : A → B be a ring homomorphism and σ∗ : Spec B →Spec A the corresponding morphism of affine schemes. Show that Ker σ is nilpotentif and only if σ∗ is dominant, that is, σ∗(Spec B) = Spec A.

Now, suppose X and Y are integral and let ξ and η be the respective genericpoints of X and Y . Let π : X → Y be a dominant morphism. This is equivalent tosaying that π maps the generic point of X to the generic point of Y . Hence in thiscase we get a field extension π∗ : K(Y ) → K(X).


Definition 12.12. π is birational if π∗ : K(Y )∼→ K(X) is an isomorphism.

Definition 12.13. A morphism of affine schemes π : Spec B → Spec A is saidto be of finite type if B is a finitely generated A-algebra. Let π : X → Y be amorphism of arbitrary schemes. π is said to be locally of finite type if the affineopen coverings {Vλ} and {Uλφ} can be chosen so that for all λ ∈ Λ, φ ∈ Φλ, themap π|Uφλ is a morphism of finite type of affine schemes. π is said to be of finitetype if, in addition, the set Φλ is finite for each λ ∈ Λ (that is, there are finitelymany Uλφ over each Vλ).

For example, any projective morphism is of finite type.

Remark 12.14. Let π : X → Y be a morphism of finite type between integralschemes. Then π is birational⇔ there exist open subsets U ⊂ X , V ⊂ Y , such thatπ induces an isomorphism U

∼→ V .Proof. Suppose π is birational. Take an affine open set V ⊂ Y and an affine

open subset U ⊂ π−1(V ), of finite type over V . Let U = Spec A, V = Spec B.Clearly, replacing X by U and Y by V does not change the problem. Since π is

birational and of finite type, we may write A = B[a1

b1, . . . ,an

bn

], where ai, bi ∈ A.

Then π∗ induces an isomorphisms of V \V (b1, . . . , bn) with its pre-image.

§13. Blowing up.

Let A be a ring, X = Spec A, I = (f1, . . . ,fn) a finitely generated ideal of A.

Definition 13.1. The blowing up of X along I is the birational projective mor-phism π : X → X , defined as follows. Consider the morphism ϕ : X\V (I)→ Pn−1

X ,which is defined precisely below, but which should intuitively be thought of asthe map which sends every ξ ∈ X \ V (I) to (ξ, (f1(ξ) : · · · : fn(ξ))) ∈ PnX . X

is defined to be the closure ϕ(X \ V (I)) ⊂ Pn−1X in the Zariski topology. More

precisely, the map ϕ is described as follows. Write X \ V (I) =n⋃i=1

Spec Afi ,

Pn−1X =

n⋃i=1

Spec A[x1

xi, . . . ,xi−1

xi, xi+1

xi, . . . ,xnxi

]. Then in each of the n affine charts

ϕ is defined by

Afi ← A

[x1

xi, . . . ,

xi−1

xi,xi+1

xi, . . . ,

xnxi

]

fjfi← xj

xi

It is easy to check that those n maps glue together to give a morphism X \V (I)→Pn−1X .

Remark 13.2.

(1) Since the blowing up X = ϕ(X \ V (I)) ⊂ Pn−1X , the natural projection

Pn−1X → Spec A induces a map X → Spec A. In particular, X is projective

over Spec A.


(2) The natural map π : X → Spec A is an isomorphism away from V (I) (the

inverse mapping is given by ϕ). This means that the map π : X → X isbirational.

As we will see later, blowing up is characterized by the properties of being birationaland projective.

Remark 13.3. For all 1 ≤ i, j ≤ n, the equation fixj − fjxi ∈ A[x1, . . . ,xn] vanishes

on X. This describes the full set of defining equations of X in Pn−1X in the case when

A is Noetherian and V (I) is a complete intersection (which means ht I = n),but not in general (a proof of this assertion is postponed). In general, the defining

equations f X in Pn−1X = Proj A[x1, . . . , xn] are not easy to describe.

If A is an integral domain, we can think of X as

(13.1) X =n⋃

i=1

(Spec A

[f1fi, . . . ,

fnfi

]).

If fact, even if A is not a domain, we can still make sense of the description (13.1)(see Example 13.13 below).Example 13.4. 1) Blowing up the plane at a point. Let A = k[x, y], I = (x, y),so that Spec A is the affine plane and I is the origin. Then P1

A = Proj A[u1, u2]should intuitively be thought of as the direct product of P1

k with the plane. Wehave the map Spec A\{0} → P1

A, described locally in each of the two coordinatecharts by A[u1

u2]→ Ay,

u1

u2→ x

yand A[u2

u1]→ Ax,

u2

u1→ y

x.

X is defined in P1A by the equation xu2 − yu1 = 0. For example, if k = R, then

X is nothing but the Mobius band.

Perhaps the most useful way of thinking about the blowing up X is that it is ascheme glued together from two coordinate charts

X = Spec k

[u1,

u2

u1

]⋃Spec k

[u2,

u1

u2

],

where the glueing is implicit in the notation.2) Similarly, we can blow up the affine n-space at the origin. Let

A = k[x1, . . . , xn], I = (x1, . . . , xn).

Then Pn−1X = Proj A[u1, . . . ,un]. X ⊂ Pn−1

X is the subscheme defined by the

equations xiuj − xjui, 1 ≤ i, j ≤ n. Again, X is covered by n coordinate charts of

the form Spec k[u1

ui, . . . ,ui−1

ui, ui,

ui+1

ui, . . . ,unui

].

3) Finally, the blowing up of Spec k[x1, . . . , xn] along (x1, . . . ,xl) for l < n is the

subset of Pl−1A defined by the equations xiuj − xjui, 1 ≤ i, j ≤ l. X is covered by

l coordinate charts of the form Spec k[u1

ui, . . . ,ui−1

ui, ui,

ui+1

ui, . . . ,ului , ul+1, . . . ,un

].


Intuitively, we may think of this last construction as first blowing up the origin inSpec k[x1, . . . ,xl] and then taking the direct product of the whole situation withSpec k[xl+1, . . . ,xn].Another characterization of blowing up. We now give yet another descriptionof the blowing up of Spec A along an ideal I, equivalent to the above, but withoutreference to any particular ideal base (f1, . . . ,fn).

First, we need a general fact about closed subschemes of ProjG for a graded al-gebra G. As mentioned earlier, such closed subschemes are defined by homogeneousideals of G. Surprisingly, while for any ring A, by definition, distinct ideals of Adefine distinct closed subschemes of Spec A, it is possible for two distinct homoge-neous ideals of G to define the same closed subscheme of ProjG. The next exercisegives an exact characterization of pairs of homogeneous ideals of G define the sameclosed subscheme of ProjG.

Let G be a graded algebra. Let {xλ}λ∈Λ be a set of homogeneous, positive degreegenerators of G over G0. Let I be a homogeneous ideal of G.

Definition 13.5. The saturation of I is the homogeneous ideal of G, defined by

I = {y ∈ G | ∀ λ ∈ Λ ∃ nλ ∈ N s.t. yxnλλ ∈ I}.

I is said to be saturated if I = I.

Exercise 13.6.

(1) Show that the definition of I does not depend on the choice of the set ofgenerators {xλ}λ∈Λ.

(2) show that I is saturated.

(3) Show that for any homogeneous ideal I,√I =√I (this shows that at least

I and I define the same closed subset of ProjG.(4) Let I and J be two homogeneous ideals of G. Show that the closed sub-

schemes V (I) and V (J) of ProjG coincide if and only if I = J . In partic-ular, I is the greatest homogeneous ideal of G which defines the same closedsubscheme as I.

We come back to our invariant characterization of blowing up. Let A be a ring,

I a finitely generated ideal, as above. Then X ∼= Proj GrIA.Indeed, let I = (f1, . . . ,fn). We have the obvious surjective homomorphism φ :

A[u1, . . . ,un]→ GrIA, given by ui 7→ fi, so that Proj GrIA is a closed subschemeof Pn−1

A . Furthermore, let J = Ker φ. Then

J ={g(u) =

∑aαu

α∣∣∣ homogeneous polynomials such that g(f1, . . . ,fm) = 0

}.

Let J ′ denote the greates homogeneous ideal having the property ϕ(Spec A\V (I)) ⊂V (J ′) (by definition, X = V (J ′).


Exercise 13.7. Show that J ′ = J , so that Proj GrIA = X.

Now let X be any scheme, not necessarily affine. We would like to define thenotion of blowing up X along an ideal. However, we run into a difficulty: it is notclear what “ideal” means in this context. The natural thing to do is to think aboutsheaves of ideals (which are a special case of sheaves of modules). This brings usto another central notion in the theory of schemes — that of quasi-coherent andcoherent sheaves of modules.

Sheaves of modules. Coherence and quasi-coherence. Roughly speaking,quasi-coherent sheaves can be thought of as a generalization of vector bundles intopology and complex analysis, with coherent sheaves corresponding to vector bun-dles of finite rank.

First, we need to define tensor product for modules over a ring. For a ring A, afree A-module is a module which is isomorphic to a (possibly infinite) direct sumof copies of A.

Definition 13.8. Let A be a ring and M1, M2 A-modules. The tensor productM1⊗AM2 is defined to be the free module generated by the symbols x1⊗x2 modulothe submodule generated by all the relations of the form (a1x1+a′1x

′1)⊗x2−a1(x1⊗

x2)− a′1(x′1 ⊗ x2) and x1 ⊗ (a2x2 + a′2x′2)− a2(x1 ⊗ x2)− a′2(x1 ⊗ x′2).

Note that if M is an A-module and B an A-algebra, then M ⊗AB has a naturalstructure of a B-module. Note also that if φ : M1 → M2 is a homomorphism ofA-modules, it makes sense to talk about the induced homomorophism φ ⊗A B :M1 ⊗A B →M2 ⊗A B.

Definition 13.9. Let A be a ring, M an A-module and S ⊂ A a multiplicative set.The localization MS of M by S is defined to be MS = {xs | x ∈ M, s ∈ S}/ ∼,where ∼ is the usual equivalence relation: x

s ∼yt whenever there exists u ∈ S such

that u(xt− ys) = 0. MS has a natural structure of an AS-module (hence also of anA-module).

Exercise 13.10. Keep the notation of Definition 12.6. Show that MS = M⊗AAS.Let X be a topological space. Let O be a sheaf of rings on X .

Definition 13.11. A sheaf of O-modules is a sheaf F of abelian groups on X ,such that for each open set U ⊂ X , FU has the structure of an OU -module, andfor any two open sets V ⊂ U , the restriction map FU → FV is a homomorphism ofOU -modules (note that OV , and hence also FV are naturally OU -modules via therestriction map OU → OV ).

Example 13.12. Let X = Spec A be an affine scheme. As usual, we denote byO the sturcture sheaf of X . Let M be an A-module. We define a sheaf M ofO-modules as follows: for each open set U , put MU = M ⊗A OU . The restrictionmaps MU →MV are induced in the obvious way from the restriction maps of O.


Definition 13.13. Let X be a scheme and let O denote the structure sheaf ofX . A sheaf F of O-modules is said to be quasi-coherent if there exists an affineopen covering X =

⋃λ∈Λ

Uλ such that the restriction of F to each Uλ is of the form

described in Example 13.12. F is said to be coherent if, in addition, FU is a finitelygenerated OU -module for each open set U .

Example 13.14. The structure sheaf O of X itself is a coherent sheaf of modules.

Remark 13.15. Let X be a scheme. There is a natural 1-1 correspondence betweenquasi-coherent ideal sheaves on X and closed subschemes of X . Namely, let I be aquasi-coherent ideal sheaf. Let X =

⋃Uλ be the affine open covering of Definition

13.13. Then V (IUλ) defines a closed subscheme of Yλ ⊂ Uλ.⋃λ

Yλ is a closed

subscheme of X , which we will denote by V (I).Let X =

⋃i

Spec Ai be a scheme, I a coherent ideal sheaf on X . We can now

define the blowing up π : X → X of X along I. Namely, let Ii denote the ideal of

sections of I on Spec Ai; Ii is an ideal of Ai. Let Xi → Spec Ai denote the blowing

up of Spec Ai along Ii. It is a routine matter to show that all the Xi can be glued

together in a natural way to give the blowing up π : X → X of X along I. Notethat there is no reason to expect π to factor through the projective space PnX forany n; we only know that such a factorization exists locally on X .

We will now give yet another characterization of blowing up, this time by auniversal mapping property.

Let π : X → Y be a morphism of schemes and I a quasi-coherent ideal sheaf onY . Let X =

⋃i,j∈Φi

Spec Bij and Y =⋃i

Spec Ai be the open coverings appearing in

the definition of morphism (Definition 12.3). For each i and each j ∈ Φi we have ahomomorphism Ai → Bij. Let π∗I denote the ideal sheaf on X whose ideal sectionsover Spec Bij is IiBij. π

∗I is a quasi-coherent ideal sheaf on X .

Let X be a scheme and I a coherent ideal sheaf on X . The idea, which we now

explain in detail, is that the blowing up π : X → X of X along I is characterizedby the universal mapping property with respect to making I invertible (Definition13.19 below).

If I = (x) is a principal ideal in a ring A, then I ∼= AAnn x as A-modules. If,

in addition, x is not a zero divisor in A, then I ∼= A as A-modules. Recall thatan ideal I is said to be locally principal if it is principal in every local ring AP ,P ∈ Spec A. The next Lemma and its Corollary show that given a locally principalideal in a ring A (not necessarily Noetherian), Spec A can be covered by finitelymany affine open charts of the form Spec Ag, such that IAg is principal for each g.

Example 13.16. Let A be a Dedekind domain (that is, a one-dimensional regular

ring, such as Z[√

5i]), and I an ideal of A of height 1. Then I is locally principal(since A is a regular 1-dimensional ring), but it need not be principal (Remark


1.15).

Lemma 13.17. Let I be a locally principal ideal in a ring A. Let f = {fλ}λ∈Λ bea set of generators of I. Then for every P ∈ Spec A, IAP = (fλ)AP for some λ.

Proof. Replacing A by AP , we may assume that A is local with maximal ideal Pand that I is principal. Let x be a generator of I. Then for every λ ∈ Λ there existsaλ ∈ A such that fλ = aλx. If aλ ∈ P for all λ ∈ Λ, then x ∈ I = fA ⊂ P (x),hence x = yx for some y ∈ P =⇒ (1− y)x = 0 =⇒ x = 0 =⇒ I = (0) and thereis nothing to prove. Hence we may assume that one of the aλ is a unit in A. Thenx = a−1

λ fλ and J = (fλ), as desired. �

Corollary 13.18. Let I be a locally principal ideal in a ring A and let f = {fλ}λ∈Λ

be a set of generators of I. Then there exists a finite subset {f1, . . . ,fn} ⊂ f and a

finite open affine cover Spec A =n⋃i=1

Spec Agi such that IAgi = (fi)Agi , 1 ≤ i ≤ n.

Proof. By Lemma 13.17,

(13.2)∑

λ∈Λ

((fλ) : I) = A

(otherwise, localizing at a maximal ideal containing∑λ∈Λ

((fλ) : I), we would find

a localization AP of A such that ((fλ) : I)AP 6= AP for any λ, that is, I is notgenerated by any of the fλ). By (13.2), 1 belongs to some finite sum of ideals ofthe form (fλ) : I, λ ∈ Λ. Hence there exists a finite subset {f1, . . . ,fn} ⊂ f and

gi ∈ (fi) : I, 1 ≤ i ≤ n, such thatn∑i=1

gi = 1. Then IAgi = (fi)Agi and no prime

of A can contain all of the gi, so that Spec A =n⋃i=1

Spec Agi is the desired open

cover. �

Definition 13.19. Let I be a locally principal ideal in a ring A. I is said to beinvertible if Spec A can be covered by affine charts of the form Ag, g ∈ A, suchthat for each g, IAg = (xg)Ag, where xg ∈ Ag is not a zero divisor.

Of course, if A is a domain, then invertible and locally principal are the samething (in that case, both properties are equivalent to saying that II−1 = A byExercise 10.21, hence the name “invertible”). Note also that if I is invertible, thenwe can strngthen Corollary 13.18 to say, in addition, that for each i, 1 ≤ i ≤ n, fi isnot a zero divisor in Agi . This fact is obtained by the same argument as Corollary13.18, if we replace (13.2) by the stronger equality

∑λ∈Λ

(Ann(Ann fλ))((fλ) : I) = A.

The point is that an element x ∈ A is not a zero divisor if and only if Ann x =(0) ⇐⇒ Ann(Ann x) = A.Definition 13.20. An ideal sheaf I on a scheme X is locally principal if thereexists an affine open cover X =

⋃i

Ui, Ui = Spec Ai, such that IUi is a principal


ideal of Ai for all i. I is said to be invertible if each of the IUi is principal andgenerated by an element which is not a zero divisor.

Again, if X is integral, then invertible and locally principal are the same thing.

Example 13.21. Let A be a ring and I = (f1, . . . ,fn) a finitely generated ideal of

A. Let X = GrIA be the blowing up of Spec A along I. By definition, X =n⋃i=1

Ui,

where Ui = {P ∈ Proj GrIA | fi /∈ P}. Then Ui = Spec ((GrIA)fi)0, where fi isviewed as an element of degree 1 in GrIA. Localization by fi annihilates Ann(fi),so that fi is not a zero divisor in ((GrIA)fi)0. Then we may identify ((GrIA)fi)0 ≡

AAnn(fi)

[f1fi, . . . , fi−1

fi, fi+1

fi, . . . , fn

fi

], where fi in the denominator makes sense since

fi is not a zero divisor in this chart. Thus

Ui = SpecA

Ann(fi)

[f1fi, . . . ,

fi−1

fi,fi+1

fi, . . . ,

fnfi

].

We have

IA

Ann(fi)

[f1fi, . . . ,

fi−1

fi,fi+1

fi, . . . ,

fnfi

]=

= (f1, . . . ,fn)A

Ann(fi)

[f1fi, . . . ,

fi−1

fi,fi+1

fi, . . . ,

fnfi

]

= (fi)A

Ann(fi)

[f1fi, . . . ,

fi−1

fi,fi+1

fi, . . . ,

fnfi

],

so that π∗I is invertible on X . Since we are dealing with a local property, this

statement remains valid even if X is not affine. In other words, if π : X → X is theblowing up of a coherent ideal sheaf I, then π∗I is invertible.

We now point out that this property is also sufficient to characterize blowing up.Namely, the blowing up π of I is the smallest (in the sense explained in Theorem13.22 below) projective morphism such that π∗I is invertible. More precisely, wehave the following theorem.

Theorem 13.22 (the universal mapping property of blowing up [2, Propo-sition II.7.14, p. 164]). Let X be a scheme, I a coherent ideal sheaf on X and

π : X → X the blowing up of I. Let ρ : Z → X be another morphism such that ρ∗Iis invertible. Then ρ factors through X in a unique way.

The Noetherian hypothesis appearing in the statement of this result in [2] issuperfluous.

Proof of Theorem 13.22. Because of the asserted uniqueness of the factorization,the problem is local on X : once we solve the problem over each affine chart of X ,uniqueness will guarantee that the resulting morphisms will patch together nicely.


Thus we may assume that X = Spec A is affine and I = (f1, . . . ,fn) ⊂ A is afinitely generated ideal. Similarly, according to Corollary 13.18, we may assumethat Z = Spec B is affine, IB = (fi)B for some i, 1 ≤ i ≤ n, and that fi is not azero divisor in B. Write

X =n⋃

j=1

SpecA

Ann(fj)

[f1fj, . . . ,

fj−1

fj,fj+1

fj, . . . ,

fnfj

].

By assumption, we are given a homomorphism ρ∗ : A→ B. It is sufficient to check

that ρ∗ factors in a unique way through AAnn(fi)

[f1fi, . . . ,fi−1

fi, fi+1

fi, . . . , fnfi

], where i

is as above. Since IB = (fi)B for some i, for each j 6= i there exists bj ∈ B suchthat fj = bjfi. Since fi is not a zero divisor in B, bj is unique. Hence the unique

homomorphism ψ : AAnn(fi)

[f1fi, . . . , fi−1

fi, fi+1

fi, . . . ,fnfi

]→ B, factoring ρ∗ is given by

(13.3) ψ

(fjfi

)= bj .

The fact that (13.3) gives a well defined homomorphism ψ follows easily from thefact that fi is not a zero divisor. �

Let X be an integral scheme. We saw that every blowing up of X → X is a bira-tional and projective morphism. Next, we show that, conversely, every birational,

projective morphism X → X is locally the blowing up of some coherent ideal sheaf,so that blowings up and birational projective morphisms are essentially the samething.

Theorem 13.23. Let X be an integral scheme and π : X → X a birational projec-tive morphism. Then there exists an open affine cover X =

⋃i

Spec Ai and finitely

generated ideals Ii ⊂ Ai such that for each i, π|π−1(Ai) is the blowing up of Ii.

Proof. Since the problem is local on X , we may assume that X = Spec A is affine

and that X is a closed subscheme of Pn−1A for some n. Let x1, . . . ,xn be the homo-

geneous coordinates on Pn−1. For 1 ≤ i ≤ n, let

Ui = Spec A

[x1

xi, . . . ,

xi−1

xi,xi+1

xi, . . . ,

xnxi

].

Since Vi = X ∩ Ui is a closed subscheme of Ui and an open subscheme of X,

Spec OVi is an affine open subscheme of X and OVi is a homomorphic image of

A[x1

xi, . . . ,xi−1

xi, xi+1

xi, . . . ,xn

xi

]. For 1 ≤ i, j ≤ n, j 6= i, let fji denote the image of

xjxi

in OVi . Let K denote the field of fractions of A. Since π is birational, all the fijare elements of K. By definition, we have the following relations among the fij, forall the allowed values of i, j, and l:

(13.4)fijfji = 1

fijfjlfli = 1.


To complete the proof of the Theorem, it is sufficient to find f1, . . . ,fn ∈ A suchthat fij = fi

fjfor all i, j (for then π is the blowing up of I = (f1, . . . ,fn)). To

construct f1, . . . ,fn, let g be a common denominator of the fij , that is, an elementof A such that gfij ∈ A for all i, j. Put f1 = g, fi = gfi1 for 2 ≤ i ≤ n. Then for

all the allowed i, j, we have fifj

= fi1fj1

= fij by (13.4), as desired. �

Let X be a scheme, I a coherent ideal sheaf on X , π : X → X the blowingup along I. The closed subscheme Y = V (I) is called the center of the blowingup π. We will sometimes refer to π as “the blowing up along Y ”. The coherent

ideal sheaf π∗I defines the closed subscheme π−1(Y ) ⊂ X . Since π∗I is invertible,

π−1(Y ) has codimension 1 in X in the case when X is Noetherian (Corollary 8.11).Codimension 1 subschemes are called divisors. The subscheme π−1(Y ) is calledthe exceptional divisor of the birational map π. By definition, π induces an

isomorphism X \ π−1(Y ) ∼= X \ Y .

Strict Transforms. Let Z be a scheme, I a coherent ideal sheaf on Z. Letι : X → Z be a closed subscheme of Z with its natural inclusion ι. Let π : Z → Zbe the blowing up of I. Let X := π−1(X\V (I)) ⊂ Z, where ” ” denotes the closurein the Zariski topology.

Definition 13.24. X is called the strict transform of X under π.

Of course, X ⊂ π−1(X) ⊂ X ∪ π−1(V (I)). To distinguish it from the stricttransform, π−1(X) is sometimes called the total transform of X under π.

Theorem 13.25. X together with the induced morphism ρ : X → X is nothingbut the blowing up of the coherent ideal sheaf ι∗I on X.

Proof. Again, the problem is local on Z, so that we may assume that Z = Spec B

is affine and I ⊂ B is a finitely generated ideal. It is sufficient to prove that X has

the universal mapping property of blowing up (that is, X satisfies the conclusionof Theorem 13.22). Here we are using a general fact about universal mappingproperties: any object satisfying the conclusion of Theorem 13.22 is necessarilyunique up to a canonical isomorphism, since given any two such objects Theorem13.22 guarantees the existence of a canonical morphism in both directions.

Let J denote the defining ideal of X in B, A = BJ , I = IA. By definition, ι∗I

is the coherent ideal sheaf I on Spec A given by the ideal I. Let ι be the inclusion

map ι : X → Z.

First, we show that ρ∗I is invertible. Let I = (f1, . . . ,fn). Let fi denote the

natural image of fi in A. Then I = (f1, . . . ,fn). Cover Z by open affine charts: Z =n⋃i=1

Spec Bi, where Bi = BAnn fi

[f1fi, . . . ,fn

fi

]. For 1 ≤ i ≤ n, let Vi = X ∩ Spec Bi.

Then Vi is a closed subscheme of Spec Bi; let Ji denote the defining ideal of Vi inBi. Let Ai = SpecBiJi , so that Vi = Spec Ai. For each i we have a commutative


diagramBi −−−−→ Aix

x

B −−−−→ A

Now, since IBi = (fi)Bi for each i, IAi = IAi = (fi)Ai is principal. To showthat fi is not a zero divisor in Ai, suppose fix = 0 for some x ∈ Ai. Let xbe any representative of x in Bi. Then fix ∈ Ji, so that x ∈ Ji(Bi)fi . Now,Spec(Bi)fi = Ui \ V (π∗I) and JiBfi is the defining ideal of Vi \ V (π∗I) in Bfi .

Since π induces an isomorphism Z \ V (π∗I) → Z \ V (I), X \ V (π∗I) ∼= X \ V (I).

Hence Vi = X ∩ Ui is the Zariski closure of Vi \ V (π∗I). Now, since x vanishes onVi \ V (π∗I) in Ui, it belongs to the ideal of its Zariski closure in Ui. This provesthat x ∈ Ji, hence x = 0, as desired. We have shown that ρ∗I is invertible.

Now, let α : Y → X be another morphism such that α∗I is invertible. It remains

to show that α has a unique factorization β : Y → X; then we can conclude that Xis the blowing up of X along I by Theorem 13.22. Since α∗I = α∗ι∗I = (ι ◦ α)∗I,ι◦α has a unique factorization γ : Y → Z by Theorem 13.22. In view of the asserteduniqueness of β, we may assume that Y = Spec C is affine and that α(Y ) is entirely

contained in one of the coordinate charts of Z, say in the first one, Spec B1. Thuswe have ring homomorphisms

(13.5)

B1γ∗−−−−→ C

xx

B −−−−→ A

It remains to show that γ∗ factors through A1, that is, γ∗(J1) = (0). By (13.5),γ∗(JB1) = (0). We have natural maps B → B1 → Bf1 , so that

(13.6) (γ∗ ⊗Bf1) (JBf1) = 0.

Since f1 is not a zero divisor in B1 and J1Bf1 = JBf1 , we have f l1J1 ⊂ JB1 forsome l ∈ N, so that γ∗(J1) = (0) by (13.6). This completes the proof of Theorem13.25. �

Example 13.26. Let k be a field, u, v independent variables. Let Z = Spec k[u, v],

I = (u, v), X = Spec k[u,v](u2−v3) ⊂ Z.

The blowing up Z of Z along I is covered by two affine charts: Z = Spec k[uv , v]∪

Spec k[u, v

u

]. Let us denote the coordinates in the first chart U1 by u1, v1, so that

v = v1, u = u1v1. Let u2, v2 be the coordinates in the second chart U2, so thatu = u2, v = u2v2.

To calculate the strict transform X of U2, we first find its full inverse image. Thisinverse image is defined by the equation u2−v3, but wirtten in the new coordinates:

u2 − v3 = u22 − u3

2v32 = u2

2(1− u2v32).


Here u2 = 0 is the equation of the exceptional divisors. To obtain the strict trans-

form X, we must factor out the maximal power of u2 out of the equation. In this

case, X ∩ U2 is defined by 1− u2v32 . In U1, we have

u2 − v3 = u21v

21 − v3

1 = v21(u2

1 − v1).

Here v1 = 0 is the equation of exceptional divisor, so that X ∩U1 = V (u21 − v3). In

particular, note that although X had a singularity at the origin, X is non-singular.Thus, in this example we started with a singular variety X with one singular point,blew up the singularity and found that the strict transform of X became non-singular. This procedure is called resolution of singularities of X .

The problem of Resolution of Singularities.

Definition 13.27. A scheme X is said to be Noetherian if there exists a finite

affine open cover X =n⋃i=1

Spec Ai, such that each Ai is a Noetherian ring.

Definition 13.28. A scheme X is said to be regular if for every ξ ∈ X , the localring OX,ξ is regular. This is equivalent to saying that X can be covered by affinecharts Spec Ai, such that each Ai is a regular ring.

Exercise 13.29. Let X = Speck[x,y,z]z2−xy denote the non-singular quadratic cone,

embedded in A3k. Construct a resolution of singularities of X by blowing up a suitable

ideal of k[x, y, z] and showing that the strict transform of X under this blowing upis regular.

Question. Let X be a reduced Noetherian scheme. Does there exist a sequence ofblowings up Xn → Xn−1 → · · · → X such that Xn is regular?

It is known, though we will not prove it here, that some mild technical conditionsmust be imposed on the ring A in order for the answer to above question to beaffirmative. We have not yet developed enough tools to state all of these conditions,but we will state one of them. This condition has to do with the closedness ofsingular locus, and we will discuss it right now.

Let A be a Noetherian ring. The regular locus of A is, by definition, Reg(A) ={P ∈ Spec A | AP is regular }. The singular locus is Sing A = Spec A\Reg(A).Using Theorem 10.13, it is possible to show that if A is an algebra of finite type over

a field, then Sing A is closed in A. For example, if A = k[x1,...,xn](f1,...,fl)

is a domain and

r = ht(f1, . . . ,fl), then Sing(A) is defined in Spec A by the ideal generated by all

the r×r minors of the Jacobian matrix(∂fi∂xj

). Similar Jacobian criteria exist in the

case of formal power series rings and complex-analytic local rings. However, thereare some pathological Noetherian rings whose singular locus is not closed. This isonly one of the pathologies which needs to be ruled out in order to have resolutionof singularities.


Definition 13.30 [3, Chapter 13, (32.B)]. A Noetherian ring A is said to be

(1) J-0 if Reg A contains a non-empty Zariski open set(2) J − 1 if Reg A is open in Spec A(3) J − 2 if any reduced finite type A-algebra is J − 1.

Note that if A is reduced, then Reg A is always non-empty since it contains all theminimal primes of A. Thus J-1 implies J-0 for reduced rings, but not in general.

As we already mentioned, finite type algebras over a field and their localizations,formal and convergent power series rings are all known to be J-2. The J-2 conditionis known to be necessary for the existence of resolution of singularities. There isone other property of Noetherian rings, called the G property (for Grothendieck),which is known to be necessary, but we will state it later, in the continuation of thiscourse.

In practice, resolution of singularities is usually achieved by locally embeddingthe given singular scheme X in a regular scheme Z, constructing a sequence ofblowings up of regular subschemes of the ambient regular scheme and studying theeffect of these transformations on the strict transform of X . The following factcomes in very useful in the above setup.

Problem 13.31. Let Z be a regular scheme, Y ⊂ Z a regular subscheme. Let

Z → Z be the blowing up along Y . Then Z is regular.

The significance of resolution of Singularities:(1) There are many constructions which can only be defined for non-singular

schemes, or at least which are much easier to define and study for non-singularschemes or varieties. These include Hodge theory, singular and etale cohomology,the canonical divisor, etc.

(2) Inasmuch as one is interested in the problem of classification of algebraicvarieties, it is natural to separate all the reduced and irreduclble varieties (or inte-gral schemes) into birational equivalence classes, that is, to group together allthe schemes having the same field of rational functions. From that point of view,resolution of singularities answers a very natural question: does every birationalequivalence class contain a non-singular scheme? Once this question has been an-swered affirmatively (as it will be in the continuation of this course), one may, onthe one hand, look for birational invariants, that is, numbers associated to the givenbirational equivalence class and defined in terms of some non-singular model, and,on the other hand, address the finer questions about the relation between differentnon-singular models in the given birational equivalence class and also what can besaid about the relation between resolution of singularities and the original singularvariety or scheme which it dominates. This is a very active area of research, knownas the Mori program; it has been the stage of some spectacular recent developments,particularly in the case of 3-dimensional varieties, or 3-folds.

(3) Embedded desingularization is a somewhat stronger form of resolutionof singularities, which is particularly useful for applications. Suppose that X isembedded in a regular scheme Z. Embedded desingularization asserts that there


exists a sequence π : Z → Z of blowings up along non-singular centers, under whichthe total transform π−1(X) of X becomes a divisor with normal crossings, which

means that locally at each point of Z, π−1(X) is defined by a monomial with respectto some regular system of parameters. Geometrically, it means that at every point of

Z there exists a local coordinate system such that π−1(X) looks locally like a unionof coordinate hyperplanes, counted with certain multiplicities. Thus divisors withnormal crossings locally have a very simple structure. There are many situations inwhich it is useful to know that any closed subscheme can be turned into a divisorwith normal crossings by blowing up. For example, this is used for compactifyingalgebraic varieties. Let X be a regular algebraic variety over a field k, embeddedin some projective space Pnk . If X is not projective over Spec k, that is, if X is notclosed in Pnk , we can always consider its Zariski closure X , which is, by definition,projective over Spec k. The problem is that even though we started with a regularX , X may well turn out to be singular. Resolution of singularities, together withits embedded version, assures us that, after blowing up closed subschemes, disjointfrom X , we may embed X in a regular projective variety X ′ such that X ′ \X is anormal crossings divisor.

(4) Finally, resolution of singularities is useful for studying the singularities them-

selves. Namely, let ξ ∈ X be a singularity and let π : X → X be a desingularization.We may adopt the following philosophy for studying the singularity ξ. All the reg-ular points are locally the same; every singular point is singular in its own way. Wemay regard resolution of singularities as a way of getting rid of the local complex-ity of the singularity ξ and turning it into global complexity of the regular scheme

X. Thus some global invariant of X may also be regarded as invariants of thesingularity ξ. For example, if X is a surface and ξ is isolated, then π−1(ξ) is a

collection of curves on the regular surface X . By embedded resolution, we mayfurther assume that π−1(ξ) is a normal crossings divisor. If {Ei}, 1 ≤ i ≤ n,are the irreducible components of π−1(ξ), then the intersection matrix (Ei.Ej) (or,

equivalently, the dual graph of the configurationn⋃i=1

Ei) is an important numeri-

cal invariant associated to the singularity ξ. A good illustration of the usefulnessof replacing local difficulties by global is Mumford’s theorem which asserts that anormal surface singularity which is topologically trivial is regular. More preicsely,given a normal surface singularity ξ ∈ X over C, one may consider its link, whichis the intersection of X with a small Euclidean sphere centered at ξ. The link is areal 3-dimensional manifold. Mumford’s theorem asserts that if the link is simplyconnected, then ξ is regular. At first, it seems unlikely that anything intelligent atall can be said about the link, until one realizes that the link is nothing but the

boundary of a tubular neighbourhood of the collectionn⋃i=1

Ei of non-singular curves

on a non-singular surface. After that the link becomes relatively easy to analyze.

We now turn to the theory of valuations, which is a very important tool in thesubject of resolution of singularities and in all of birational geometry.


§14. Valuations.

We will start with the algebraic definition of valuations and will then move onto their beautiful geometric interpretation in birational geometry.Definition 14.1. An ordered group is an abelian group Γ together with a subsetP ⊂ Γ (here P stands for “positive elements”) which is closed under addition andsuch that Γ = P

∐{0}∐(−P ).Remark 14.2. P induces a total ordering on Γ: a < b ⇐⇒ b − a ∈ P . Thus anequivalent way to define an ordered group would be “a group with a total ordering,which respects addition, that is, a > 0, b > 0 =⇒ a+ b > 0”.

Note that an ordered group is necessarily torsion-free.Examples 14.3. Z,R with the usual ordering are ordered groups. Any subgroupΓ ⊂ R is an oredred group with the induced ordering (more generally, any subgroupof an ordered group is an ordered group). Zn with the lexicographical ordering (cf.Definition 14.4) are ordered groups.Definition 14.4. Let Γ1, . . . ,Γr be ordered groups. The lexicographical ordering

on their direct sumr⊕i=1

Γi is defined by

(a1, . . . ,an) < (b1, . . . ,bn) ⇐⇒ ∃ i, 1 ≤ i ≤ n such that

a1 = b1

a2 = b2

...

ai−1 = bi−1

ai < bi.

All the ordered groups which will appear in this course will be of the formr⊕i=1

Γi,

where Γi ⊂ R for all i and the total order is lexicographic.We are now ready to define valuations. Let K be a field, Γ an ordered group.

Let K∗ denote the multiplicative group of K.Definition 14.5. A valuation of K with value group Γ is a surjective grouphomomorphism ν : K∗ → Γ, such that for any x, y ∈ K∗,

(14.1) ν(x+ y) ≥ min{ν(x), ν(y)} and

Exercise 14.6. Let K be a field, ν a valuation of K and x, y non-zero elements ofK such that ν(x) 6= ν(y). Show that in this case equality must hold in (14.1), thatis,

(13.2) ν(x+ y) = min{ν(x), ν(y)}.

Example 14.7. Let X be an integral scheme, K = K(X), ξ ∈ X such that OX,ξis a regular local ring. As usual, let mX,ξ be the maximal ideal of OX,ξ.


Define νξ : k(x)∗ → Z by

νξ(f) = max{n | f ∈ mnX,ξ}, f ∈ OX,ξ.

νξ extends from OX,ξ to all of K in the obvious way by additivity: νξ(fg) = νξ(f)−

νξ(g). νξ induces a group homomorphism because⊕ mnX,ξ

mn+1X,ξ

is an integral domain.

In the above example, note that ξ is any scheme-theoretic point; for example, itcould stand for the generic point of an integral codimension 1 subscheme. In thatcase, the condition that OX,ξ is non-singular holds automatically whenever X isnormal (Theorem 10.19).Remark 14.8. Let A be a domain, K its field of fractions, I ⊂ A an ideal. Wecan generalize the above example as follows. Define

νI(f) = max{n | f ∈ In}, for f ∈ A.

In general, νI is a pseudo-valuation, which means that the condition of additivity inthe definition of valuation is replaced by the inequality νI(xy) ≥ νI(x) + νI(y). νIis a valuation if and only if ⊕ In

In+1 is an integral domain (a condition which alwaysholds if I is maximal and AI is regular).

Valuations of the form νI , I ∈ Spec A are called divisorial. The reason for thisname is that even if dim AI > 1, we can always blow up Spec A along I. Let

π : X → Spec A be the blowing up along I. Then K(X) = K(X).

The property that ⊕ In

In+1 is a domain means that D := V (π∗I) = Proj ⊕ In

In+1

is irreducible. Then OX,D

is a regular local ring of dim 1 and νI = νD. This

example illustrates an important philosophical point about valuations: a valuationis an object associated to the field K, that is, to an entire birational equivalenceclass, not to a particular model in that birational equivalence class. Thus to studya given valuation, one is free to perform blowings up until one arrives at a modelwhich is particularly convenient for understanding the given valuation.

The next several examples will be located on the plane. Let k be a field, K =k(u, v) a pure transcendental extension of k of degree 2. Surprisingly, valuationsof K are already highly non-trivial and interesting to study. They display almostall of the same properties as valuations of arbitrary fields, at least those arising inalgebraic geometry.Example 14.9. Let Γ = (1, π) = {a+ bπ | a, b ∈ Z} ⊂ R. Let

(14.3)ν(v) = 1,

ν(u) = π.

A very important fact for us is the observation that (14.3) determines ν completely.Indeed, by additivity, specifying ν(u) and ν(v) determines the value of every

monomial: for any α, β ∈ N0, we have ν(uαvβ) = απ+β. Now, let f =∑cαβu

αvβ.


By (14.2), ν(f) = min{πα+ β | cαβ 6=0}. Now it is clear that ν extends uniquely toall of K by additivity.

Example 14.10. Let K be as in the previous example. Let Γ = Z2 with lexico-graphical ordering. Let ν(v) = (0, 1), ν(u) = (1, 0).

As above, this data determines ν completely, that is, ν(uαvβ) = (α, β) andν(∑cα,βu

αvβ) = min{(α, β) | cα,β 6= 0}.

Exercise 14.11. Generalize Examples 14.7 and 14.10 as follows. Let X be anintegral scheme. Consider a collection X1 $ X2 $ · · · $ Xn $ X of closed integralsubschemes such that the local ring OXi,Xi−1

is regular for each i. Define a valuationν : K(X)∗ → Zn with lexicographical ordering in a way which yields Example 14.7if n = 1 and Example 14.10 in the case n = 2, X = Spec k[u, v], X1 = {(u, v)} andX2 = V (u).

Since an abelian group is automatically a Z-module, it makes sense to talk aboutelements of Γ being linearly independent over Z. In view of the above examples,it is very easy to show that any valuation of k(u, v) such that ν(u) and ν(v) arelinearly independent is completely determined by ν(v) and ν(u). We now give anexample of a valuation with ν(u) and ν(v) linearly dependent over Z, such that νis not determined by ν(u) and ν(v).

Example 14.12. Take K = k(u, v), as before. We define a valuation ν : K∗ → Qas follows.

Let Q−1 = v, Q0 = u. Let ν(Q−1) = 1, ν(Q0) = 32 .

Let pn be the n-th prime number, assume that Qn is defined and that ν(Qn) =βn = qn

pnfor some qn ∈ N.

Put Qn+1 = Qpnn −vqn and define ν(Qn+1) = qn+ 1pn+1

. We define a valuation ν

with value group Q as follows. Let f ∈ k[u, v]. For α ∈ Q, let Pα denote the ideal

of k[u, v] generated by all the products of the formn∏i=1

Qγii withn∑i=1

γiβi ≥ α.

Put ν(f) = max{α | f ∈ Pα}. It is not too hard to show that this valuationν has the following alternative description. Let t be a new variable. Let k[[tQ+ ]]denote the ring of formal power series in t with exponents in the set of non-negativerational numbers, such that the set of exponents appearing in any given series is wellordered. There is an infinite increasing sequence α1 < α2 < α3 < . . . of rationalnumbers and an infinite sequence of constants cn ∈ k, such that the denominatorof αn is pn and such that ν is induced from the obvious t-adic valuation on k[[tQ+ ]]via the injection

v → t

u→∞∑

i=0

citαi .


Example 14.13. The next example may be regarded as a special case of Example14.12, in which αi ∈ Z for all i ∈ N0. Let K = k(u, v), as before.

Let K = k((u, v)) denote the field of fractions of k[[u, v]]. Choose an element

t ∈ k[[u, v]] \ k(u, v) of the form t = u−∞∑i=1

civi, ci ∈ k∗.

Let ν : K → Z2 be given by ν(t) = (1, 0), ν(v) = (0, 1), where we take thelexicographical order on Z2. Put ν := ν|K . Since ν(K∗) ⊂ (0) ⊕ Z, we obtainν : K∗ → Z. By definition, ν(v) = 1. Since ν(u−v(c1+c2v+. . . )) = (1, 0) > (0, 1) =ν(v(c1+c2v+ . . . )), we have ν(u) = ν(v) = 1. Since ν(u−c1v−v2(c2+c3v+ . . . )) =(1, 0) > (0, 2) = ν(v2(c2 + c3v + . . . )), we have ν(u − c1v) = 2. Continuing in thisway, we obtain

ν(u) = ν(v) = 1

ν(u− c1v) = 2

ν(u− c1v − c2v2) = 3

...

ν(u− c1v − c2v2 − · · · − cnvn) = n+ 1.

§15. Valuation rings.

Let K be a field, Γ an ordered group, ν : K∗ ։ Γ a valuation of K. Associatedto ν is a local subring (Rν , mν) of K, having K as its field of fractions:

Rν = {x ∈ K | ν(x) ≥ 0}mν = {x ∈ K | ν(x) > 0}.

Example 15.1 (Divisorial valuations). Let X be an integral scheme, D ⊂ X aclosed integral subscheme, ξ the generic point of D.

Assume that OX,ξ is a regular local ring of dimension 1. Let t be a generator ofmX,ξ. Then K = (OX,ξ)t. Indeed, any element f ∈ OX,ξ can be written as f = tnu,where n ∈ N and u is invertible. For each f = tnu as above, we have νD(f) = n.Then Rν = OX,D.

In this example, Rν is Noetherian; in fact, it is a regular local ring of dimension1. We already know (Theorem 10.19) that if (R,m, k) is a 1-dimensional local noe-therian ring, R is regular⇐⇒ R is normal. In other words, regular one-dimensionallocal rings and normal one-dimensional (Noetherian) local rings are the same thing.Below, we show that Example 15.1 is typical in two ways:

(1) Any valuation ring Rν is normal (whether or not ν is divisorial).(2) A local domain (R,M, k) is a valuation ring for a valuation with value group

Z if and only if it is noetherian, regular and of dimension 1.

(2) says, in other words, that every valuation with value group Z arises in theway described in Example 15.1; regular 1-dimensional local noetherian rings and


valuation rings Rν for ν with value group Z are one and the same thing. Later(Proposition 15.8), we will also prove that for any field K and any valuation ν ofK, Rν is Noetherian if and only if the value group of ν is Z.

Theorem 15.2.. Let K be a field and ν : K → Γ any valuation of K. Let Rν bethe valuation ring of ν. Then Rν is normal.

Proof. Take any x ∈ K. Suppose x is integral over Rν , that is,

(14.1) xn + c1xn−1 + · · ·+ cn = 0

for some ci ∈ Rν . Then ν(ci) ≥ 0 for all i. By (14.1), ν(xn) = ν(c1xn−1+· · ·+cn) ≥

min1≤i≤n

{ν(cixn−i)}. Hence there exists i, 1 ≤ i ≤ n, such that ν(xn) ≥ ν(cixn−i) =

ν(ci) + (n− i)ν(x). Then ν(x) ≥ ν(ci)i≥ 0, so that that x ∈ Rν . �

Theorem 15.3. Let (R,m, k) be a local domain, K the field of fractions of R. Thefollowing conditions are equivalent:

(1) R is Noetherian, regular of dimension 1.(2) R = Rν for some valuation ν : K∗ → Z.(3) There exists t ∈ m such that every ideal of R is of the form (tn), n ≥ 0.(4) Every ideal of R is a power of m.

Proof of Theorem 15.3. (1) =⇒ (2). Take X = Spec R, D = {m} in Example15.1.

(2) =⇒ (3). Let t be an element of R such that ν(t) = 1. Let I be any idealof R. Let n = min{ν(x) | x ∈ I} (the minimum exists because all the ν(x) arenon-negative integers; here we are using very strongly that Γ = Z and not, say, Q).For any x ∈ I, ν

(xtn

)≥ 0, hence

(15.1)x

tn∈ R.

On the other hand, there exists y ∈ I such that ν(y) = n. Then ν(tn

y

)= 0, so

that tn

yis a unit of R. Thus

(15.2) tn ∈ (y) ⊂ I.

(15.1) and (15.2) prove that I = (tn).(3) ⇒ (1) is immediate, because by (3), m = (t).We have proved that (1)–(3) are all equivalent. It remains to show that (4) is

equivalent to (1)–(3). We will show that (4) ⇐⇒ (3).(3) ⇒ (4). Clearly, t ∈ m. Then m = (t) and (4) follows.(4) ⇒ (3). Take any t ∈ m \m2. Then (t) = mn for some n. Hence (t) = m and

(3) follows. This completes the proof of Theorem 15.3. �.


Rings R satisfying (1)–(4) of Theorem 15.3 are called discrete valuation rings.Indeed, Theorem 15.3 says that these are precisely the valuation rings correspondingto valuations with value group Z.

Now that we understand the case of discrete valuation rings, we describe thevaluation rings in Examples 14.9, 14.10 and 14.12.Examples 15.4.

1. The valuation ν : k(u, v)∗ → (1, π) of Example 14.9. Here

Rν =(k[uαvβ ]α,β∈Z; απ+β≥0

)m,

where m is the maximal generated by all the monomials uαvβ, απ + β > 0.2. The valuation ν : k(u, v)∗ → Z2 (with lexicographical order) of Example 14.10.

Rν =(k[uαvβ]α>0 or α=0,β≥0

)m

, where m is the maximal ideal generated by all

the monomials uαvβ such that ν(uαvβ) ≥ 0.3. The valuation ν : k(u, v)∗ → Q of Example 14.12. Let Q−1 = v, Q0 = u,

Q1, . . . ,Qn, . . . be as in Example 14.12. Let β−1 = 1, β0 = 32, βn = ν(Qn) for

n ∈ N. Then Rν = (k[Qγ])m, where Qγ ranges over all the products of the formn∏i=1

Qγii , with n ∈ N, γi ∈ Z and∑γi

βi ≥ 0, and where m is the maximal ideal

generated by all thosen∏i=1

Qγii such thatn∑i=1

γiβi > 0.

Notice that none of the valuation rings in these three examples are Noetherian.Theorem 15.3 gives several equivalent characterizations of discrete valuation

rings. Next, we would like to characterize arbitrary valuation rings. In other words,we pose the questions: given a local domain (R,m) with field of fractions K, whendoes there exist a valuation ν of K such that R = Rν?

In the next theorem, we will give three criteria, each of which is equivalent tothe existence of such a ν. For geometric applications, the most useful one is thecriterion in terms of birational domination, which we will now define.

Definition 15.5. Let (R1, m1), (R2, m2) be two local domains with the same fieldof fractions K. We say that R2 birationally dominates R1, denoted R1 < R2, ifR1 ⊂ R2 and m1 = m2 ∩R1.

Another way of phrasing Definition 15.5 is to say that there is a natural mapπ : Spec R2 → Spec R1, such that π(m2) = m1. For example, let X be an integralscheme and π : X ′ → X a blowing up of X . Let ξ ∈ X , ξ′ ∈ X ′ be such thatξ = π(ξ′). Then OX,ξ < OX′,ξ′ .Theorem 15.6. Let (R,m) be a local domain with field of fractions K. Thefollowing conditions are equivalent:

(1) R = Rν for some valuation ν : K∗ ։ Γ(2) for any x ∈ K∗, either x ∈ R or 1

x∈ R (or both)

(3) the ideals of R are totally ordered by inclusion(4) (R,m) is maximal (among all the local subrings of K) with respect to birational

domination.


Proof. (1) ⇒ (2). Suppose R = Rν for some ν. Take any x ∈ K∗. Thenν(x) + ν

(1x

)= ν

(x · 1

x

)= ν(1) = 0. Hence either ν(x) ≥ 0 or ν

(1x

)≥ 0. This

proves that either x ∈ Rν or 1x∈ Rν , as desired.

(2) ⇒ (3). Suppose the ideals of R are not totally ordered by inclusion. Thenthere exist ideals I1, I2 ⊂ R such that I1 6⊂ I2 and I2 6⊂ I1. Take a ∈ I1 \ I2,b ∈ I2 \ I1. By assumption, either a

b ∈ R or ba ∈ R. Say, a

b ∈ R. Then b ∈ I2,ab ∈ R⇒ a = b · ab ∈ I2, which gives the desired contradiction.

(3) ⇒ (1) We are given that the ideals of R are totally ordered by inclusion andwe must construct a valuation ν of K, such that R = Rν . First, we describe thevalue group of ν.

Let Φ denote the set of all principal ideals of R (other than R itself), with theoperation of multiplication.

By assumption Φ is totally ordered by the order relation I1 < I2 if and onlyif I2 ⊂ I1. Moreover, if α, β ∈ Φ and α < β, then there exists γ ∈ Φ such thatγ + α = β: indeed, write α = (a), β = (b), with a, b ∈ R. α < β means (b) ⊂ (a),that is, there exists x ∈ R such that b = ax. If we set γ = (x), we have γ + α = βin Φ.

Let Γ = Φ∐{0}∐(−Φ). It is easy to check that the operation of Φ extends to

Γ and that Γ becomes an ordered group under this operation. For x ∈ R, defineν(x) = (x) ∈ Φ. This defines a map ν : R∗ → Φ. Extend ν to all of K by additivity.This gives the desired valuation ν : K∗ → Γ. We leave details to the reader.

(2) ⇒ (4). Let (R′, m′) be a local domain with field of fractions K such that(R,m) < (R′, m′). Suppose R $ R′. Take x ∈ R′ \ R. Then, by assumption,1x∈ R⇒ 1

x∈ m⇒ 1

x∈ m′ (since R < R′) ⇒ x 6∈ R′, which is a contradiction.

(4) ⇒ (2). Suppose there is x ∈ K∗ such that x 6∈ R and 1x /∈ R.

Consider the birational extension R ⊂ R[x]. If there exists a maximal idealm′ in R[x], containing mR[x], we would have R < R[x]m′, which contradicts thehypothesis.

Hence we may assume that mR[x] is not contained in any maximal ideal of R[x],that is,

(15.3) mR[x] = R[x].

Applying the same reasoning to 1x

instead of x, we obtain

(15.4) mR

[1

x

]= R

[1

x

].

(15.3)–(15.4) mean that 1 ∈ mR[x] and 1 ∈ mR[ 1x], that is,

1 = a0 + a1x+ · · ·+ anxn for some ai ∈ m(15.5)

1 = b0 + b11

x+ · · ·+ bl

1

xlfor some bi ∈ m.(15.6)


Choose expressions (15.5)–(15.6) so that both n and l are the smallest possible.Say, l ≤ n. Multiply (15.6) by xl. We obtain:

xl = b0xl + b1x

l−1 + · · ·+ bl,

so that

(15.7) 0 = xl +b1

b0 − 1xl−1 + · · ·+ bl

b0 − 1.

Multiply (15.7) by anxn−l and subtract it from (15.5). We obtain a relation of the

form (15.5), but with a smaller n. This is a contradiction, and the proof of Theorem15.6 is complete. �

Remark 15.7. It is clear from the proof of (3) =⇒ (1) that the valuation ringRν completely determines ν.

Proposition 15.8. Let ν : K∗ → Γ be a valuation. Then Rν is Noetherian ⇔Γ ≃ Z.

Proof. ⇐ is contained in the (2) =⇒ (1) of Theorem 15.3.⇒ Since the ideals of Rν are totally ordered by inclusion, saying that Rν is

Noetherian is the same as saying that the set of ideals of Rν is well ordered. If Rνis Noetherian, then the set of ideals of Rν is well ordered ⇒ the set of principalideals of Rν is well ordered.

By (3) =⇒ (1) of Theorem 15.6, the set of (proper) principal ideals of Rν canbe identified with the semi-group Φ of positive elements of Γ. Thus to complete theproof of the Proposition, it remains to prove the following lemma.

Lemma 15.9. Let Γ be an ordered group, Φ the set of positive elements of Γ. IfΦ is well ordered, then Φ ≃ N.

Proof. Let 1 := min{α : α ∈ Φ}. For n ∈ N let n := n ·1. This gives an inclusionι : N → Φ. We want to show that ι is an isomorphism.

Suppose not. Take α ∈ Φ \ N. For any n ∈ N, we cannot have n < α < n + 1:indeed, if n < α < n+ 1, then 0 < α− n < 1, which contradicts the definition of 1.

Hence α > n for any n ∈ N. Let β = min{α ∈ Φ | α > n for all n ∈ N}. Thenβ− 1 > n ∀ n. Since β− 1 < β, we have a contradiction. This completes the proofof Lemma 15.9 and with it of Proposition 15.8. �.

§16. Numerical invariants of valuations.

In this section we define three important integer invariants associated with val-uations and investigate relations between them.

Let Γ be an ordered group.Definition 16.1. An isolated subgroup of Γ is a subgroup ∆ ⊂ Γ, which is asegment in the ordering. This means: for any a ∈ Γ, b ∈ ∆ such that −b < a < b,we have b ∈ ∆.


Example 16.2. (0)⊕Z is an isolated subgroup of Z2 with lexicographical ordering.

Remark 16.3. Let ν : K∗ → Z2 be a valuation. Then there is a one-one corre-spondence between isolated subgroups of Γ and prime ideals of Rν , described asfollows:

∆ ⊂ Γ←→ P = {x ∈ Rν | ν(x) /∈ ∆}.In particular, the set of isolated subgroups of Γ is totally ordered by inclusion (thisis also very easy to show directly for any ordered group Γ).Definition 16.4. For a valuation ν : K∗ → Γ, the rank of ν, denoted rk ν, isdefined to be the number of distinct isolated subgroups of Γ (not counting (0), butcounting Γ itself). In other words, if (0) $ ∆1 $ ∆2 $ · · · $ ∆n = Γ is the full listof isolated subgroups of Γ, then rk ν = n.

Remark 16.5. rk ν = dim Rν .

Examples 16.6. If Γ = Z2 with lexicographical ordering, then rk ν = 2. Moregenerally, if Γ = Zn with lexicographical ordering, then rk ν = n.

In all the other examples we have encountered so far, we had rk ν = 1. In fact,

rk ν = 1 if and only if there exists an injection Γ ⊂ R. If Γ =r⊕i=1

Γi with Γi ⊂ R

and the total order on Γ is lexicographical, then rk ν = r.

Definition 16.7. The rational rank of ν, denoted rat. rk ν, is defined byrat. rk ν = dimQ Γ⊗Z Q.

rat. rk ν is nothing but the maximal number of elements of Γ, which are linearlyindependent over Z.

Remark 16.8. rk ν ≤ rat. rk ν for all valuations ν.Indeed, let (0) $ ∆1 $ ∆2 $ · · · $ ∆n = Γ be the isolated subgroups of Γ. Pick

xi ∈ ∆i \∆i−1, i = 1, 2, . . . ,n. Then x1, x2, . . . ,xn are linearly independent over Zin Γ.

Example 16.9. Let Γ = (1, π) ⊂ R. Then rk ν = 1 < 2 = rat. rk ν, since 1, π arelinearly independent over Z.

Definition 16.10. Let (R,m, k) be a local domain with field of fractions K and νa valuation of K. We say that ν is centered in R if R < Rν (this is equivalent tosaying that ν(R) ≥ 0 and ν(m) > 0).

If ν is centered at R, then we have a natural injection k → Rνmν

.

Definition 16.11. The transcendence degree of ν over k is

tr. degkν := tr. deg

(Rνmν

/k

).

Examples:


References

1. M. Atiyah and MacDonald, Introduction to Commutative Algebra, Addison-Wesley, 1969.

2. R. Hartshorne, Algebraic Geometry, Springer-Verlag, 1977.3. H.Matsumura, Commutative Algebra, Benjamin/Cummings Publishing Co.,

Reading, Mass., 1970.4. M. Nagata, Local Rings, Krieger Publishing Co., Huntington, N.Y., 1975.5. O. Zariski and P. Samuel, Commutative Algebra, Vol. II, Springer-Verlag,

1960.

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