ca lecture notes

7/30/2019 CA Lecture Notes

1/59

Commutative Algebra

Lecture Notes

Anand Prabhakar Sawant

Research Scholar

School of Mathematics

Tata Institute of Fundamental Research, Mumbai

January 2009


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Contents

1 Basic Definitions and Tools 1

1.1 Basic Concepts and Definitions . . . . . . . . . . . . . . . . . . . 1

1.2 Prime Avoidance Lemma . . . . . . . . . . . . . . . . . . . . . . 3

1.3 Nakayamas Lemma . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Localization of Rings and Modules 72.1 Localization of Rings . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Localization of Modules . . . . . . . . . . . . . . . . . . . . . . 10

2.3 Localization at a Prime Ideal . . . . . . . . . . . . . . . . . . . . 12

2.4 An Interesting Application . . . . . . . . . . . . . . . . . . . . . 13

3 Noetherian Rings and Modules 15

3.1 Definitions and Basic Properties . . . . . . . . . . . . . . . . . . 15

3.2 The Hilbert Basis Theorem . . . . . . . . . . . . . . . . . . . . . 17

3.3 Cohens Theorem and Applications . . . . . . . . . . . . . . . . . 19

3.4 Primary Decomposition of Radicals . . . . . . . . . . . . . . . . 22

4 Artinian Rings and Modules 25

4.1 Definitions and Basic Properties . . . . . . . . . . . . . . . . . . 25

4.2 Equivalent Characterizations of Artinian Rings and Modules . . . 28

4.3 Jordan-Holder Series . . . . . . . . . . . . . . . . . . . . . . . . 30

4.4 Artinian Local Rings . . . . . . . . . . . . . . . . . . . . . . . . 31

5 Integral Extensions 33

5.1 Integral Elements and Integral Extensions . . . . . . . . . . . . . 33

5.2 Properties of Integral Extensions . . . . . . . . . . . . . . . . . . 36

5.3 Noether Normalization . . . . . . . . . . . . . . . . . . . . . . . 385.4 The Going-up Theorem . . . . . . . . . . . . . . . . . . . . . . . 41

5.5 The Going-down Theorem . . . . . . . . . . . . . . . . . . . . . 44

6 The Krull Dimension of a Ring 47

6.1 The Krull Principal Ideal Theorem and its Converse . . . . . . . . 47

6.2 Dimension of Polynomial Algebras . . . . . . . . . . . . . . . . . 53

6.3 Dimension of Affine Algebras . . . . . . . . . . . . . . . . . . . 56

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Chapter 1

Basic Definitions and Tools

1.1 Basic Concepts and Definitions

Throughout the course, every ring appearing is assumed to be commutative withunity (always denoted by 1) which is not equal to zero (always denoted by 0).

Definition 1.1.1 (Multiplicative set) A subset S of a ring A is called multiplica-

tive if1 S, 0 S, and ab S whenever a, b S .

Definition 1.1.2 (Prime ideal) An ideal p of a ring A is called a prime ideal if

A\p is a multiplicative set.

Note that our definition automatically implies that a prime ideal p of A does not

contain 1, so it is proper. Equivalently, an idealp

ofA

is prime ifp

is a propersubset ofA and ifab p implies a p or b p.

Definition 1.1.3 (Maximal ideal) A proper ideal m of A is said to be a maximal

ideal ifm is not properly contained in any proper ideal of A.

Remarks. 1. An ideal p ofA is prime if and only ifA/p is an integral domain.

2. An ideal m of A is maximal if and only if A/m is a field. In particular, a

maximal ideal is prime.

3. Ifp is a prime ideal of a ring A and I, Jare ideals ofA such that I Jp, then

I p or J p. For, if I p and J p, then we can find a I and b Jsuch that ab p. This is a contradiction since I J p.

Definition 1.1.4 (Spectrum of a ring) Let A be a commutative ring. The set of

all prime ideals of A is called the spectrum of A and is denoted by Spec(A).

Definition 1.1.5 (Maximum spectrum of a ring) The set of maximal ideals in

A is a subset of Spec(A), called the maximum spectrum of A and is denoted by

Max(A).

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Zariski Topology. We shall now give a topology on the set Spec(A). For any

subset E A, let

V(E) = {p Spec(A) | E p}.

Note that V(0) = Spec(A), V(A) = . For any family {Ei}iI of subsets of A, wehave

iIV(E) = V(iIEi).Now, let a, b be ideals of A. Ifp Spec(A), then a b p if and only ifa p orb p. (For, by the above remark, ab a b p a p or b p. The otherimplication is obvious.) Thus, the set

{V(I) | I is an ideal of A}satisfies the axioms for the topology defined by closed sets. This topology is called

the Zariski Topology on Spec(A).

Examples. 1. Spec(Z) = {0} {p | p is a prime number }. Any ideal ofZis of the form nZ. Then V(nZ) = {p | p is a prime number dividing n }.Thus, all closed sets except V(0) in the Zariski topology ofZ are finite.

2. Let k be a field. Then Spec(k) = {0}. Now consider the polynomial ringk[x]. Since k[x] is a PID, we see that

Spec(k[x]) = {0} { p(x) | p(x) is an irreducible polynomial }

Any ideal Iofk[x] is generated by a polynomial q(x)

I of minimal degree;

hence, it follows that

V(I) = {p(x) | p(x) is an irreducible polynomial dividing q(x) }

3. Let k be an algebraically closed field. Then the irreducible polynomials in

k[x] are precisely the linear polynomials. Hence, in this case,

Spec(k[x]) = {0} { x a | a k}

Thus, for any ideal I = (q(x)), we have

V(I) = {x a | a is a root of q(x) }Definition 1.1.6 (Local ring, Semilocal ring) A ring A is said to be local if A

has a unique maximal ideal. A ring A is said to be semilocal if A has only finitely

many maximal ideals.

Let A be a ring. By a chain of prime ideals of A we shall mean a finite strictly

increasing sequence of prime ideals

p0 p1 pn

ofA. The integer n will be called the length of the chain.

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Definition 1.1.7 (Krull dimension) The supremum of the lengths of all chains of

prime ideals of A is called the Krull dimension of A; we denote it by dim(A).

Definition 1.1.8 (Height of an ideal) Let A be a ring. Ifp Spec(A), then theheight of p, denoted by ht(p), is defined to be the supremum of the lenghts of

chains of prime ideals of A

p0 p1 pn = p

ending atp. For any ideal I of A, we define

ht(I) = inf{ht(p) | p I }.

Definition 1.1.9 (Minimal prime over an ideal) Let A be a ring and I A anideal. p Spec(A) is said to be minimal over I if I p and if for any prime idealp of A such that I

p

p, we have p = p. A prime ideal p is said to be minimal

if it is minimal over the ideal 0.

Remark. The only prime ideal minimal over a prime ideal p is p itself.

1.2 Prime Avoidance Lemma

We shall now prove that an ideal contained in a finite union of prime ideals is

contained in at least one of them. The name prime avoidance comes from the fol-

lowing typical application: If an ideal I is not contained in any of a finite number

of prime ideals p1, . . ., pn, then there exists an element a

I that avoids being

contained in any of the pjs.

Lemma 1.2.1 (Prime avoidance lemma) Let A be a ring and I be an ideal of A.

Letp1, . . . , pn be prime ideals of A such that I ni=1pi. Then I pj for some j.

Proof We shall prove the contrapositive: I pj for all j I ni=1pi. We provethis by induction on n, the number of prime ideals. For n=1, the proof is clear.

Assume the result for n 1 prime ideals.Now, suppose that p1, p1, . . ., pn, are prime ideals of A such that I pj for

j = 1, . . . , n. For each i, 1 i n, we have

I pj, for j = 1, . . . , i 1, i + 1, . . . , n,

so by the induction hypothesis, for each i, 1 i n, we can find an element xi I such that xi jipi. If for some i we have xi pi, we are through. Otherwise,we have xi pi for all i. Then it is easy to see that y =

ni=1 x1 xi1xi+1 xn is

an element of I which is not in any pi. This completes the proof.

We shall now prove that if I is a finitely generated ideal of a ring A satisfy-

ing the hypotheses of the above lemma, then there is a linear combination of the

generators ofI which avoids n

i=1pi.

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Lemma 1.2.2 Let A be a ring and I = a1, . . . ar be a finitely generated ideal ofA. Letp1, . . . , pn be prime ideals of A such that I pi for each i, 1 i n. Then b2, . . . br A such that .

c = a1 + b2a2 +

+ brar

ni=1pi

Proof Without loss of generality, assume that pi pj for i j. We prove the

lemma by induction on n. The case n = 1 is clear. Assume the result for n 1prime ideals. Then c2, . . . cr A such that .

d = a1 + c2a2 + + crar n1i=1 piIf d pn, we are through. Assume d pn. If a2, . . . ar pn, then from theexpression for d we see that a1 pn, contradicting that I pn. So for some i,ai pn. Without loss of generality, let i = 2. Since pi pj for i j, we can find

x

n1i=1

pi such that x pn. Then

c = a1 + (c2 + x)a2 + + crar ni=1pi,

completing the proof.

Remark. The lemmas need not hold if I is contained in a finite union of arbitrary

(not necessary prime) ideals. For example, let k be a field and consider the ring

A = k[x,y]/x,y2 = { a+bx +cy | a, b, c k }. Let k = Z/2Z. x,y are zerodivisorsin A. It is easy to see that

x,y = x y x +y,

but x,y is not contained in any of them. (In fact, A is an artinian local ring withunique maximal ideal x,y.)

1.3 Nakayamas Lemma

Definition 1.3.1 Let A be a ring. The intersection of all maximal ideals of A is

called the Jacobson radical of A and is denoted by Jac(A).

Remark. It is easy to see that Jac(A) = { x A | 1 ax is a unit for all a A}.For, if x Jac(A) and 1 ax is not a unit for some a A, then there exists amaximal ideal m of A that contains 1 ax; but since x m, this gives 1 m,a contradiction. On the other hand, if x m for some maximal ideal m, then

m + Ax = A so for some a A we have 1 ax m, whence 1 ax cannot be aunit of A.

Lemma 1.3.2 (Nakayamas lemma, first version) Let A be a ring, M a finitely

generated A-module and I an ideal of A such that IM = M. Then we can find an

a

I such that(1

a)M = 0.

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Proof If M = 0, there is nothing to prove. So let M 0 be generated by

m1, mr. Since I M = M, there exist i j I for 1 i, j rsuch that

mi =

r

j=1

i jmj,

for i = 1, . . . , r. Thus, we have

1 11 12 1r21 1 22 2r...

.... . .

...

r1 r2 1 rr

m1m2...

mr

=

0

0...

0

.

If we write

T =

1 11 12 1r21 1 22 2r... ... . . . ...r1 r2 1 rr

,

then multiplying the above relation on the left by Adj(T), we get

(det T)

m1m2...

mr

=

0

0...

0

,

so (det T)M = 0. But since T Idr (mod I), we have det T 1(mod I), sodet T = 1 a for some a I. This completes the proof.

Corollary 1.3.3 (Nakayamas lemma, second version) Let A be a ring and M

be a finitely generated A-module. Suppose that I is an ideal of A such that I Jac(A). If IM= M, then M = 0.

Proof From the first version of Nakayamas lemma, I M = M for some a I, (1 a)M = 0. But since a lies in Jac(A), 1 a is a unit of A. Hence, it follows

thatM =

0.

Corollary 1.3.4 (Nakayamas lemma, third version) Let A be a ring and sup-

pose that I is an ideal of A such that I Jac(A). Let M be a finitely generatedA-module and N an A-submodule of M. If M = N + I M, then M = N.

Proof Put M = M/N. Since Mis finitely generated, so is Mand clearly, M = I M.

Since I Jac(A), by the second version of Nakayamas lemma, M = 0, that is,M = N.

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An application of Nakayamas lemma. Suppose that W is a set of generators

of an A-module M which is minimal in the sense that no proper subset of W

generates M; then W is said to be a minimal basis of M. Two minimal bases

of an A-module M do not necessarily have the same number of elements; for

example, when M = A, if there exist nonunits x,y

A such that x + y = 1, then

both {1} and {x,y} are minimal bases of A. (For example, A[x] has minimal bases{1} and {x, 1 x}.) However, if A is a local ring, then for any finitely generatedA-module M, any two minimal bases have the same number of elements.

Proposition 1.3.5 Let A be a local ring with unique maximal ideal m. Let M be

a finitely generated A-module. Then M = M/mM is a finite dimensional A/m-

vector space. Let the vector space dimension of M be n.

(i) If B = {v1, . . . , vn} is a basis of M over A/m, and if vi is a lift in M of vi, thenB = {v1, . . . , vn} is a minimal basis of M.

(ii) Conversely, if B = {v1, . . . , vm} is a minimal basis of M, then B = {v1, . . . , vm}is a basis of M over A/m.

Since dimension of a vector space is well defined, we have m = n.

Proof (i) We have

M = Av1 + + Avn + mM.Since M is finitely generated, and Jac(A)=m, by Nakayamas lemma, we

have M = Av1 + + Avn. If a proper subset B B generates M, then theset of images ofB in Mis proper subset ofB generating M, a contradiction.

(ii) Clearly, B is a basis of M. Note that B is a linearly independent set. Other-

wise, some proper subset B B would be a basis of M and using (i), wewould get a generating set of Mwhich is a proper subset of B, contradicting

minimality ofB.

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Chapter 2

Localization of Rings and Modules

2.1 Localization of Rings

Recall the definition of a multiplicative set: A subset S of a ring A is called mul-tiplicative if 1 S, 0 S, and ab S whenever a, b S.

Examples of typical multiplicative sets.

1. Let A be a ring and a A be such that an 0 for all n N. Then S ={1, a, a2, . . . , an, . . .} is a multiplicative set.

2. Ifp A is a prime ideal, then S = A\p is a multiplicative set.3. For any family {pi}iI of prime iedeals ofA, the set S\iIpi is mutiplicative.4. For any ring A, the set of units of A is a multiplicative set, so is the set of

elements of A that are not zerodivisors.

5. IfI is an ideal of a ring A, then S = 1 + I is a multiplicative set.

Definition 2.1.1 (Localization of a ring) Let A be a ring and S A be a muti-plicative set. Define a relation on the set A S by

(a, s) (b, t) u S such that u(at bs) = 0.

It is easy to see that

is an equivalence relation. Denote the equivalence class

of (a, s) by as

. Let us denote the set of all equivalence classes under by S1A.Define binary operations of addition and multiplication on S1A as follows:

a

s+

b

t=

at+ bs

st,

a

s b

t=

ab

st.

It is easy to verify that these operations are well-defined and make S1A into a

ring with zero

0

1 and unity

1

1 . Since A is commutative, so is S1

A.

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We shall now see some basic facts on localization.

Lemma 2.1.2 Let A be a ring and S A a multiplicative set. There is a ringhomomorphism f : A S1A given by f(x) = x

1. Then f is injective if and only

if S contains no zerodivisors.

Proof It is easy to see that f is a ring homomorphism. If f is injective and S

contains a zerodivisor u, then a A such that a 0 and ua = 0. But then wehave

a

1=

ua

u=

0

u=

0

1,

contradicting that f is injective. Hence, S does not contain any zerodivisor.

Conversely, if S contains no zerodivisors, thena

1=

b

1 u S such that

u(a b) = 0, which implies a = b. Thus, f is injective.

Throughout the chapter, the letter f will denote the canonocal homomorphism

f : A S1A of Lemma 2.1.2.

Example. In general, the homomorphism f is not injective. Let k be a field and

consider the ring A = k[x,y]/x,y and the multiplicative set S = {1, x, x2, . . .}.Not that in this case, f is not injective:

f(y) =y

1=

xy

x=

0

1

and f(A) = k[x] S1A = k[x, x1].We now prove the following universal property of localization.

Lemma 2.1.3 Let A be a ring and S A a multiplicative set. Let g : A B be aring homomorphism such that g(s) is a unit in B for all s S . Then there exists aunique ring homomorphism h : S1A B such that g = h f .

Proof Define h : S1A B by

h a

s = g(a)g(s)

1 for all a

A and s

S.

Then h is a ring homomorphism and clearly, for any a A, we have

h f(a) = h

a

1

= g(a)g(1)1 = g(a).

Moreover, if h : S1A B is such that g = h f, then for any a A we haveh

a

1

= g(a). Since h is a ring homomorphism, for any s S, h

s

1

h

1

s

= 1,

so h

1

s

= g(s)1. Thus, h = h, proving the uniqueness ofh.

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Remark. From the above lemma, g cannot be injective if f is not injective. Thus,

in view of Lemma 2.1.2, if S contains zerodivisors, then no homomorphism of

S1A is injective.

The following properties of localization of rings are routine verifications and

are hence left as exercises.

Proposition 2.1.4 Let A be a ring, I, J be ideals of A and S A be a multiplica-tive set. Then:

1. S1(I + J) = S1I + S1J.

2. S1(I J) = S1I.S1J.

3. S1(I J) = S1I S1J.

4. S1

I = S1

A if and only if I S

.We shall now see that the ideals ofS1A can described in terms of ideals of A.

Then we characterize the prime ideals ofS1A in terms of prime ideals ofA.

Proposition 2.1.5 Let I be an ideal of a ring A. Then

S1I :=

i

s| i I, s S

is an ideal of S1A. Moreover, any ideal of S1A is of the form S1I for someideal I of A.

Proof It is easy to see that S1I is an ideal of S1A. Let J be any ideal of S1A.Let f denote the canonical homomorphism of Lemma 2.1.2. Put I = f1(J).

Clearly, I is an ideal of A. We claim that J = S1I. Leta

s J. Since J is an ideal,

s

1.a

s=

a

1 J so a I. This implies that J S1(I). The other containment is

clear as f(I) J. Therefore, J = S1I.

We now prove the most important property of localization of rings.

Theorem 2.1.6 The only prime ideals of S1A are S1p, where p is a prime idealof A such thatp S = . Thus, prime ideals of S1A are in bijective correspon-dence with the prime ideals of A that do not intersect S .

Proof First, we prove that S1p is a prime ideal of S1A for any prime ideal p

in A not intersecting S. Let p Spec(A) be such that p S = . Then as.b

t

S1p abst

=c

ufor some c p and u S. This implies v(abu cst) = 0 for

some v S, so ab(uv) = cstv p, which gives ab p, as p S = . But this

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gives a p or b p, p being prime. Hence as

S1p or bt

S1p, showing thatS1p Spec(S1(A)).

Now, let q Spec(S1(A)). Then by Lemma 2.1.6, we have q = S1p, wherep = f1(q). Since the inverse image of a prime ideal under a ring homomorphism

is a prime ideal, p Spec(A). (Reason: Any homomorphism g : A1 A2 of ringsinduces an injective homomorphism g : A1/I1 A2/I2, where I2 Spec(A2)and I1 = g

1(I2). Since A2/I2 is a domain, it follows that A1/I1 is a domain, soI1 Spec(A1).)

Thus, it follows that prime ideals of S1(A) are in bijective correspondencewith prime ideals ofA not intersecting S, under the correspondence

{p Spec(A); p S = } {S1p Spec(S1(A))}.

2.2 Localization of Modules

Definition 2.2.1 (Localization of a module) Let A be a ring, S A be a muti-plicative set and M be an A-module. Define a relation on the set M S by

(m, s) (m, s) u S such that u(sm sm) = 0.It is easy to see that is an equivalence relation. Denote the equivalence classof(m, s) by

m

s. Let us denote the set of all equivalence classes under by S1M.

Define the binary operation of addition on S

1M as follows:

m

s+

m

s=

sm + sm

ss,

for allm

s,

m

s S1M. Define an action of S1A on S1M by

a

t m

s=

am

ts,

for alla

t S1A, m

s S1M. It is easy to verify that these operations are well-

defined and make S1M into an S1A-module.

Let M,Nbe two A-modules and let f HomA(M,N). Then f naturally inducesa homomorphism of S1A-modules S1M S1N denoted by S1f, which isdefined by

S1f

m

s

=

f(m)

s.

Now, (m, s) = (m, s) u S such that u(sm sm) = 0; hence, u(sf(m) s f(m)) = 0, so

f(m)

s=

f(m)s

. Thus, S1f is well-defined. It is easy to verify

that S1f HomS1A(S1M, S1N).The following property of localization of modules is a routine verification and

is hence left as an exercise.

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Proposition 2.2.2 Let A be a ring, M,N be A-modules and S A be a multiplica-tive set. Then S1(M+ N) = S1M+ S1N. In particular, (S1An) = (S1A)n.

We now prove that localization behaves well with exact sequences of modules.

Definition 2.2.3 (Exact sequence) A sequence

Mf M g M

of A-modules M, M, M and A-homomorphisms f, g is said to be an exact se-quence ifker f = Im g.

Definition 2.2.4 (Short exact sequence) A sequence

0 M f M g M 0of A-modules M, M, M and A-homomorphisms f, g is said to be a short exactsequence if f is surjective, g is injective, andker f = Im g.

Theorem 2.2.5 Let A be a ring and let

Mf M g M

be an exact sequence of A-modules. Then the sequence

S1MS1f S1M S

1g S1Mof S1A-modules is also exact.

Proof We have ker f = Im g, so g f = 0. Now, for any ms

S1M, we have

S1g S1f ms

= S1g

f(m)

s

= g f(m)

s= 0

1,

so Im S1f ker S1g. Now, let ms

ker S1g. Then S1g

m

s

=

g(m)

s= 0, so

u S such that ug(m) = g(um) = 0, whence um ker g = Im f. Therefore,f(m) = um for some m M. Thus,

m

s=

f(m)us

= S1f

m

us

Im S1f,

so ker S1g Im S1f. This proves the theorem.

A particular case of this theorem which is useful is that of short exact se-

quences.

Corollary 2.2.6 Let A be a ring and let

0 M f M g M 0be an exact sequence of A-modules. Then the sequence

S1MS1f S1M S

1g S1Mof S

1A-modules is also short exact.

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2.3 Localization at a Prime Ideal

The technique of localization at a prime ideal is useful because it sufficient to ver-

ify some properties of rings/modules locally in the sense that a property holds for a

ring/module R if and only if it holds for localization ofR at all the prime/maximal

ideals.

Definition 2.3.1 Let A be a ring andp be a prime ideal of A. Then S = A\p is amultiplicative set. Then the ring S1A is called the localization of A atp and isdenoted by Ap.

We shall now see that Ap is a local ring; that is why this process is called

localization.

Theorem 2.3.2 Let A be a ring andp be a prime ideal of A. Thena

s Ap is a unit

if and only if a p. Therefore, Ap ia local ring with unique maximal ideal pAp.

Proof Supposea

s Ap is a unit. Then there exists b

t Ap such that a

s.b

t=

1

1, so

we for some u S ( here S = A\p) we have u(ab st) = 0, whence uab = ust S,therefore, a p. Conversely, if a p, then a S, so s

a Ap for any s S. Then

a

s.

s

a=

1

1, so

a

sis a unit, and we are through.

Thus, it follows that every element outside the ideal pAp ofAp is a unit. Hence,

pAp is the unique maximal ideal of Ap and the ring Ap is local. This completes the

proof.

Examples. 1. Zp =

a

b Q | a, b Z, p b

, where p is a prime number, is

an example of localization ofZ at the prime ideal p.2. Let k be a field. Then localization of k[x] at the prime ideal generated by

x a where a k is

k[x]xa =

f(x)

g(x) k(x) | a is not a root of g(x)

Theorem 2.3.3 Let A be a ring and letp be a prime ideal of A. Then we have

ht(p

)=

dim(Ap)

.

Proof By Theorem 2.1.6, there is a one to one correspondence between prime

ideals ofAp and prime ideals q ofA that do not intersect A\p, that is, are containedin p. Hence, chains of prime ideals of Ap correspond to chains of prime ideals of

A that are contained in p. Hence, it follows that ht(p) = dim(Ap).

Remark. We can use localization to construct rings having any desired number

of maximal ideals. Let A be a ring and let a collection {pi}iI of prime ideals of Abe given. Put S = A\ iI pi. S is a multiplicative set. It is an easy exercise tocheck that S

1

A is a ring with Max(S1

A) = {S1

pi | i I}.12


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The following result illustrates the use of localization at prime ideals to obtain

a property of modules.

Lemma 2.3.4 Let A be a ring and M an A-module. Then M = 0 if and only if

Mp = 0 for all maximal ideals p of A.

Proof () follows trivially. We shall prove the nontrivial implication (). Sup-pose M 0. Choose any nonzero x M. Let I = AnnA(x) := {a A | ax = 0}.Clearly, 1 I, so I is proper. Hence, I is contained in a maximal ideal p ofA. We

claim that Mp 0. For, if Mp = 0, we would havex

1=

0

1in Mp, so then we would

find an element u A\p with ux = 0, giving that u I. But this contradicts thatI p. Hence the lemma follows.

2.4 An Interesting Application

Let A be a ring. In this section, we shall use the localization technique to obtain a

bound on the Krull dimension of A[x] in terms of the Krull dimension of A.

We say that an ideal q of A[x] is lying over an ideal p of A ifq A = p. Weclaim the following: there cannot be more than two prime ideals of A[x] lying

over the same prime ideal of A. Suppose, if possible, that we can find a chain

p1 p2 p3

of prime ideals of A[x] lying over a single prime ideal p ofA:

p1 A = p2 A = p3 A = p.

Going modulo p, we may assume that A is an integral domain and

p1 A = p2 A = p3 A = 0.

Let S = A\{0} and let K = S1A, the field of fractions of A. It is easy to checkthat (S1A)[x] = S1(A[x]). Now, prime ideals of S1A[x] are in bijective cor-respondence with the prime ideals of A[x] that do not intersect S. Hence, we

have

S1p1 S1p2 S1p3.

But K = S1A is a field, whence S1A[x] is a PID, from which we conclude thatdim(S1A[x])=1, a contradiction. Hence, we cannot have more than two primeideals of A[x] lying over the same prime ideal of A. This means, given a prime

ideal p ofA, there are at most two prime ideals p1, p2 ofA[x] such that

p1 p2 and p1 A = p2 A = p.

Therefore, given any chain of prime ideals of A[x], say, of length n,

p0 p1 pn,13


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the length of the corresponding chain

p0 A p1 A pn A,

of prime ideals in A cannot have length less than n/2. From this we get a bound

on the Krull dimension of A[x]:

dimA dimA[x] 2dimA + 1.

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Chapter 3

Noetherian Rings and Modules

3.1 Definitions and Basic Properties

Theorem 3.1.1 Let A be a ring and M be an A-module. Then the following areequivalent:

1. Any nonempty collection of submodules of M has a maximal element.

2. Any ascending chain of A-submodules of M is stationary.

3. Every A-submodule of M is finitely generated.

Proof Exercise.

Definition 3.1.2 (Noetherian modules, Noetherian rings) Let A be a ring. An

A-module M is called noetherian if M satisfies any of the equivalent conditions

of Theorem 3.1.1. A ring A is called a noetherian ring if A is noetherian as an

A-module.

Remark. A subring of a noetherian ring need not be noetherian. For example,

let k be a field and consider B = k[x1, x2, . . .] be the polynomial ring in infinitely

many variables over k. Let A denote the field of fractions of B. A is noetherian

being a field but B is not; for, there exists a non-terminating strictly increasing

chain of ideals of B:

x1 x1, x2 x1, . . . , xn .

Proposition 3.1.3 Let A be a ring, M be an A-module and N be an A-submodule

of M. Then M is noetherian if and only if N and M/N are noetherian.

Proof Suppose M is noetherian. An ascending chain of A-submodules of N is

also an ascending chain of A-submodules of M. Also, an ascending chain of A-

submodules of M/N corresponds to an ascending chain of A-submodules of M

containing N. Hence it follows that N and M/N are noetherian.

Now assume that N and M/N are noetherian. Let K be any A-submodule of

M. We shall prove that K is finitely generated. Since M/N is noetherian, the

15


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submodule (N + K)/N is finitely generated, say, by the set {k1, . . . , kn}, whereki K for all i. Also, the submodule K N of the noetherian A-module N isfinitely generated, say by the set {q1, . . . , qm}, where qj K N for all j. Weclaim that K is generated by the finite set {k1, . . . , kn, q1, . . . , qm}. Let x K bearbitrary. Then there exist i

A such that

x =

ni=1

iki (N + K)/N K/(K N).

Therefore, x ni=1 iki K N, so we can find j A such thatx

ni=1

iki =

mj=1

jqj

. The proof now follows from this.

Corollary 3.1.4 Let A be a ring and M1, . . . , Mn be A-modules. Then the direct

sum ni=1

Mi is noetherian if and only if each of M1, . . . , Mn is noetherian.

Proof Exercise.

Corollary 3.1.5 Let A be a noetherian ring and M be a finitely generated A-

module. Then M is noetherian.

Proof Let Mbe generated by {a1, . . . , an}, aM. We shall prove that M is noethe-rian by induction on n. For n = 1, note that a1A A/ AnnA a1 and since A is

noetherian, by Proposition 3.1.3, M = a1A is noetherian. By induction, assume

that B = a1A + + an1A is noetherian. Let M be generated by n elementsa1, . . . , an. Since

M/B = (anA + B)/B anA/(anA B),

M/B is noetherian; hence B is noetherian by Proposition 3.1.3.

Corollary 3.1.6 A homomorphic image of a noetherian ring is noetherian.

Proof Let f : M Nbe a homomorphism ofA-modules, where Mis noetherian.Then f(M) M/ ker f, so by Proposition 3.1.3, f(M) is noetherian.

Corollary 3.1.7 Let A be a noetherian ring and f : A B be a homomorphismof rings such that B is a finitely generated ring extension over f(A). Then B is

noetherian. Thus, in particular, any finitely generated algebra overZ or over a

field k is noetherian.

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Proof Note that the hypotheses imply that B is a finitely generated f(A)-module.

The result now follows from corollaries 3.1.6 and 3.1.5.

We shall now see an interesting property of certain endomorphisms of noethe-

rian modules.

Lemma 3.1.8 Let A be a ring and M be a noetherian A-module. Then any sur-

jective endomorphism f : M M is injective.

Proof Consider the following increasing chain of submodules of M:

ker f ker f2 ker fn .

Since M is noetherian, for some n N, we have ker fn = ker fn+1. That is,fn(a) = 0 if and only if f(fn(a)) = 0. Now, let b ker f be arbitrary. Bysurjectivity of f, we can find b1 Msuch that f(b1) = b. Continuing, we can findbn M such that fn(bn) = b. Then we see that fn+1(b) = f(fn(bn)) = 0 if andonly if fn(bn) = b = 0. Thus, ker f = 0, so f is injective.

Finally, we see that a noetherian ring remains noetherian after localization.

Lemma 3.1.9 Let A be a noetherian ring and S A be a multiplicative set. ThenS1A is noetherian.

Proof We know from Proposition 2.1.5 that any ideal of S1A is of the form S1I

for some ideal I ofA. Given any nonempty collection {S1I | } of ideals ofS1A, consider the corresponding collection {I | } of ideals of A. Since Ais noetherian, this has a maximal element, say I0. It is now easy to see that S

1I0is a maximal element of {S1I | }. Thus, S1A is noetherian.

3.2 The Hilbert Basis Theorem

Theorem 3.2.1 (Hilberts basis theorem) Let A be a noetherian ring. Then the

polynomial ring A[x] is also a noetherian ring.

Proof (Sarges, 1976) Let I be an ideal of A[x]. Since the zero ideal 0 is finitely

generated, assume I 0. Suppose, if possible, that I is not finitely generated.

Clearly, I contains nonconstant polynomials. Choose a polynomial f0(x) Iof minimal positive degree. For each n > 0, inductively choose a polynomial

fn(x) I\f0(x), . . . , fn1(x) of minimal positive degree. Let kn = deg fn(x) foreach n. Clearly, kn kn+1 for each n. (For, kn > kn+1 would contradict the choiceof fn(x).) For each n, let an denote the leading coefficient of fn(x). Since A is

noetherian, the increasing sequence of ideals of A:

a0 a0, a1 a0, . . . , an 17


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is stationary, so for some n we have a0, . . . , an = a0, . . . , an+1. Hence, thereexist i A such that an+1 =

ni=1 iai. Since fn+1(x) f0(x), . . . , fn(x), we have

g(x) = fn+1(x) n

i=1

ifi(x)xkn+1ki f0(x), . . . , fn(x).

But deg g(x) < deg fn+1(x), a contradiction to the choice of fn+1. Thus, I must be

finitely generated and the theorem follows.

Corollary 3.2.2 If A is a noetherian ring, then for any n N, A[x1, . . .xn] is anoetherian ring.

Proof This follows from Theorem 3.2.1 and induction.

Let A be a ring. We define the power series ring A[[x]] to be the set of all

formal power series in x, that is, elements of the form

n=0 anxn, which becomes

a ring with addition and multiplication by

n=0

anxn +

n=0

bnxn =

n=0

(an + bn)xn,

n=0

anxn

n=0

bnxn

=

n=0

i+j=n

aibj

xn.

We define the degree of a power series f(x) =

n=0 anxn to be the least non-

negative integer n such that an 0 and ai = 0 for all i < n; we shall denote it by

Deg(f(x)) and call an the starting coefficient of f(x). It is easy to see that A[[x]]

is commutative if A is commutative. We shall now prove that A[[x]] is noetherian

if A is noetherian.

Theorem 3.2.3 If A is a noetherian ring, then the power series ring A[[x]] is also

a noetherian ring.

Proof We shall prove that A[[x]] is noetherian by proving that its every ideal is

finitely generated. Let J be any ideal of A[[x]]. Define

In = {a A | f(x) =

n=0

anxn J such that Deg(f(x)) = n and an = a }.

Note that each In is an ideal ofA. Since x f(x) J for any f(x) J, we have

I0 I1 In .

Since A is noetherian, r N such that In = Ir for all n r. Since each In isfinitely generated, we can find an,1, . . . , an,kn

A such that In =

an,1, . . . , an,kn

.

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Let fi,j(x) J denote a polynomial whose degree is i and the starting coefficientis ai,j. We claim that J is generated by the finite set

{fi,j(x) | 1 i rand 1 j ki}.

Let f(x) J be arbitrary. If Deg(f(x)) = n r, and a is the starting coefficient off(x), then there exist n,j A such that a =

knj=1

n,jan,j. Then

f(x) kn

j=1

n,jfn,j(x)

is a power series of degree at least n + 1. We can procced inductively to assume

that Deg(f(x)) r. If Deg(f(x)) = n > r, then In = Ir and with the same notationas above,

f(x) kr

j=1

r,jx

n

r

fr,j(x)

is a power series of degree at least n + 1. Continuing in the same manner, we can

find power series

gj(x) =

l=n

l,jxlr, for 1 j kr

such that

f(x) =

krj=1

gj(x)fr,j(x).

Thus, it follows that J is finitely generated, and A[[x]] is noetherian.

3.3 Cohens Theorem and Applications

We now give a characterization of noetherian rings which allows us to conclude

that a ring is noetherian by only proving that all its prime ideals are finitely gen-

erated.

Theorem 3.3.1 (Cohen) A ring A is noetherian if and only if all its prime ideals

are finitely generated.

Proof One implication is clear. Let us prove the nontrivial implication. Suppose

all prime ideals of A are finitely generated. Suppose, if possible, that A is not

noetherian. Then the collection

= {I A | I is an ideal which is not finitely generated }

is nonempty and partially ordered by inclusion. By Zorns lemma, it follows that

has a maximal element, say I. By hypothesis, I cannot be a prime ideal, so we

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can find x,y I such that xy I. By maximality of I in , the ideals I, x andI,y must be finitely generated. Let

I,y = u1, . . . , un.

Write ui = ci +ydi for each i, where ci I and di A. Note that the set(I : y) = {a A | ay I }

is an ideal of A which contains both I and x. Hence, (I : y) and is finitely

generated, say

(I : y) = v1, . . . , vm.Now, let z I be arbitrary. Considering z as an element of I,y, we can findi A such that

z =

n

i=1

iui =

n

i=1

ici +

n

i=1

iydi.

Since z I and ci I, we have y(n

i=1 idi) I, so we have (n

i=1 idi) (I : y).Hence, we can find j A such that

ni=1

idi =

mj=1

jvj,

which immediately gives

z =

n

i=1

ici +

m

j=1

jyvj,

whence it follows that I = c1, . . . , cn,yv1, . . .yvm is finitely generated, a contra-diction to our hypothesis that I . Therefore, our assumption is wrong and A isnoetherian. This completes the proof.

Recall that if M is an A-module, then the annihilatorof M in A is defined to

be

AnnA(M) := {a A | aM = 0 }.

It is easy to see that AnnA(M/N) is an ideal ofA. We shall now see a generalizationof Cohens theorem for modules:

Theorem 3.3.2 Let A be a commutative ring and M be a finitely generated A-

module. Then M is a noetherian A-module if and only if pM is finitely generated

for every prime ideal p of A.

Proof One implication () is obvious; we prove the nontrivial implication. Sup-pose, if possible, that pM is finitely generated for every prime ideal p of A but M

is not noetherian. Then the collection

= {N M | N is an A-submodule of M which is not finitely generated }20


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is nonempty and partially ordered by inclusion. By Zorns lemma, it follows that

has a maximal element, say N. Consider M/Nwith natural A- module structure.

Claim : The annihilator of M/N in A, AnnA(M/N), is a prime ideal of A.

Suppose it is not prime. Then we can find a, b AnnA(M/N) such that ab AnnA(M/N). By maximality ofN, both N+aM and N+ bM are finitely generated.

Let

N + bM = Au1 + + Aun.Write ui = ci + bdi for each i, where ci N and di M. Note that the set

(N : b) := {m M | bm N }

is an A-submodule of M which contains both N and aM. Hence, (N : b) and

is finitely generated, say

(N : b) = Av1 +

+ Avm.

Now, let z N be arbitrary. Considering z as an element of N + bM, we can findi A such that

z =

ni=1

iui =

ni=1

ici +

ni=1

ibdi.

Since z Nand ci N, we have b(n

i=1 idi) N, so we have (n

i=1 idi) (N : b).Hence, we can find j A such that

n

i=1idi =

m

j=1jvj,

which immediately gives

z =

ni=1

ici +

mj=1

jbvj,

whence it follows that N = Ac1 + +Acn + bAv1 + + bAvm is finitely generated,a contradiction to our hypothesis that N . Therefore, AnnA(M/N) must be aprime ideal of A. Let p = AnnA(M/N).

Note that M is a finitely generated A-module; hence, so is M/N. If x1, . . . , xr

generate M as an A-module, then M/N = Ax1 + + Axr. Then it is easy to seethat

p = AnnA(M/N) = ri=1AnnA(Axi).Since p is a prime ideal, it follows that p = AnnA(Axi) for some i. Now, the

submodule N+Axi must be finitely generated. Suppose that N+ Axi is generated

by the finite set {y1 + r1xi, . . . ,ys + rsxi}. By an argument similar to the one above,it follows that N = Ay1 + +Ays + pxi. But observe that pM N, which impliesthat N is generated by finitely many elements y1, . . . ,ys, contradicting that N .Thus, M must be a noetherian A-module. This proves the theorem.

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Recall that we proved in Corollary 3.1.5 that any finitely generated module B

over a noetherian ring A is noetherian. As a consequence of the above general-

ization of Cohens theorem, we obtain the following important theorem by Eakin

and Nagata which gives a converse to Corollary 3.1.5 in a certain case.

Theorem 3.3.3 (Eakin-Nagata) Let A be a ring. Let B be a ring such that A Band B is a finitely generated A-module. If B is noetherian, then A is noetherian.

Proof Let p be any prime ideal of A. Then pB is a prime ideal of B and is finitely

generated as a B-module since B is noetherian. Hence there exist x1, . . . , xm psuch that pB = x1B + + xmB. Now, B is finitely generated as an A- module,say B = Ay1 + + Ayn for some yj B. Then it follows that pB is generatedby the finite set {xiyj | 1 i m, 1 i m} over A. Using Theorem 3.3.2we conclude that B is a noetherian A-module. Observe that since B is an over-

ring of A, AnnA(B) = 0 and that A is an A-submodule of B. Therefore, A is a

noetherian A-module and consequently, a noetherian ring. (Alternatively, definef : A n

j=1B by f(a) = (ay1, , ayn). Since AnnA(B) = 0, it is easy to see that

f is injective and A is isomorphic to f(A) as an A-module. Since B is a noetherian

A-module, so is nj=1

B. Hence, f(A) is a noetherian A-module and consequently,

so is A. Therefore, A is a noetherian ring.)

An immediate generalization of the Eakin-Nagata theorem is the following.

Corollary 3.3.4 Let A be a ring and M be a finitely generated faithful (that is,

AnnA(M) = 0) A-module. If M is a notherian A-module, then A is a noetherian

ring.

Proof Suppose M = Ax1 + + Axn, for some xi M. Define f : A ni=1Mby f(a) = (ax1, , axn). Since AnnA(M) = 0, it is easy to see that f is injectiveand A is isomorphic to f(A) as an A-module. Since M is a noetherian A-module,

so is ni=1

M. Hence, f(A) is a noetherian A-module and consequently, so is A.

Therefore, A is a noetherian ring.

3.4 Primary Decomposition of RadicalsDefinition 3.4.1 (Radical, Nilradical) Let A be a ring and I be an ideal of A.

The set of elements {x A | xn I for some n N } is an ideal of A, called theradical of I; we shall denote it by

I. The ideal

0 of all nipotent elements of A

is called the nilradical of A and is sometimes denoted by nil(A).

Remark. If A is a noetherian ring, then for any ideal I of A,

I is finitely gen-

erated, say

I = a1, . . . , ak. For each i, ni N such that anii I. Putn = n1 + + nk. Then it is easy to see that (

I)n I.

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Lemma 3.4.2 Let A be a ring and S A be a mutiplicative set. Let I be an idealof A which is maximal with respect to the property that I S = . Then I is aprime ideal.

Proof Consider the collection

= {I A | I is an ideal of A such that I S = }. is nonempty as 0 . I is a maximal element of . Suppose, if possible, that Iis not prime. Then a, b I such that ab I. By maximality ofI in , I, a andI, b intersect S, so we can find x I, a and y I, b such that x,y S. Write

x = + ra, y = + sb, where , I, r, s A.Since ab I, we have xy I. But xy S, S being a multiplicative set, acontradiction as I S = . Thus, I must be a prime ideal.

Lemma 3.4.3 Let A be a ring and I A be an ideal. Then

I =

pSpec(A),pI

p.

Proof Ifa

I, then an I for some n N. Hence, an p for every prime idealp that contains I. Hence a p for every prime ideal p I. On the other hand,assume a p for every prime ideal p I. If a

I, then S = {1, a, a2, . . .} is

a multiplicative set with S I . By Lemma 3.4.2, there exists a prime idealp

I such that p

S =

. But then a

p

S, a contradiction. Hence, a

I and

we are through.

Proposition 3.4.4 Let A be a noetherian ring and I be an ideal of A. Then

I is

a finite intersection of prime ideals of A.

Proof Assume the contrary. Then the collection

= {I A| I is an ideal ofA,

I is not a finite intersection of prime ideals of A}is nonempty, and partially ordered by inclusion. An application of Zorns lemma

gives a maximal element I . I cannot be prime; for, otherwise we would haveI = I, contradicting that I . Thus, we can find elements x,y I such that

xy I. By maximality of I in , we have I, x, I,y . So there exist finitelymany prime ideals p1, . . . , pn, q1, . . . qm ofA such that

I, x = ni=1pi and

I,y = mj=1qj.

We shall obtain a contradiction by proving that

I =I, x

I,y is a finiteintersection of prime ideals. Clearly, we have

I I, x

I,y. Let a I, x

I,y. There exist k, l N such that ak I, x and al I,y. Write

ak

= + rx, al

= + sy where , I, r, s A.23


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Then ak+l I as xy I. Hence, it follows that

I =I, x

I,y and wearrive at a contradiction. Therefore, the proposition holds.

Theorem 3.4.5 Let A be a noetherian ring and I be an ideal of A. Suppose n

N

is the least such that I can be written as an intersection of n prime ideals of Acontaining I:

I = n

i=1pi. Then { pi | 1 i n} is precisely the set of all prime

ideals of A minimal over I.

Proof IfIis itself a prime ideal, then there is nothing to prove as

I = I. Suppose

I is not prime. We have I = ni=1pi. We claim that each pi is minimal over I. For,

if some pi is not minimal over I, say p1, then q Spec(A) such that I q p1.Then it is easy to see by taking radicals that

I = ni=1pi q p1.

From Remark 3 following Definition 1.1.5, there exists r 2 such that pr q p1. But then we have

I = ijpi, contradicting the minimality of n. Thus, each

pi must be minimal over I.

Now, let p Spec(A) be minimal over I. Since I is not prime, I p. Takingradicals, we get

I = ni=1pi p,so there exists 1 r n such that pr p. But since p is minimal over I, pr = p.Thus, the theorem follows.

Finally, we note a fact which immediately follows from Proposition 3.4.4 and

Theorem 3.4.5.

Corollary 3.4.6 For any ideal I of a noetherian ring A, there are only finitely

many prime ideals of A minimal over I.

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Chapter 4

Artinian Rings and Modules

4.1 Definitions and Basic Properties

Theorem 4.1.1 Let A be a ring and M be an A-module. Then the following areequivalent:

1. Any nonempty collection of submodules of M has a minimal element.

2. Any descending chain of A-submodules of M is stationary.

Proof Exercise.

Definition 4.1.2 (Artinian modules, Artinian rings) Let A be a ring and M an

A-module. Then M is called artinian if M satisfies any of the equivalent conditionsof Theorem 4.1.1. A ring A is called an artinian ring if A is artinian as an A-

module.

Examples. 1. A field kis an artinian ring.

2. A finite dimensional vector space over a field kis an artinian k-module.

3. A finite ring is artinian. In particular, if p is a prime number and n N, thenZ/(pnZ) is a Z-module.

4. Ifkis a field and A = k[x], then

x x2 xn .

is an infinite strictly descending chain of ideals of A. Thus A is not artinian.

Remark. Example 4 above shows that an analogue of Hilberts Basis Theorem is

not true for artinian rings.

Proposition 4.1.3 Let A be a ring, M be an A-module and N be an A-submodule

of M. Then M is artinian if and only if both N and M/N are artinian.

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Proof If M is artinian, it os easy to see that both N and M/N are artinian (since

chains of submodules of N and M/N correspond to chains of submodules of M).

Conversely, suppose N and M/N are artinian. Consider a descendng chain of

submodules of M:

K1

K2

Kn

.

Since the descending chain

(K1 + N)/N (K2 + N)/N (Kn + N)/N of submodules of M/N must terminate, n N such that

(Kn + N)/N = (Kn + N)/N,

for all n n. For each n N we have (Kn + N)/N K/(Kn N). But thedescending chain

K1

N

K2

N

Kn

N

of submodules of N must also terminate, n N such that

(Kn + N)/N = (Kn + N)/N,

for all n n. Taking n0 = max{n, n}, it is easy to see that Kn = Kn0 for alln n0. Therefore, it follows that M is artinian. This completes the proof.

As immediate corollaries of this, we have the following results, which can be

proved along similar lines of corresponding results in the noetherian case.

Corollary 4.1.4 Let A be a ring and M1, . . . , Mn be A-modules. Then the directsum n

i=1Mi is artinian if and only if each of M1, . . . , Mn is artinian.

Corollary 4.1.5 Let A be an artinian ring and M a finitely generated A-module.

Then M is artinian.

Corollary 4.1.6 A homomorphic image of an artinian A-module is artinian.

We shall now see an interesting property of certain endomorphisms of artinian

modules, similar to the one we saw in the previous chapter (See Lemma 3.1.8).

Lemma 4.1.7 Let A be a ring and M be an artinian A-module. Then any injective

endomorphism f : M M is surjective.Proof Consider the following decreasing chain of submodules of M:

Im f Im f2 Im fn .Since M is artinian, for some n N, we have Im fn = Im fn+1. Let y M bearbitrary. Since fn(y) Im fn = Im fn+1, we can find an element x M suchthat fn(y) = fn+1(x), that is, f(fn1(y) = f(fn(x)), so injectivity of f impliesthat fn1(y) = fn(x). Proceeding inductively, we see that y = f(x). Thus, f issurjective.

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Proposition 4.1.8 An artinian domain is a field.

Proof Let A be an artinian domain and let x A be nonzero. Since the decreasingsequence of ideals

x

x2

xn

must be stationary, there exists n N such that xn = xn+1. Since xn xn+1,we have

xn = xn+1y,

for some y A. Since A is a domain and x 0, we get xy = 1. It follows that A isa field.

Corollary 4.1.9 In an artinian ring, every prime ideal is maximal.

Proof Let Let A be an artinian ring and p be a prime ideal of A. Then A/p is an

artinian domain, which has to be a field by Proposition 4.1.8. Hence p must be amaximal ideal of A.

Remark. IfA is an artinian ring, then the nilradical of A is the same as the Jacob-

son radical of A:

nil(A) = pSpec(A)p = pMax(A)p = Jac(A).

Proposition 4.1.10 An artinian ring is semilocal.

Proof Let A be an artinian ring. Suppose, if possible that A is not semilocal; thenwe can find an infinite set {mn | n N} of distinct maximal ideals of A. Since thedescending sequence of ideals

m1 m1 m2

must terminate, for some n N we have

ni=1mi = n+1i=1 mi.

whence ni=1mi mn+1. Since mn+1 is a prime ideal, by Remark 3 following

Definition 1.1.3, we see that mi mn+1 for some 1 i n, so mi = mn+1,mi being maximal; a contradiction to our assumption. Thus, A must have only

finitely many maximal ideals.

Proposition 4.1.11 Let A be an artinian ring. Then the nilradical of A (which is

the same as the Jacobson radical of A) is nilpotent.

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Proof For brevity, write N= nil(A). The descending chain

N N2 terminates, so for some k N, we have Nk = Nl for all l k. We claim thatN

k

= 0. Assume the contrary. Then the collection

C = {I A | I is an ideal such that INk 0}is nonempty (for, Nl C for all l), and has a minimal element, say I, A beingartinian. Since INk 0, there exists an element x I such that

xNk 0.The minimality of I in C implies that I = x. Also,

(

x

Nk)Nk =

x

N2k =

x

Nk 0,

and xNk I, so the minimality of I again implies xNk = I. Hence, xNk =x. Hence, x = xy for some y Nk N. Since y is nilpotent, yr = 0 for some r,so

x = xy = xy2 = = xyr = 0,whence I = x = 0, a contradiction. Thus, Nk = 0 and N is nilpotent.

4.2 Equivalent Characterizations of Artinian Rings

and Modules

Proposition 4.2.1 For a vector space V over a fieldk, the following are equiva-

lent:

1. V is a finite dimensional vector space over k.

2. V is a noetherian k-module.

3. V is an artinian k-module.

Proof (1)

(2): If V is a finite dimensional vector space over k, then every k-

submodule ofV is a subspace ofV which is finite dimensional and hence, finitely

generated. Therefore, V is a noetherian k-module.

(2) (3): IfV is a noetherian k-module, then V is finitely generated, so it hasa finite basis. Given any nonempty collection C of k-submodules (that is, vectorsubspaces) of V, we can choose a subspace of least dimension, which serves as a

minimal element ofC. Thus, V os artinian.(3) (1): Suppose V is artinian but not a finite dimensional vector subspace

over k. Then we can find an infinite subset {en | N N} of linearly independentvectors in V. Then

i1

kei

i2kei

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is an infinite stictly decresing chain of k-submodules ofV, a contradiction.

This completes the proof.

Lemma 4.2.2 Let A be a ring. Let

0 = M0 M1 Mn = Mbe a filtration of A-modules. Then

1. M is noetherian if and only if Mi+1/Mi are noetherian for each i.

2. M is artinian if and only if Mi+1/Mi are artinian for each i.

Proof This is just a repeated application of Propositions 3.1.3 and 4.1.3.

Lemma 4.2.3 Let A be a ring andm1, . . . ,mn be maximal ideals of A (not neces-sarily distinct). Suppose m1 mnM = 0. Then M is noetherian if and onle if Mis artinian.

Proof Suppose M is noetherian. Consider the filtration

0 m1 mnM m1 mn1M m1M MPut Mi = m1 miM. Since M is noetherian, Mi/Mi+1 is noetherian for all i. ButMi/Mi+1 is an A/mi+1 module, that is, a vector space over the field A/mi+1. Thus,

M is artinian by Lemma 4.2.1. The other implication can be proved similarly.

We are now set to prove the main equivalent characterization of artinian rings.

A noetherian ring may not be artinian; for example, Z is noetherian but not ar-

tinian. The next theorem will imply that an artinian ring is always noetherian,

so the descending chain condition (DCC) is stronger than the ascending chain

condition (ACC).

Theorem 4.2.4 A ring A is artinian if and only if A is noetherian and dim(A)=0.

Proof Suppose A is an artinian ring. Then every prime ideal of A is maximal

(see Corollary 4.1.9); hence dim(A)=0. By Proposition 4.1.10, A has only finitelymany maximal ideals, say m1, ,mn. Proposition 4.1.11 implies that there existsk N such that

nil(A)k = Jac(A)k = (ni=1mi)k = (m1 mn)k = 0.This along with Lemma 4.2.3 implies that A is noetherian.

Conversely, assume that A is noetherian and dim(A)=0. Hence, it follows that

every prime ideal in A is maximal. Also, it follows that all the prime ideals of A

are minimal (that is, minimal over 0). Hence, by Theorem 3.4.5, we have

nil(A) = n

i=1mi = m1 mn.29


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But the since A is noetherian, the nilradical of A is nilpotent, so for some r N,we have

(m1 mn)r = 0.Again, this along with Lemma 4.2.3 implies that A is artinian. This proves the

theorem.

Remark. Using this characterization artinian rings, conclusions of Lemmas 3.1.8

and 4.1.7 can be put together to see the following interesting and useful fact sabout

endomorphisms of artinian rings: An endomorphism f : A A of an artinian ringA is surjective if and only if it is injective.

4.3 Jordan-Holder Series

Definition 4.3.1 (Simple module) Let A be a ring. An A-module M is said to besimple if M 0 and the only A-submodules of M are 0 and M.

Remark. A simple A-module is both artinian and noetherian.

Definition 4.3.2 (Jordan-Holder series) Let A be a ring and M an A-module.

Then M is said to be of finite length if there exixts a filtration

0 = M0 M1 Mn = M

such that Mi+1/Mi is simple for all i. The length of M is then defined to be n;

it is denoted by lA(M). Such a fitration is called a Jordan-Holder series or acomposition series for M.

We now state two well-known results which we shall use.

Theorem 4.3.3 Let A be a ring. Suppose that an A-module M has a Jordan-

Holder series of length n. Then every Jordan-Holder series of M has length n.

Proposition 4.3.4 Let A be a ring, M an A-module and N an A-submodule of M.

Then M has a finite length if and only if N and M/N have finite length. In this

case, we have lA(M) = lA(N) + lA(M/N).

Theorem 4.3.5 An A-module M has a Jordan-Holder series if and only if M is

both noetherian and artinian as an A-module.

Proof Suppose that M has a Jordan-Holder series:

0 = M0 M1 Mn = M.

Then each Mi+1/Mi is simple, whence it follows that M is both noetherian and

artinian as an A-module (See Lemma 4.2.2).

To prove the converse, assume that M is both noetherian and artinian as an

A-module. Since M is noetherian, it is easy to see that M has a maximal proper

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submodule, say M1. If M1 = 0, we are through. Otherwise, M1 is noetherian and

hence, has a maximal proper submodule, say M2. If M2 = 0, we are through.

Otherwise, we proceed inductively as before. If Mn 0 for all n N, then we getan infinite descending chain:

M M1 M2 Mn ,a contradiction as M is artinian. Thus, for some n N, we must have Mn = 0. Itis easy to see that

0 = Mn M1 Mis a Jordan-Holder series for M, Mi/Mi+1 being simple for each i. This completes

the proof.

4.4 Artinian Local RingsAs we shall see in this section, every artinian ring can be written uniquely as a

product of artinian local rings. Therefore, to study artinian rings, it is enough to

study artinian local rings.

Remark. IfA is an artinian ring with unique maximal ideal m, then m is the only

prime ideal of A, and we have nil(A) = Jac(A) = m. Hence, every element ofm is

nilpotent and so is m. Therefore, every element of A is either a unit or nilpotent.

An example of such a ring is Z/(pnZ), where p is a prime number and n N.Theorem 4.4.1 (Structure theorem for artinian rings) Every artinian ring is a

finite direct product of artinian local rings.

Proof Let A be an artinian ring. We konw that A is semilocal; let m1, mn be allthe distinct maximal ideals of A. As in the proof of Theorem 4.2.4, we have for

some k N(ni=1mi)k = mk1 mkn = 0.

Consider the natural map f : A ni=1

(A/mki ), which is pojection onto every

co-ordinate. Since, mki s are also pairwise coprime, by chinese remainder theorem

it follows that f is an isomorphism of rings. Clearly, every A/mki is artinian. It

remains to show that each A/mki is local. Any maximal ideal ofA/mki corresponds

to a maximal ideal m of A such that m mki . Since mi is maximal in A, it isalso prime in A, and hence, it follows that mi m. Maximality ofmi implies thatmi = m. Now, it is easy to see that the unique maximal idael of A/m

ki is mi/m

ki .

Thus, each A/mki is local and the proof follows.

Example. A ring with only one prime ideal need not be noetherian (and hence,

not artinian). Let k be a field and A = k[x1, x2, . . .] be a polynomial ring with in-

finitely many variables. Let I = x1, x22, x33, . . .; then the ring B = A/I has a singleprime ideal, namely, x1, x2, . . .. Hence, B is a local ring with Krull dimension 0.But B is not noetherian as the ideal x1, x2, . . . is not finitely generated.

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IfA is a local ring, we shall always denote by m its unique maximal ideal. The

field k = A/m is called the residue fieldof A. The A-module m/m2 is annihilated

by m and hence, is a vector space over k. If A is noetherian (or artinian), m is

finitely generated, so dimk(m/m2) is finite.

Proposition 4.4.2 Let A be an artinian local ring. Then the following are equiv-alent:

1. Every ideal in A is principal.

2. The maximal ideal m of A is principal.

3. dimk(m/m2) 1.

Proof (1) (2): Trivial.(2) (3): This follows from the fact that the image of the generator ofm in m/m2generates m/m

2

.(3) (1): If dimk(m/m2) = 0, then m = m2, so by Nakayamas lemma, m =0, whence A is a field. Hence, every ideal in A is principal. Now suppose

dimk(m/m2) = 1. By Proposition 1.3.5, it follows that m is principal, say m = x.

Let I A be any ideal. Since m = Jac(A) is nilpotent, mn = 0 for some n. IfI is nontrivial, then we can find an r such that I mr but I mr+1. Hence, y I such that for some a A, y = axr but y mr+1. Therefore, it follows thata x = m. Thus, a is a unit ofA. Hence, xr = a1y I, so we have Ixr. Thus,I is principal. This completes the proof.

Example. Not all artinian rings satisfy the conditions of the above Proposition.

For example, the rings Z/(pnZ) and k[x]/f(x)n (f irreducible) satisfy the con-ditions. On the other hand, the artinian local ring k[x2, x3]/x4 does not: Herem = x2, x3, so m2 = 0 and dimk(m/m2) = 2.

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Chapter 5

Integral Extensions

5.1 Integral Elements and Integral Extensions

In some sense, integral extensions are a generalization of field extensions. Studyof diophantine equations in number theory led to the conept of algebraic integers,

while in classical algebraic geometry, algebraic curves were frequently studied by

projecting them onto a line and regarding the curve as a ramified covering of the

line. This is very much analogous to the relation between the rings of integers of

a number field and the ring Q, and the common algebraic feature is the notion of

integral dependence, which we shall study in this chapter.

Recall that a polynomial f(x) with coefficients in a ring A is called monic if its

leading coefficint is 1, the unity of A. If a ring A is a subring of a ring B (where

the unity elements of A and B coincide), B will be called a ring extension or just

an extension ofA.

Definition 5.1.1 (Integral element) Let A B be an extension of rings. An ele-ment x B is said to be integral over A if there exists a monic polynomial withcoefficints in A ssatisfied by x.

Examples. 1. For any ring A, every element a A is integral over A.2. Let K/F be a field extension and a K algebraic over F. Then a is integral

over F.

3. ConsiderZ Z+ iZ. 1 + i Z+ iZ is integral over Z as it satisfies the monicpolynomial x

2

+ 2x 2 = 0.4. Consider Z Q. Then p

q Q, where q 1 and (p, q) = 1, is not integral

over Z. For, if there exist a1, . . . , an Z such thatp

q

n+ a1

p

q

n1+ + an = 0,

then we would have

pn + a1pn1q + + anqn = 0,

which would imply that p and q have a common factor, a contradiction.

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5. Generalizing the above example, let A be a UFD and B be the field of frac-

tions of A. Thenp

q B, where q is not a unit and (p, q) = 1, is not integral

over A.

Proposition 5.1.2 Let B be a ring and A a subring of B. Then the following areequivalent:

1. x B is integral over A.2. A[x] is a finitely generated A-module.

3. A[x] is contained in a subring C of B such that C is a finitely generated

A-module.

Proof (1) (2): Suppose x B is integral over A. Then n N and a1, . . . , an A such that

xn + a1xn1 + + an = 0.Hence, xn 1, x, x2, . . . , xn1, from which it follows that A[x] is a finitely gener-ated A-module (generated by the set {1, x, x2, . . . , xn1}).

(2) (3): Trivial (Take C = A[x]).(3) (1): Suppose A[x] C B, where C is a finitely generated A-module,

say C = c1, . . . , cn. Then we can find i j A for 1 i, j n such that

xci =

nj=1

i jcj, for i = 1, . . . , n.

Writing

Tx =

x 11 12 1n21 x 22 2n...

.... . .

...

n1 n2 x nn

,

we get

Tx

c1c2...

cn

=

0

0...

0

.

Multiplying the above relation on the left by Adj(Tx), we get

(det Tx)

c1c2...

cr

=

0

0...

0

so (det Tx)C = 0. In particular, det Tx.1 = 0, so det Tx = 0. Expanding the

determinant, we get a monic polynomial with coefficients in A satisfied by x.

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Corollary 5.1.3 Let A be a subring of a ring B. If x1, . . . , xn B are integral overA, then A[x1, . . . , xn] is a finitely generated A-module.

Proof This follws from Proposition 5.1.2(3) by induction on n.

Proposition 5.1.4 Let A B be a ring extension. Then the setC = {x B | x is integral over A}

is a subring of B containing A.

Proof Clearly, C A, so C is nonempty. Let x,y C. By Corollary 5.1.3,A[x,y] B is a finitely generated A-module. Note that A[x y],A[xy] A[x,y].By Proposition 5.1.2(3), it follows that xy, xy are integral overA, that is xy, xy C. Therefore, C is a subring of B containing A.

Definition 5.1.5 (Integral closure) Let A B be a ring extension. Then the ringAB := {x B | x is integral over A}

is called the integral closure of A in B.

If B = AB, that is, every element of B is integral over A, then B is said to be

integral over A.

If A = AB, that is, the only elements of B is integral over A are those in A, then

A is said to be integrally closed in B.

If A is a domain and B is the field of fractions of A, then A is said to beintegrally closed or normal if A = AB.

Integral closure of a domain A in its field of fractions is called the normaliza-

tion of A.

Examples. 1. Z is normal (integrally closed in Q).

2. In general, any UFD is normal.

3. An important particular case of the above definition is the ring of integers

in algebraic number theory: A number field K is a finite field extension of

Q. We have ring extensions ZQ

K. The ring of integers

OK ofK is the

ring

OK = {x K | x is integral over Z},the integral closure ofZ in K.

Let n be a square-free integer and consider Z Q Q( n). The integralclosure ofZ in Q(

n) is Z[], where

=

1 +

n

2ifn 1(mod 4),

n ifn

2 or 3(mod 4).

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4. Consider A = k[x,y]/y2 x3, where k is a field. It is not difficult to seethat A k[t2, t3]. It is easy to see that the field of fractions of A is k(t).

Then t is integral over A but t A, so A is not normal. The integral closure

ofA in k(t) is k[t], which is normal, being a UFD.

5.2 Properties of Integral Extensions

We first prove a very important fact, namely, the transitivity of integral extensions.

Proposition 5.2.1 Let A B C be ring extensions. If C is integral over B andB is integral over A, then C is integral over A.

Proof Let x C. Since C is integral over B, n N and b1, . . . , bn B such that

xn + b1xn1 +

+ bn = 0.

Since b1, . . . , bn are integral over A, A[b1, . . . , bn] is finitely generated. Now, x is

also integral over A[b1, . . . , bn]; so A[b1, . . . , bn][x] = A[b1, . . . , bn, x] is a finitely

generated A-module. Since A[x] A[b1, . . . , bn, x], it follows that x is integralover A. This proves that C is integral over A.

Proposition 5.2.2 Let A B be an integral extension. Let S A be a multiplica-tive set. Then S1A S1B is an integral extension.

Proof Let

x

s S1

B, where x B and s S. Since B is integral over A, n Nand a1, . . . , an A such that

xn + a1xn1 + + an = 0.

Hence, it follows that

x

s

n+

a1

s

x

s

n1+ + an

sn= 0,

whereai

s

i

S1A for all i. Thus,

x

s

is integral over S1A.

Proposition 5.2.3 Let A B be an integral extension. Let I be an ideal of B andput J = I A. Then B/I is an integral extension of A/J.

Proof Let x B/I. Since B is integral over A, n N and a1, . . . , an A suchthat

xn + a1xn1 + + an = 0.

Reducing modulo J (and noting that J I),we see that x is integral over A/J.

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Corollary 5.2.4 Let A be a ring and I A[x] be an ideal containing a monicpolynomial. Put J = I A. Then A[x]/I is an integral extension of A/J.

Proof This is easily seen by observing that a monic polynomial in Igives a monic

polynomial satisfied by x after reduction modulo J.

If in Proposition 5.2.3 we have I Spec(A) and J Spec(B), then A/I B/Jis an integral extension of domains. Hence, it is important to study integral exten-

sions of domains. We now see a very important result about integral extension of

domains.

Lemma 5.2.5 Let A B be an integral extension of domains. Then A is a field ifand only if B is a field.

Proof Suppose A is a field. Let b

B be nonzero. Let f(x)

A[x] be a monic

polynomial of minimal degree satisfied by b, say

f(x) = xn + a1xn1 + + an,

where ai A. We havef(b) = bn + a1b

n1 + + an = 0.Ifan = 0, since B is a domain, we would have

bn1 + a1bn2 + + an1 = 0,

contradicting the choice of f(x). Thus, an 0 and we have

b1 = a1n (bn1 + a1bn2 + + an1).Therefore, it follows that B is a field.

Conversely, let that B be a field and a A be nonzero. a has an inverse in B,say b. But B is integral over A, so m N and a1, . . . , am A such that

bm + a1bm1 + + am = 0.

Multiplying by am, we get

1 + a1a + + amam

= 0,

so a1 = (a1 + + amam1) A, whence it follows that A is a field.

Corollary 5.2.6 Let A B be an integral extension of rings. Letq Spec(B) andp = q A. Then p is a maximal ideal of A if and only ifq is a maximal ideal of B.

Proof Since B is integral over A, B/q is integral over A/p. This is an integral

extension of domains, so A/p is a field if and only if B/q is a field. The result

follows from this.

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5.3 Noether Normalization

The Noether normalization theorem says that any finitely generated k-algebra is

integral over a suitably chosen polynomial subalgebra over k, where kis a field.

Definition 5.3.1 (Algebra over a ring) Let A be a ring. An A-algebra is a ringB (not necessarily commutative) with a ring homomorphism f : A B such thatf(A) is contained in the centre of B.

With the same notation of the above definition, B has a natural A-module struc-

ture given by

a b = f(a)b = b a,for all a A, b B.

Definition 5.3.2 (Affine algebra over a field) Let k be a field and I be an ideal

of k[x1, . . . , xn]. Then A := k[x1, . . . , xn]/I is called an affine k- algebra.

In other words, A is a finitely generated k-algebra. If A as above is also a

domain, it is called an affine k-domain.

Lemma 5.3.3 Let k be a field and f(x1, . . . , xn) k[x1, . . . , xn] be a nonconstantpolynomial. Then there exists a ring automorphism of k[x1, . . . , xn] such that

a (f(x1, . . . , xn)) is monic in xn for some a k.

Proof We claim that there exist d1, . . . , dn1 N such that the ring automorphism ofk[x1, . . . , xn] determined by

(1) = 1, (xn) = xn, and (xi) = xi + xdin for 1 i n 1

is such that (f(x1, . . . , xn)) is monic in xn, after multiplication by a suitable

nonzero constant.

Note that for a monomial x11

xnn , we have

(x11

xnn ) = (x1 + xd1n )1 (xn1 + xdn1n )n1xnn= x1d1++n1dn1+nn + lower degree terms in xn.

Now, we show that dis can be chosen so that for any two distinct monomials

x11

xnn and x11 xnn in f, we have

1d1 + +n1dn1 +n 1d1 + + n1dn1 + n.

Choose t N such that t > max{1, . . . , n} for any monomial x11 xnn in f.If we set di = t

i for each i, then by uniqueness of t-adic expansion of a natural

number it follows that di N, 1 i n are the required ones. Thus, no twodistict monomials in (f(x1, . . . , xn)) cancel each other and it follows that the term

in (f(x1, . . . , xn)) of highest degree is purely in xn. Then by composing with

multiplication by a suitable constant (if required), we are done.

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We now the prove the main theorem of the section, namely, the Noether nor-

malization lemma.

Theorem 5.3.4 (Noether normalization lemma) Let k be a field and I an ideal

of k[x1, . . . , xn]. Consider the affine k-algebra A = k[x1, . . . , xn]/I. Then there

exist z1, . . . ,zm A algebarically independent over k such that A is an integralextension of the polynomial subring B = k[z1, . . . ,zm].

Proof We prove the theorem by induction on n. The cases I = 0 and I =

k[x1, . . . , xn] are trivial, so we assume that 0 I k[x1, . . . , xn]. Consider the

case n = 1. Note that I k = since I is a proper ideal and I contains a monicpolynomial; therefore, k k[x1]/I is an integral extension and the result followsby taking B = k. Assume the result for all k< n.

Let A = k[x1, . . . , xn]/I be an affine algebra over k in n variables and let

f(x1, . . . , xn) be a nonzero polynomial in I. By Lemma 5.3.3, there exists an

automorphism of k[x1, . . . , xn] such that (f(x1, . . . , xn)) is monic in xn. PutJ = (I). Then A = k[x1, . . . , xn]/I (k[x1, . . . , xn])/(I) = k[x1, . . . , xn]/J.

Now, J contains a monic polynomial, namely, (f). Therefore, it follows that

C :=k[x1, . . . , xn1]

J k[x1, . . . , xn] k[x1, . . . , xn]

J A

is an integral extension. But C is an affine k-algebra in n 1 variables, so byinduction hypothesis, there exists a polynomial subalgebra B = k[z1, . . . ,zm] with

z1, . . . ,zm algebraically indepenxdent over k such that C is integral over B. Since

A is integral over C, it follows that A is integral over B. This proves the theorem.

Remark. With the notation of the Noether normalization lemma above ( Theorem

5.3.4), the minimal number m N such that A is integral over B = k[z1, . . . ,zm] isthe transcendence degree ofA over k.

As corollaries of the Noether normalization lemma, we obtain the weak and

the strong form of Hilberts nullstellensatz.

Lemma 5.3.5 (Hilberts nullstellensatz, weak form) Let k be an algebraically

closed field. Then any maximal ideal m of k[x1, . . . , xn] is of the form

m = x1 a1, . . . , xn an,

for some a1, . . . , an k.

Proof We claim that k[x1, . . . , xn]/m is a finite extension of k. k[x1, , xn]/m isa finitely generated k-algebra, so by the Noether normalization lemma, there exit

algebraically independent elements y1, . . . ,yr such that k[x1, . . . , xn]/m is integral

over k[y1, . . . ,yr]. Since k[x1, . . . , xn]/m is a field, by Lemma 5.2.5, k[y1, ,yr] isa field. Therefore, we must have r = 0 and it follows that k[x1, . . . , xn]/m is a finite

extension of k, and hence, an algebraic extension of k. Since k is algebraically

closed, k[x1, . . . , xn]/m k. Let ai denote the image of xi + m in k under this

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isomorphism. Then clearly x1 a1, . . . , xn an m. But it is easy to see thatx1a1, . . . , xnan is a maximal ideal ofk[x1, . . . , xn], som = x1a1, . . . , xnan.

Theorem 5.3.6 (Hilberts nullstellensatz) Let k be an algebraically closed fieldand I be an ideal of k[x1, . . . , xn]. Define

V(I) = { (a1, . . . , an) kn | f(a1, . . . , an) = 0 for all f I }

Then for any proper ideal I k[x1, . . . , xn], V(I) is an ideal of kn and

i(V(I)) := { h k[x1, . . . , xn] | h(a1, . . . , an) = 0 for all (a1, . . . , an) V(I) }

is an ideal of k[x1, . . . , xn] satisfying i(V(I)) =

I.

Proof If I is a proper ideal ofk[x1, . . . , xn], then I is contained in some maximalideal m ofk[x1, . . . , xn]. Note that v(I) V(m). From the weak form of the Hilbertnullstellensatz it follows that m = x1 a1, . . . , xn an for some (a1, . . . , an) kn,so (a1, . . . , an) V(m). Therefore, V(m) is nonempty, and hence, so is V(I). Thestatements that V(I) and i(V(I)) are ideals ofkn and k[x1, . . . , xn] respectively, are

easy verifications and are left as exercises.

One inclusion

I i(V(I)) is clear from the definition. We prove the nontriv-ial inclusion: Let f = f(x1, . . . , xn) i(V(I)). Consider the ideal J = 1 y f,I inthe polynomial ring k[x1, . . . , xn,y] in n + 1 variables. Note that (a1, . . . , an, b) V(J) if and only if (a1, . . . , an) V(I) and b f(a1, . . . , an) = 1. Therefore, V(J) =; for, otherwise, for some (a1, . . . , an, b) k

n+1

we would have

0 = 1 b f(a1, . . . , an) = 1,

a contradiction. Hence, it follows that J = k[x1, . . . , xn,y]. Since 1 J, we canfind polynomials gi k[x1, . . . , xn,y] and hi I such that

1 =

gi(x1, . . . , xn,y)hi(x1, . . . , xn) + g0(x1, . . . , xn,y)(y f(x1, . . . , xn) 1).

Since there are only finitely many gis in the above sum, we can find m Nsufficiently large such that after multiplying by fm both the sides of hte above

equation we can write

fm =

Gi(x1, . . . , xn,y f)hi(x1, . . . , xn) + G0(x1, . . . , xn,y f)(y f(x1, . . . , xn) 1).

Now, look at this polynomial identity in the localization of the ring k[x1, . . . , xn,y]

with respect to the multiplicative set S = {1, f, f2, . . .}. Note that f is a unit inS1k[x1, . . . , xn,y], so substituting y =

1

fin the above equation we get fm S1I,

so it follows that some power of f lies in I, that is, f

I. The theorem now

follows from this.

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5.4 The Going-up Theorem

Suppose that A B is a ring extension. We say that a prime ideal q of B is lyingover a prime ideal p ofA ifq A = p. Given a prime ideal p ofA, can we alwaysfind a prime ideal q of B lying over p? We prove that this is the case when B is

integral over A. If such a q exists, then q pB and q (A\p) = . We first provethat pB (A\p) = and use this to find a required q Spec(B).

Lemma 5.4.1 Let A B be an integral extension of rings. Letp Spec(A). Thenfor any y pB, there is a monic polynomial f(x) p[x] such that f(y) = 0.

Proof Let y pB; then y = ni=1 pibi for some pi p, bi B. Since B is integralover A, S = A[b1, . . . bn] is a finitely generated A-module, say

S = Aw1 + + Awr.

Then we can find i jk A such that

biwj =

rk=1

i jkwk, for 1 i n, 1 j r.

Now, y =n

i=1 pibi, so for each j, 1 j r, we have

ywj =

ni=1

pi(biwj) =

ni=1

pi

r

k=1

i jkwk

=r

k=1

jkwk,

where jkp as pi

p. Write

T =

y 11 12 1n21 y 22 2n...

.... . .

...

n1 n2 y nn

,

so we have

T

w1w2...

wn

=

0

0...

0

.

Multiplying the above relation on the left by Adj(T), we get

(det T)

w1w2...

wr

=

0

0...

0

.

so (det T)S = 0. In particular, det T.1 = 0, so det T = 0. Expanding the determi-

nant, we get a monic polynomial with coefficients in p satisfied by y.

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Lemma 5.4.2 Let A B be an integral extension of rings. Ifp Spec(A), thenpB (A\p) = .

Proof On contrary, assume that there exists an element x pB (A\p). Then byLemma 5.4.1, there exist a1, . . . , an

p such that

xn + a1xn1 + + an = 0.

But this implies xn p, so x p, p being a prime ideal. This is a contradictionsince x A\p. Hence, the lemma holds.

Theorem 5.4.3 (Lying-over theorem) Let A B be an integral extension ofrings and letp Spec(A). Then there exists q Spec(B) such thatq A = p.

Proof From the Lemma 5.4.1 and Lemma 5.4.2, the collection

= {I B | I is an ideal such that I A\p = }

is nonempty. An application of Zorns lemma gives us a maximal element q pB,which has to be a prime ideal of B by Lemma 3.4.2. Since q A\p = , we haveq A p; but we also have pB q, so p q A. Thus, p = q A. This provesthe lying over theorem.

The question now is the following: if we have p1, p2 Spec(A) and q1 Spec(B) such that p

1 p

2and q

1lies over p

1, can we find a prime ideal q

2of B

lying over p2 such that q1 q2?

Theorem 5.4.4 Let A B be an integral extension of rings. Letp1, p2 be primeideals of A such thatp1 p2 and letq1 be a prime ideal of B such thatq1 A = p1.Then there exists q2 Spec(B) such thatq1 q2 andq2 A = p2.

Proof Since B is integral over A, B/q is integral over A/p. We now have a com-

mutative diagram:

A //

B

A/p1 // B/q1

where the the horizontal maps are natural inclusions and , are natural projec-

tions. By the lying over theorem (Theorm 5.4.3), there exists a prime ideal q2 of

B/q1 such that q2 A/p1 = p2 = (p2). Then it follows from the commutativity ofthe diagram that q2 :=

1(q2) is a prime ideal ofB such that q2A = 1(p2) = p2.

Theorem 5.4.4 and induction immediately give the important Going-up theo-

rem of Cohen-Seidenberg.

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Theorem 5.4.5 (Going-up theorem of Cohen-Seidenberg) Let A B be an in-tegral extension of rings. Let m, n N be such that m n. Letp1 pn be achain of prime ideals of A andq1 qm be a chain of prime ideals of B suchthatqi A = pi for each 1 i m. Then there exist prime ideals qm+1 qnof B such thatqi

A = pi for each 1

i

n.

The going-up theorem has many important applications. A very useful fact

which follows from the going-up theorem is the following:

Proposition 5.4.6 If A B is an integral extension of rings, then dim(A) =dim(B).

Proof From the lying-over and going-up theorems (Theorem 5.4.3 and Theorem

5.4.5), given any chain

p0 p1 pmof prime ideals of A, there is a chain

q0 q1 qmof prime ideals of B such that qi A = pi for each 1 i n. Thus, clearlydim(A) dim(B). If dim(A) = , we are through, so assume that dim(A) is finite.Consider a maximal chain

p0 p1 pnof prime ideals of A and let

q0 q1 qnbe a corresponding chain of prime ideals of B such that qi lies over pi for each i.

Since pn is maximal in A, by Corollary 5.2.6, qn is maximal in B. To prove that

q0 q1 qnis a maximal chain of prime ideals of B, by going modulo a suitable prime ideal, it

suffices to prove the following: If A B an integral extension of domains, thenany nonzero prime ideal q of B satisfies q A 0.

We now prove this assertion: Let b

q be a nonzero element and let

xr + c1xr1 + + cr

be a minimal monic polynomial satisfied by b, with coefficients in A. Since B isa domain, it follows that cn 0 and that cn b A q A = p, so p 0.The result follows from this.

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5.5 The Going-down Theorem

Let A B be an integral extension of rings. We now consider the followingquestion: if we have p0, p1 Spec(A) and q0 Spec(B) such that p1 p0 and q0lies over p0, can we find a prime ideal q1 of B lying over p1 such that q1

q0? If

such a prime ideal q1 exists, then we must have q1 p1B, q1 (B\q0) = andq1(A\p1) = . In that case, q1(B\q0)(A\p1) = and hence p1B(B\q0)(A\p1) =. We shall see that this need not hold in the case of a general integral extensionA B; we show that this holds when A and B are domains and A is normal (thatis, integrally closed in its field of fractions).

Lemma 5.5.1 Let A B be an integral extension of domains with A integrallyclosed in its field of fractions k. Let b B. Then the minimal satisfied by b over kis an element of A[x] (that is, has coefficients in A).

Proof Let f(x) = xn

+ a1xn

1

+ + an be the monic polynomial in k[x] ofminimal degree satisfied by b. Since B is integral over A, the field of fractions of

B, say k, is an algebraic extension ofk. Hence, there exists an algebraic extensionL k kcontaining all the roots, say d1, . . . , dn of f(x). Since b is integral overA, there exists a monic polynomial g(x) A[x] such that g(b) = 0. Since g(x) canalso be thought of as an element of k[x], f(x)|g(x) in k[x], so g(di) = 0 for all i.Thus, all the dis are integral over A. But since integral elements over A form a

ring and

(x d1) (x dn) = xn + a1xn1 + + an,it follows that each ai k is integral over A. Since A is integrally closed, ai Afor all 1 i n, proving the lemma.

Lemma 5.5.2 Let A B be an integral extension of domains with A integrallyclosed in its field of fractions k. Letp be a prime ideal of A and b pB. Letf(x) = xm + c1x

m1 + + cm be the monic polynomial in k[x] of minimal degreesatisfied by b. Then ci p for all i.

Proof Since B is integral over A and b pB, by Lemma 5.4.1, there exists amonic polynomial

g(x) = xn + a1xn1 + + an p[x]

such that g(b) = 0. From Lemma 5.5.1 above, we have f(x) A[x]. since f(x) ismonic and of minimal degree satisfying f(b) = 0, it follows that f(x)|g(x) in A[x],so there exists h(x) A[x] such that g(x) = f(x)h(x). Reducing modulo p, we get

g(x) = xm = f(x) h(x),

so it follows that f(x) = xm, A/p being a domain. Thus, ci p for each i.

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Lemma 5.5.3 Let A B be an integral extension of dom

ca lecture notes

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