A Course In Commutative Algebra

Robert B. Ash

Preface

This is a text for a basic course in commutative algebra, written in accordance with the

following objectives.

The course should be accessible to those who have studied algebra at the beginning

graduate level. For general algebraic background, see my online text “Abstract Algebra:

The Basic Graduate Year”, which can be downloaded from my web site

www.math.uiuc.edu/∼ r-ash

This text will be referred to as TBGY.

The idea is to help the student reach an advanced level as quickly and e¬ciently as

possible. In Chapter 1, the theory of primary decomposition is developed so as to apply to

modules as well as ideals. In Chapter 2, integral extensions are treated in detail, including

the lying over, going up and going down theorems. The proof of the going down theorem

does not require advanced ¬eld theory. Valuation rings are studied in Chapter 3, and

the characterization theorem for discrete valuation rings is proved. Chapter 4 discusses

completion, and covers the Artin-Rees lemma and the Krull intersection theorem. Chapter

5 begins with a brief digression into the calculus of ¬nite di¬erences, which clari¬es some

of the manipulations involving Hilbert and Hilbert-Samuel polynomials. The main result

is the dimension theorem for ¬nitely generated modules over Noetherian local rings. A

corollary is Krull™s principal ideal theorem. Some connections with algebraic geometry

are established via the study of a¬ne algebras. Chapter 6 introduces the fundamental

notions of depth, systems of parameters, and M -sequences. Chapter 7 develops enough

homological algebra to prove, under approprate hypotheses, that all maximal M -sequences

have the same length. The brief Chapter 8 develops enough theory to prove that a regular

local ring is an integral domain as well as a Cohen-Macaulay ring. After completing

the course, the student should be equipped to meet the Koszul complex, the Auslander-

Buchsbaum theorems, and further properties of Cohen-Macaulay rings in a more advanced

course.

Bibliography

Atiyah, M.F. and Macdonald, I.G., Introduction to Commtative Algebra, Addison-Wesley

1969

Balcerzyk, S. and Joze¬ak, T., Commutative Noetherian and Krull Rings, Wiley 1989

Balcerzyk, S. and Joze¬ak, T., Commutative Rings: Dimension, Multiplicity and Homo-

logical Methods, Wiley 1989

Eisenbud, D., Commutative Algebra with a view toward algebraic geometry, Springer-

Verlag 1995

Gopalakrishnan, N.S., Commutatilve Algebra, Oxonian Press (New Delhi) 1984

Kaplansky, I., Commutative Rings, Allyn and Bacon 1970

2

Kunz, E., Introduction to Commutative Algebra and Algebraic Geometry, Birkh¨user

a

1985

Matsumura, H., Commutatlive Ring Theory, Cambridge 1986

Raghavan, S., Singh, B., and Sridharan, S., Homological Methods in Commutative Alge-

bra, Oxford 1975

Serre, J-P., Local Albegra, Springer-Verlag 2000

Sharp, R.Y., Steps in Commutative Algebra, Cambridge 2000

c copyright 2003, by Robert B. Ash. Paper or electronic copies for noncommercial use

may be made freely without explicit permission of the author. All other rights are reserved.

Table of Contents

Chapter 0 Ring Theory Background

0.1 Prime Avoidance

0.2 Jacobson Radicals, Local Rings, and Other Miscellaneous Results

0.3 Nakayama™s Lemma

Chapter 1 Primary Decomposition and Associated Primes

1.1 Primary Submodules and Ideals

1.2 Primary Decomposition

1.3 Associated Primes

1.4 Associated Primes and Localization

1.5 The Support of a Module

1.6 Artinian Rings

Chapter 2 Integral Extensions

2.1 Integral Elements

2.2 Integrality and Localization

2.3 Going Down

Chapter 3 Valuation Rings

3.1 Extension Theorems

3.2 Properties of Valuation Rings

3.3 Discrete Valuation Rings

Chapter 4 Completion

4.1 Graded Rings and Modules

4.2 Completion of a Module

4.3 The Krull Intersection Theorem

1

2

Chapter 5 Dimension Theory

5.1 The Calculus of Finite Di¬erences

5.2 Hilbert and Hilbert-Samuel Polynomials

5.3 The Dimension Theorem

5.4 Consequences of the Dimension Theorem

5.5 Strengthening of Noether™s Normalization Lemma

5.6 Properties of A¬ne k-Algebras

Chapter 6 Depth

6.1 Systems of Parameters

6.2 Regular Sequences

Chapter 7 Homological Methods

7.1 Homological Dimension: Projective and Global

7.2 Injective Dimension

7.3 Tor and Dimension

7.4 Application

Chapter 8 Regular Local Rings

8.1 Basic De¬nitions and Examples

Exercises

Solutions

Chapter 0

Ring Theory Background

We collect here some useful results that might not be covered in a basic graduate algebra

course.

0.1 Prime Avoidance

Let P1 , P2 , . . . , Ps , s ≥ 2, be ideals in a ring R, with P1 and P2 not necessarily prime,

but P3 , . . . , Ps prime (if s ≥ 3). Let I be any ideal of R. The idea is that if we can avoid

the Pj individually, in other words, for each j we can ¬nd an element in I but not in Pj ,

then we can avoid all the Pj simultaneously, that is, we can ¬nd a single element in I that

is in none of the Pj . We will state and prove the contrapositive.

0.1.1 Prime Avoidance Lemma

With I and the Pi as above, if I ⊆ ∪s Pi , then for some i we have I ⊆ Pi .

i=1

Proof. Suppose the result is false. We may assume that I is not contained in the union

of any collection of s ’ 1 of the Pi ™s. (If so, we can simply replace s by s ’ 1.) Thus

for each i we can ¬nd an element ai ∈ I with ai ∈ P1 ∪ · · · ∪ Pi’1 ∪ Pi+1 ∪ · · · ∪ Ps . By

/

hypothesis, I is contained in the union of all the P ™s, so ai ∈ Pi . First assume s = 2, with

I ⊆ P1 and I ⊆ P2 . Then a1 ∈ P1 , a2 ∈ P1 , so a1 + a2 ∈ P1 . Similarly, a1 ∈ P2 , a2 ∈ P2 ,

/ / /

so a1 + a2 ∈ P2 . Thus a1 + a2 ∈ I ⊆ P1 ∪ P2 , contradicting a1 , a2 ∈ I. Note that P1

/ /

and P2 need not be prime for this argument to work. Now assume s > 2, and observe

that a1 a2 · · · as’1 ∈ P1 © · · · © Ps’1 , but as ∈ P1 ∪ · · · ∪ Ps’1 . Let a = (a1 · · · as’1 ) + as ,

/

which does not belong to P1 ∪ · · · ∪ Ps’1 , else as would belong to this set. Now for all

i = 1, . . . , s ’ 1 we have ai ∈ Ps , hence a1 · · · as’1 ∈ Ps because Ps is prime. But as ∈ Ps ,

/ /

so a cannot be in Ps . Thus a ∈ I and a ∈ P1 ∪ · · · ∪ Ps , contradicting the hypothesis. ™

/

It may appear that we only used the primeness of Ps , but after the preliminary reduc-

tion (see the beginning of the proof), it may very well happen that one of the other Pi ™s

now occupies the slot that previously housed Ps .

1

2 CHAPTER 0. RING THEORY BACKGROUND

0.2 Jacobson Radicals, Local Rings, and Other Mis-

cellaneous Results

0.2.1 Lemma

Let J(R) be the Jacobson radical of the ring R, that is, the intersection of all maximal

ideals of R. Then a ∈ J(R) i¬ 1 + ax is a unit for every x ∈ R.

Proof. Assume a ∈ J(R). If 1 + ax is not a unit, then it generates a proper ideal, hence

1 + ax belongs to some maximal ideal M. But then a ∈ M, hence ax ∈ M, and therefore

1 ∈ M, a contradiction. Conversely, if a fails to belong to a maximal ideal M, then

M + Ra = R. Thus for some b ∈ M and y ∈ R we have b + ay = 1. If x = ’y, then

1 + ax = b ∈ M, so 1 + ax cannot be a unit (else 1 ∈ M). ™

0.2.2 Lemma

Let M be a maximal ideal of the ring R. Then R is a local ring (a ring with a unique

maximal ideal, necessarily M) if and only if every element of 1 + M is a unit.

Proof. Suppose R is a local ring, and let a ∈ M. If 1 + a is not a unit, then it must

belong to M, which is the ideal of nonunits. But then 1 ∈ M, a contradiction. Conversely,

assume that every element of 1+M is a unit. We claim that M ⊆ J(R), hence M = J(R).

If a ∈ M, then ax ∈ M for every x ∈ R, so 1 + ax is a unit. By (0.2.1), a ∈ J(R), proving

the claim. If N is another maximal ideal, then M = J(R) ⊆ M © N . Thus M ⊆ N , and

since both ideals are maximal, they must be equal. Therefore R is a local ring. ™

0.2.3 Lemma

Let S be any subset of R, and let I be the ideal generated by S. Then I = R i¬ for every

maximal ideal M, there is an element x ∈ S \ M.

Proof. We have I ‚ R i¬ I, equivalently S, is contained in some maximal ideal M. In

other words, I ‚ R i¬ ∃M such that ∀x ∈ S we have x ∈ M. The contrapositive says

that I = R i¬ ∀M ∃x ∈ S such that x ∈ M. ™

/

0.2.4 Lemma

√ √

Let I and J be ideals of the ring R. Then I + J = R i¬ I+ J = R.

Proof. The “only if” part holds because any ideal is contained in its radical. Thus assume

that 1 = a + b with am ∈ I and bn ∈ J. Then

m+n i j

1 = (a + b)m+n = ab.

i

i+j=m+n

Now if i + j = m + n, then either i ≥ m or j ≥ n. Thus every term in the sum belongs

either to I or to J, hence to I + J. Consequently, 1 ∈ I + J. ™

0.3. NAKAYAMA™S LEMMA 3

0.3 Nakayama™s Lemma

First, we give an example of the determinant trick ; see (2.1.2) for another illustration.

0.3.1 Theorem

Let M be a ¬nitely generated R-module, and I an ideal of R such that IM = M . Then

there exists a ∈ I such that (1 + a)M = 0.

Proof. Let x1 , . . . , xn generate M . Since IM = M , we have equations of the form

n n

xi = j=1 aij xj , with aij ∈ I. The equations may be written as j=1 (δij ’ aij )xj = 0.

If In is the n by n identity matrix, we have (In ’ A)x = 0, where A = (aij ) and x is a

column vector whose coe¬cients are the xi . Premultiplying by the adjoint of (In ’ A),

we obtain ∆x = 0, where ∆ is the determinant of (In ’ A). Thus ∆xi = 0 for all i, hence

∆M = 0. But if we look at the determinant of In ’ A, we see that it is of the form 1 + a

for some element a ∈ I. ™

Here is a generalization of a familiar property of linear transformations on ¬nite-

dimensional vector spaces.

0.3.2 Theorem

If M is a ¬nitely generated R-module and f : M ’ M is a surjective homomorphism,

then f is an isomorphism.

Proof. We can make M into an R[X]-module via Xx = f (x), x ∈ M . (Thus X 2 x =

f (f (x)), etc.) Let I = (X); we claim that IM = M . For if m ∈ M , then by the

hypothesis that f is surjective, m = f (x) for some x ∈ M , and therefore Xx = f (x) = m.

But X ∈ I, so m ∈ IM . By (0.3.1), there exists g = g(X) ∈ I such that (1 + g)M = 0.

But by de¬nition of I, g must be of the form Xh(X) with h(X) ∈ R[X]. Thus (1+g)M =

[1 + Xh(X)]M = 0.

We can now prove that f is injective. Suppose that x ∈ M and f (x) = 0. Then

0 = [1 + Xh(X)]x = [1 + h(X)X]x = x + h(X)f (x) = x + 0 = x. ™

In (0.3.2), we cannot replace “surjective” by “injective”. For example, let f (x) = nx on

the integers. If n ≥ 2, then f is injective but not surjective.

The next result is usually referred to as Nakayama™s lemma. Sometimes, Akizuki and

Krull are given some credit, and as a result, a popular abbreviation for the lemma is

NAK.

0.3.3 NAK

(a) If M is a ¬nitely generated R-module, I an ideal of R contained in the Jacobson

radical J(R), and IM = M , then M = 0.

(b) If N is a submodule of the ¬nitely generated R-module M , I an ideal of R contained

in the Jacobson radical J(R), and M = N + IM , then M = N .

4 CHAPTER 0. RING THEORY BACKGROUND

Proof.

(a) By (0.3.1), (1 + a)M = 0 for some a ∈ I. Since I ⊆ J(R), 1 + a is a unit by (0.2.1).

Multiplying the equation (1 + a)M = 0 by the inverse of 1 + a, we get M = 0.

(b) By hypothesis, M/N = I(M/N ), and the result follows from (a). ™

Here is an application of NAK.

0.3.4 Proposition

Let R be a local ring with maximal ideal J. Let M be a ¬nitely generated R-module, and

let V = M/JM . Then

(i) V is a ¬nite-dimensional vector space over the residue ¬eld k = R/J.

(ii) If {x1 + JM, . . . , xn + JM } is a basis for V over k, then {x1 , . . . , xn } is a minimal

set of generators for M .

(iii) Any two minimal generating sets for M have the same cardinality.

Proof.

(i) Since J annihilates M/JM , V is a k-module, that is, a vector space over k. Since M

is ¬nitely generated over R, V is a ¬nite-dimensional vector space over k.

n

(ii) Let N = i=1 Rxi . Since the xi + JM generate V = M/JM , we have M = N + JM .

By NAK, M = N , so the xi generate M . If a proper subset of the xi were to generate

M , then the corresponding subset of the xi + JM would generate V , contradicting the

assumption that V is n-dimensional.

(iii) A generating set S for M with more than n elements determines a spanning set for

V , which must contain a basis with exactly n elements. By (ii), S cannot be minimal. ™

0.4 Localization

Let S be a subset of the ring R, and assume that S is multiplicative, in other words,

0 ∈ S, 1 ∈ S, and if a and b belong to S, so does ab. In the case of interest to us, S will

/

be the complement of a prime ideal. We would like to divide elements of R by elements

of S to form the localized ring S ’1 R, also called the ring of fractions of R by S. There

is no di¬culty when R is an integral domain, because in this case all division takes place

in the fraction ¬eld of R. We will sketch the general construction for arbitrary rings R.

For full details, see TBGY, Section 2.8.

0.4.1 Construction of the Localized Ring

If S is a multiplicative subset of the ring R, we de¬ne an equivalence relation on R — S

by (a, b) ∼ (c, d) i¬ for some s ∈ S we have s(ad ’ bc) = 0. If a ∈ R and b ∈ S, we de¬ne

the fraction a/b as the equivalence class of (a, b). We make the set of fractions into a ring

in a natural way. The sum of a/b and c/d is de¬ned as (ad + bc)/bd, and the product of

a/b and c/d is de¬ned as ac/bd. The additive identity is 0/1, which coincides with 0/s for

every s ∈ S. The additive inverse of a/b is ’(a/b) = (’a)/b. The multiplicative identity

is 1/1, which coincides with s/s for every s ∈ S. To summarize:

S ’1 R is a ring. If R is an integral domain, so is S ’1 R. If R is an integral domain and

S = R \ {0}, then S ’1 R is a ¬eld, the fraction ¬eld of R.

0.4. LOCALIZATION 5

There is a natural ring homomorphism h : R ’ S ’1 R given by h(a) = a/1. If S

has no zero-divisors, then h is a monomorphism, so R can be embedded in S ’1 R. In

particular, a ring R can be embedded in its full ring of fractions S ’1 R, where S consists

of all non-divisors of 0 in R. An integral domain can be embedded in its fraction ¬eld.

Our goal is to study the relation between prime ideals of R and prime ideals of S ’1 R.

0.4.2 Lemma

If X is any subset of R, de¬ne S ’1 X = {x/s : x ∈ X, s ∈ S}. If I is an ideal of R, then

S ’1 I is an ideal of S ’1 R. If J is another ideal of R, then

(i) S ’1 (I + J) = S ’1 I + S ’1 J;

(ii) S ’1 (IJ) = (S ’1 I)(S ’1 J);

(iii) S ’1 (I © J) = (S ’1 I) © (S ’1 J);

(iv) S ’1 I is a proper ideal i¬ S © I = ….

Proof. The de¬nitions of addition and multiplication in S ’1 R imply that S ’1 R is an

ideal, and that in (i), (ii) and (iii), the left side is contained in the right side. The reverse

inclusions in (i) and (ii) follow from

ab at + bs a b ab

+= , =.

s t st st st

To prove (iii), let a/s = b/t, where a ∈ I, b ∈ J, s, t ∈ S. There exists u ∈ S such that

u(at ’ bs) = 0. Then a/s = uat/ust = ubs/ust ∈ S ’1 (I © J).

Finally, if s ∈ S © I, then 1/1 = s/s ∈ S ’1 I, so S ’1 I = S ’1 R. Conversely, if

S ’1 I = S ’1 R, then 1/1 = a/s for some a ∈ I, s ∈ S. There exists t ∈ S such that

t(s ’ a) = 0, so at = st ∈ S © I. ™

Ideals in S ’1 R must be of a special form.

0.4.3 Lemma

Let h be the natural homomorphism from R to S ’1 R [see (0.4.1)]. If J is an ideal of

S ’1 R and I = h’1 (J), then I is an ideal of R and S ’1 I = J.

Proof. I is an ideal by the basic properties of preimages of sets. Let a/s ∈ S ’1 I, with

a ∈ I and s ∈ S. Then a/1 = h(a) ∈ J, so a/s = (a/1)(1/s) ∈ J. Conversely, let a/s ∈ J,

with a ∈ R, s ∈ S. Then h(a) = a/1 = (a/s)(s/1) ∈ J, so a ∈ I and a/s ∈ S ’1 I. ™

Prime ideals yield sharper results.

0.4.4 Lemma

If I is any ideal of R, then I ⊆ h’1 (S ’1 I). There will be equality if I is prime and disjoint

from S.

Proof. If a ∈ I, then h(a) = a/1 ∈ S ’1 I. Thus assume that I is prime and disjoint from

S, and let a ∈ h’1 (S ’1 I). Then h(a) = a/1 ∈ S ’1 I, so a/1 = b/s for some b ∈ I, s ∈ S.

There exists t ∈ S such that t(as ’ b) = 0. Thus ast = bt ∈ I, with st ∈ I because

/

S © I = …. Since I is prime, we have a ∈ I. ™

6 CHAPTER 0. RING THEORY BACKGROUND

0.4.5 Lemma

If I is a prime ideal of R disjoint from S, then S ’1 I is a prime ideal of S ’1 R.

Proof. By part (iv) of (0.4.2), S ’1 I is a proper ideal. Let (a/s)(b/t) = ab/st ∈ S ’1 I,

with a, b ∈ R, s, t ∈ S. Then ab/st = c/u for some c ∈ I, u ∈ S. There exists v ∈ S such

that v(abu ’ cst) = 0. Thus abuv = cstv ∈ I, and uv ∈ I because S © I = …. Since I is

/

prime, ab ∈ I, hence a ∈ I or b ∈ I. Therefore either a/s or b/t belongs to S ’1 I. ™

The sequence of lemmas can be assembled to give a precise conclusion.

0.4.6 Theorem

There is a one-to-one correspondence between prime ideals P of R that are disjoint from

S and prime ideals Q of S ’1 R, given by

P ’ S ’1 P and Q ’ h’1 (Q).

Proof. By (0.4.3), S ’1 (h’1 (Q)) = Q, and by (0.4.4), h’1 (S ’1 P ) = P . By (0.4.5), S ’1 P

is a prime ideal, and h’1 (Q) is a prime ideal by the basic properties of preimages of sets.

If h’1 (Q) meets S, then by (0.4.2) part (iv), Q = S ’1 (h’1 (Q)) = S ’1 R, a contradiction.

Thus the maps P ’ S ’1 P and Q ’ h’1 (Q) are inverses of each other, and the result

follows. ™

0.4.7 De¬nitions and Comments

If P is a prime ideal of R, then S = R \ P is a multiplicative set. In this case, we write

RP for S ’1 R, and call it the localization of R at P . We are going to show that RP is

a local ring, that is, a ring with a unique maximal ideal. First, we give some conditions

equivalent to the de¬nition of a local ring.

0.4.8 Proposition

For a ring R, the following conditions are equivalent.

(i) R is a local ring;

(ii) There is a proper ideal I of R that contains all nonunits of R;

(iii) The set of nonunits of R is an ideal.

Proof.

(i) implies (ii): If a is a nonunit, then (a) is a proper ideal, hence is contained in the

unique maximal ideal I.

(ii) implies (iii): If a and b are nonunits, so are a + b and ra. If not, then I contains a

unit, so I = R, contradicting the hypothesis.

(iii) implies (i): If I is the ideal of nonunits, then I is maximal, because any larger ideal J

would have to contain a unit, so J = R. If H is any proper ideal, then H cannot contain

a unit, so H ⊆ I. Therefore I is the unique maximal ideal. ™

0.4. LOCALIZATION 7

0.4.9 Theorem

RP is a local ring.

Proof. Let Q be a maximal ideal of RP . Then Q is prime, so by (0.4.6), Q = S ’1 I

for some prime ideal I of R that is disjoint from S = R \ P . In other words, I ⊆ P .

Consequently, Q = S ’1 I ⊆ S ’1 P . If S ’1 P = RP = S ’1 R, then by (0.4.2) part (iv), P

is not disjoint from S = R \ P , which is impossible. Therefore S ’1 P is a proper ideal

containing every maximal ideal, so it must be the unique maximal ideal. ™

0.4.10 Remark

It is convenient to write the ideal S ’1 I as IRP . There is no ambiguity, because the

product of an element of I and an arbitrary element of R belongs to I.

0.4.11 Localization of Modules

If M is an R-module and S a multiplicative subset of R, we can essentially repeat the

construction of (0.4.1) to form the localization of M by S, and thereby divide elements

of M by elements of S. If x, y ∈ M and s, t ∈ S, we call (x, s) and (y, t) equivalent if for

some u ∈ S, we have u(tx ’ sy) = 0. The equivalence class of (x, s) is denoted by x/s,

and addition is de¬ned by

xy tx + sy

+= .

s t st

If a/s ∈ S ’1 R and x/t ∈ S ’1 M , we de¬ne

ax ax

= .

st st

In this way, S ’1 M becomes an S ’1 R-module. Exactly as in (0.4.2), if M and N are

submodules of an R-module L, then

S ’1 (M + N ) = S ’1 M + S ’1 N and S ’1 (M © N ) = (S ’1 M ) © (S ’1 N ).

Chapter 1

Primary Decomposition and

Associated Primes

1.1 Primary Submodules and Ideals

1.1.1 De¬nitions and Comments

If N is a submodule of the R-module M , and a ∈ R, let »a : M/N ’ M/N be mul-

tiplication by a. We say that N is a primary submodule of M if N is proper and for

every a, »a is either injective or nilpotent. Injectivity means that for all x ∈ M , we have

ax ∈ N ’ x ∈ N . Nilpotence means that for some positive integer n, an M ⊆ N , that is,

an belongs to the annihilator of M/N , denoted by ann(M/N ). Equivalently, a belongs to

the radical of the annihilator of M/N , denoted by rM (N ).

Note that »a cannot be both injective and nilpotent. If so, nilpotence gives an M =

a(an’1 M ) ⊆ N , and injectivity gives an’1 M ⊆ N . Inductively, M ⊆ N , so M = N ,

contradicting the assumption that N is proper. Thus if N is a primary submodule of M ,

then rM (N ) is the set of all a ∈ R such that »a is not injective. Since rM (N ) is the radical

of an ideal, it is an ideal of R, and in fact it is a prime ideal. For if »a and »b fail to be

injective, so does »ab = »a —¦ »b . (Note that rM (N ) is proper because »1 is injective.) If

P = rM (N ), we say that N is P -primary.

√

If I is any ideal of R, then rR (I) = I, because ann(R/I) = I. (Note that a ∈

ann(R/I) i¬ aR ⊆ I i¬ a = a1 ∈ I.)

Specializing to M = R and replacing a by y, we de¬ne a primary ideal in a ring R

as a proper ideal Q such that if xy ∈ Q, then either x ∈ Q or y n ∈ Q for some n ≥ 1.

Equivalently, R/Q = 0 and every zero-divisor in R/Q is nilpotent.

√

A useful observation is that if P is a prime ideal, then P n = P for all n ≥ 1. (The

radical of P n is the intersection of all prime ideals containing P n , one of which is P . Thus

√ √

P n ⊆ P . Conversely, if x ∈ P , then xn ∈ P n , so x ∈ P n .)

1

2 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

1.1.2 Lemma

√

I is a maximal ideal M, then I is M-primary.

If

√

Proof. Suppose that ab ∈ I and b does not belong to I = M. Then by maximality of

M, it follows that M + Rb = R, so for some m ∈ M and r ∈ R we have m + rb = 1. Now

√

m ∈ M = I, hence mk ∈ I for some k ≥ 1. Thus 1 = 1k = (m + rb)k = mk + sb for

some s ∈ R. Multiply by a to get a = amk + sab ∈ I. ™

1.1.3 Corollary

If M is a maximal ideal, then Mn is M-primary for every n ≥ 1.

√

Proof. As we observed in (1.1.1), Mn = M, and the result follows from (1.1.2). ™

1.2 Primary Decomposition

1.2.1 De¬nitions and Comments

A primary decomposition of the submodule N of M is given by N = ©r Ni , where the

i=1

Ni are Pi -primary submodules. The decomposition is reduced if the Pi are distinct and

N cannot be expressed as the intersection of a proper subcollection of the Ni .

We can always extract a reduced primary decomposition from an unreduced one, by

discarding those Ni that contain ©j=i Nj and intersecting those Ni that are P -primary

for the same P . The following result justi¬es this process.

1.2.2 Lemma

If N1 , . . . , Nk are P -primary, then ©k Ni is P -primary.

i=1

Proof. We may assume that k = 2; an induction argument takes care of larger values.

Let N = N1 © N2 and rM (N1 ) = rM (N2 ) = P . Assume for the moment that rM (N ) = P .

If a ∈ R, x ∈ M, ax ∈ N , and a ∈ rM (N ), then since N1 and N2 are P -primary, we have

/

x ∈ N1 © N2 = N . It remains to show that rM (N ) = P . If a ∈ P , then there are positive

integers n1 and n2 such that an1 M ⊆ N1 and an2 M ⊆ N2 . Therefore an1 +n2 M ⊆ N , so

a ∈ rM (N ). Conversely, if a ∈ rM (N ) then a belongs to rM (Ni ) for i = 1, 2, and therefore

a ∈ P. ™

We now prepare to prove that every submodule of a Noetherian module has a primary

decomposition.

1.2.3 De¬nition

The proper submodule N of M is irreducible if N cannot be expressed as N1 © N2 with

N properly contained in the submodules Ni , i = 1, 2.

1.3. ASSOCIATED PRIMES 3

1.2.4 Proposition

If N is an irreducible submodule of the Noetherian module M , then N is primary.

Proof. If not, then for some a ∈ R, »a : M/N ’ M/N is neither injective nor nilpotent.

The chain ker »a ⊆ ker »2 ⊆ ker »3 ⊆ · · · terminates by the ascending chain condition, say

a a

at ker »a . Let • = »a ; then ker • = ker •2 and we claim that ker • © im • = 0. Suppose

i i

x ∈ ker • © im •, and let x = •(y). Then 0 = •(x) = •2 (y), so y ∈ ker •2 = ker •, so

x = •(y) = 0.

Now »a is not injective, so ker • = 0, and »a is not nilpotent, so »i can™t be 0 (because

a

ai M ⊆ N ). Consequently, im • = 0.

Let p : M ’ M/N be the canonical epimorphism, and set N1 = p’1 (ker •), N2 =

p’1 (im •). We will prove that N = N1 © N2 . If x ∈ N1 © N2 , then p(x) belongs to

both ker • and im •, so p(x) = 0, in other words, x ∈ N . Conversely, if x ∈ N , then

p(x) = 0 ∈ ker • © im •, so x ∈ N1 © N2 .

Finally, we will show that N is properly contained in both N1 and N2 , so N is reducible,

a contradiction. Choose a nonzero element y ∈ ker •. Since p is surjective, there exists

x ∈ M such that p(x) = y. Thus x ∈ p’1 (ker •) = N1 (because y = p(x) ∈ ker •), but

x ∈ N (because p(x) = y = 0). Similarly, N ‚ N2 (with 0 = y ∈ im •), and the result

/

follows. ™

1.2.5 Theorem

If N is a proper submodule of the Noetherian module M , then N has a primary decom-

position, hence a reduced primary decomposition.

Proof. We will show that N can be expressed as a ¬nite intersection of irreducible sub-

modules of M , so that (1.2.4) applies. Let S be the collection of all submodules of M

that cannot be expressed in this form. If S is nonempty, then S has a maximal element

N (because M is Noetherian). By de¬nition of S, N must be reducible, so we can write

N = N1 © N2 , N ‚ N1 , N ‚ N2 . By maximality of N , N1 and N2 can be expressed

as ¬nite intersections of irreducible submodules, hence so can N , contradicting N ∈ S.

Thus S is empty. ™

1.3 Associated Primes

1.3.1 De¬nitions and Comments

Let M be an R-module, and P a prime ideal of R. We say that P is an associated prime

of M (or that P is associated to M ) if P is the annihilator of some nonzero x ∈ M . The

set of associated primes of M is denoted by AP(M ). (The standard notation is Ass(M).

Please do not use this regrettable terminology.)

Here is a useful characterization of associated primes.

1.3.2 Proposition

The prime ideal P is associated to M if and only if there is an injective R-module homo-

morphism from R/P to M . Therefore if N is a submodule of M , then AP(N ) ⊆ AP(M ).

4 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

Proof. If P is the annihilator of x = 0, the desired homomorphism is given by r +P ’ rx.

Conversely, if an injective R-homomorphism from R/P to M exists, let x be the image of

1 + P , which is nonzero in R/P . By injectivity, x = 0. We will show that P = annR (x),

the set of elements r ∈ R such that rx = 0. If r ∈ P , then r + P = 0, so rx = 0, and

therefore r ∈ annR (x). If rx = 0, then by injectivity, r + P = 0, so r ∈ P . ™

Associated primes exist under wide conditions, and are sometimes unique.

1.3.3 Proposition

If M = 0, then AP(M ) is empty. The converse holds if R is a Noetherian ring.

Proof. There are no nonzero elements in the zero module, hence no associated primes.

Assuming that M = 0 and R is Noetherian, there is a maximal element I = annR x in

the collection of all annihilators of nonzero elements of M . The ideal I must be proper,

for if I = R, then x = 1x = 0, a contradiction. If we can show that I is prime, we have

I ∈ AP(M ), as desired. Let ab ∈ I with a ∈ I. Then abx = 0 but ax = 0, so b ∈ ann(ax).

/

But I = ann x ⊆ ann(ax), and the maximality of I gives I = ann(ax). Consequently,

b ∈ I. ™

1.3.4 Proposition

For any prime ideal P , AP(R/P ) = {P }.

Proof. By (1.3.2), P is an associated prime of R/P because there certainly is an R-

monomorphism from R/P to itself. If Q ∈ AP(R/P ), we must show that Q = P .

Suppose that Q = ann(r + P ) with r ∈ P . Then s ∈ Q i¬ sr ∈ P i¬ s ∈ P (because P is

/

prime). ™

1.3.5 Remark

Proposition 1.3.4 shows that the annihilator of any nonzero element of R/P is P .

The next result gives us considerable information about the elements that belong to

associated primes.

1.3.6 Theorem

Let z(M ) be the set of zero-divisors of M , that is, the set of all r ∈ R such that rx = 0

for some nonzero x ∈ M . Then ∪{P : P ∈ AP(M )} ⊆ z(M ), with equality if R is

Noetherian.

Proof. The inclusion follows from the de¬nition of associated prime; see (1.3.1). Thus

assume a ∈ z(M ), with ax = 0, x ∈ M, x = 0. Then Rx = 0, so by (1.3.3) [assuming R

Noetherian], Rx has an associated prime P = ann(bx). Since ax = 0 we have abx = 0, so

a ∈ P . But P ∈ AP(Rx) ⊆ AP(M ) by (1.3.2). Therefore a ∈ ∪{P : P ∈ AP(M )}. ™

Now we prove a companion result to (1.3.2).

1.3. ASSOCIATED PRIMES 5

1.3.7 Proposition

If N is a submodule of M , then AP(M ) ⊆ AP(N ) ∪ AP(M/N ).

Proof. Let P ∈ AP(M ), and let h : R/P ’ M be a monomorphism. Set H = h(R/P )

and L = H © N .

Case 1: L = 0. Then the map from H to M/N given by h(r + P ) ’ h(r + P ) + N is

a monomorphism. (If h(r + P ) belongs to N , it must belong to H © N = 0.) Thus H

is isomorphic to a submodule of M/N , so by de¬nition of H, there is a monomorphism

from R/P to M/N . Thus P ∈ AP(M/N ).

Case 2: L = 0. If L has a nonzero element x, then x must belong to both H and N , and

H is isomorphic to R/P via h. Thus x ∈ N and the annihilator of x coincides with the

annihilator of some nonzero element of R/P . By (1.3.5), ann x = P , so P ∈ AP(N ). ™

1.3.8 Corollary

AP( Mj = AP(Mj ).

j∈J j∈J

Proof. By (1.3.2), the right side is contained in the left side. The result follows from

(1.3.7) when the index set is ¬nite. For example,

AP(M1 • M2 • M3 ) ⊆ AP(M1 ) ∪ AP(M/M1 )

= AP(M1 ) ∪ AP(M2 • M3 )

⊆ AP(M1 ) ∪ AP(M2 ) ∪ AP(M3 ).

In general, if P is an associated prime of the direct sum, then there is a monomorphism

from R/P to •Mj . The image of the monomorphism is contained in the direct sum of

¬nitely many components, as R/P is generated as an R-module by the single element

1 + P . This takes us back to the ¬nite case. ™

We now establish the connection between associated primes and primary decomposi-

tion, and show that under wide conditions, there are only ¬nitely many associated primes.

1.3.9 Theorem

Let M be a nonzero ¬nitely generated module over the Noetherian ring R, so that by

(1.2.5), every proper submodule of M has a reduced primary decomposition. In particular,

the zero module can be expressed as ©r Ni , where Ni is Pi -primary. Then AP(M ) =

i=1

{P1 , . . . , Pr }, a ¬nite set.

Proof. Let P be an associated prime of M , so that P = ann(x), x = 0, x ∈ M . Renumber

the Ni so that x ∈ Ni for 1 ¤ i ¤ j and x ∈ Ni for j + 1 ¤ i ¤ r. Since Ni is Pi -primary,

/

we have Pi = rM (Ni ) (see (1.1.1)). Since Pi is ¬nitely generated, Pini M ⊆ Ni for some

ni ≥ 1. Therefore

j r

⊆

Pini )x

( Ni = (0)

i=1 i=1

6 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

so ©j Pini ⊆ ann(x) = P . (By our renumbering, there is a j rather than an r on the left

i=1

side of the inclusion.) Since P is prime, Pi ⊆ P for some i ¤ j. We claim that Pi = P ,

so that every associated prime must be one of the Pi . To verify this, let a ∈ P . Then

ax = 0 and x ∈ Ni , so »a is not injective and therefore must be nilpotent. Consequently,

/

a ∈ rM (Ni ) = Pi , as claimed.

Conversely, we show that each Pi is an associated prime. Without loss of generality, we

may take i = 1. Since the decomposition is reduced, N1 does not contain the intersection

of the other Ni ™s, so we can choose x ∈ N2 ©· · ·©Nr with x ∈ N1 . Now N1 is P1 -primary, so

/

as in the preceding paragraph, for some n ≥ 1 we have P1 x ⊆ N1 but P1 x ⊆ N1 . (Take

n’1

n

P1 x = Rx and recall that x ∈ N1 .) If we choose y ∈ P1 x \ N1 (hence y = 0), the proof

n’1

0

/

will be complete upon showing that P1 is the annihilator of y. We have P1 y ⊆ P1 x ⊆ N1

n

and x ∈ ©r Ni , so P1 x ⊆ ©r Ni . Thus P1 y ⊆ ©r Ni = (0), so P1 ⊆ ann y. On the

n

i=2 i=2 i=1

other hand, if a ∈ R and ay = 0, then ay ∈ N1 but y ∈ N1 , so »a : M/N1 ’ M/N1 is not

/

injective and is therefore nilpotent. Thus a ∈ rM (N1 ) = P1 . ™

We can now say something about uniqueness in primary decompositions.

1.3.10 First Uniqueness Theorem

Let M be a ¬nitely generated module over the Noetherian ring R. If N = ©r Ni is a

i=1

reduced primary decomposition of the submodule N , and Ni is Pi -primary, i = 1, . . . , r,

then (regarding M and R as ¬xed) the Pi are uniquely determined by N .

Proof. By the correspondence theorem, a reduced primary decomposition of (0) in M/N

is given by (0) = ©r Ni /N , and Ni /N is Pi -primary, 1 ¤ i ¤ r. By (1.3.9),

i=1

AP(M/N ) = {P1 , . . . , Pr }.

But [see (1.3.1)] the associated primes of M/N are determined by N . ™

1.3.11 Corollary

Let N be a submodule of M (¬nitely generated over the Noetherian ring R). Then N is

P -primary i¬ AP(M/N ) = {P }.

Proof. The “only if” part follows from the displayed equation above. Conversely, if P is

the only associated prime of M/N , then N coincides with a P -primary submodule N ,

and hence N (= N ) is P -primary. ™

1.3.12 De¬nitions and Comments

Let N = ©r Ni be a reduced primary decomposition, with associated primes P1 , . . . , Pr .

i=1

We say that Ni is an isolated (or minimal ) component if Pi is minimal, that is Pi does not

properly contain any Pj , j = i. Otherwise, Ni is an embedded component (see Exercise 5

for an example). Embedded components arise in algebraic geometry in situations where

one irreducible algebraic set is properly contained in another.

1.4. ASSOCIATED PRIMES AND LOCALIZATION 7

1.4 Associated Primes and Localization

To get more information about uniqueness in primary decompositions, we need to look at

associated primes in localized rings and modules. In this section, S will be a multiplicative

subset of the Noetherian ring R, RS the localization of R by S, and MS the localization

of the R-module M by S. Recall that P ’ PS = P RS is a bijection of C, the set of prime

ideals of R not meeting S, and the set of all prime ideals of RS .

The set of associated primes of the R-module M will be denoted by APR (M ). We

need a subscript to distinguish this set from APRS (MS ), the set of associated primes of

the RS -module MS .

1.4.1 Lemma

Let P be a prime ideal not meeting S. If P ∈ APR (M ), then PS = P RS ∈ APRS (MS ).

(By the above discussion, the map P ’ PS is the restriction of a bijection and therefore

must be injective.)

Proof. If P is the annihilator of the nonzero element x ∈ M , then PS is the annihilator

of the nonzero element x/1 ∈ MS . (By (1.3.6), no element of S can be a zero-divisor,

so x/1 is indeed nonzero.) For if a ∈ P and a/s ∈ PS , then (a/s)(x/1) = ax/s = 0.

Conversely, if (a/s)(x/1) = 0, then there exists t ∈ S such that tax = 0, and it follows

that a/s = at/st ∈ PS . ™

1.4.2 Lemma

The map of (1.4.1) is surjective, hence is a bijection of APR (M ) © C and APRS (MS ).

Proof. Let P be generated by a1 , . . . , an . Suppose that PS is the annihilator of the

nonzero element x/t ∈ MS . Then (ai /1)(x/t) = 0, 1 ¤ i ¤ n. For each i there exists

si ∈ S such that si ai x = 0. If s is the product of the si , then sai x = 0 for all i, hence

sax = 0 for all a ∈ P . Thus P ⊆ ann(sx). On the other hand, suppose b annihilates sx.

Then (b/1)(x/t) = bsx/st = 0, so b/1 ∈ PS , and consequently b/1 = b /s for some b ∈ P

and s ∈ S. This means that for some u ∈ S we have u(bs ’ b ) = 0. Now b , hence ub ,

belongs to P , and therefore so does ubs . But us ∈ P (because S © P = …). We conclude

/

that b ∈ P , so P = ann(sx). As in (1.4.1), s cannot be a zero-divisor, so sx = 0 and the

proof is complete. ™

1.4.3 Lemma

Let M be a ¬nitely generated module over the Noetherian ring R, and N a P -primary

submodule of M . Let P be any prime ideal of R, and set M = MP , N = NP . If

P ⊆ P , then N = M .

Proof. By (1.4.1) and (1.4.2), there is a bijection between APRP (M/N )P (which coin-

cides with APRP (M /N )) and the intersection APR (M/N ) © C, where C is the set of

prime ideals contained in P (in other words, not meeting S = R \ P ). By (1.3.11), there

is only one associated prime of M/N over R, namely P , which is not contained in P by

hypothesis. Thus APR (M/N ) © C is empty, so by (1.3.3), M /N = 0, and the result

follows. ™

8 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

At the beginning of the proof of (1.4.3), we have taken advantage of the isomorphism

between (M/N )P and M /N . The result comes from the exactness of the localization

functor. If this is unfamiliar, look ahead to the proof of (1.5.3), where the technique is

spelled out. See also TBGY, Section 8.5, Problem 5.

1.4.4 Lemma

In (1.4.3), if P ⊆ P , then N = f ’1 (N ), where f is the natural map from M to M .

Proof. As in (1.4.3), APR (M/N ) = {P }. Since P ⊆ P , we have R \ P ⊆ R \ P . By

(1.3.6), R \ P contains no zero-divisors of M/N , because all such zero-divisors belong to

P . Thus the natural map g : x ’ x/1 of M/N to (M/N )P ∼ M /N is injective. (If

=

x/1 = 0, then sx = 0 for some s ∈ S = R \ P , and since s is not a zero-divisor, we have

x = 0.)

If x ∈ N , then f (x) ∈ N by de¬nition of f , so assume x ∈ f ’1 (N ). Then f (x) ∈ N ,

so f (x) + N is 0 in M /N . By injectivity of g, x + N is 0 in M/N , in other words, x ∈ N ,

and the result follows. ™

1.4.5 Second Uniqueness Theorem

Let M be a ¬nitely generated module over the Noetherian ring R. Suppose that N =

©r Ni is a reduced primary decomposition of the submodule N , and Ni is Pi -primary,

i=1

i = 1, . . . , r. If Pi is minimal, then (regarding M and R as ¬xed) Ni is uniquely determined

by N .

Proof. Suppose that P1 is minimal, so that P1 ⊇ Pi , i > 1. By (1.4.3) with P =

Pi , P = P1 , we have (Ni )P1 = MP1 for i > 1. By (1.4.4) with P = P = P1 , we have

N1 = f ’1 [(N1 )P1 ], where f is the natural map from M to MP1 . Now

NP1 = (N1 )P1 © ©r (Ni )P1 = (N1 )P1 © MP1 = (N1 )P1 .

i=2

Thus N1 = f ’1 [(N1 )P1 ] = f ’1 (NP1 ) depends only on N and P1 , and since P1 depends

on the ¬xed ring R, it follows that N1 depends only on N . ™

1.5 The Support of a Module

The support of a module M is closely related to the set of associated primes of M . We

will need the following result in order to proceed.

1.5.1 Proposition

M is the zero module if and only if MP = 0 for every prime ideal P , if and only if MM = 0

for every maximal ideal M.

Proof. It su¬ces to show that if MM = 0 for all maximal ideals M, then M = 0.

Choose a nonzero element x ∈ M , and let I be the annihilator of x. Then 1 ∈ I (because

/

1x = x = 0), so I is a proper ideal and is therefore contained in a maximal ideal M. By

hypothesis, x/1 is 0 in MM , hence there exists a ∈ M (so a ∈ I) such that ax = 0. But

/ /

then by de¬nition of I we have a ∈ I, a contradiction. ™

1.5. THE SUPPORT OF A MODULE 9

1.5.2 De¬nitions and Comments

The support of an R-module M (notation Supp M ) is the set of prime ideals P of R such

that MP = 0. Thus Supp M = … i¬ MP = 0 for all prime ideals P . By (1.5.1), this is

equivalent to M = 0.

If I is any ideal of R, we de¬ne V (I) as the set of prime ideals containing I. In

algebraic geometry, the Zariski topology on Spec R has the sets V (I) as its closed sets.

1.5.3 Proposition

Supp R/I = V (I).

Proof. We apply the localization functor to the exact sequence 0 ’ I ’ R ’ R/I ’ 0 to

get the exact sequence 0 ’ IP ’ RP ’ (R/I)P ’ 0. Consequently, (R/I)P ∼ RP /IP .=

Thus P ∈ Supp R/I i¬ RP ⊃ IP i¬ IP is contained in a maximal ideal, necessarily P RP .

But this is equivalent to I ⊆ P . To see this, suppose a ∈ I, with a/1 ∈ IP ⊆ P RP . Then

a/1 = b/s for some b ∈ P, s ∈ P . There exists c ∈ P such that c(as ’ b) = 0. We have

/ /

cas = cb ∈ P , a prime ideal, and cs ∈ P . We conclude that a ∈ P . ™

/

1.5.4 Proposition

Let 0 ’ M ’ M ’ M ’ 0 be exact, hence 0 ’ MP ’ MP ’ MP ’ 0 is exact.

Then

Supp M = Supp M ∪ Supp M .

Proof. Let P belong to Supp M \ Supp M . Then MP = 0, so the map MP ’ MP is

injective as well as surjective, hence is an isomorphism. But MP = 0 by assumption, so

MP = 0, and therefore P ∈ Supp M . On the other hand, since MP is isomorphic to a

submodule of MP , it follows that Supp M ⊆ Supp M . If MP = 0, then MP = 0 (because

MP ’ MP is surjective). Thus Supp M ⊆ Supp M . ™

Supports and annihilators are connected by the following basic result.

1.5.5 Theorem

If M is a ¬nitely generated R-module, then Supp M = V (ann M ).

Proof. Let M = Rx1 + · · · + Rxn , so that MP = (Rx1 )P + · · · + (Rxn )P . Then Supp M =

∪n Supp Rxi , and by the ¬rst isomorphism theorem, Rxi ∼ R/ ann xi . By (1.5.3),

=

i=1

Supp Rxi = V (ann xi ). Therefore Supp M = ∪i=1 V (ann xi ) = V (ann M ). To justify

n

the last equality, note that if P ∈ V (ann xi ), then P ⊇ ann xi ⊇ ann M . Conversely, if

P ⊇ ann M = ©n ann xi , then P ⊇ ann xi for some i. ™

i=1

And now we connect associated primes and annihilators.

10 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

1.5.6 Proposition

If M is a ¬nitely generated module over the Noetherian ring R, then

√

P = ann M .

P ∈AP(M )

Proof. If M = 0, then by (1.3.3), AP(M ) = …, and the result to be proved is R = R.

Thus assume M = 0, so that (0) is a proper submodule. By (1.2.5) and (1.3.9), there is

a reduced primary decomposition (0) = ©r Ni , where for each i, Ni is Pi -primary and

i=1

AP(M ) =√ 1 , . . . , Pr }.

{P

If a ∈ ann M , then for some n ≥ 1 we have an M = 0. Thus for each i, »a : M/Ni ’

M/Ni is nilpotent [see (1.1.1)]. Consequently, a ∈ ©r rM (Ni ) = ©r Pi . Conversely, if

i=1 i=1

a belongs to this intersection, then √ all i there exists ni ≥ 1 such that ani M ⊆ Ni . If

for

n = max ni , then an M = 0, so a ∈ ann M . ™

1.5.7 Corollary

If R is a Noetherian ring, then the nilradical of R is the intersection of all associated

primes of R.

√

Proof. Take M = R in (1.5.6). Since ann R = 0, ann R is the nilradical. ™

And now, a connection between supports, associated primes and annihilators.

1.5.8 Proposition

Let M be a ¬nitely generated module over the Noetherian ring R, and let P be any prime

ideal of R. The following conditions are equivalent:

(1) P ∈ Supp M ;

(2) P ⊇ P for some P ∈ AP(M );

(3) P ⊇ ann M .

Proof. Conditions (1) and (3) are equivalent by (1.5.5). To prove that (1) implies (2),

let P ∈ Supp M . If P does not contain any associated prime of M , then P does not

contain the intersection of all associated primes (because P is prime). By (1.5.6), P does

√

not contain ann M , hence P cannot contain the smaller ideal ann M . This contradicts

(1.5.5). To prove that (2) implies (3), let Q be the intersection of all associated primes.

√

Then P ⊇ P ⊇ Q = [by (1.5.6)] ann M ⊇ ann M . ™

Here is the most important connection between supports and associated primes.

1.5.9 Theorem

Let M be a ¬nitely generated module over the Noetherian ring R. Then AP(M ) ⊆

Supp M , and the minimal elements of AP(M ) and Supp M are the same.

Proof. We have AP(M ) ⊆ Supp M by (2) implies (1) in (1.5.8), with P = P . If P is

minimal in Supp M , then by (1) implies (2) in (1.5.8), P contains some P ∈ AP(M ) ⊆

1.6. ARTINIAN RINGS 11

Supp M . By minimality, P = P . Thus P ∈ AP(M ), and in fact, P must be a minimal

associated prime. Otherwise, P ⊃ Q ∈ AP(M ) ⊆ Supp M , so that P is not minimal

in Supp M , a contradiction. Finally, let P be minimal among associated primes but not

minimal in Supp M . If P ⊃ Q ∈ Supp M , then by (1) implies (2) in (1.5.8), Q ⊇ P ∈

AP(M ). By minimality, P = P , contradicting P ⊃ Q ⊇ P . ™

Here is another way to show that there are only ¬nitely many associated primes.

1.5.10 Theorem

Let M be a nonzero ¬nitely generated module over the Noetherian ring R. Then there is

a chain of submodules 0 = M0 < M1 < · · · < Mn = M such that for each j = 1, . . . , n,

Mj /Mj’1 ∼ R/Pj , where the Pj are prime ideals of R. For any such chain, AP(M ) ⊆

=

{P1 , . . . , Pn }.

Proof. By (1.3.3), M has an associated prime P1 = ann x1 , with x1 a nonzero element

of M . Take M1 = Rx1 ∼ R/P1 (apply the ¬rst isomorphism theorem). If M = M1 ,

=

then the quotient module M/M1 is nonzero, hence [again by (1.3.3)] has an associated

prime P2 = ann(x2 + M1 ), x2 ∈ M1 . Let M2 = M1 + Rx2 . Now map R onto M2 /M1 by

/

r ’ rx2 + M1 . By the ¬rst isomorphism theorem, M2 /M1 ∼ R/P2 . Continue inductively

=

to produce the desired chain. (Since M is Noetherian, the process terminates in a ¬nite

number of steps.) For each j = 1, . . . , n, we have AP(Mj ) ⊆ AP(Mj’1 ) ∪ {Pj } by (1.3.4)

and (1.3.7). Another inductive argument shows that AP(M ) ⊆ {P1 , . . . , Pn }. ™

1.5.11 Proposition

In (1.5.10), each Pj belongs to Supp M . Thus (replacing AP(M ) by {P1 , . . . , Pn } in the

proof of (1.5.9)), the minimal elements of all three sets AP(M ), {P1 , . . . , Pn } and Supp M

are the same.

Proof. By (1.3.4) and (1.5.9), Pj ∈ Supp R/Pj , so by (1.5.10), Pj ∈ Supp Mj /Mj’1 . By

(1.5.4), Supp Mj /Mj’1 ⊆ Supp Mj , and ¬nally Supp Mj ⊆ Supp M because Mj ⊆ M . ™

1.6 Artinian Rings

1.6.1 De¬nitions and Comments

Recall that an R-module is Artinian if it satis¬es the descending chain condition on

submodules. If the ring R is Artinian as a module over itself, in other words, R satis¬es

the dcc on ideals, then R is said to be an Artinian ring. Note that Z is a Noetherian ring

that is not Artinian. Any ¬nite ring, for example Zn , is both Noetherian and Artinian,

and in fact we will prove later in the section that an Artinian ring must be Noetherian.

The theory of associated primes and supports will help us to analyze Artinian rings.

1.6.2 Lemma

If I is an ideal in the Artinian ring R, then R/I is an Artinian ring.

12 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

Proof. Since R/I is a quotient of an Artinian R-module, it is also an Artinian R-module.

In fact it is an R/I module via (r + I)(x + I) = rx + I, and the R-submodules are

identical to the R/I-submodules. Thus R/I is an Artinian R/I-module, in other words,

an Artinian ring. ™

1.6.3 Lemma

An Artinian integral domain is a ¬eld.

Proof. Let a be a nonzero element of the Artinian domain R. We must produce a

multiplicative inverse of a. The chain of ideals (a) ⊇ (a2 ) ⊇ (a3 ) ⊇ · · · stabilizes, so for

some t we have (at ) = (at+1 ). If at = bat+1 , then since R is a domain, ba = 1. ™

1.6.4 Proposition

If R is an Artinian ring, then every prime ideal of R is maximal. Therefore, the nilradical

N (R) coincides with the Jacobson radical J(R).

Proof. Let P be a prime ideal of R, so that R/I is an integral domain, Artinian by (1.6.2).

By (1.6.3), R/P is a ¬eld, hence P is maximal. ™

One gets the impression that the Artinian property puts strong constraints on a ring.

The following two results reinforce this conclusion.

1.6.5 Proposition

An Artinian ring has only ¬nitely many maximal ideals.

Proof. Let Σ be the collection of all ¬nite intersections of maximal ideals. Then Σ is

nonempty and has a minimal element I = M1 © · · · © Mr (by the Artinian property). If

M is any maximal ideal, then M ⊇ M © I ∈ Σ, so by minimality of I we have M © I = I.

But then M must contain one of the Mi (because M is prime), hence M = Mi (because

M and Mi are maximal). ™

1.6.6 Proposition

If R is Artinian, then the nilradical N (R) is nilpotent, hence by (1.6.4), the Jacobson

radical J(R) is nilpotent.

Proof. Let I = N (R). The chain I ⊇ I 2 ⊇ I 3 ⊇ · · · stabilizes, so for some i we have

I i = I i+1 = · · · = L. If L = 0 we are ¬nished, so assume L = 0. Let Σ be the collection of

all ideals K of R such that KL = 0. Then Σ is nonempty, since L (as well as R) belongs

to Σ. Let K0 be a minimal element of Σ, and choose a ∈ K0 such that aL = 0. Then

Ra ⊆ K0 (because K0 is an ideal), and RaL = aL = 0, hence Ra ∈ Σ. By minimality of

K0 we have Ra = K0 .

We will show that the principal ideal (a) = Ra coincides with aL. We have aL ⊆ Ra =

K0 , and (aL)L = aL2 = aL = 0, so aL ∈ Σ. By minimality of K0 we have aL = K0 = Ra.

From (a) = aL we get a = ab for some b ∈ L ⊆ N (R), so bn = 0 for some n ≥ 1.

Therefore a = ab = (ab)b = ab2 = · · · = abn = 0, contradicting our choice of a. Since the

1.6. ARTINIAN RINGS 13

assumption L = 0 has led to a contradiction, we must have L = 0. But L is a power of

the nilradical I, and the result follows. ™

We now prove a fundamental structure theorem for Artinian rings.

1.6.7 Theorem

Every Artinian ring R is isomorphic to a ¬nite direct product of Artinian local rings Ri .

Proof. By (1.6.5), R has only ¬nitely many maximal ideals M1 , . . . , Mr . The intersection

of the Mi is the Jacobson radical J(R), which is nilpotent by (1.6.6). By the Chinese

remainder theorem, the intersection of the Mi coincides with their product. Thus for

r r

some k ≥ 1 we have ( 1 Mi )k = 1 Mk = 0. Powers of the Mi still satisfy the

i

r

hypothesis of the Chinese remainder theorem, so the natural map from R to 1 R/Mk i

is an isomorphism. By (1.6.2), R/Mk is Artinian, and we must show that it is local. A

i

maximal ideal of R/Mk corresponds to a maximal ideal M of R with M ⊇ Mk , hence

i i

M ⊇ Mi (because M is prime). By maximality, M = Mi . Thus the unique maximal

ideal of R/Mk is Mi /Mk . ™

i i

1.6.8 Remarks

A ¬nite direct product of Artinian rings, in particular, a ¬nite direct product of ¬elds,

is Artinian. To see this, project a descending chain of ideals onto one of the coordinate

rings. At some point, all projections will stabilize, so the original chain will stabilize.

A sequence of exercises will establish the uniqueness of the Artinian local rings in the

decomposition (1.6.7).

It is a standard result that an R-module M has ¬nite length lR (M ) if and only if M

is both Artinian and Noetherian. We can relate this condition to associated primes and

supports.

1.6.9 Proposition

Let M be a ¬nitely generated module over the Noetherian ring R. The following conditions

are equivalent:

(1) lR (M ) < ∞;

(2) Every associated prime ideal of M is maximal;

(3) Every prime ideal in the support of M is maximal.

Proof.

(1) ’ (2): As in (1.5.10), there is a chain of submodules 0 = M0 < · · · < Mn = M ,

with Mi /Mi’1 ∼ R/Pi . Since Mi /Mi’1 is a submodule of a quotient M/Mi’1 of M , the

=

hypothesis (1) implies that R/Pi has ¬nite length for all i. Thus R/Pi is an Artinian

R-module, hence an Artinian R/Pi -module (note that Pi annihilates R/Pi ). In other

words, R/Pi is an Artinian ring. But Pi is prime, so R/Pi is an integral domain, hence

a ¬eld by (1.6.3). Therefore each Pi is a maximal ideal. Since every associated prime is

one of the Pi ™s [see (1.5.10)], the result follows.

14 CHAPTER 1. PRIMARY DECOMPOSITION AND ASSOCIATED PRIMES

(2) ’ (3): If P ∈ Supp M , then by (1.5.8), P contains some associated prime Q. By

hypothesis, Q is maximal, hence so is P .

(3) ’ (1): By (1.5.11) and the hypothesis (3), every Pi is maximal, so R/Pi is a ¬eld.

Consequently, lR (Mi /Mi’1 ) = lR (R/Pi ) = 1 for all i. But length is additive, that is, if

N is a submodule of M , then l(M ) = l(N ) + l(M/N ). Summing on i from 1 to n, we get

lR (M ) = n < ∞. ™

1.6.10 Corollary

Let M be ¬nitely generated over the Noetherian ring R. If lR (M ) < ∞, then AP(M ) =

Supp M .

Proof. By (1.5.9), AP(M ) ⊆ Supp M , so let P ∈ Supp M . By (1.5.8), P ⊇ P for some

P ∈ AP(M ). By (1.6.9), P and P are both maximal, so P = P ∈ AP(M ). ™

We can now characterize Artinian rings in several ways.

1.6.11 Theorem

Let R be a Noetherian ring. The following conditions are equivalent:

(1) R is Artinian;

(2) Every prime ideal of R is maximal;

(3) Every associated prime ideal of R is maximal.

Proof. (1) implies (2) by (1.6.4), and (2) implies (3) is immediate. To prove that (3)

implies (1), note that by (1.6.9), lR (R) < ∞, hence R is Artinian. ™

1.6.12 Theorem

The ring R is Artinian if and only if lR (R) < ∞.

Proof. The “if” part follows because any module of ¬nite length is Artinian and Noethe-

rian. Thus assume R Artinian. As in (1.6.7), the zero ideal is a ¬nite product M1 · · · Mk

of not necessarily distinct maximal ideals. Now consider the chain

R = M0 ⊇ M1 ⊇ M1 M2 ⊇ · · · ⊇ M1 · · · Mk’1 ⊇ M1 · · · Mk = 0.

Since any submodule or quotient module of an Artinian module is Artinian, it follows

that Ti = M1 · · · Mi’1 /M1 · · · Mi is an Artinian R-module, hence an Artinian R/Mi -

module. (Note that Mi annihilates M1 · · · Mi’1 /M1 · · · Mi .) Thus Ti is a vector space

over the ¬eld R/Mi , and this vector space is ¬nite-dimensional by the descending chain

condition. Thus Ti has ¬nite length as an R/Mi -module, hence as an R-module. By

additivity of length [as in (3) implies (1) in (1.6.9)], we conclude that lR (R) < ∞. ™

1.6.13 Theorem

The ring R is Artinian if and only if R is Noetherian and every prime ideal of R is

maximal.

Proof. The “if” part follows from (1.6.11). If R is Artinian, then lR (R) < ∞ by (1.6.12),

hence R is Noetherian. By (1.6.4) or (1.6.11), every prime ideal of R is maximal. ™

1.6. ARTINIAN RINGS 15

1.6.14 Corollary

Let M be ¬nitely generated over the Artinian ring R. Then lR (M ) < ∞.

Proof. By (1.6.13), R is Noetherian, hence the module M is both Artinian and Noetherian.

Consequently, M has ¬nite length. ™

Chapter 2

Integral Extensions

2.1 Integral Elements

2.1.1 De¬nitions and Comments

Let R be a subring of the ring S, and let ± ∈ S. We say that ± is integral over R if ±

is a root of a monic polynomial with coe¬cients in R. If R is a ¬eld and S an extension

¬eld of R, then ± is integral over R i¬ ± is algebraic over R, so we are generalizing a

familiar notion. If ± is a complex number that is integral over Z, then ± is said to be an

√

algebraic integer For example, if d is any integer, then d is an algebraic integer, because

it is a root of x2 ’ d. Notice that 2/3 is a root of the polynomial f (x) = 3x ’ 2, but f

is not monic, so we cannot conclude that 2/3 is an algebraic integer. In a ¬rst course in

algebraic number theory, one proves that a rational number that is an algebraic integer

must belong to Z, so 2/3 is not an algebraic integer.

There are several conditions equivalent to integrality of ± over R, and a key step is

the following result, sometimes called the determinant trick.

2.1.2 Lemma

Let R, S and ± be as above, and recall that a module is faithful if its annihilator is 0. Let

M be a ¬nitely generated R-module that is faithful as an R[±]-module. Let I be an ideal

of R such that ±M ⊆ IM . Then ± is a root of a monic polynomial with coe¬cients in I.

Proof. let x1 , . . . , xn generate M over R. Then ±xi ∈ IM , so we may write ±xi =

n

j=1 cij xj with cij ∈ I. Thus

n

(δij ± ’ cij )xj = 0, 1 ¤ i ¤ n.

j=1

In matrix form, we have Ax = 0, where A is a matrix with entries ± ’ cii on the main

diagonal, and ’cij elsewhere. Multiplying on the left by the adjoint matrix, we get

∆xi = 0 for all i, where ∆ is the determinant of A. But then ∆ annihilates all of M , so

∆ = 0. Expanding the determinant yields the desired monic polynomial. ™

1

2 CHAPTER 2. INTEGRAL EXTENSIONS

2.1.3 Remark

If ±M ⊆ IM , then in particular, ± stabilizes M , in other words, ±M ⊆ M .

2.1.4 Theorem

Let R be a subring of S, with ± ∈ S. The following conditions are equivalent:

(1) ± is integral over R;

(2) R[±] is a ¬nitely generated R-module;

(3) R[±] is contained in a subring R of S that is a ¬nitely generated R-module;

(4) There is a faithful R[±]-module M that is ¬nitely generated as an R-module.

Proof.

(1) implies (2): If ± is a root of a monic polynomial over R of degree n, then ±n and all

higher powers of ± can be expressed as linear combinations of lower powers of ±. Thus

1, ±, ±2 , . . . , ±n’1 generate R[±] over R.

(2) implies (3): Take R = R[±].

(3) implies (4): Take M = R . If y ∈ R[±] and yM = 0, then y = y1 = 0.

(4) implies (1): Apply (2.1.2) with I = R. ™

We are going to prove a transitivity property for integral extensions, and the following

result will be helpful.

2.1.5 Lemma

Let R be a subring of S, with ±1 , . . . , ±n ∈ S. If ±1 is integral over R, ±2 is integral

over R[±1 ], . . . , and ±n is integral over R[±1 , . . . , ±n’1 ], then R[±1 , . . . , ±n ] is a ¬nitely

generated R-module.

Proof. The n = 1 case follows from (2.1.4), part (2). Going from n ’ 1 to n amounts

to proving that if A, B and C are rings, with C a ¬nitely generated B-module and B a

¬nitely generated A-module, then C is a ¬nitely generated A-module. This follows by a

brief computation:

r s r s

Ayj xk . ™

C= Byj , B = Axk , so C =

j=1 j=1 k=1

k=1

2.1.6 Transitivity of Integral Extensions

Let A, B and C be subrings of R. If C is integral over B, that is, every element of C is

integral over B, and B is integral over A, then C is integral over A.

Proof. Let x ∈ C, with xn + bn’1 xn’1 + · · · + b1 x + b0 = 0. Then x is integral over

A[b0 , . . . , bn’1 ]. Each bi is integral over A, hence over A[b0 , . . . , bi’1 ]. By (2.1.5),

A[b0 , . . . , bn’1 , x] is a ¬nitely generated A-module. By (2.1.4), part (3), x is integral

over A. ™

2.1. INTEGRAL ELEMENTS 3

2.1.7 De¬nitions and Comments

If R is a subring of S, the integral closure of R in S is the set Rc of elements of S that

are integral over R. Note that R ⊆ Rc because each a ∈ R is a root of x ’ a. We say that

R is integrally closed in S if Rc = R. If we simply say that R is integrally closed without

reference to S, we assume that R is an integral domain with fraction ¬eld K, and R is

integrally closed in K.

If the elements x and y of S are integral over R, then just as in the proof of (2.1.6), it

follows from (2.1.5) that R[x, y] is a ¬nitely generated R-module. Since x + y, x ’ y and

xy belong to this module, they are integral over R by (2.1.4), part (3). The important

conclusion is that

Rc is a subring of S containing R.

If we take the integral closure of the integral closure, we get nothing new.

2.1.8 Proposition

The integral closure Rc of R in S is integrally closed in S.

Proof. By de¬nition, Rc ⊆ (Rc )c . Thus let x ∈ (Rc )c , so that x is integral over Rc . As in

the proof of (2.1.6), x is integral over R. Thus x ∈ Rc . ™

We can identify a large class of integrally closed rings.

2.1.9 Proposition

If R is a UFD, then R is integrally closed.

Proof. Let x belong to the fraction ¬eld K of R. Write x = a/b where a, b ∈ R and a and

b are relatively prime. If x is integral over R, there is an equation of the form

(a/b)n + an’1 (a/b)n’1 + · · · + a1 (a/b) + a0 = 0

with ai ∈ R. Multiplying by bn , we have an + bc = 0, with c ∈ R. Thus b divides an ,

which cannot happen for relatively prime a and b unless b has no prime factors at all, in

other words, b is a unit. But then x = ab’1 ∈ R. ™

A domain that is an integral extension of a ¬eld must be a ¬eld, as the next result

shows.

2.1.10 Proposition

Let R be a subring of the integral domain S, with S integral over R. Then R is a ¬eld if

and only if S is a ¬eld.

Proof. Assume that S is a ¬eld, and let a be a nonzero element of R. Since a’1 ∈ S,

there is an equation of the form

(a’1 )n + cn’1 (a’1 )n’1 + · · · + c1 a’1 + c0 = 0

with ci ∈ R. Multiply the equation by an’1 to get

a’1 = ’(cn’1 + · · · + c1 an’2 + c0 an’1 ) ∈ R.

4 CHAPTER 2. INTEGRAL EXTENSIONS

Now assume that R is a ¬eld, and let b be a nonzero element of S. By (2.1.4) part (2),

R[b] is a ¬nite-dimensional vector space over R. Let f be the R-linear transformation on

this vector space given by multiplication by b, in other words, f (z) = bz, z ∈ R[b]. Since

R[b] is a subring of S, it is an integral domain. Thus if bz = 0 (with b = 0 by choice of b),

we have z = 0 and f is injective. But any linear transformation on a ¬nite-dimensional

vector space is injective i¬ it is surjective. Therefore if b ∈ S and b = 0, there is an

element c ∈ R[b] ⊆ S such that bc = 1. Consequently, S is a ¬eld. ™

2.1.11 Preview

Let S be integral over the subring R. We will analyze in great detail the relation between

prime ideals of R and those of S. Suppose that Q is a prime ideal of S, and let P = Q © R.

(We say that Q lies over P .) Then P is a prime ideal of R, because it is the preimage

of Q under the inclusion map from R into S. The map a + P ’ a + Q is a well-de¬ned

injection of R/P into S/Q, because P = Q © R. Thus we can regard R/P as a subring of

S/Q. Moreover, S/Q is integral over R/P . To see this, let b + Q ∈ S/Q. Then b satis¬es

an equation of the form

xn + an’1 xn’1 + · · · + a1 x + a0 = 0

with ai ∈ R. But b + Q satis¬es the same equation with ai replaced by ai + P for all i,

proving integrality of S/Q over R/P . We can now invoke (2.1.10) to prove the following

result.

2.1.12 Proposition

Let S be integral over the subring R, and let Q be a prime ideal of S, lying over the prime

ideal P = Q © R of R. Then P is a maximal ideal of R if and only if Q is a maximal ideal

of S.

Proof. By (2.1.10), R/P is a ¬eld i¬ S/Q is a ¬eld. ™

2.1.13 Remarks

Some results discussed in (2.1.11) work for arbitrary ideals, not necessarily prime. If R

is a subring of S and J is an ideal of S, then I = J © R is an ideal of R. As in (2.1.11),

R/I can be regarded as a subring of S/J, and if S is integral over R, then S/J is integral

over R/I. Similarly, if S is integral over R and T is a multiplicative subset of R, then

ST is integral over RT . To prove this, let ±/t ∈ ST , with ± ∈ S, t ∈ T . Then there is an

equation of the form ±n + cn’1 ±n’1 + · · · + c1 ± + c0 = 0, with ci ∈ R. Thus

± cn’1 ± n’1 c1 ± c0

+ · · · + ( n’1 ) + n = 0

( )n + ( )( )

t t t t t t

with cn’j /tj ∈ RT .

2.2. INTEGRALITY AND LOCALIZATION 5

2.2 Integrality and Localization

Results that hold for maximal ideals can sometimes be extended to prime ideals by the

technique of localization. A good illustration follows.

2.2.1 Proposition

Let S be integral over the subring R, and let P1 and P2 be prime ideals of S that lie over

the prime ideal P of R, that is, P1 © R = P2 © R = P . If P1 ⊆ P2 , then P1 = P2 .

Proof. If P is maximal, then by (2.1.12), so are P1 and P2 , and the result follows. In the

general case, we localize with respect to P . Let T = R\P , a multiplicative subset of R ⊆ S.

The prime ideals Pi , i = 1, 2, do not meet T , because if x ∈ T © Pi , then x ∈ R © Pi = P ,

contradicting the de¬nition of T . By the basic correspondence between prime ideals in a

ring and prime ideals in its localization, it su¬ces to show that P1 ST = P2 ST . We claim

that

P RT ⊆ (P1 ST ) © RT ‚ RT .

The ¬rst inclusion holds because P ⊆ P1 and RT ⊆ ST . The second inclusion is proper,

for otherwise RT ⊆ P1 ST and therefore 1 ∈ P1 ST , contradicting the fact that P1 ST is a

prime ideal.

But P RT is a maximal ideal of RT , so by the above claim,

(P1 ST ) © RT = P RT , and similarly (P2 ST ) © RT = P RT .

Thus P1 ST and P2 ST lie over P RT . By (2.1.13), ST is integral over RT . As at the

beginning of the proof, P1 ST and P2 ST are maximal by (2.1.12), hence P1 ST = P2 ST . ™

If S/R is an integral extension, then prime ideals of R can be lifted to prime ideals of

S, as the next result demonstrates. Theorem 2.2.2 is also a good example of localization

technique.

2.2.2 Lying Over Theorem

If S is integral over R and P is a prime ideal of R, there is a prime ideal Q of S such that

Q © R = P.

Proof. First assume that R is a local ring with unique maximal ideal P . If Q is any

maximal ideal of S, then Q © R is maximal by (2.1.12), so Q © R must be P . In general,

let T be the multiplicative set R \ P . We have the following commutative diagram.

R ’’’ S

’’

¦ ¦

¦ ¦g

f

R T ’ ’ ’ ST

’’

The horizontal maps are inclusions, and the vertical maps are canonical (f (r) = r/1 and

g(s) = s/1). Recall that ST is integral over RT by (2.1.13). If Q is any maximal ideal

of ST , then as at the beginning of the proof, Q © RT must be the unique maximal ideal

6 CHAPTER 2. INTEGRAL EXTENSIONS

of RT , namely P RT . By commutativity of the diagram, f ’1 (Q © RT ) = g ’1 (Q ) © R.

(Note that if r ∈ R, then f (r) ∈ Q © RT i¬ g(r) ∈ Q .) If Q = g ’1 (Q ), we have

f ’1 (P RT ) = Q©R. By the basic localization correspondence [cf.(2.2.1)], f ’1 (P RT ) = P ,

and the result follows. ™

2.2.3 Going Up Theorem

Let S be integral over R, and suppose we have a chain of prime ideals P1 ⊆ · · · ⊆ Pn

of R, and a chain of prime ideals Q1 ⊆ · · · ⊆ Qm of S, where m < n. If Qi lies

over Pi for i = 1, . . . , m, then there are prime ideals Qm+1 , . . . , Qn of S such that

Qm ⊆ Qm+1 ⊆ · · · ⊆ Qn and Qi lies over Pi for every i = 1, . . . , n.

Proof. By induction, it su¬ces to consider the case n = 2, m = 1. Thus assume P1 ⊆ P2

and Q1 © R = P1 . By (2.1.11), S/Q1 is integral over R/P1 . Since P2 /P1 is a prime ideal

of R/P1 , we may apply the lying over theorem (2.2.2) to produce a prime ideal Q2 /Q1 of

S/Q1 such that

(Q2 /Q1 ) © R/P1 = P2 /P1 ,

where Q2 is a prime ideal of S and Q1 ⊆ Q2 . We claim that Q2 © R = P2 , which gives the

desired extension of the Q-chain. To verify this, let x2 ∈ Q2 © R. By (2.1.11), we have

an embedding of R/P1 into S/Q1 , so x2 + P1 = x2 + Q1 ∈ (Q2 /Q1 ) © R/P1 = P2 /P1 .

Thus x2 + P1 = y2 + P1 for some y2 ∈ P2 , so x2 ’ y2 ∈ P1 ⊆ P2 . Consequently, x2 ∈ P2 .

Conversely, if x2 ∈ P2 then x2 + P1 ∈ Q2 /Q1 , hence x2 + P1 = y2 + Q1 for some y2 ∈ Q2 .

But as above, x2 + P1 = x2 + Q1 , so x2 ’ y2 ∈ Q1 , and therefore x2 ∈ Q2 . ™

It is a standard result of ¬eld theory that an embedding of a ¬eld F in an algebraically

closed ¬eld can be extended to an algebraic extension of F . There is an analogous result

for ring extensions.

2.2.4 Theorem

Let S be integral over R, and let f be a ring homomorphism from R into an algebraically

closed ¬eld C. Then f can be extended to a ring homomorphism g : S ’ C.

Proof. Let P be the kernel of f . Since f maps into a ¬eld, P is a prime ideal of R. By

(2.2.2), there is a prime ideal Q of S such that Q © R = P . By the factor theorem, f

induces an injective ring homomorphism f : R/P ’ C, which extends in the natural way

to the fraction ¬eld K of R/P . Let L be the fraction ¬eld of S/Q. By (2.1.11), S/Q is

integral over R/P , hence L is an algebraic extension of K. Since C is algebraically closed,

f extends to a monomorphism g : L ’ C. If p : S ’ S/Q is the canonical epimorphism

and g = g —¦ p, then g is the desired extension of f , because g extends f and f —¦ p|R = f .

™

In the next section, we will prove the companion result to (2.2.3), the going down

theorem. There will be extra hypotheses, including the assumption that R is integrally

closed. So it will be useful to get some practice with the idea of integral closure.

2.3. GOING DOWN 7

2.2.5 Lemma

Let R be a subring of S, and denote by R the integral closure of R in S. If T is a

multiplicative subset of R, then (R)T is the integral closure of RT in ST .

Proof. Since R is integral over R, it follows from (2.1.13) that (R)T is integral over RT .

If ±/t ∈ ST (± ∈ S, t ∈ T ) and ±/t is integral over RT , we must show that ±/t ∈ (R)T .

There is an equation of the form

± a1 ± an

( )n + ( )( )n’1 + · · · + =0

t t1 t tn

n

with ai ∈ R and ti , t ∈ T . Let t0 = i=1 ti , and multiply the equation by (tt0 )n to

conclude that t0 ± is integral over R. Therefore t0 ± ∈ R, so ±/t = t0 ±/t0 t ∈ (R)T . ™

2.2.6 Corollary

If T is a multiplicative subset of the integrally closed domain R, then RT is integrally

closed.

Proof. Apply (2.2.5) with R = R and S = K, the fraction ¬eld of R (and of RT ). Then

RT is the integral closure of RT in ST . But ST = K, so RT is integrally closed. ™

Additional results on localization and integral closure will be developed in the exercises.

The following result will be useful. (The same result was proved in (1.5.1), but a slightly