THE PRIME SPECTRUM OF A RING: A SURVEY

by James S. Fernandez

A Thesis Submitted to the Faculty of the College of Science in Partial Fulfillment of the Requirements for the Degree of Master of Science in Mathematics

Florida Atlantic University Boca Raton, Florida December 1991 THE PRIME SPECTRUM OF A RING: A SURVEY

by

James S. Fernandez

This thesis was prepared under the direction of the candidate's thesis advisor, Dr.

Lee Klingler, Department of Mathematics, and has been approved by the members of his supervisory committee. It was submitted to the faculty of the College of

Science and was accepted in partial fulfillment of the requirements for the degree of

Master of Science in Mathematics.

Supervisory Committee

nt of Mathematics

Date

ii ACKNOWLEDGEMENTS

I wish to express my sincere appreciation to my thesis advisor, Dr. Lee Klingler, for his patience and insight, and for lending assistance far beyond that which one normally assumes upon accepting the role of thesis advisor.

I also wish to thank Dr. J . Brewer for his direction in getting this endeavor started and Dr. F. Richman for his insight into fine tuning the nuances during the final revision.

I further wish to thank B. Broer, R. Pelava, and R. Ebel for their assistance and cooperation in getting this thesis printed.

iii ABSTRACT

AUTHOR James S. Fernandez TITLE The Prime Spectrum of a Ring: A Survey INSTITUTION Florida Atlantic University

THESIS ADVISOR Dr. Lee Klingler DEGREE Master of Science in Mathematics YEAR 1991

This thesis has as its motivation the exploration, on an informal level, of a correspondence between Algebra and Topology. Specifically, it considers the prime spectrum of a ring, that is, the set of prime ideals, endowed with the Zariski topology. Questions posed by M. Atiyah and I. MacDonald in their book, "Introduction to Commutative Algebra", serve as a guideline through most of this work. The final section, however, follows R. Heitmann's paper, "Generating Non-Noetherian Modules Efficiently". This section examines the patch topology on the prime spectrum of a ring where the patch topology has as a closed subbasis the

Zariski closed and Zariski quasi-compact open sets. It is proven that the prime spectrum of a ring with the patch topology is a compact Hausdorff space, and several relationships between the patch and Zariski topologies are established. The final section concludes with a technical theorem having a number of interesting corollaries, among which are a stable range theorem and a theorem of Kronecker, both generalized to the non-Noetherian setting.

iv TABLE OF CONTENTS

ACKNOWLEDGEMENTS ...... iii ABSTRACT ...... iv

Section

1 Preliminaries: Some Properties of Prime Ideals ...... 1

2 The Topology of the Prime Spectrum of a Ring ...... 4

3 Irreducible Spaces ...... 8

4 Homomorphisms and Ring Spectra ...... 12

5 Localization and Ring Spectra ...... 22

6 Integral Dependence ...... 25

7 Noetherian Spaces ...... 32

8 The Patch Topology ...... 40

References ...... 53

v THE PRIME SPECTRUM OF A RING: A SURVEY

§1 Preliminaries: Some Properties of Prime Ideals

We begin our discussion by reviewing several definitions and propositions concerning the prime and radical ideals of a ring. The results proved in our discussion depend heavily on the fact that the rings involved are commutative and possess a multiplicative unity. Henceforth, all rings mentioned in the text are assumed to be commutative and possess a multiplicative unity.

DEFINITION 1.1. The nilradical of a ring A is the subset consisting of all the nilpotent elements of A.

For the sake of completeness we now recall, without proof, some standard properties of the prime and radical ideals of a ring A.

PROPOSITION 1.2. Let A be a ring. 1) Any intersection of ideals of A, is an ideal of A; 2) The nilradical N of A is the intersection of all prime ideals contained in

A. In particular, N is an ideal;

3) Let I be an ideal of A. The radical of I, denoted r(I), is defined to be

the set {a E A I an E I for some n E IN}. The radical of I equals the intersection of all prime ideals in A which contain I. In particular, r(I) is an ideal of A.

1 4) Let P , ... , P n be prime ideals of A and let I be an ideal contained in 1 their union. Then I c Pi for some i. 5) Let I , ... , In be ideals of A and let P be a prime ideal containing their 1 intersection. Then Ii c P for some i.

PROPOSITION 1.3. Let A be a ring, A 1 0. Let !fJ denote the set of all

prime ideals of A. Then !fJ contains minimal elements with respect to inclusion.

Proof Note that !fJ 1 ¢ since 0 is contained in some maximal, hence prime, ideal of A. Let C = { Pi I i E I} be a chain in !fJ where I is some indexing set. By invoking Zorn's lemma it suffices to show that i~ri E C. Since i~ri is the intersection of ideals of A, it is an ideal. We need only show i~ri is a prime ideal. Let a,b E A and suppose ab E i~ri" Without loss of generality, if b E Pi for all iEI then we're done, so suppose there exists kEI with b i Pk. Let j E I be such that P. c Pk. Then we have ab E P. and b; P . so that a E P. since P. J J J J J is a prime ideal. In particular, a E Pk. Hence for any Pi E I we have either Pi c Pk, in which case a E Pi by the above argument, or Pk c Pi in which case a E Pk n n c Pi. Thus a E iEli so that iEipi E C. Therefore, !fJ has a minimal element. C

PROPOSITION 1.4. Let A be a ring, A 1 ¢, and let N be its nilradical.

The following are equivalent: 1) A has exactly one prime ideal;

2) Every element of A is either a unit or a nilpotent element; 3) ~ is a field.

2 Proof We show {1) ~ {2) ~ {3) ~ {1).

{1) ~ {2) : Suppose P is the sole prime ideal of A. Then P = N (prop. 1.3) and since a maximal ideal is also prime we know that P is the only maximal ideal of

A. Hence, given x E A we have either x E P = N so that x is nilpotent; or x ;. P implying x is not contained in any maximal ideal of A so that x is a unit. {2) ~ {3): Assume {2) holds. Let I :/: (0) be an ideal of ~· Then there exists x;.

N such that x+N E I. Since x ;. N, there exists y E A such that xy = 1. Thus

(x+N)(y+N) = xy+N = 1+N E I. Hence I = {1) so that the only ideals of ~ are {0) and {1) . Thus ~ is a field. (3) ~ {1): Suppose ~ is a field. Then N is maximal ideal of A. Let P be any prime ideal of A. Then N c P. But N is maximal so N = P. Therefore, N is the only prime ideal of A. C

DEFINITION 1.5. A ring having only one maximal ideal is called a local ring.

PROPOSITION 1.6. Let A be a local ring with maximal ideal M. Then 1) Every element of A-M is a unit; 2) The only idempotents of A are 0 and 1.

Proof 1) Since every non-unit of A is contained in some maximal ideal, (1) follows immediately.

2) Let a E A be an idempotent so that a(a-1) = 0 E M. Since M is a prime ideal, a E M or a-1 E M. If a E M then, since 1 ;. M, a-1 ;. M. Hence, a-1 is a unit so that a= 0. Similarly, a-1 E M ~ a-1 = 0. Therefore, a E {0,1 }. C

3 §2 The Tooology of the Prime Spectrum of a Ring

Let A be a ring and let X denote the set of all prime ideals of A. Our goal is to endow X with a topology. To this end, for each subset E c A, let

V(E) = { p E X I E ( p }. Suppose I is the ideal generated by E c A. Then since E c I c r(I), where r(I) denotes the radical of I, we clearly have V(r(I)) c V(I) c V(E). However, I is the smaUest ideal containing E so that P E V(E) ~ P E V(I). Hence, V(E) = V(I) . Also, r(I) equals the intersection of all prime ideals containing I so that V(r(I)) = V(I). Therefore, for any subset E c A we have V(E) = V(I) = V(r(I)). Consider the cases when E = (0) or E = (1). Since 0 E P, for all P EX, we certainly have V(O) = X. Now, if P E X, then 1 ;. P so that V(1) = ¢. Let (Ei)iE/ be any family of subsets of A (I some indexing set). Let P E V(i~/Ei) . Then i~.zEi c P so that Ei c P, for each i E I. Thus P E V(Ei), each i E I and hence P E i~/V(Ei). Therefore, V(i~.zEi) c i~/V(Ei). Inspection shows this argument can be reversed to give i~/V(Ei) c V(i~/Ei). Therefore, i~/V(Ei) = V(i~.zEi). Now suppose I and J are ideals of A. Since IJ c I n J c I (or c J), it is clear that V(I) U V(J) c V(I n J) c V(IJ). We wish to show that, in fact, equality holds throughout. It suffices then to show that V(IJ) c V(I) u V(J). Let P E

V(IJ). Then IJ c P . If I c P, then P E V(I) and we're done so suppose I { P. We show JcP,thatis,PE V(J). Let xEI-P. PickaEJ. Then xaEIJCP so x E P or a E P. Hence we must have a E P by the choice of x so that P E V( J) .

Thus V(IJ) c V(I) U V(J). Therefore, V(I) U V(J) = V(I n J) = V(IJ).

Summarizing, we have proved

4 PROPOSITION 2.1. Let A be a ring and let X denote the set of all prime ideals of A. For each subset E c A, let V(E) denote the set of all prime ideals of A which contain E. Then, 1) If I is the ideal generated by E, then V(E) = V(I) = V(r(I));

2) V(O) =X; V(l) = ¢; n u 3) If (Ei)iEI is any family of subsets of A, then iEIV(Ei) = V(iE~J; 4) V(I) u V{J) = V(I n J) = V(IJ) for any ideals I and J of A.

If we let 'I = { V{E) I E c A }, then by 2), 3), and 4) of the above proposition we see that 'I satisfies the axioms for the closed sets of a unique topology on X. The topology 'I on X is called the Zariski topology. The topological space (X,'I), denoted Spec( A), is called the prime spectrum of A. Let A be a ring and let X= Spec( A). We know any closed subset of X is of the form V(E) for some E c X. But now the question arises as to what the open subsets of X look like. To say that any open subset of X is of the form X- V(E) for some E c A, though true, is not very helpful. For f E A, define

Xf =X - V(f). We claim { Xf I f E A } is a basis for the Zariski topology. Let U c X be an open, non-void set in X. We need to show U = f~ I Xf, for some subset I c A. U open in X implies U =X- V(E) for some E cA. Hence,

U =X- V{f~E{f}) =X- f~E V(f) = f~E(X- V(f)) = f~E Xf Therefore, since U c X was arbitrary, { Xf I f E A} is a basis for X. We will make frequent use of this basis for Spec(A) as our discussion progresses. Aside from the obvious, it has several other interesting properties. To get a better feel for some of these properties we have the following.

5 PROPOSITION 2.2. Let A be a ring. For f E A let Xf =X- V(f) .

1) xf n xg = xfg' v f, g e A;

2) Xf = ¢=> f is nilpotent;

3) Xf = X ¢:::> f is a unit;

4) Xf = Xg ¢:::> r((f)) = r((g)); 5) X is quasi-compact; 6) Xf is quasi-compact for all f E A;

7) An open subset G of X is quasi-compact ¢:::> G is a finite union of sets

of the form xf

Proof 1) Let f,g E A. P E Xf n Xg ¢=> P is a prime ideal such that f; P and g ¢ P ¢=> fg ¢ P ¢=> P E xfg· Hence, xf n xg = xfg·

2) Xf = ¢=> V(f) =X¢=> f E P, V P EX¢=> f is nilpotent.

3) f is a unit ¢:::> f is not contained in any prime ideal of A ¢:::> V(f) = ¢:::> Xf = X.

4) Xf = Xg ¢:::> V(f) = V(g) ¢:::> for P E X, (f E P ¢:::> g E P) ¢:::> r((f)) = r((g)) by proposition 1.2. 5) Let { Gi I i E I} be an open cover of X. Since each Gi can be written as a union of sets Xf we may assume without loss of generality Gi = Xr. for all i E I. I

Now, X= .u xf = .uJX- V(f.)) =X- .niV(f.) =X-V({f. 1 iEI}) IE 1 i IE.f\ I IE I I so that V( {fi I iEI}) = ¢. Thus, no prime ideal contains ( {fi I i E 1} ), so that ( {fi I i E 1}) = (1). There exists then a finite subset J c I such that 1 = j~J a/j where aj E A, for all j E J. Now, working backwards, V( {fj I jEJ}) = so that

x = x- V( {f. 1 jEJ}) = x- .n v(f.) = .u;x- V(f.)) = .U xf . J JE 1 J JE J JE 1 j Hence, X can be covered by finitely many of the Xf. so that X is quasi-compact. I

6 6) Let f E A and let {X I i E I} be an open cover of Xf, where a. E A, for all a. I I i E I (we may assume we have a "basic" open covering of Xf). Thus we have

Xf= X- V(f) c . u =. u JX- Vl(a.)) =X- . n Via. ( ) =X- V( .uftE a. }) IE 1x a. IE .f\ 1 1E 1 1 1 1 1 so that V(i~ft a)) c V(f) = V( r(f) ). Hence, r(f) c r(i~/{ ai} ). Thus there exists n E IN such that fn E ( {a. IiE.l} ). This in turn implies there exists a finite subset 1 J c I such that fn = .EEJ z.a. where z. E A, for all j E J. Whence, since P E V(fn) J J J J ~ P E V(f), we have

(fn) c ( {ajljEJ}) ~ V( {ajljEJ}) c V(fn) = V(f) n ~ jEJV(aj) c V(f)

~ j~jX- V(aj)) J X- V(f) u ~xfcjEJxa : J Therefore, since J is finite, Xf is quasi-compact. 7) Let G c X be open. ( ~) Suppose G can be written G = f~IXf where I c A is finite. Let 'e= {Xg I gEJ} be a "basic" open cover of G. Then '8 is also an open cover for each Xf so each Xf has a finite subcover { Xgf I gf E Jf } consisting of u u elements from 'e where Jf c J is finite for each f E I. Thus fEI gflf{Xg/ is a

finite subset of 'e which covers G. Therefore, G is quasi-compact. (~)Let G c X be open and quasi-compact. G is open so G =X- V(I) = f~IXf, for some ideal I of A. Therefore, by quasi-compactness, G = f~JXf for some

finite J c I. C

7 §3 Irreducible Spaces

DEFINITION 3.1. A topological space X is said to be irreducible if X "f and if every pair of non-void open sets in X intersect.

Let X be a topological space. A subset A c X is said to be dense in X if and only if A n G "f for every non-void open subset G c X. Thus we see X is irreducible ¢::::} every non-void open subset of X is dense. Consider now the special case where X= Spec(A) for some ring A. Let N denote the nilradical of A. Suppose N is a prime ideal of A. Let U, V c X be non-void open subsets.

Pick P u E U and P v E V. Now, U =X- V(E) for some E cA. Thus P u E U:::}

E ( P and N c P :::} E ( N so that N E U. Similarly, N E V. Hence, N E U n V so that U n V f . Therefore, X is irreducible. Conversely, if N is not a prime ideal of A there exists a,b E A-N such that ab E N. Now, at N:::} V(a) "f X so Xa "f

. Similarly, Xb 1= . Both X a and Xb are open in X, however

xa n xb = xab =X- V(ab) c X- V(N) = . Therefore, X is not irreducible. This proves the following:

LEMMA 3.2. Spec(A) is irreducible ¢::::} N is a prime ideal of A.

We now prove some basic properties of irreducible spaces.

PROPOSITION 3.3. Let X be a topological space.

1) If Y is an irreducible subspace of X then the closure Y of Y is irreducible.

8 2) Every irreducible subspace of X is contained in a maximal irreducible subspace. 3) The maximal irreducible subspaces of X are closed and cover X.

Proof {1) Let Y be an irreducible subspace of X and let U and V arbitrary non-void open subsets of Y. Then U n Y, V n Y are non-void subsets of Y since Y is dense in Y. Now U open in Y means there exists open G c X such that U = G n Y. Thus Y n U = Y n (Y nG) = Y n G so that Y n U is open in Y. Similarly, Y n V is open in Y. Thus, {Y n U) n {Y n V) :f ¢>, so in particular, U n V :f ¢> . Therefore, Y is irreducible since U,V were arbitrary.

{2) Let E c X be an irreducible subspace. Let n = { SeX I S is an irreducible subspace, E c S}. Since E E 11, 1l :f ¢> . Let <& be a chain in 1l and let M = SE ~ S. If we can show M E n then the existence of a maximal element in n is assured by Zorn's Lemma. To this end, let U, V c M be arbitrary non-void open subsets of M. Then there exists S1, S2 E <& such that S1 n U :f ¢> and S2 n V :f ¢>. <8 is a chain so we may assume s1 c s2. Thus s2 n U, s2 n V are non-void open subsets of . Since is irreducible, s2 s2 ¢> :f {S2 n U) n (s2 n V) c u n v. Hence, M is an irreducible subspace of X since U, V were arbitrary. Clearly, E c M so that M E 11. {3) Let M c X be a maximal irreducible subspace. Well, M c M and by {1) M is an irreducible subspace. Thus we must have M = M. Now, pick x E X. Since the only open subsets of {x} are itself and ¢>, {x} is an irreducible subspace of X. Thus by {2) above, there exists a maximal irreducible subspace M of X such that {x} c M. C

9 COROLLARY 3.4. If X is a Hausdorff space, X :f. ¢, then the maximal irreducible subspaces are the singletons.

DEFINITION 3.5. A maximal irreducible subspace of a topological space

X is called an irreducible component of X.

PROPOSITION 3.6. Let A be a ring, X = Spec(A). The irreducible components of X are the closed sets V(P) where P is a minimal prime

ideal of A.

Proof Let Y = { S c X I Sis a closed irreducible subspace}. The proof has several parts.

1) P E X~ V(P) E Y. Let U, V c V(P) be arbitrary non-void open subsets of V(P) . Then there exists ideals E and F of A such that U = V(P) n (X- V(E)) and V = V(P) n (X-V(F)). Thus,

U n V = V(P) n [X- ( V(E) U V(F))] = V(P) n [X- V(E n F)]. Now U :f. ¢ implies V(P) { V(E) so that E { P. Similarly, F { P. But P is prime, so En F { P. Hence V(P) { V(E n F). Therefore, U n V :f. ¢ so that V(P) is a closed irreducible subspace, that is, V(P) E Y. We now wish to show the map V:X-+ Y is a bijection.

2) If S E Y, then there exists P E X with V(P) = S, that is, V is a surjection. Here we will prove the contrapositive. Let E be an ideal of A which is not prime.

We may assume E = r(E). There exist a,b E A-E such that abE E. Now, (a) {

E implies there exists Q E V(E) with a¢ Q since PE ~(E)p = E. Thus we have

10 V{E) ( V(a). Similarly, V(E) ( V{(b)). Hence, Xa n V(E) and Xb n V{E) are both open and non-void in V(E). However,

(XanV{E)) n (XbnV(E)) = V(E) n Xab = V(E) n (X-V(ab)) and since (ab) c E, V(E) n (X-V(ab)) = ¢. Therefore, V(E) is reducible. 3) V is injective.

For P,Q EX clearly P = Q ~ V(P) = V(Q). Therefore, V:X ..... Y is a bijection. The maximal irreducible subspaces of X correspond to the maximal elements of Y. Since the action of V reverses inclusions, (i.e., P c Q in X ~ V(Q) c V{P) in Y) the maximal elements of Y correspond, under V, to the minimal elements of X. Specifically, M c X is an irreducible component if and only if M = V(P) for some minimal prime ideal

P of X. C

11 §4 Homomorphisms and Ring Spectra

In our discussion so far we have concerned ourselves with the prime spectrum of but one ring at any given time. A natural question to ask is what relationships on their respective prime spectra are induced by a homomorphism between two rings. We now address this question. Let A, B be rings and let rp:A-+ B be a ring homomorphism The inverse image under a homomorphism of an ideal J of B is an ideal of A, in particular, if 1 J is a prime ideal then so is rp- (J). {Note, this requires us to have rp{1) = 1.

Henceforth, all ring homomorphisms mentioned will have this property). If J is an ideal of B we write rp- 1(J) = Jc, the contraction of J to A. Let X = Spec{ A) and Y = Spec{B). Thus if P E Y, then Pc E X. So we see that rp induces a map rp *:Y-+ X defined by rp *(P) = Pc, for all P E Y = Spec{B). Before continuing we introduce a more explicit notation: P E VA {E) means E c A and E c P E X. Similarly, P E VB{F) means F c B and F c P E Y. If E ={a} for some a E A, we write V{a) instead of VA{{a}). Consider the following: for E c A and P E Y, 1 rp{E) c P ¢:::> E c rp- {P)

¢:::> rp *{P) E VA {E) 1 ¢:::> PErp-{VA(E)).*

Hence, if f E A and P E Y, then

P E Y rp{ f) ¢:::> rp{ f) i P

¢:::> p ;.rp*-1{VA{f))

12 1 1 ~ P E cp*- (X)- cp*- (VA(f)) *-1 *-1 ~ P E cp (X- VA(f)) = cp (Xf).

Therefore, cp*-1(Xf) = Y cp(f)· In particular, if U c X is open (i.e., if there exists 1 EcA such that U = f~EXf) then cp*- (U) is open in Y. Hence, cp* is continuous. Let F c X be a closed set. Then we know that cp *-\F) is a closed set in

Y since cp * is continuous. However, we can do much better than this. F closed in

X implies F = VA (I) for some ideal I of A. (If I is an ideal of A we write Bcp(I) = Ie = the ideal in B generated by cp(I), the extension of I to B). From the discussion above we see that if P E Y, then 1 P E cp*- (VA(I)) ~ cp(I) c P

~ PE VB(cp(I))

~ P E VB(Ie) *-1 *-1 e Therefore, cp (F) = cp (VA (I)) = VB(I ).

PROPOSITION 4.1. Let cp:A -i B be a ring homomorphism. Let X=

Spec(A), Y = Spec(B). Let cp *:Y -i X be the induced map.

1) cp* is continuous (In fact, the inverse image of a basis element is a basis

element); *-1 e 2) If I is an ideal of A, then cp (VA(I)) = VB(I );

3) If J is an ideal of B, then cp ( VB(J)) = VA (Jc); 4) If cp is surjective, then cp * is a homeomorphism of Y onto the closed

subset VA (ker( cp)) of X;

5) If cp is injective, then cp *(Y) is dense in X. More generally, cp *(Y) is

dense in X ~ ker( cp) c N, where N is the nilradical of A; * * * 6) Let ,P:B -i C be another ring homomorphism. Then ( ,Pocp) = cp o,P ;

13 Proof (1) and (2) have already been proven above. Before moving on we note the following: if I is an ideal of B, then r(Ic) = r(I)c. [y E r(I)c ~ cp(y) E r(I) ~ there exists n E IN such that cp(y)n = cp(yn) E I ~ yn E cp- 1(I) = Ic ~ y E r(Ic)] .

3) To show cp ( VB(I)) = VA (Ic) it suffices to show that cp *( VB(I)) is a dense subset of VA (Ic). Let f E A and suppose Xf n VA (Ic) 4: ¢>, that is, Xf n VA (Ic) is open and non-void in VA (Ic). We will show cp *( VB (I)) n [Xf n VA (Ic)] 4: ¢>. Let

P E Xf n VA{Ic). Then f;. P and Icc P. We claim there exists Q E VB(I) such that cp(f) E Q. Suppose this were not so. Then cp(f) E n{Q I Q E VB(I)} = r(I). Hence, f E cp- 1( cp(f)) c cp- 1(r(I)) = r(I)c.

But r(I)c = r(Ic) so that

VA (Ic) = VA (r(Ic)) = VA (r(I)c) and so r(I)c c P. Thus f E r(I)c c P which contradicts f;. P. Therefore, pick Q E

VB(I) such that cp(f) '/. Q. Then cp *(Q) E cp *( VB(I)) c VA (Ic) and f '/. cp *(Q). Hence cp *(Q) E cp *(VB(I)) n [Xf n VA(Ic)] . Therefore, since Xf n VA(Ic) was arbitrary, cp *( VB(I)) intersects every non-void open set in VA (Ic).

Therefore, cp ( VB(I)) = VA (Ic).

4) Suppose cp is surjective. Then there exists a one-to-one correspondence between ideals of B and ideals of A containing ker( cp) under which prime ideals of B correspond to prime ideals of A containing ker( cp) . Thus, in lieu of this correspondence, cp *: Y ---1 VA (ker( f)) is a bijection. We know from ( 1) that cp * is continuous. It remains to show cp * is an open map. Let U c Y be an open set.

Note we may assume without loss of generality U = Yf for some fEB. Then,

cp *(Yf) = { cp *(P) I P E Y and f;. P } 1 = { Q E VA(ker(cp)) I cp- (f);. Q}

= Xf n VA (ker( cp))

14 Therefore, cp *(Yf) is open in VA (ker( cp)) so that cp * is an open map.

Thus, cp *:Y -1 VA (ker( cp)) is a homeomorphism.

5) Recall from (3) if I is an ideal of B, then cp ( VB(I)) = VA (Ic). Since cp *(Y) = cp *( VB(O)) we have

cp *(Y) =X ¢::> VA (Oc) = X, where 0 denotes the 0-ideal

¢::> VA (ker( cp)) = X

¢::> ker( cp) c P for all P E X

¢::> ker( cp) c N.

* ~.-- Therefore, cp (Y) is dense in X ¢::> cp (Y) = X ¢::> ker( cp) c N. In particular, if cp is injective, then ker( cp) = {0} c N so that cp *(Y) is dense in X.

6) Let P E Spec(C). Then, (1Pocp)*(P) = ('I/Jocp)-1(P) = (cp-lo'I/J-l)(P) = (cp\'1/J*)(P). * * * Therefore, (1Pocp) = cp o'I/J since P E Spec(C) was arbitrary. C

COROLLARY 4.2. Let A be a ring, N the nil-radical of A. Then Spec( A) and Spec(~) are naturally homeomorphic.

Proof The natural quotient map cp:A -1 ~ is surjective so that, by ( 4), cp* is a homeomorphism of Spec(~) onto V(ker( cp)). However, V(ker( cp)) = V(N) = Spec( A). C

We continue our discussion with some examples. Let A be an integral domain having but one non-zero prime ideal P. Let A K be the field of quotients of A and let B = P" x K. Define cp:A -1 B by

15 c,o(a) = (a,a), where a= a+P, the image of a in ~ · Now,~ is a field so that A Spec(B) = { ((1,0)) , ((0,1))} = { F x (0) , (0) x K }. Thus, we have cp *((1,0)) = cp-1((1,0)) = (0) and cp *((0,1)) = cp-1 ((0,1)) = P. Hence cp * is a bijection. Note VB((1,0)) = { ((1,0)) } is closed in Spec(B), however, (cp*-1)-1[(1,0)] = cp-1[(1,0)] = (0) which is not closed in Spec(A) since *-1 * VA (0) = Spec( A). Thus cp is not continuous. Therefore, cp is not a homeomorphism.

Let A = IIi~ 1Ai the direct product of rings Ai . We claim that Spec( A) is the disjoint union of open and closed subspaces Xi, where X. is canonically 1 homeomorphic with Spec(Ai). For i = 1,2, ... ,n define

x. = { rr .n1s. I s. =A. if j#, s. E Spec(A.) }, 1 J= J J J 1 1 and let cpi :Spec(Ai) -1 Xi be the canonical injection. Note that if Q E Xi for some i, then Q is a prime ideal of A. The point however is that every prime ideal of A is of this form. Suppose I is a proper ideal of A, I ¢ X. for any i. We show I is 1 not a prime ideal of A. There are two cases.

1) I= E1 x Ili~ 2 Ai where Ei is a non-prime ideal of A1 (without loss of generality we assume the "proper factor ideal" occurs in the first factor) . Clearly then I is not a prime ideal of A.

2) I= E1 x E2 x Ili~ 3 Ai where E1,E2 are any proper ideals of A1,A2 respec­ tively. Let a1 E E1, a2 E E2, x1 = (a1,1,1, ... ,1), x2 = (1 ,a2,1, .. . ,1) . Then x1~ E I, but x1 ,x2 E A-1. Thus I is not a prime ideal of A.

16 Therefore, Q E Spec(A) {:::::} Q E Xi for some i = 1,2, ... ,n. Hence, if we endow Xi 1 with the subspace topology of Spec( A) we see that both cpi and cpj are continuous.

Now consider x1 as a subspace of Spec(A). Let e1 = {1,0, ... ,0) E A. Then, V( e1) = { P E Spec( A) I cpi"\P) = A1 } = { P E Spec( A) I the first factor ideal = A1 }. Now, Spec(A)- V{e1) = X1 so that X1 is open. However we also have X1 = V({0,1, ... 1)) so that X1 is closed. A similar argument shows that Xi is both open and closed in Spec(A) for all i = 1,2, ... ,n. Therefore, Spec(A) is the disjoint union of the open and closed subspaces Xi of Spec( A), each Xi being canonically homeomorphic to Spec( Ai).

Suppose now we were given an arbitrary ring A and asked whether or not Spec{A) is disconnected. Generally this question may be answered with a straight­ forward set-theoretic argument. However, as the following proposition shows, this can often be determined without resorting to the topological space Spec{ A) .

PROPOSITION 4.3. Let A be any ring. The following are equivalent: 1) X = Spec( A) is disconnected:

2) A~ A1 x A2 where neither of the rings is the 0-ring; 3) A contains an idempotent a 1 0,1.

Proof {1) ~ {2): Suppose X is disconnected. Then there are proper radical ideals

I,J of A such that X = V{I) U V{ J) and V{I) n V( J) = ¢;. We first consider the case where the nilradical N ={0).

17 Hence, X= V(I) U V(J) = V(InJ) :} In J = (0) and ¢ = V(I) n V(J) = V(IUJ) :} (IUJ) = (1) =A. A A A h · h f h . A A . th 0 . Thus TifJ ~ T x J w ere ne1t er o t e nngs T' J 1s e -nng. Now suppose N f (0). Consider the ring B = ~- Clearly then if Spec(A) is disconnected then so is Spec(B). In fact, VB(~) and VB(~) serve as the disconnection of Spec(B). Applying the above argument to B we can find non­ zero rings B1 and B2 with B ~ B1 x B2. Let e E B be the element which corresponds to the idempotent (1,0) E B1 X B2 . Then e f 0, 1 is an idempotent of B. But now we can "lift" e to an idempotent e E A with e :1 0, 1 [3, §3 .6 prop. 1] Let a EA. We can write a= a1 = a(e + (1--e)) = ae + a(l-e).

Since (e) n (1--e) = ¢ we see that A~ (e) x (1--e). However, (e) and (1--e) are not only ideals of A but are non-zero rings in their own right having e and 1--e as their respective multiplicative identities. (2) :} (3): Suppose ¢:A-+ A x A is a ring isomorphism where neither A nor 1 2 1 ~ is the 0-ring. Then (1,0) E A1 x A2 is an idempotent and (1,0) :1 0, 1 E 1 A1 x A2. Then ¢- ((1,0)) E A is an idempotent which is not the 0 or 1 of A. (3):} (1): Let a E A-{0,1} be an idempotent. Then a2 =a so that a(a-1) = 0 is in every prime ideal of A. That is, P E Spec( A) :} a E P or a-1 E P. Hence, X=

V(a) U V(a-1). Since a+ (a-1) = 1, if P E Spec(A), then a E P ~ a-1 i P.

Thus V(a) n V(a-1) = ¢. Therefore, X is disconnected. C

18 We wish to end this section with a discussion about a very important example. Let A be a ring. Consider the subset of Spec(A) consisting of all

maximal ideals of A. This set, endowed with the induced (subspace) topology is called the maximal spectrum of A and is denoted Max( A). We now discuss some of the properties of Max( A) with regard to a very specific, though important ring A.

Let X be a compact Hausdorff space and let C(X) denote the ring of all real valued continuous functions on X (here the addition and multiplication of ring elements are the pointwise addition and multiplication of functions). Let x E X and denote by Mx the set of all f E C(X) such that f(x) = 0. If we let 1,0:C(X)-+ IR be the (surjective) homomorphism defined by I,O(f) = f(x) for all f E C(X), we see that Mx = ker( 1,0) so that Mx is an ideal of C(X). Furthermore ~ ~ IR, so that X A Mx is a maximal ideal. If X denotes Max(C(X)), we have then defined a map

A J.£ : X -+ X by J.£( x) = M , for all x E X. X We will show that J.£ is a homeomorphism of X onto X.

1) J.1. is surjective. Let M be any maximal ideal of C(X). Let <& = {xEX I

f(x)=O, VfEM}, that is, <& is the set of common zeros for elements of M. If

xE ~ then M c M so that M = M since M is maximal. Thus it suffices to X X show <& t ¢>. Suppose <& = ¢> . Pick x E X, then there exists f E M such X that f (x) t 0. Hence, there exists an open set G c IR with 0 G and X X t X 1 f (x) E G . Since f is continuous, H = (G ) is open in X with X E X X X X r X X Hx. Thus { Hx I xEX} is an open cover of X. X is compact so there are n x , x , ... , xn E X with X= U H . Define 1 2 . 1 x. 1= 1 f = f 2 + f 2 + ... + f 2 x1 x2 xn Then f E M and f(x) t 0 for all x E X. The function f is then continuous

19 and defined for all x E X so that f is a unit in C(X). Thus M = (1),

contradicting the choice of M. Therefore, ~~ ¢. Hence, J.L is surjective.

2) J.L is injective. To prove this we first need the following lemma

URYSOHN'S LEMMA Let 'I' be a locally compact Hausdorff space.

Let V and K be subsets of 'I', K c V, K compact, V open. Then

there exists a continuous real valued function f on 'I' such that

f(k) = 1, Vk E K and f(v) = 0, Vv i V. [5, (prop.) 2.12]

Let x, y E X, x f. y. Then { x} and {y} are disjoint and closed in X so by

Urysohn's lemma there exits fEC(X) with f(x) = 1 and f(y) = 0. Thus f E

M but f i M . In particular M f. M . Therefore J.L is injective. y X X y

3) J.L is bicontinuous. Fix an element f E C(X). Define

uf = {X E X I f( X) f. 0 } and

uf = {Me x 1 f t M } .

A For x E X, x E U f <=> f( x) f. 0 <=> f i Mx <=> Mx = J.L( x) E U f Hence, 1 J.L(Uf) = Uf Since Uf = 1 (1R-{O}) we see that Uf is open in X. Also,

A A Uf = X n (Spec[C(X)]- V(f))

where Spec[C(X)]-V(f) is open in Spec(C(X)), so that Uf is open in X.

We now claim that the set { Uf I f E C(X) } is a basis for X. Let G c X be a non-void open subset and let x E G. We wish to show there exists an

f E C(X) such that x E Uf c G. Now {x} and X-G are both closed in X

so that there exists an f E C(X) with f(x) = 1 and f(X-G) = 0 by

Urysohn's lemma. Hence, X E uf ( G. Thus { uf I f E C(X) } is a basis

20 A for X. Consider the set { Uf I f E C(X) }. Let B be an arbitrary

A non-void open set in X. Then there is an ideal E of C(X) such that

A A B =X n (Spec[C(X)]- V(E)) = X- V(E).

Let Mx E B (for some x E X), and let f E E be such that f(x) "f 0. Then

A A M E Uf. Now for y E X we have M E Uf => f(y) -j 0 => f ¢ M => My E A X A y A y X- V(E) = B, so Uf c B. Hence, Mx E Uf c B. We have shown that the set

A { uf I f E C(X) } is a basis for X. In particular then, the map J.L : X -+ X

A maps the basis elements of X onto the basis elements of X.

Therefore, J.L is bicontinuous.

Therefore, by 1), 2), and 3), J.L is a homeomorphism.

21 §5 Localization and Ring Spectra

Let A be a ring and let S c A be a multiplicatively closed subset of A such that 1 E A. Define an equivalence relation = on Ax S by

(a,s) =(b,t) ¢=>(at- bs)u = 0 for some u E S. Let afs denote the equivalence class of (a,s), and let As denote the set of all equivalence classes. If we put a ring structure on As in the natural way, that is

afs + b/t =(at+ bs)/st, (a/s)(b/t) = abfst, then it can be shown (with a long but routine argument) that As is in fact a ring. We formalize this with the following definition.

DEFINITION 5.1. Let A be a ring and let S c A be a multiplicatively

closed subset of A such that 1 E S. Then As is called the ring of fractions of A with respect to S.

As an immediate consequence of our definition we have As = 0 (the 0-ring)

¢=> 0 E S. To avoid this highly uninteresting scenario we henceforth assume 0 ;. S.

We also have the canonical ring homomorphism f:A-+ As defined by f(x) = x/1. Though in general f is not injective (i.e., an imbedding) we do have the following.

LEMMA 5.2. f:A-+ As is injective if and only if S contains no 0-divisors.

22 Proof(~) Let yES. Suppose there exists a E A with ya = 0. Then f(a) =I= ¥ so that a E ker(f) = {0}. Hence, y is not a 0-divisor.

( ~) Let x E ker(f). Then f(x) = x/1 = 0/1 so there exists y E S with xy = 0.

Since we are assuming S contains no 0--divisors, x = 0. Hence, ker(f) = {0}. 0

Let r.p:A -1 B be a ring homomorphism, r.p *:Spec( A) -1 Spec(B) the associ- ated mapping. We claim every prime ideal of A is a contracted ideal ~ r.p * is surjective. Suppose every prime ideal of A is a contracted ideal and let P E Spec(A). Then P = Pee by [1,prop. 1.17(iii)]. Hence, P is the contraction of a prime ideal Q of B [1, prop. 3.16]. Thus P = Q c = r.p *(Q) so that r.p * is surjective. Conversely, if r.p * is surjective the result is immediate. Now suppose that every prime ideal of B is an extended ideal. Let P 1, P 2 E Spec( B). Then * there exists ideals 11, 12 of A such that Pi= lie ,i = 1,2. Suppose r.p (P1) = * . I ec I ec h b [ 7] r.p (P 2). We claim P 1 = P 2. Smce 1 = 2 , we ave, y 1, prop. 1.1 e ece ece e p 1 = 11 = 11 = 12 = 12 = p 2" Therefore, r.p * is injective. However, if r.p * is injective it does not necessarily follow that every prime ideal of B is an extended ideal. Consider the canonical injection r.p:U. -1 U.[i], where U.[i] denotes the ring of Gaussian integers. In U., (2) is a prime ideal. However, (2) = (1+i) 2 in U.[i] where 1+i is an irreducible element of U.[i]. Hence, r.p *((1+i)) = (2) but (1+i) is not an extended ideal. Consider now the rings "0.( 2) and U.[i]( 1+i)' each having only one non-zero prime ideal. Since U.

and U.[i] are integral domains and (2) c (1+i) in U.[i], we can think of u.( 2) as a * subring of U.[i]( 1+i)" Now if we let 7/J:U.( 2) -1 U.[i](l+i) be our imbedding, then 7/J is injective but (1+i) is not the extension of any ideal of "0.( 2).

23 PROPOSITION 5.3. Let A be a ring, S c A a multiplicatively closed

subset, and let t.p:A -+ As be the cannonical homomorphism. Then

t.p *:Spec(As) -+ Spec(A) is a homeomorphism of Spec(As) onto its image Xs in X= Spec( A).

Proof. That 1.fJ* is onto Xs is clear. Since every (prime) ideal of As is an extended ideal [1, prop. 3.1l(i)], t.p * is 1-1 by the preceding argument. Such a map t.p * was shown always to be continuous. Thus it remains to show t.p * is an open map. Let G c Spec( As) be a "basic" open set. Then G = Spec( As)- V((f)) for some f E As. Since every ideal of As is an extended ideal [1, prop. 3.11(i)], there exists an ideal F of A such that Fe = (f) . To show l.fJ * is an open map, it suffices to prove t.p *(G) = Xs- V(F). Let Q E G. Then Fe = (f) { Q which implies F { Qc = 1.fJ *(Q) [else, F c Qc ~ Fe = (f) c Qce c Q, a contradiction].

Thus 1.fJ *(Q) c Xs- V(F). Now let H E Xs- V(F). Then H = 1.fJ *(P) for some

P E Spec(As)· If P E V((f)), then F c Fee = (f)c c Pc = t.p *(P) = H which contradicts the choice of H. Hence P E Spec(A ) - V((f)) = G and so H E 5 1.fJ* (G). Therefore, Xs- V(F) = 1.fJ* (G) so that t.p * is an open map. C

We now consider a particular case of the above proposition. Let A be a ring, f E A. Let S = { fn I n=0,1,2, ... }. In this case, As is denoted by Af Now, the prime ideals of Af are in one-to-one correspondence with the prime ideals of A having void intersection with S. Hence, for Xf = Spec(A)- V(f) we have

P E Xf ¢:::> P E Spec(A) and P n S = ¢ ¢:::> P = t.p *(Q) where Q E Spec(Af)

Therefore, t.p *(Spec(Af)) = Xf

24 §6 Integral Dependence

DEFINITION 6.1. Let A and B be rings with A c B. An element x E B is said to be integral over A if x satisfies a monic polynomial with

coefficients in A. If every element of B is integral over A then B is said

to be integral over A.

We now wish to recall the following two important theorems. The proof for each may be found in the reference stated.

THEOREM 6.2. ("Lying Over")

Let A c B be rings with B integral over A, and let P E Spec(A). Then

there exists Q E Spec(B) such that P = Q n A. [1, thm. 5.10)

THEOREM 6.3. ("Going-up theorem")

Let A c B be rings with B integral over A. Let P 1 c P 2 c · · · c P n be a chain of prime ideals of A and Q 1 c Q2 c · · · c Qk (k < n) a chain of rime ideals of B such that P. = Q. A for 1 i k. Then the chain P 1 1 n <- -<

Q1 c Q2 c · · · c Qk can be extended to a chain Q1 c Q2 c · · · c Qn such that Pi= Qi n A for 1 ~ i ~ n. [1, thm. 5.11)

Let cp:A -+ B be a ring homomorphism such that B is integral over its subring R = cp(A) (such a ring homomorphism is said to be integral). What can then be said about the induced map cp *:Spec(B) -+ Spec(A)? If we let 1/J:A -+ R be the canonical surjection then we see that 1/J *:Spec(R) -+ Spec(A) is a

25 homeomorphism of Spec{R) onto the closed subset VA {ker( cp)) of Spec{ A) (prop.

4.1]. Hence, '1/J * is a closed map. Consider the induced map f *:Spec{B) -+ Spec{R)

where f:R-+ B is the canonical injection. Let VB{I) be a closed set in Spec{B)

for some ideal I of B. Let J = I n R. Clearly f * ( VB(I)) c VR(J). Now, TB is

integral over ~ (1, prop. 5.6]. Let P E VR(J). The image J> of P in ~ is a prime ideal so by lying over there exists a prime ideal q in ¥ such that J> = q n ~- Let Q be the unique prime ideal of B having q as its image in ¥- Then f *(Q) = P so that f * ( VB(I)) = VR(J). Thus f * is a closed map. However, cp * =

7/J *of * and since the composition of closed maps is closed we see that cp * is a closed

map.

LEMMA 6.4. Let A and B be rings, A c B with B integral over A.

Let f:A -+ F be a homomorphism of A into an algebraically closed field

f. Then f can be extended to a homomorphism f:B-+ f.

Proof Let P = ker(f). f has no zer~ivisors so that P is a prime ideal of A. By lying over there exists a prime ideal Q of B such that P = Q n A. q is integral over ~ [1, prop. 5.6] and both are integral domains. Also, f induces an injective homomorphism -+ f. Let F , F be the quotient fields of and g:~ 1 2 ~ q respectively. Let E F where a,b E and b j 0. Each of a and b satisfies a E 2 q monic polynomial with coefficients in hence with coefficients in F . Thus, both ~, 1 a and b are algebraic over F so that is algebraic over F . In particular, 1 E 1 since E F was arbitrary, F is algebraic over F . Since f is a field, we can E 2 2 1 factor g into g = 1 o'Y where ~---+ F ---+f. F is algebraic over F so that 2 1 'Y1 1 'Y2 2 1 1 can be extended to a (injective) homomorphism :F -+ f since f is 2 ;y2 2

26 algebraically closed [4, thm. 7.2.8]. If we let q:B -1 q and rp:q -1 F 2 be the canonical quotient map and imbedding respectively, then f = .:Y2 o rp o q is the desired extension of f. C

Let A c B be rings with B integral over A. We know if P is a prime ideal of A then Bp is integral over Ap [1, prop. 5.6]. Now let Q be a prime ideal of B such that P = Q n A. A natural question to ask is whether or not BQ is integral over Ap. We will show this is generally not the case. Let A= k[x2-1] 2 2 c B = k[x], where k is a field. Note x - (x -1) - 1 = 0 so that x E B is integral over A. Hence A[x] = k[x2-l,x] = B so that B is integral over A [1, prop. 5.1]. Let N be the maximal ideal in B generated by x-1 and let M = N n 2 A= {q E Alx -1 dividies q}. Form the local rings BN, AM. The denominators in BN are polynomials p(x) E k[x]-{0} such that x-1 does not divide p(x); the 2 2 denominators in AM are polynomials q(x) E k[x -1)-{0} such that x -1 doesn't divide q(x). Suppose x! 1 E BN is integral over AM. Then there exists ~~, , ... , fn elements of AM such that i1q 1 qn P(t) = tn+l + fn

0 = qoql···qn + fnql·· ·qn -l(x+l) + .. . + f lqoq2···qn (x+ 1 )n+l + foql·· .qn (x+ 1)n+l . Thus qoql···qn = (x+l) ·("other stuff'') so that, in particular, x+l divides 2 2 qoq t· · ·qn = am(x -l)m + · · · +at(x -1) + ao, where ai E k for all i and ao f. 0 since qoq 1• • • qn E A - M. Hence, x+ 1 divides a0 establishing a contradiction.

Thus x!1 is not integral over AM. Therefore, BN is not integral over AM.

27 We continue our discussion with the following.

LEMMA 6.5. Let A c B be rings with B integral over A.

1) If x E A is a unit in B, then x is a unit in A;

2) The Jacobson radical of A is the contraction of the Jacobson radical of B.

Proof (1) Using the lying over property, given a prime ideal P of A, we may find a prime ideal Q of B with P = Q n A. If x E P, then x E Q which can't be since x is a unit in B. Hence, x is not contained in any prime ideal of A.

Therefore, x is a unit in A. (2) First note, if P and Q are prime ideals of A and B respectively with P =

Q n A, then P is maximal ~ Q is maximal [1, cor. 5.8]. Let M and N be the Jacobson radicals of A and B respectively. In lieu of [1, cor. 5.8] and lying over we then have:

x EM ~ xis contained in every maximal ideal of A

~ x E A and x is contained in every maximal ideal of B

~ xEAnN

Therefore, the Jacobson radical of A is the contraction of the Jacobson radical of B C

DEFINITION 6.6. A ring homomorphism f:A -+ B is said to have the

goinrrup property if the conclusion of the going-up theorem holds between

B and its subring f(A).

28 Let cp:A -.. B be a ring homomorphism, cp *:Spec(B) -.. Spec(A) the associated mapping. Consider the following:

1) cp * is a closed map;

2) cp has the going-up property; c B A 3) for Q E Spec(B), P Q E Spec(A) the map cp*:Spec(q)-.. Spec(p-) is = 0 surjective, where cp*(q) u = cp *(U)/ cp *(Q) = cp *(U)/P for all 0 U B q E Spec(q)· We claim that (1) ::} (2) {::::} (3)

First we remark that the map cp * is certainly well defined. 0 (2) ::} (3) Suppose cp has the going-up property. Let P and Q be prime ideals of A and B respectively with P = Qc. Then ker(cp) c P. Let U E Spec(~) and let U E Spec(A) be the unique prime ideal of A having U as its image in ~ (ie., U = ¥). Making use of the isomorphism between ke ~( cp) and cp(A), we may consider cp(U) as a prime ideal in cp(A) since ker( cp) c U [note then Q n cp(A) c cp(U)]. Since cp has the going-up property we can find V E Spec(B) so that Q c V and V n cp(A) = cp(U). But the image ~ of V in ~ remains a prime ideal. 1 rn*(V) _ w*(y) _ cp- (Vncp(A)) _ U _ UA Hence ""oQ" -~- P -p-- .

Therefore, cp * is surjective. 0 (3) ::} (2) Follows immediately by a reversal of the above argument.

(1) ::} (3) Suppose cp * is a closed map. Let Q E Spec(B) and let P = Qc.

Recall that

(0) * - A where the last equality holds since cp is a closed map. Pick U E Spec( F) and let - u U E VA (P) be such'that U = p-· Now, by (0) there exists V E VB(Q) with

29 c/(v) = U. Hence, qis a prime ideal of Spec(~) with

PROPOSITION 6.7. Let G be a finite group of automorphisms of a ring

A and let A G = {a E AI u(a) =a, Va E A}. Then A is integral over A G. In addition, if S is a multiplicatively closed subset of A such that u(S) c S

for all u E G, and we let sG = S n A G, then the action of G on A

extends to an action on As and (A G)sG ~(As) G.

Proof First we show A G is a subring of A. Certainly 0,1 E A G. Let x,y E A G. u(x+y) = u(x) + u(y) = x + y, for all u E G so that x+y E A G. Similar reasoning shows xy E A G and -x E AG. Since the associative and distributive properties are inherited from A, we see that A G is in fact a subring of A.

Let G = { u1, u2, .. . ,un} where the ui are distinct and u1 = identity map on A. Fix a E A, define the polynomial

P a(t) = (t-u1(a))·(t-u2(a))· · ·(t-un(a)). Since u1(a) =a, we see that P a(a) = 0 for all a EA. If we multiply out P a(t) and collect terms we get n n-1 ( ) P a t = t + sn_1t + ... + s1t + s0, where the si are the elementary symmetric functions of u (a), .. . ,un(a) up to sign. 1 Hence, the coefficients si are invariant under G so that si E A G for i=0,1, ... ,n.

Therefore, A is integral over A G since a E A was arbitrary.

Let S and sG be as stated above. For each u E G define &-:As -+ As by

A (a) uf E a a b a b (] s = (] a~s lOr all s E As. Suppose s' T E As and s = r· Then there exists u E

30 S such that ( at-bs )u = 0. Pick u E G. Then

u[(at-bs)u] = [u(a)u(t)-u(b)u(s)]u(u) = u(O) = 0.

Since u(u) E S, we have ~~:~ = ~~ ~ so that u(i) = ;.(¥). Thus, u is well defined for all u E G. Hence, u H u extends the action of G on A to an action on As.

Let (As)G = { i E As I u(i) = i, Vu E G }. Note that sG is a multiplicatively closed subset of A. We first show A(SG) ~ As. Define tp:A ~As by tp(a) = r· 1) Let t E SG. Then tp(t) = f is a unit in As; 2) Let a E A and suppose tp(a) = 0. Then there exists t E S such that at = 0.

Let t, t1'" .. tk denote the distinct elements in {u1(t), ... ,un(t)}. Then

attl·. ·tk = 0 and ttl·. ·tk E sG. Hence, if X E A and tp(x) = 0, then there

exists s E SG with xs = 0.

3) Let a E A, t E S. Then

tp(a)tp(t)-l = j·} = (at1· · ·tk/1)·(1/tt1· · ·tk) = tp(at 1· · ·tk)tp(tt1· · ·tk)-l where tt1· · ·tk E SG. Thus every element of As is of the form tp(a)tp(s)-l where s E SG.

Therefore, by (1), (2), (3), there exists a unique isomorphism 1/J:A(SG) ~ As [1, corl. 3.2]. We certainly have then [A(SG)]G ~ (As)G. It suffices to show [A(SG)]G ~ (A G)(SG). We will show that in fact [A(SG)]G = (A G)(SG). Clearly [A(SG)]G J (A G)(SG). Let i E [A(SG)]G. Then ~(i) = ~f:J = i for all u E G, where, since s E SG, u(s) = s. Now, u(:) = i :::} there exists u E SG such that asu = tp(a)su = u(a)u(s)u(u) = u(asu) since s,u E SG. Hence, asu E A G. But then asu E (A G)(SG). This shows ~s = ssu G G G G [A(SG)] c [(A )(SG)] so that [A(SG)] = [(A )(SG)). G G Therefore [(A )(SG)] ~ (As) . C

31 §7 Noetherian Spaces

PROPOSITION 7.1. Let }; be a set partially ordered with respect to ~- Then the following are equivalent:

1) Any ascending sequence x1 ~ x2 ~ x3 ~ · · · in }; finitely terminates (ie., there exists n E IN such that xn = xn+ 1 = · · · ); 2) Every non-void subset of }.; contains a maximal element. Remarks: (1) is called the ascending chain condition (a.c.c.); (2) the maximal condition. If the partial order is with respect to "~", (1) is then called the descending chain condition; (2) the minimal condition.

DEFINITION 7.2. A topological space 'I' is said to be Noetherian if the

open subsets of satisfy the ascending chain condition. By

complementation, this is equivalent to the closed sets of 'I' satisfying the descending chain condition.

Armed solely with the above definition we now attack the following.

PROPOSITION 7.3. Let 'I' be a Noetherian space. Then 1) Every subspace of 'I' is Noetherian; 2) 'I' is quasi-compact.

Proof (1) Let A be a (non-void) subspace of 'I'. Let U U be an 1 c u2 c 3 c · · · ascending chain of open sets in A. For each i E IN, there exists G. open in 'I' with 1

Ui = Gi n 'I'. We may assume without loss of generality G1 c G2 c G3 c · · · for all

32 i E IN (if not, define Gj+ = Gj U Gi+ for all i E IN). Since 'I' is Noetherian, there 1 1 exists n E IN such that G1 c G2 c · · · c Gn = Gn+1 = · · ·. Hence,

u1 = G1 n'I'c ··· c Un = Gn n'I'= Un+1 = Gn+1 n'I'= ···. Thus, A is Noetherian since {Ui}iEIN was arbitrary. (2) Let ~1 = { Gi I i E I} be an open cover for 'I. We will construct a finite subcover. Let M be a maximal element of ~ . Define ~ = { G E ~ G ~ M } 1 1 2 1 I 1 and pick M maximal in ~ . Continuing inductively, define 2 2 n -1 ~n={GE ~n-11 G~ i~tM) and pick Mn maximal in ~n.

n M eM UM c ···cUM c ··· 1 1 2 i=l i is an ascending chain of open sets in 'I, hence, there exists k E IN such that k k+r u M = u M. i=l i i = 1 1 k k for any rEIN. Then 'I= i~lMi, for if not, ~+ 1 :f

Suppose 'I is a topological space such that every open subspace is quasi­ compact. Let G c G c G c · · · be an ascending chain of non-void open sets in 1 2 3 'I. Then i~INGi is an open subspace of 'I' with open cover {Gi}iEIN" Hence, there

U n exists n E IN such that . EING. = U G. = G . In particular, 1 1 i=l 1 n G1 c G2 c G3 c · · · c Gn = Gn+1 = · · ·. Thus we see that 'I' is a Noetherian space. This fact, combined with the above proposition, yields the following lemma.

33 LEMMA 7.4. The following are equivalent: 1) '.r is a Noetherian space; 2) Every open subspace of '.r is quasi---compact; 3) Every subspace of '.r is quasi---compact.

Let '.r be a Noetherian space and let }.; be the set of all closed subsets of '.r which are not finite unions of closed irreducible subspaces. If ~ :/= ¢ then, since '.r is Noetherian, ~ contains a minimal element U. Note that U is reducible, hence there exists closed proper subsets V 1, V2 of U such that U = V 1 u V 2. By the rninimality of U, neither V 1 E }.; nor V 2 E }.;. Thus, both V 1 and V 2 can be written as a finite union of closed irreducible subspaces, hence, so can U which contradicts U E ~- Therefore, }.; = ¢. In particular, '.r can be written as a finite union of closed irreducible subspaces. Consider now the set .At of irreducible components of '.r. By prop. 3.3 the elements of .At are closed and cover '.r.

n Thus, there exists M , M , ... ,M E .At such that '.r = .u M. . We wish to show 1 2 n 1 = 1 1 that in fact .At= {M1, M2, ... ,Mn} . If '.r E .At then .At= {'.r} and we're done. Suppose '.r t .At. Let Y c '.r be a proper subspace such that Y ~ M. for i=1,2, ... ,n. 1

Let {M1, M2, .. . ,Mk} c {M1, M2, ... ,Mn} be a minimal covering for Y, that is, no subset of {M1, M2, ... ,Mn} having fewer than k elements will cover Y. By the n rninimality of our cover, Y n M and Y n i~ Mi are closed proper subsets of Y 1 2 whose union is Y. Thus Y is reducible. We have shown Y c '.r, Y irreducible ::} Y c Mi for some i=1,2, ... ,n.

Therefore, .At= {M1, M2, ... ,Mn}. We formalize the above with the following proposition.

34 PROPOSITION 7.5. 1) A Noetherian space is a finite union of closed irreducible subspaces;

2) The set of irreducible components of a Noetherian space is finite.

Let us now return to our discussion of the prime spectrum of a ring.

LEMMA 7.6. If A is a Noetherian ring, then X Spec(A) is a

Noetherian topological space.

Proof Let V(I ) J V(I ) J V(I ) J · · · be an arbitrary descending chain of closed 1 2 3 subsets of X. We may assume I. = r(I.) for all i E IN . Then we have 1 1 I1 ( 12 ( 13 ( ....

A is Noetherian so there exists n E IN such that 11 c 12 c · · · C In = In+1 =

Hence, V(I1) J V(I2) J V(I 3) J · · · J V(In) = V(ln + 1) = · · · . Therefore, X is Noetherian. C

Note however that the converse to the lemma is false. Let k be a field.

Consider the following chain of integral extensions

i where t is an indeterminate. If for each i E IN we let Mi + = ( the sole 1 Vf),

maximal ideal of Ai, we see that M1 c M2 c · · · c Mn c · · ·. Define the ring A= i~INAi. Since each stage is an integral extension we see that A is integral over Ai for all i E IN. Let M = i~INMi . Certainly M is an ideal of A and M f A since 1 ¢ M. However, we claim that in fact M is a prime ideal. Let a,b E A and

suppose abE M. Then a,b E Ai, some i, and so ab E Ai n M c Mi. Thus, we may

35 assume without loss of generality a E Mi c M. Therefore, M is a prime ideal. Now, let P be a prime ideal of A. If P n A1 = (0) 1 then, since (0) c P, (0)1 n A = (0), and (0) is a prime ideal of A, we have P = (0) [1, corl. 5.9]. Now suppose

P :f (0) is a prime ideal of A. By the above then P n Ai :f (0) for all i E IN. Hence P n Ai = Mi since the only prime ideals in Ai are (O)i and Mi. Thus

Mi c P for all i implies M = i~INMi c P. However, again by corl 5.9 M n A1 = M1 = P n A1 implies P = M. Therefore, the only prime ideals of A are (0) and M. Clearly then Spec(A) = {(O),M} is Noetherian. However, A is not Noetherian since (t) c (v'i) c (4v'i) c · · · is an ascending chain of ideals in A which does not finitely terminate.

COROLLARY 7. 7. The set of minimal prime ideals in a Noetherian ring A is finite.

Proof Let X = Spec(A). By lemma 7.6 X is a Noetherian space. By prop. 7.5 the set of irreducible components of X is finite. Prop. 3.6 tells us there is a one-to-one correspondence between the irreducible components of X and the minimal prime ideals of A. Therefore, the number of minimal prime ideals of A is finite. C

PROPOSITION 7.8. Let cp:A -+ B be a ring homomorphism and suppose

Spec(B) is a Noetherian space. Then cp *:Spec(B) -+ Spec(A) is a closed

mapping ~ cp has the going-up property.

36 Proof (:::}) In section 6 it was shown that this holds generally.

( ~) Recall by prop. 4.1, if f:R -+ S is a surjective ring homomorphism, then f *:Spec(S) -+ Spec(R) is a homeomorphism of Spec(S) onto the closed subset VR(ker(f)) of Spec(R). Thus for purposes of this proof, we may assume A c B and our map cp is simply the canonical injection (so for P E Spec(B), Pc = PnA).

Suppose cp has the going-up property. We first show cp *( VB(P)) is a closed set in Spec(A) for all P E Spec(B). Pick P E Spec(B). By prop. 4.1(3) it suffices to show VA(Pc) c cp *(VB(P)). Let Q' E VA(Pc). Since cp has the going-up property, there exists Q E Spec(B) with P c Q and Qc = Q'. In particular,

Q' = Qc E cp *( VB(P)).

Hence, cp *( VB(P)) = VA (P c) for all P E Spec(B). Now let VB(I) be any closed set in Spec(B) and we may assume I = r(I). Because Spec(B) is Noetherian, the one-t()--{)ne correspondence between ideals of ¥ and ideals of B which contain I shows us that Spec(¥) is also Noetherian. Let q, denote the set of all minimal prime ideals of ¥, and let w denote the set of primes in B corresponding to q, . Since Spec(¥) is Noetherian, q, is a finite set by cor. 7. 7. Thus w is a finite set.

Also, since I = r(I), we have I = n{P E Spec(B) I P E w}.

Hence, cp *( VB(I)) = cp *( VB(n{P 1 P E w})

= cp *( u { vB(P) 1 P E w} )

= u { cp *( VB(P)) I p E w } = u { vA (PC) I p E w}. The last expression is a finite union of closed sets in Spec(A) hence is closed.

Thus, cp *( VB(I)) is closed in Spec( A). Therefore, since I was arbitrary, cp * is a closed map. C

37 We have already seen if A is a Noetherian ring, then Spec(A) is a Noetherian space but that the converse does not hold. What then can be said, if anything, of a ring A given that Spec(A) is Noetherian? The following proposition addresses this question.

PROPOSITION 7.9. Let A be a ring such that Spec(A) is a Noetherian space. Then the set of prime ideals of A satisfies the ascending chain condition.

Proof Let P 1 c P 2 c P 3 c · · · be an ascending chain of prime ideals of A. Then V(P ) J V(P ) J V(P ) J • • • is a descending chain of closed subsets of Spec( A) . 1 2 3 Hence, there exists n E IN such that

V(P 1) J V(P2) J • • • J V(P n) = V(P n+1) = · · · so that P 1 c P 2 c · · · c P n = P n+ 1 = · · ·. Therefore, the prime ideals of A satisfy the ascending chain condition. C

It would be somewhat of a comforting thought if the converse to the proposition above were true but alas, such comfort must be found elsewhere. For a counterexample, consider the Boolean ring A = 21N where addition and multiplication are, respectively, the symmetric difference and intersection of sets. Note, elements of A are subsets of IN . Recall, a Boolean ring is a commutative ring with unit where every element is idempotent. Some properties of A are: a+ a = 0 and a·(a+1) = 0 V a E A, where 0 =¢and 1 =IN.

Let P be a prime ideal of A and let a E A- P. Since an(1+a) = 0 and a¢ P, we must have a+1 E P (recall "n" is our ring multiplication). However

38 (a+ 1) + a = 1. Thus, if we adjoin to P any element not already in P we generate the unit ideal. Hence, every prime ideal of A is maximal. In particular, the set of prime ideals of A satisfy the ascending chain condition trivially. Now, for n E IN, define P n = 21N -{n}, that is, P n is the "power set" of IN -{n}. Clearly then P n is an ideal of A for each n E IN . Fix n E IN . Let a, b E A and suppose anb E P n· Then n t anb so that n t a or n t b, that is, a E P n or b E P n·

Thus, P n is a prime ideal of A for each n E IN. Consider the following descending chain of closed sets in Spec( A) :

V({l}) J V({1,2}) J V({1,2,3}) J · • · Since { Pk I k > n } c V( {1,2, ... ,n}) we see that the above chain does not finitely terminate. Hence, Spec(A) is not Noetherian.

39 §8 The Patch Topology

DEFINITION 8.1. Let A be a ring. The patch topology on the set Spec(A) is the topology having as a closed subbasis the Zariski-closed and Zariski quasi-compact open subsets of Spec{A). A closed set in the patch topology is called a patch.

Throughout the remainder of our discussion, given a ring A, we will be making frequent use of both the Zariski and patch topologies on Spec( A). So as to avoid confusion, we introduce the following notation.

NOTATION Let A be a ring.

1) Spec{ A) refers to the set of all prime ideals of A;

2) Z = Spec(A) endowed with the Zariski topology; 3) 1' = Spec(A) endowed with the patch topology; 4) For U c A, D{E) = Spec{A)- V(E); 5) For E c Spec(A), to say that E is Z-open (resp. Z-closed, Z-compact) means that E is open (resp. closed, compact) in the Zariski topology; 6) For E c Spec{A), to say that E is 7'-open (resp. 7'-closed, 7'-compact) means that E is open (resp. closed, compact) in the patch topology.

7) For f E A, Xf = Spec( A)- V(f) (the same notation as before).

Let A be a ring. The closed basis elements of 1' are formed by taking all possible finite unions of subbasis elements. Thus we see that the closed basis

40 elements of 'P are of the form V(I), D(J), or V(I) U D(J), where I and J are ideals of A with J finitely generated. Now, by complementation, we see that the open basis elements of 'P are sets of the form D(E), V(F), or D(E) n V(F), where

E and F are ideals of A with F finitely generated.

PROPOSITION 8.2. Let A be a ring. The patch topology 'P on

Spec(A) is compact.

Proof We first show 'P is a Hausdorff space. Let P, Q E Spec(A) with P -f Q.

We may assume without loss of generality there exists f E P-Q. Then V(f) and

D(f) are both open in 'P with P E V(f) and Q E D(f). Hence, 'P is Hausdorff.

Let C = {Pi}iE/ be a collection of patches having the finite intersection property.

To show that 'P is a compact space we wish show i~/ Pi f ¢. Note that each patch P. E C can be written as an intersection of basic closed sets 1 P. =.n B . . 1 JE 1i 1,J where the B. . are of the form V, D, or V U D, where V is Z-closed, D is l,J Z--Qpen and Z-quasi-compact. We claim the collection

c· = . u [.U .{B. ·}] = {B .. 1 i E I, j E Ji} IE 1 JE 11 l,J l,J also has the finite intersection property. Pick n E IN and B .. , ... ,B. . E C'. lt,Jl ln,Jn Then there exists P , ... , P E C such that B. . J P , ... , B. . J P . Hence, 1 n lt,Jl 1 In,Jn n n n ¢ -f n Pk c n B. . k=l k=l lk,Jk so we see that C' does in fact have the finite intersection property. In addition, it is clear that Thus we may as well assume our collection C consists entirely of basic open sets so that

41 for some indexing sets I, J, and K. Also, we may assume C is such a collection which is maximal with respect to the finite intersection property. Let V = i~IVi . .Z is quasi-compact (prop. 2.2) so {Vi}iEI having the finite intersection property implies V I= ¢>. Pick V k U Dk E C (some k E K). V k is closed in the quasi-compact space .Z hence is quasi- compact. Thus Vk U Dk is

.Z-quasi--compact for all k E K. Now pick V 1 U D1, V 2 U D2 E C- {Vi}iEI, where possibly one or both of V 1, V 2 = ¢>. Then ¢>I= (V1 U D1) n (V2 U D2) = (V1 n V2) U (V2 n D1) U (V1 n D2) U (D 1 n D2). Note that V 1 n D2 is a .Z--closed subset of the .Z-quasi--compact space D2, hence is .Z-quasi--compact. Similarly for V 2 n D1 . Thus, each intersection on the right hand side is a .Z-quasi--compact set so that their union is as well . Inductively we

n see if U , U , ... ,U E C- {V. }·El' then .n U. I= ¢> and is .Z-quasi--compact. Let 1 2 n 1 1 1=t 1

ul, u2, ... ,un E c- {Vi}iEI" For each i E I define yi = vi n L~~uJ 0 Then given i E I, Yi is non-void, being a finite intersection of elements from C, and is closed in the .Z-quasi--compact set U1 n U2 n···n Un. Note that {Yi}iEI has the finite intersection property since any finite intersection of such elements may be considered as a finite intersection of elements from C. Thus ¢> I= i~Iyi c V. Hence, since U1, ... , Un E C were arbitrary, for any C1, C2, ... , Cm E C we have V n (C1n···nCm) I=¢> . Therefore, by the maximality of C, V =Vi for some i E I. Suppose now that there exists proper closed subsets E and F of V such that

V = E U F. Since neither F E C nor E E C, there exists G1, ... , Gn, H1, ... , Hm E C suchthat En(G1 n···nGn)=¢> and Fn(H1 n···nHm)=¢>. Thus, v n (G1 n .. ·n Gn) n (H1 n .. ·n Hm) = ¢>. This contradicts the finite intersection property. Therefore, no such sets E, F exist so that V is an irreducible closed subspace of .Z. Hence, V = V(P) for some P E

42 Spec( A) (proof of prop. 3.6). Claim: P E C~CC. We already know P E Vi for all i E I. Pick Dj E C (some j E.!). Let Q E V n Dj" Since Dj = Spec(A)-V(E) for

some E c A, P c Q ~ P E Dj" Thus P E D a' for all a E J. Now let Vk U Dk E C

(some k E K). V n (Vk U Dk) -f ¢ implies we have two cases:

V n Dk -f ¢ or V n Dk = ¢.

1) V n Dk -f ¢ in which case P E Dk c Vk u Dk by the above argument.

2) V n Dk = ¢. Then for any U E C we have ¢ -f V n (Vk u Dk) n U = V n V k n

U so that, by the max:imality of C and since V k is closed, V k = Vi, some i E I.

Hence, P E V k c V k U Dk so that P E V a U D a for all a E K.

By 1) and 2) we see that in any event we have P E C~CC. Therefore, .9 is compact. C

COROLLARY 8.3. Let F be a patch in 'P. Then the Zariski closure F of

F is the set F = P~F V(P).

Proof Note that F = V(P~FP). Thus P~F V(P) c F. Let Q E F. Consider the collection .U = {Xa I a'/. Q} of Z-open, Z-quasi-compact neighborhoods of Q in

Z. Each Xa E is closed in and for a ,a , ... ,a '/. Q, X X X = .U .9 1 2 n n· · ·n n a1 a2 an

Xg E .U where g = a1 a2 ···an '/. Q. By the definition of F, F n Xa /= ¢ for all Xa E .U . Hence, {F} u {Xa I a '/. Q} is a collection of patches with the finite

intersection property. Since 'P is compact, we can pick P E F n (a~QXa)· Note

that a~QXa = {G E Spec(A) I G c Q} by the definition of the sets Xa. Hence,

Q E V(P) so that F c P~F V(P). Therefore, F = P~F V(P). c

Adopting the terminology of [2) we have the following.

43 DEFINITION 8.4. The ~spectrum of a ring A is the closure of Max( A) in the patch topology. For E c Spec(A), j-spec(E) is the patch closure of the set of prime ideals in A which are maximal in E.

PROPOSITION 8.5. Let A be a ring. Let w be the set of all primes in Spec(A) which are the intersection of maximal ideals. If Max(A) is a

Noetherian space, then j-spec(A) = w.

Proof Let P = {3~BM{3 E w, M/3 E Max(A) for all {3 E B. To show that P E j-spec(A) it suffices to show that any open basis element G of 'P which contains

P also contains some maximal ideal. We recall that the open basis elements of 'P are sets of the form D(I), V(J), or D(I) n V(J), where I and J are ideals of A with J finitely generated. We have then three cases.

1) P E D(I) for some ideal I of A. Then I { P = {3~BM{3 so that there exists {3 E B for which I { M . Therefore, M {3 E D(I) 13 n 2) P E V(a ,a , ... ,a ) = .n V(a.) for some a ,a , ... ,a E A. Then a. E P for each 1 2 n 1=l 1 1 2 n 1 i so that ai E M/3 for 1 ~ i ~ n and for all {3 E B. Hence, M/3 E V(a1,a2, ... ,an) for all {3 E B.

3) P E D(I) n V(a1,a2, .. . ,an) for some ideal I of A and some al'a2, ... ,an E A.

By invoking (1) and (2) above we can find {3 E B with M/3 E D(I) n V(a1,a2, ... ,an). Hence, in any event, we have P E j-spec(A). Therefore, w c j-spec(A). Now let

P E j-spec(A). There exists {ak}kEK c A with P = ( {ak}kEK) and we may assume K is a well-ordered set. For any Q E Spec(A) let VM(Q) = V(Q) n

Max( A), a Z-dosed subset of Max(A). Consider the following,

VM(a1) J VM(al, a2) J ••. J VM(a1, ... , an) J ••.

44 is a descending chain of closed, non-void subsets of Max(A). Since Max(A) is Noetherian the chain terminates at VM(a1' ... , at) for some t E IN . Define

R = n{M I M E VM(a1, ... , at)}. We wish to show P = R. Certainly we have P c R. Suppose b E R- P. Note that D(b) n V{a1, ... , at) is a 'P-open set containing P. Thus, since P E j-spec(A) there exists ~E Max(A) with ~E D(b) n VM(a1' ... , at). But ME

VM(a1' ... , at) implies ~ E VM(P) so that P c ~ E D(b ). This contradicts VM(P) = VM(R) . Thus R- P = ¢> and so P = R. Therefore P E 'Ill completing the proof. C

DEFINITIONS 8.6. Let A be a ring, E c Spec(A). 1) The dimension of E, denoted dim(E), is the maximal length of a

chain of primes in E (If E = Spec( A) we write dim( A)); 2) The j-dimension of E, denoted j-dim(E), is the maximal length of a

chain of primes in j-spec(E) (If E = Spec( A) we write j-dim(A)); 1 3) For P E Spec( A), o(P) = sup dim(j-spec(A[a- ]) n V(P)] ; aEA 4) For U c Spec( A), 8-dim(U) = sup o(P), if U = Spec( A) we write PEU 8-dim(U) = 8-dim(A).

As a consequence of our definitions we have the following. LEMMA 8. 7. Let A be a ring. 1) For P E Spec(A), o(P) S dim(~); 2) For a E A, 8-dim(A[a- 1]) S 8-dim(A).

45 Proof (1) Let P E Spec(A) and let a E A. Since the set of prime ideals which 1 survive as primes in A[a- ] is the set D(a), we may consider 1 j--spec(A[a- ] n V(P)) c j--spec( V(P)) = j--spec(~) c Spec(~). The result now follows.

1 A A b {2) Let a E A, A[a- ] = B, and let bE B. Then b =an for some bE A and 1 some n E IN U {0}. Hence we see that Spec(B[b- ]) is in one-to-one correspon­ 1 1 1 dence with the set D{ab). In other words, (A[a- ])[b- ] = A[(ab)- ]. The result now follows since a E A was arbitrary. C

One might wonder if we truly need a definition such as that for 8-dim.

Unfortunately, we do. The reason is that a statement for ~mension analogous to that of lemma 8. 7(2) does not hold. That is, if A is a ring, and a E A, it does not 1 necessarily follow that j-dim(A[a- ]) ~ j-dim(A), a fact which we must contend with later in our discussion. For an example of such a ring consider the following.

Let k be a field and let A = k[x,y]( x,y ) where x and y are indeterminates. If we let M denote the maximal ideal of A then we see that {M} = j--spec(A) so that ~m(A) = 0. Note that prime ideals in A are generated by irreducible polynomials in k[x,y] having 0 constant term. Now consider the ring B = 1 A[y - ]. In B, (y) now generates the unit ideal. However, now primes in A of the form (xn + y), where n E IN, become maximal in B. Since we saw in the proof of proposition 8.5 that primes in B which are the intersections of maximal ideals are elements j--spec(B), we must have n~IN(xn + y) = (0) E j--spec(B). Hence, we have

~m(B) = 1 > 0 = j-dim(A). We now state and prove two lemmas which will be needed in the proof of the main theorem of this section.

46 LEMMA 8.8. Let A be a ring, U c Spec( A). If F 1, F 2 are disjoint patches, F , F U Spec(A), then either F F U =¢,or 1 2 c c 1 n 2 n dim(F1 n F 2 n U) < dim(U). (Here the bar denotes the Zariski closure).

Proof Recall, F = P~F V(P) and F = P~F V(P) . Let P E F n F . Then 1 1 2 2 1 2 there exists Q E F and Q E F such that Q , Q P. Since F F = ¢, Q 1 1 2 2 1 2 c 1 n 2 1 and Q2 can't both be equal to P. Suppose Q1 f P. Q1 E U ~ P is not a minimal element of U. Hence F 1 n F 2 contains no minimal elements of U so that dim(F1 n F2 n U) < dim(U). c

LEMMA 8.9. Let A be a ring. If F 1, F 2 are disjoint Z-closed sets in

Spec( A), then there exists a E A such that F 1 c V(a) and F 2 c D(a).

Proof Let F 1 = V(11) and F 2 = V(12). Then V(I1) n V(12) = V(11 +12) = ¢ <==> there exists a unit u E 11+1 2. Thus we may write 1 = a1 + a2 where ai E li. Then a = a1 has the desired properties. C

1 THEOREM 8.10. Let A be a ring and let s E A, d = j-dim(A[s- ]).

Suppose a1, ... ,ak E A are such that k > d+1 and D(s) c D(a1' ... ,ak). Then there exists elements b2, ... bk E A with bi E A(s-a) such that D(s) c D(a2 + a1 b2se2, ... ' ak + a1 bksek)

for any choices of e2, ... ek E IN U { 0} .

47 1 Proof First consider the ring A[s- ] = As. If x E A, let i denote its image in

A . Then D( s) and D( s) correspond to the same set of primes but is now all of s Spec(As). Suppose the result holds for As. Then for each hi E As(s-ai) we can write = b.J/i where b. E A(s-a.). Note, the only ideals surviving in As are b.1 1 1 1 those which miss s. Since we are assuming

A) D(A A bA Ae 2+f2 A A bA Aek+fk) D( s c a2 + a1 2s , ... , ak + a1 ks we then get

D(s) c D(a2 + a1b2se2, ... 'ak + a1bksek). Hence, we may assume s is a unit in A and that D(s) = Spec( A).

[Note we have then d = j-dim(A)].

Let X= j-spec(A). Partition X as follows: let x1 = D(ak) n X and x2 = V(ak) n x. Note that X and X are disjoint patches. Let Y = X X , where the bar 1 2 1 n 2 denotes the Zariski closure. We proceed by induction on dim(X) = j-dim(A). By lemma 8.8 we have Y n X = ¢ or dim(Y n X) < dim(X). Suppose Y n X = ¢. Y is Z-closed so Y = V(J) for some ideal J of A. However, X contains all maximal ideals of A and Y n X = ¢. Thus Y = ¢ whence, by lemma 8.9, there exists r E A such that X1 c V(r) and X2 c D(r). Let bk = r(s-ak). We want to show D(a , ... , ak_ , ak + a bksek) = Spec( A) = or equivalently, if we let 2 1 1 D(s) I = A(a , ... , ak_ , ak + a bksek), then we wish to show I = A. It suffices then to 1 1 1 show I is not contained in any maximal ideal of A. Pick M E Max(A) and let A I+M A A A. A A I = M c A= y where y 1s a field. We show I = A (for then I ~ M). There are two cases: M E or M E . x1 x2

48 1) M E x1 :::} ak ~ M so that ak E A is a unit (where .. ~.. denotes the corresponding image in ~). Also, in this case we have M E V(r) so r E M and hence r = 0 E A which in turn implies bk = 0. Now consider i = A(a2, ... ak-1'

~ ek A A A A A ak + a1bks ) where a1bk = 0 and (0) -1 Aak c I. Therefore, I = A.

2) ME X2 :::} ME V(ak):::} ak = 0 EA. However, ME D(r):::} r f 0 so both r, s are units. Thus bk = r(s-ak) = rs f 0. Also, A = A(al'"""l ak) but Aa.k = 0, hence IA- AA(~ ~ ~ ~ bA ~ek)- AA(~ ~ ~ ~~ek+1) .J. 0 A AA - a2, ... ak_1, ak + a1 ks - a2, ... , ak_1, a1 rs .,. so I = . Therefore in any case, i =A and since ME Max:( A) was arbitrary, I= A. Now suppose dim(Y n X) < dim(X). Assume the result holds for all rings

R with j-dim(R) < d. Y is Z-dosed so Y = V(I), some radical ideal I of A.

We may then consider Y = Spec(*)· Note that

j-spec(*) = Max(*) = V(l) n Max(A) c V(I) n Max(A) = V(I) n j-spec(A)

=YnX. Thus k - 1 1 + Let E Y. Then E V(P) for some P E x > j-dim(~). Q Q 2 [recall X2 = P~X V(P) c Y). But P E X :::} ak E P c Q. Thus if we mod out by 2 2 I we see that ak makes no contribution (ie. since I = r(I), ak +I = 0). Let "~" denote the corresponding image in ~ - Hence we see that

D(s) =Spec(*)= D(a1, ... , a.k_1).

A A A A Applying our induction hypothesis to T' there exist b2, ... , bk_1 E T' where hi E ~8-ai), such that D(s) = D(a2 + a1 b2§e2, ... ,ak-1 + a1 bk-1 sek-1) . Thus Y c D( a + a b se2, ... , ak_ + a bk_ sek-1), that is, every prime ideal 2 1 2 1 1 1

49 containing I does not contain a . + a b.sej for some j = 2, ... , k-1. It remains to J 1 J find bk of the appropriate form so that D(s) = Spec( A) = D(a2 + a1 b2se2, ... ' ak-1 + a1 bk-1 sek-1, ak + a1 bksek) Let Z = V(a + a b se2, ... , ak_ + a bk_ sek-1) and define the Z-closed sets 2 1 2 1 1 1 z1 = Z n X1 and z2 = Z n X2. Then z1 n z2 = Z n Y = ¢J so by lemma 8.8 there exists r E A such that z1 c V(r) and z2 c D(r). We claim bk = r(s-ak) 2 1 is of the appropriate form. Let J = A(a2 + a1b 2se , ... , ak_1 + a1bk_ 1sek- , ak + a1bksek). Then it suffices to show J = A. By repeating the argument above for the case Y n X = ¢, we see that for bk so chosen we do indeed have J = A.

This completes the induction and hence the proof. C

We wish to conclude our discussion on the prime spectrum of a ring with several important corollaries of the above theorem.

COROLLARY 8.11. Let d = j-dim(A) and k > d+l. If (a1, ... , ak) = A, then there exists b2, .. . , bk E A such that (a2 + a1b 2, ... , ak + a1bk) =A. (This is commonly known as the Stable Range Theorem.)

1 COROLLARY 8.12. Let sEA and d = j-dim(A[s- ]). Suppose a1, ... , ak E A are such that k > d+l and D(a1) c D(s) c D(a1, ... , ak). Then there exists b2, ... , bk E (s) such that D(a2 + a1b2, ... , ak + a1bk) = D(a1, .. . , ak).

Proof By theorem 8.10 we can find b2, ... , bk E (s) so that D(s) c D(a2 + a1b 2, ... , ak + a1bk). Let I= (a2 + a1b2, ... , ak + a1bk) and J = (a2, .. . ak). Then

50 D(I) = D(a2 + a1b 2, ... , ak + a1bk) = D(s) u [D(I) n V(s)]. Now consider the following, suppose P E Spec( A) and s E P , then since a1bi E ( s), for i=2, ... ,k we must have a1bi E P, for i=2, ... ,k. Thus P E D(I) n V(s) ~ s E P and a. + a b . ¢ P, some j = 2, ... , k. But then a b. E P so that we must have s E J 1 J 1 J P and aj ¢ P. Hence we see that P E D(I) n V(s) ~ P E D(J) n V(s). Thus

D(I) = D(s) u [D(J) n V(s)] = D(s) u D(J) J D(a1) U D(J) = D(a1, ... , ak). Thus D(a1' ... ak) c D(I). To show the other inclusion we show that D(s) U D(J) c

D(a1) U D(J) so that we have in fact D(s) U D(J) = D(a1) U D(J) . If P E D(J) this is clear so suppose P E D(s). Then s ¢ P so that P E D(a1, ... , ak) by our hypothesis. But P E D(a1, ... , ak) implies P E D(a1) or P E D(a2, .. . , ak) = D(J). Thus D(s) u D(J) c D(a1) U D(J). Therefore, D(I) = D(a2 + a1b2, ... , ak + a1bk) = D(a1'" .. , ak) c

1 COROLLARY 8.13. Suppose j-dim(A[ai ]) = d and k > d+l. Then

there exists b2, ... , bk E (a1) with D(al, .. . , ak) = D(a2 + al b2, ... , ak + al bk). (This is a generalization of a theorem of Kronecker.)

COROLLARY 8.14. Let d = 6--dim(A). If I = r(t1, ... , tn, a1, ... , ad+l) !: then I= r(c , ... , cd+l) where c.= a.+ . b . . t .. (here "r()" denotes the 1 1 1 J J,1 J corresponding radical ideal).

Proof We use induction on n. For n = 1 note that by corollary 8.13 we have

51 So now suppose the result holds for all n < a. Again by corollary 8.13 we have

V(t 1, ... , ta, a1, ... , ad+l) = V(t2 + tl b1,2'"'' t a+ tl bl,a ,al + tl bl,a+l'"'' ad+l + tl bl,a+d+l). Now we apply our induction hypothesis to the right hand side to get

V(t 1, ... , ta, a1, ... , ad)= V(c1, ... , cd+l) a where c. = a. + (t b +') + ._~ {J . . (t. + t b .) which a moments reflection will 1 1 1 1 , a 1 J-2 J,I J 1 1 ,J show is of the appropriate form. C

52 REFERENCES

1 M. F. Atiyah and I. G. MacDonald, Introduction to Commutative Algebra, Addison-Wesley Publishing Co., 1969.

2 R. Heitmann, Generating Non-Noetherian Modules Efficiently, Michigan Math. J ., 31 (1984), no. 2, 167-180.

3 J. Lambek, Lectures on Rings and Modules, Blaisdell Publishing Co., 1966.

4 S. Lang, Algebra, Addison-Wesley Publishing Co., 1984.

5 W. Rudin, Real and Complex Analysis, McGraw-Hill Book Co., 1987.

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