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Substructures of Spec R[X]
188, 184]202 Ž1997. JA966803
JOURNAL OF ALGEBRA ARTICLE NO.
Substructures of Spec Rw X x Stephen McAdam Department of Mathematics, Colgate Uni¨ ersity, Hamilton, New York 13346
and Chandni Shah Department of Mathematics, Uni¨ ersity of California, Ri¨ erside, California 92521 Communicated by Mel¨ in Hochster Received August 2, 1994
1. INTRODUCTION This work is part of a loosely related series of papers attempting to understand the poset structure of Spec Rw X x, with R a Noetherian ring. An underlying difficulty in this endeavor is that the nature of R itself has very subtle influences on the poset structure of Spec Rw X x. That fact is illustrated in wWx Žin which Wiegand characterizes Spec Zw X x, with Z the integers., and also plays a large role in the present work. Because of this, it appears to be too difficult to understand all of Spec Rw X x except in cases in which Spec R is very simple Žsee wHWx and wSh2x., and Wiegand’s work in wWx seems to be as complicated a case in which understanding all of Spec Rw X x is feasible. Therefore this paper, like wM1x and wM3x before it, only attempts to understand a finite chunk of Spec Rw X x. We begin with three definitions. DEFINITION. If P is a prime ideal in the ring R and if p is a prime ideal in the polynomial ring Rw X x, then we call p an upper to P in Rw X x if p l R s P, but p / PRw X x. ŽSee wK, Sect. 1-5x for basic facts about uppers.. DEFINITION. Let R be a domain, let P be a nonzero prime ideal in R, and let K be an upper to 0 in Rw X x. We use UŽ P, K . to denote the set p ¬ p is an upper to P in Rw X x, and K ; p4 . 184 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
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DEFINITION. Let 0 ; P ; M be prime ideals in the domain R. Let U be a finite set of uppers to P in Rw X x, and assume that for each p g U, p MRw X x. We will say that M overwhelms U if for every upper K to 0 in Rw X x with UŽ P, K . s U, we have K ; MRw X x. The question raised in this paper is Žfor R a Noetherian but not a local Henselian domain. what conditions on R, P, M, and U are required to assure that U is not overwhelmed by M ? Let us briefly outline what considerations lead us to that question. We need a fourth definition. DEFINITION. Let P1 , . . . , Pn be distinct nonzero prime ideals in the domain R and, for 1 F i F n, let Ui be a finite, possibly empty, set of uppers to Pi in Rw X x. We will say that U1 , . . . , Un are algebraically discernible if there is an upper K to 0 in Rw X x such that for 1 F i F n, UŽ Pi , K . s Ui . The question we originally asked was what sets of uppers U1 , . . . , Un are algebraically discernible? We felt that if a satisfactory answer to that question was available, then we could strengthen the results reported in wSex, wSh1x, wM1x, and wM3x. Exploring this original question led us down some unexpected paths. Let us describe that journey in the easiest possible setting Žwhich nonetheless encompasses all the main ideas of the situation.. Let Ž R, M . be an integrally closed, local non-Henselian domain and let P be a prime ideal of R with 0 / P / M. Let U and V be finite sets of uppers to P and M, respectively, such that for each p g U, p MRw X x. We hoped to show that U, V are algebraically discernible. We first proved the following theorem. THEOREM A. There are infinitely many uppers K to 0 in Rw X x such that UŽ P, K . s U. That was a good start. We only needed to find one such K which also satisfied UŽ M, K . s V. However, we then showed that this was impossible for all but one choice of V. THEOREM B. With notation as before, if U, V are algebraically discernible, then V s q ¬ q is an upper to M in Rw X x and there is a p g U with p ; q4 . Remark. Since for p g U we are assuming p MRw X x, it follows that p can be contained in at most finitely many uppers to M. Thus, since U is finite, the set V of Theorem B is also finite Žwhether or not U, V are algebraically discernible .. Somewhat naively, we then hoped to show for the special choice of V in Theorem B, that U, V are algebraically discernible. Instead, we discovered that the best we could do was the following theorem.
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THEOREM C. With notation as before, if V s q ¬ q is an upper to M in Rw X x and there is a p g U with p ; q4 , then U, V are algebraically discernible if and only if M does not o¨ erwhelm U. Theorem C is interesting. It says that if K is an upper to 0 with UŽ P, K . s U, then either UŽ M, K . s V Žas originally hoped. or K : MRw X x. ŽIn the latter case UŽ M, K . consists of all uppers to M, and so is as far from equaling V as possible.. We set about looking for an example in which M overwhelmed U. Some examples appear in Section 3. The easiest to describe Žbut the hardest to prove. is as follows. With ŽT, N . a non-Henselian Discrete Valuation Ring ŽDVR., let R s T ww Y xx, P s Ž Y . R, and M s Ž N, Y . R. If U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 , then M overwhelms U. ŽSee Example 3.12 and wM2, Ž3.6.x.. Thus, our original desire to understand when U, V are algebraically discernible was frustrated by the fact that M might overwhelm U, and so our goal became Žand still is. to determine under what circumstances M overwhelms U. ŽThe examples in Section 3 show that the question of when M overwhelms U is rather subtle. Those examples do not hint at any clear pattern.. In Section 2, we discuss Theorems A, B, and C, though in more generality than stated here. ŽHowever, we defer some of the more technical arguments until Section 4.. In Section 3 we give examples of situations in which M does and does not overwhelm U. 2. ALGEBRAICALLY DISCERNIBLE UPPERS Notation. Throughout this section, R will be a domain and P1 , . . . , Pn will be distinct nonzero prime ideals of R. For 1 F i F n, Ui will be a finite set of uppers to Pi in Rw X x. While we are primarily interested in the case that R is Noetherian, the ideas are easily extended to the case that R lies between some Noetherian domain D and the integral closure of D. R will never be a field. Also, we will assume that R is not a quasilocal Henselian domain. Our first theorem shows that when P1 , . . . , Pn are pairwise incomparable, U1 , . . . , Un are always algebraically discernible. ŽAs the proof of Theorem 2.1 is highly technical, and contributes little to the understanding of the main ideas of this paper, we defer the proof until Section 4.. Ž2.1. THEOREM. Suppose there is a Noetherian domain D with integral closure D9 such that D : R : D9. Suppose R is not Henselian. If P1 , . . . , Pn are pairwise incomparable, then U1 , . . . , Un are algebraically discernible. In fact, there are < R < uppers K to 0 in Rw X x such that for 1 F i F n, UŽ Pi , K . s Ui . Ž Here, < R < denotes the cardinality of R..
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Remarks. Ža. Theorem A of the Introduction is a special case of Theorem 2.1. Žb. In the case that R s Z, the integers, Theorem 2.1 also follows from Wiegand’s characterization of Zw X x in wWx. The question of which U1 , . . . , Un are algebraically discernible, is similar to the question of which rings satisfy Wiegand’s final axiom characterizing Spec Zw X x. We now turn to the general case. The next lemma is crucial. Ž2.2. LEMMA. Let 0 ; Q1 ; Q2 be prime ideals in an integrally closed domain R. Suppose that L ; q2 are uppers in Rw X x to 0 and Q2 , respecti¨ ely. Suppose that L Q2 Rw X x. Then there is an upper q1 to Q1 with L ; q1 ; q2 . We defer the proof of Lemma 2.2 until Section 4. Ž2.3. PROPOSITION. In the Ž arbitrary. domain R, if U1 , . . . , Un are algebraically discernible, then the following two conditions must hold. GU: If p ; q, where p g U1 j ??? j Un and q is an upper to Pi Žsome i, 1 F i F n., then q g Ui . Minimality: Suppose that p g Ui Žsome i, 1 F i F n. and p is minimal in the set U1 j ??? j Un . Then Žwith R9 the integral closure of R . there is a Q g Spec R9 with Q l R s Pi such that Q is minimal in the set Q9 g Spec R9 ¬ Q9 l R g P1 , . . . , Pn44 . Proof. To prove GU, suppose that p ; q with p g Uj Žsome j, 1 F j F . n and with q an upper to Pi Žsome i, 1 F i F n.. As we are assuming that U1 , . . . , Un are algebraically discernible, there is an upper K to 0 in Rw x x with UŽ Ph , K . s Uh for 1 F h F n. Since p g Uj , K ; p ; q. Thus q g UŽ Pi , K . s Ui . This proves GU. As for minimality, let p g Ui and suppose that p is minimal in the set U1 j ??? j Un . Since p g Ui s UŽ Pi , K ., K ; p. By lying over, there is a prime L in R9w X x with L l Rw X x s K Žso that L is an upper to 0 in R9w X x., and by going up, there is a prime q of R9w X x with L ; q and with q l Rw X x s p. Let q l R9 s Q. Thus Q l R s Pi and q is an upper to Q in R9w X x. We will show that Q is minimal in the set Q9 g Spec R9 ¬ Q9 l R g P1 , . . . , Pn44 . If not, then for some 1 F j F n there is some Q9 g Spec R9 with Q9 ; Q and Q9 l R s Pj . Since K is contained in only finitely many uppers to Pi Žnamely, those uppers in Ui ., clearly K Pi Rw X x. It follows that L QR9w X x. We have 0 ; Q9 ; Q, L ; q, and L QR9w X x. By Lemma 2.2, there is an upper q9 to Q9 in R9w X x with L ; q9 ; q. Intersecting with Rw X x Žand letting p9 s q9 l Rw X x., we have K ; p9 ; p. Since p9 is an upper to Pj , we have p9 g UŽ Pj , K . s Uj . This contradicts that p is minimal in U1 j ??? j Un , and so completes the proof.
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Ž2.4. LEMMA. Let R be integrally closed, and suppose that both GU and minimality hold. Also suppose that P1 , . . . , Pr are the minimal members of P1 , . . . , Pn4 . Then for r q 1 F i F n, Ui s q ¬ q is an upper to Pi and there exists a p g U1 j ??? j Ur with p ; q4 . Proof. Suppose that q is an upper to Pi Ž r q 1 F i F n. and there exists a p g U1 j ??? j Ur with p ; q. By GU, q g Ui . Conversely, suppose that q g Ui . Surely there is a p which is minimal in U1 j ??? j Un with p : q. Suppose p g Uj , so that p is an upper to Pj . Since minimality holds and since R is integrally closed, we see that Pj is minimal in P1 , . . . , Pn4 . Thus 1 F j F r. This shows that p g U1 j ??? j Ur , as desired. Since i ) r G j, it also shows that p / q, so that p ; q, as desired. Remark. Theorem B of the Introduction is a special case of Lemma 2.4 Žcombined with Proposition 2.3.. The converse of Proposition 2.3 does not hold, as examples in the next section show. The best we can get is a partial converse. We consider that first in the integrally closed case. Ž2.5. THEOREM. Let R be integrally closed and suppose that U1 , . . . , Un satisfy GU and minimality. Suppose that P1 , . . . , Pr are the minimal members of P1 , . . . , Pn4 . Also, suppose that K is an upper to 0 in Rw X x such that for 1 F i F r, UŽ Pi , K . s Ui . Ž By Theorem 2.1, if R is Noetherian and nonHenselian, then < R < such K exist.. Then for r q 1 F j F n, either UŽ Pj , K . s Uj or K ; Pj Rw X x. Proof. Suppose K Pj Rw X x Žfor some j, r q 1 F j F n.. Let UŽ Pj , K . s Vj . ŽWe will show that Vj s Uj .. Since the intersection of any infinite set of uppers to Pj is Pj Rw X x Žwhich does not contain K ., we see that Vj is finite. Consider the primes P1 , . . . , Pr , Pj and the corresponding finite sets of uppers U1 , . . . , Ur , Vj . Since UŽ Pi , K . s Ui for 1 F i F r and UŽ Pj , K . s Vj , we see that U1 , . . . , Ur , Vj are algebraically discernible. By Proposition 2.3, U1 , . . . , Ur , Vj satisfy GU and minimality. By Lemma 2.4, Vj s q ¬ q is an upper to Pj and there exists a p g U1 j ??? j Ur with p ; q4 . On the other hand, U1 , . . . , Un also satisfy GU and minimality Žby hypothesis., and so Lemma 2.4 shows that Uj equals that same set. Thus UŽ Pj , K . s Vj s Uj . Remark. Theorem C of the Introduction follows easily from Theorem 2.5. We now drop the assumption that R is integrally closed. The next theorem is proved in Section 4. Ž2.6. THEOREM. Let D be a Noetherian domain with integral closure D9 and suppose that D : R : D9. Suppose R is not Henselian. Let P1 , . . . , Pr , Prq1 , . . . , Pn be distinct nonzero prime ideals in R, ordered such that P1 , . . . , Pr are the minimal members of the set P1 , . . . , Pn4 . Suppose that
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GU and minimality hold. Then there exist < R < uppers K to 0 in Rw X x such that for 1 F i F r, UŽ Pi , K . s Ui , while for r q 1 F i F n, either UŽ Pi , K . s Ui or K ; Pi Rw X x.
3. EXAMPLES Notation. In this section, we restrict our attention to the special case that 0 ; P ; M are prime ideals in the Noetherian domain R, and U is a finite set of uppers to P such that for all p g U, p MRw X x. Suppose Žmomentarily. that R is integrally closed. Also suppose that V is a finite set of uppers to M such that U, V satisfy GU and minimality. Then Theorem 2.5 shows that either U, V are algebraically discernible or M overwhelms U. ŽSee the Introduction.. Therefore, in this section we explore through examples when M does or does not overwhelm U. ŽThus, we do not consider V, just U. Also, not all the examples have R integrally closed.. We can detect no pattern in our examples, and so consider the question of when M overwhelms U a subtle one. Let us begin by considering the case that U consists of a single upper to P. Thus, U s p4 , where p is an upper to P and p MRw X x. Our first three examples discuss this case. Ž3.1. EXAMPLE. With notation as before, suppose that RrP is integrally closed. Also suppose U s p4 , with p an upper to P such that p MRw X x. Then M does not overwhelm U. Proof. Let overbars denote reduction modulo P. By wT, Theorem Bx, p is generated by the set of polynomials of minimal degree which are contained in p. Note that each such polynomial is in no other upper to 0 in Rw X x except p. ŽTo see this, note that the analogous statement is easily seen to be true in F w X x, with F the quotient field of R, and so our statement follows from the fact that F w X x is a localization of Rw X x.. Therefore, since p MRw X x, there is a nonconstant polynomial b Ž X . g p y MRw X x with b Ž X . not contained in any upper to 0 in Rw X x except p. Pick f Ž X . g Rw X x such that f Ž X . s b Ž X .. Thus f Ž X . is contained in p y MRw X x, and f Ž X . is not contained in any other upper to P except p. Let K be a prime in Rw X x with f Ž X . g K : p and with K minimal over f Ž X . Rw X x. By the principal ideal theorem, height K s 1. Let K l R s Q. We must have either Q s 0 and K is an upper to 0, or height Q s 1 and K s QRw X x. However, the latter case is impossible, since if it held, we would have f Ž X . g K s QRw X x s Ž K l R . Rw X x : Ž p l R . Rw X x s PRw X x : MRw X x, which is a contradiction. Thus K is an upper to 0. Obviously K is contained in p by construction and is not contained in any
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other upper to P wsince that is true of f Ž X . g K x. Therefore, UŽ P, K . s p4 s U. Now since f Ž X . f MRw X x, we see that K MRw X x. Thus M does not overwhelm U. Ž3.2. Question. With notation as at the start of this section, is it true that R M rPM is integrally closed if and only if for each upper p to P with p MRw X x, M does not overwhelm U s p4 ? Concerning the preceding question, if R M rPM is integrally closed, then M does not overwhelm U s p4 for any upper p to P with p MRw X x. This follows easily from Example 3.1. The next two examples lend credence to the converse, since in the next two examples, R M rPM is not integrally closed and M does overwhelm U s p4 , for carefully chosen p. ŽNonetheless, we suspect that the converse is false and that a counterexample can be constructed which avoids the assumptions of either of the next two examples.. Ž3.3. LEMMA. Let Ž R, M . be a quasilocal domain containing a field of characteristic 0 and let F be the quotient field of R. Suppose that a g F and that a f R and ay1 f R. Let p be the kernel of the map Rw X x ª Rw a x. If f Ž X . is a polynomial contained in p, but not contained in any other upper to 0 in Rw X x except p, then f Ž X . g MRw X x. Proof. Obviously p s Ž X y a . F w X x l Rw X x. Since contraction gives a one-to-one map from the uppers to 0 in F w X x to the uppers to 0 in Rw X x, we see that f Ž X . is not contained in any upper to 0 in F w X x except Ž X y a . F w X x. Thus, in the Unique Factorization Domain ŽUFD. F w X x, the only prime polynomial dividing f Ž X . is X y a , so that f Ž X . s aŽ X y a . k , for some a g F and some integer k G 1. For some h with 0 F h F k, consider the degree k y h coefficient of this polynomial. That coefficient is in R Žsince f Ž X . g Rw X x. and has the form ma a h for some nonzero integer m. We will show that each a a h Ž0 F h F k . is in M. This will show that f Ž X . g MRw X x as desired. Since R contains a field of characteristic 0, we see that m represents a unit in R. Thus a a h is in R for 0 F h F k. Letting h s 0, we see that a g R. Letting h s 1 and recalling that a f R, we see that a is not a unit in R, so that a g M. Now consider an h with 0 - h F k. If a a h f M, then Žsince a a h g R . we have that a a h is a unit of R, so that ay1 s a a hy 1ra a h g R. This contradiction shows that a a h g M for all 0 F h F k, and completes the argument that f Ž X . g MRw X x. Ž3.4. EXAMPLE. Let P be a nonzero prime ideal in the local UFD Ž R, M .. Suppose that RrP is not integrally closed, but does contain a field of characteristic 0. Then there is an upper p to P in Rw X x with p MRw X x, such that M overwhelms U s p4 .
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Proof. Let R s RrP and let R9 be the integral closure of R. Pick any a g R9 y R. ŽA well known and easy exercise shows that ay1 f R.. Let p be the kernel of the map Rw X x ª Rw a x. Since a is integral over R, p contains a monic polynomial. Thus p MRw X x. Let p be the inverse image of p under the map Rw X x ª Rw X x. Now p is easily seen to be an upper to P in Rw X x and p MRw X x. We will show that M overwhelms p4 . Assume that K is an upper to 0 in Rw X x and that UŽ P, K . s p4 . We must show that K ; MRw X x. Since R and hence Rw X x is a UFD, we see that K is principal. Let K s f Ž X . Rw X x. Thus f Ž X . is contained in p, but is not contained in any other upper to P except p Žsince the same is true of K .. Since p is the image of p in Rw X x, we have that f Ž X . is in p, but is not in any other upper to 0 in Rw X x except p. By Lemma 3.3, f Ž X . g MRw X x. Thus f Ž X . g MRw X x, so that K s f Ž X . Rw; MRw X x. Ž3.5. LEMMA. Let P ; Q ; M be prime ideals in the integrally closed Ž not necessarily Noetherian. domain R and let U be a finite set of uppers to P such that for each p g U, p MRw X x. Let W s q ¬ q is an upper to Q and there is a p g U with p ; q4 , V1 s m ¬ m is an upper to M and there is a p g U with p ; m4 , and V2 s m ¬ m is an upper to M and there is a q g W with q ; m4 . If M does not o¨ erwhelm U, then V1 s V2 . Proof. Suppose M does not overwhelm U. Then there is an upper K to 0 in Rw X x with UŽ P, K . s U such that K MRw X x. Since no p g U is contained in MRw X x, V1 is finite. We easily see that U, V1 satisfy GU and minimality. By Theorem 2.5, we see that UŽ M, K . s V1. Obviously K QRw X x. Thus W is finite and we easily see that U, W satisfy GU and minimality. By Theorem 2.5, UŽ Q, K . s W. Since no prime in U is contained in MRw X x, no prime in W is contained in MRw X x. Thus, V2 is finite. As W, V2 satisfy GU and minimality, Theorem 2.5 shows that UŽ M, K . s V2 . Thus V1 s UŽ M, K . s V2 . Ž3.6. EXAMPLE. Let P be a nonzero prime ideal in the integrally closed local domain Ž R, M .. Let T s RrP and N s MrP. Assume that height N G 2. Also assume that in the integral closure of T, N1X and N2X are distinct primes lying over N, with height N1X G 2. ŽThus, T s RrP is not integrally closed.. Pick a g N1X y N2X and let p be the kernel of the map Rw X x ª T w X x ª T w a x. Then p is an upper to P in Rw X x, p MRw X x, and M overwhelms U s p4 . Proof. Since T w a x is a simple integral extension of T s RrP, clearly p is an upper to P in Rw X x, and p contains a monic polynomial, so that p MRw X x. In order to show that M overwhelms U s p4 , we will use Lemma 3.5. Thus we will now find a special prime Q with P ; Q ; M.
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Let N1 s N1X l T w a x and N2 s N2X l T w a x. Note that N1 / N2 and height N1 G 2. Let S be the set of height 1 primes of T w a x which are contained in N1 and let SX s Q9 g S ¬ QX N2 4 . The principal ideal theorem shows that N1 s D Q9 g S4 . Thus N1 : D Q9 g S94 j N2 . If S9 is finite, the prime avoidance lemma shows that N1 is contained in either N2 or in some Žheight 1. prime in S9. As neither of these is true, we see that S9 is infinite. Since T w a x is a finite T-module Žand contained in the quotient field of . T , ŽT :T w a x. s b g T ¬ bT w a x : T 4 / 0. Thus, only finitely many height 1 primes of T w a x can contain ŽT :T w a x.. Therefore, we may select Q9 g S9 such that ŽT :T w a x. Q9. Since Q9 g S9, we have that Q9 ; N1 and Q9 N2 . Since ŽT :T w a x. Q9, an easy exercise shows that Q9 is the unique prime of T w a x lying over Q9 l T. Note that since Q9 ; N1 , Q9 l T ; N. Letting Q be the inverse image in R of Q9 l T, we see that P ; Q ; M. Let W, V1 , and V2 be as in Lemma 3.5. We will show that V1 / V2 . Since the prime N2 of T w a x lies over N in T, under the isomorphism w R X xrp , T w a x, N2 corresponds to a prime mrp, where m is an upper to M in Rw X x. By definition, m g V1. We need only show that m f V2 . Suppose m g V2 . Then there is a q g W with q ; m. Since q g W, we have p ; q. Since qrp ; mrp, under the isomorphism Rw X xrp , T w a x, qrp corresponds to a prime q9 of T w a x contained in N2 . Since Žbeing in W . q is an upper to Q Žwhich taken modulo P is Q9 l T ., we see that q9 in T w a x lies over Q9 l T in T. Since Q9 is the only prime of T w a x lying over Q9 l T, we see that q9 s Q9. This contradicts that Q9 is not contained in N2 . Therefore, V1 / V2 , and so by Lemma 3.5, M overwhelms p4 . This completes our study of the case that U consists of a single upper to P. We now consider more general U. Ž3.7. LEMMA. Let P be a prime ideal in the Noetherian domain R, with P not contained in the Jacobson radical of R. Let g Ž X . be a nonconstant polynomial in Rw X x. Then there is a polynomial hŽ X . g Rw X x such that hŽ X . ' g Ž X . modulo P and hŽ X . cannot be factored into the product of two nonconstant polynomials. Proof. Let N be a maximal ideal of R not containing P, so that P and N 2 are comaximal. By wK, Theorem 77x, N 2 / N. Suppose that g Ž X . s a d X d q ??? qa0 . Pick bd f N, b 0 g N y N 2 , and, for 0 - i - d, bi g N. Now for 0 F i F d, use the Chinese remainder theorem to pick c i ' a i mod P and c i ' bi mod N 2 . Let hŽ X . s c d X d q ??? qc0 . Clearly hŽ X . ' g Ž X . mod P. Also, we clearly have c d f N, c 0 g N y N 2 , and, for 0 - i d, c i g N. The argument used to prove Eisenstein’s criterion shows that hŽ X . cannot be factored into the product of two nonconstant polynomials.
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The next example strengthens both the assumption and the conclusion of Example 3.1. Ž3.8. EXAMPLE. Let 0 ; P ; M be prime ideals in the Noetherian UFD R and suppose that RrP is integrally closed. Also suppose that P is not contained in the Jacobson radical of R. If U is any finite set of uppers to P in Rw X x such that p MRw X x for all p g U, then M does not overwhelm U. Proof. Let overbars denote reduction modulo P. For each p g U, p is an upper to 0 in the integrally closed domain Rw X x, and so Žas in the proof of Example 3.1., wT, Theorem Bx shows there is a nonconstant polynomial b p Ž X . g p y MRw X x with b p Ž X . not contained in any upper to 0 in Rw X x except p. Let b Ž X . s P b p Ž X . over all p g U. Clearly b Ž X . f MRw X x, and the only uppers to 0 in Rw X x which contain b Ž X . are the p for p g U. Pick g Ž X . g Rw X x with g Ž X . s b Ž X .. By Lemma 3.7, pick hŽ X . g Rw X x such that hw X x s g w X x s b Ž X . and hŽ X . cannot be factored into the product of two nonconstant polynomials. Obviously hŽ X . f MRw X x, and the only uppers to P in Rw X x which contain hŽ X . are the uppers p g U Žsince these statements are true modulo P .. Since R is a UFD, we can write hŽ X . s cf Ž X . with c g R and with the greatest common divisor of the coefficients of f Ž X . equaling 1. Thus, the only constant factors if f Ž X . are units in R. Since also cf Ž X . s hŽ X . cannot be factored into the product of two nonconstant polynomials, the same is true of f Ž X .. Therefore, f Ž X . is irreducible in the UFD Rw X x. Since cf Ž X . s hŽ X . f MRw X x, we have f Ž X . f MRw X x and c f M. Thus c f P, and so c is not contained in any upper to P. Since we already know that the set of uppers to P in Rw X x which contain cf Ž X . s hŽ X . is exactly the set U, we see that the set of uppers to P in Rw X x which contain f Ž X . is also the set U. Let K s f Ž X . Rw X x. Since f Ž X . is irreducible Žand nonconstant. in Rw X x, K is an upper to 0 in Rw X x. Since f Ž X . f MRw X x, K MRw X x. Since the set of uppers to P which contain f Ž X . is exactly U, UŽ P, K . s U. This shows that M does not overwhelm U. Before presenting the next example Žour most interesting one., we must introduce a concept. DEFINITION. The nonzero prime ideal P in the Noetherian domain R is called strongly comaximizable if for every positive integer m, there is a finitely generated integral extension domain T of R such that there are exactly m primes in T which lie over P, and those m primes in T are pairwise comaximal.
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Strongly comaximizable primes are studied in wM2x. In particular, wM2, Ž1.4.x says that if P is a prime not contained in the Jacobson radical of the Noetherian domain R, then P is strongly comaximizable. Thus strongly comaximizable primes are plentiful. ŽAlso, since strong comaximizability is easily seen to be preserved under localization, it is easy to construct strongly comaximizable primes in local domains.. While it is easy to find strongly comaximizable primes, finding primes which are not strongly comaximizable is significantly harder. Here is an example of one. Let T be an integrally closed Noetherian domain, let R s T ww Y xx, and let P s Ž Y . R. P is not strongly comaximizable wM2, Ž3.6.x. We need the following results from wM2x. Ž3.9. LEMMA. Suppose that the prime P is not a strongly comaximizable prime ideal in the Noetherian domain R. Then the following holds. If N is a maximal ideal in R9, the integral closure of R, and if P : N l R, then there is a prime ideal Q of R9 with Q : N and Q l R s P. Proof. This follows from wM2, Ž1.12.Ža.x. ŽWe leave it to the reader to add the little argument necessary to compensate for the fact that R9 might not be finitely generated over R, the key ingredient of that argument being the fact that only finitely many primes of R9 lie over P wN, Ž33.10.x.. Ž3.10. LEMMA. Let P be a prime ideal in the Noetherian domain R. The following are equi¨ alent. Ži. P is strongly comaximizable. Žii. There is a finitely generated integral extension domain T of R in which all of the primes which lie o¨ er P can be ordered P1 , . . . , Pr , Prq1 , . . . , Pn , with 0 - r - n, in such a way that P1 l ??? l Pr and Prq1 l ??? l Pn are comaximal. Proof. This is wM2, Ž1.5.Ži. m Žiii.x. The next lemma is similar to Lemma 2.4, but instead of assuming that R is integrally closed, we assume that P is not strongly comaximizable. Ž3.11. LEMMA. Suppose that P is not a strongly comaximizable prime in the local domain Ž R, M ., with P / M. Suppose that U and V are a finite sets of uppers to P and M, respecti¨ ely, such that U, V satisfy GU and minimality. Then V s q ¬ q is an upper to M, and there is a p g U with p ; q4 . Proof. If p g U and if p ; q with q an upper to M, then q g V by GU. Conversely, suppose that q g V. We must show there is a p g U with p ; q. If not, then q is minimal in U j V, and so by minimality, in the integral closure R9 of R, there is a prime N lying over M such that N does not contain any prime of R9 lying over P. This contradicts Lemma 3.9.
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Ž3.12. EXAMPLE. With notation as at the start of this section, and with Ž R, M . local, suppose that P is a prime of R which is not strongly comaximizable. Let U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 . Then M overwhelms U. Proof. Let K be an upper to 0 in Rw X x such that UŽ P, K . s U. In order to show that M overwhelms U, we must show that K ; MRw X x. Suppose that K MRw X x Žand we will derive a contradiction.. Let UŽ M, K . s V. ŽSince the intersection of any infinite set of uppers to M is MRw X x, we see that V is finite.. By definition, we have that U, V are algebraically discernible. By Proposition 2.3, U, V satisfy GU and minimality. Using Lemma 3.11, it is now trivial to see that we must have UŽ M, K . s V s Ž M, X . Rw X x, Ž M, X q 1. Rw X x4 . Now Rw X xrK , Rw u x for some u algebraic over R. Since we know exactly which two uppers to M contain K, we see that in Rw u x there are exactly two primes lying over M, namely, Ž M, u. Rw u x and Ž M, u q 1. Rw u x. Similarly, the only two primes lying over P are Ž P, u. Rw u x and Ž P, u q 1. Rw u x. Let T be the integral closure of R in Rw u x and let L be the kernel of w T X x ª T w u x s Rw u x. Let Q s Ž P, u. Rw u x l T and let qrL be the preimage of Ž P, u. Rw u x under the isomorphism T w X xrL , Rw u x. Now q is an upper to Q in T w X x, and since u g Ž P, u. Rw u x, we see that X g q. Thus q s Ž Q, X .T w X x. Similarly, if N s Ž M, u. Rw u x l T, then the pre-image of Ž M, u. Rw u x in T w X xrL is Ž N, X .T w X xrL. Obviously Q : N. We now claim that N is the only maximal ideal of T which contains Q. Suppose that N9 / N is another maximal ideal of T with Q : N9. Then since L ; Ž Q, X .T w X x : Ž N 9, X .T w X x, under T w X xrL , R w u x, Ž N9, X .T w X xrL would correspond to a prime in Rw u x which lies over M in R and which contains u. This is impossible since we already know that the only prime in Rw u x which lies over M and contains u is Ž M, u. Rw u x Žwhich corresponds to Ž N, X .T w X xrL.. This proves the claim that N is the only maximal ideal of T containing Q. We next claim that Q is the only prime of T which lies over P and is contained in N. Note that Lemma 4.3 shows TN s Ž Rw u x.Ž M , u. Rw ux. Since Ž P, u. Rw u x is the only prime in Rw u x which is contained in Ž M, u. Rw u x and lies over P, we see that there is only one prime in T which is contained in N and lies over P. Clearly Q is that prime. We claim that Ž P, u q 1. Rw u x l T / Q. If equality held, then we would have Q ; Ž M, u q 1. Rw u x l T. Since we already know that N is the only maximal ideal of T containing Q, we have Ž M, u q 1. Rw u x l T s N. By Lemma 4.3 Žused twice. we get that Ž Rw u x.Ž M, u. Rw ux s Ž Rw u x.Ž M , uq1. Rw ux. The unique maximal ideal of this ring is seen to contain both u and u q 1, a contradiction.
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We wish to use Lemma 3.10, which refers to a finitely generated extension of R. Since T might not be finitely generated over R, in this paragraph and the next, we will cut T down a bit. For now, we claim that only finitely many primes in T lie over P. Since u is in the quotient field of T, write u s arb with a and b in T. Then T lies between Rw a, b x and its integral closure. Since Rw a, b x is Noetherian and integral over R, only finitely many primes of Rw a, b x can lie over P, and so the same is true of T, by wN, Ž33.10.x. Let Q s Q1 , Q2 , . . . , Q h be all of the primes of T lying over P. Since we have already seen that Ž P, u q 1. Rw u x l T / Q s Q1 , we see that h G 2. We have also previously seen that none of Q2 , . . . , Q h is contained in N. Pick an element b g Ž Q2 l ??? l Q h . y N. The primes of Rw b x which lie over P are Q1 l Rw b x, Q2 l Rw b x, . . . , Q h l Rw b x. The first of these is contained in N l Rw b x and is not contained in any other maximal ideal of Rw b x Žsince Q1 is contained in N but not in any other maximal ideal of T .. On the other hand, none of Q2 l Rw b x, . . . , Q h l Rw b x is contained in N l Rw b x Žsince b f N .. Therefore, we see that the two ideals Q1 l Rw b x and F Q i l Rw b x ¬ 2 F i F h4 are comaximal. Now Lemma 3.10 shows that P is strongly comaximizable. This contradicts the hypothesis and completes the proof. Remark. Let T be a non-Henselain DVR, let R s T ww Y xx, and let P s Ž Y . R. By wM2, Ž3.6.x, P is not strongly comaximizable. However, RrP , T is integrally closed. Thus Example 3.12 shows that the conclusion of Example 3.1 cannot be extended to a set U consisting of two uppers to P. Furthermore, since T is a DVR, R is a UFD. Thus Example 3.12 shows that in Example 3.8, the assumption that P is not contained in the Jacobson radical is important Žalthough we wonder if it can be weakened to assuming that P is strongly comaximizable.. Before giving our final example, we need another concept. DEFINITION. The nonzero prime ideal P in the Noetherian domain R is called integrally discerning if for every finite set U of uppers to P such that every p g U contains a monic polynomial, there is an upper K to 0 in Rw X x such that K contains a monic polynomial and UŽ P, K . s U. Integrally discerning primes are studied in wMSx, which shows, among other things, that if P is not contained in the Jacobson radical of R and if RrP is integrally closed, then P is integrally discerning. Thus integrally discerning primes are plentiful. Ž3.13. EXAMPLE. With notation as at the start of this section, suppose that P is integrally discerning. If every p g U contains a monic polynomial, then M does not overwhelm U.
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Proof. Since P is integrally discerning, there is an upper K to 0 in Rw X x such that UŽ P, K . s U and K contains a monic polynomial. Obviously K MRw X x, and so M does not overwhelm U.
4. TECHNICAL ARGUMENTS In order to prove Theorem 2.1, we need the following easy consequence of wHW, Ž1.1.x. ŽFor a strengthening of this, see wHMx.. Ž4.1. LEMMA. Let P be a nonzero prime ideal in the Noetherian domain R. Furthermore, suppose that either R is not Henselian or R is Henselian but P is not the maximal ideal of R. Then for any positi¨ e integer m, there is a finitely generated integral extension domain of R in which at least m primes lie o¨ er P. The next corollary follows pleasantly from Lemma 4.1. Ž4.2. COROLLARY. Let R be a Noetherian domain and let R S be a localization of R. If R S is a local Henselian domain, but not the quotient field of R, then R S s R Ž and R is a local Henselian domain.. Proof of Theorem 2.1. We first reduce to the case that R is Noetherian. Since Dw X x : Rw X x : D9w X x and D is Noetherian, any prime in Dw X x has at most finitely many primes of Rw X x lying over it by wN, Ž33.10.x. It is therefore an easy exercise to find a ring T, finitely generated over D Žand hence Noetherian., with D : T : R, such that for 1 F i F n, Pi is the only prime of R lying over Pi l T, and for each p g U1 j ??? j Un , p is the only prime of Rw X x lying over p l T w X x. Let Wi s p l T w X x ¬ p g Ui 4 . Suppose that we can find < T < s < R < uppers L to 0 in T w x x such that for each i, UŽ Pi l T, L. s Wi . Then it is easily seen that for any such L, if K is a prime in Rw X x lying over L, then K is an upper to 0 and UŽ Pi , K . s Ui . ŽSince D : T : R : D9 and R is not Henselian, T is not Henselian.. Therefore, we may replace R by T, Pi by the Pi l T, and Ui by Wi . In other words, we may simply assume that R is Noetherian. Let S s R y Ž P1 j ??? j Pn .. We leave to the reader the easy verification of the fact that we can replace R by R S , Pi by the Ž Pi .S , and Ui by pS ¬ p g Ui 4 Žhere simply using that uppers to 0 in R S w X x have the form K S , where K is an upper to 0 in Rw X x.. Since P1 , . . . , Pn are pairwise incomparable, we see that the primes Ž Pi .S Ž1 F i F n. are all maximal in R S . Also, by Corollary 4.2, R S is not Henselian. The upshot is that we may now assume that R is a Noetherian, non-Henselian domain and that P1 , . . . , Pn are all maximal ideals. For 1 F i F n, suppose Ui s pi1 , . . . , pi m i 4 . By repeated applications of Lemma 4.1, we can find a finite integral extension domain R* of R such that there are at least m i distinct primes of R* lying over Pi . Now pick any
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p g U1 j ??? j Un . Say p g Ui . Since p is an upper to the maximal ideal Pi , p has the form p s Ž Pi , f p Ž X .. Rw X x, where f p Ž X . is a monic polynomial. Suppose to R* we adjoin all of the roots of f p Ž X . for all p g U1 j ??? j Un , and call the resulting ring R** Žan integral extension domain of R .. Note that in R**, there are at least m i primes lying over Pi Ž1 F i F n. and, for any p g U1 j ??? j Un , the polynomial f p Ž X . factors into a product of linear polynomials in R**w X x. For i s 1, 2, . . . , n, suppose that Q i1 , . . . , Q i r i are all of the distinct Žmaximal. prime ideals of R** lying over Pi , and note that we already have ri G m i . Pick any p g Ui , so that according to our earlier notation, p s pi j for some j Ž1 F j F m i .. By going up, there is an upper qi j to Q i j in R**w X x with qi j l Rw X x s pi j . Since f p Ž X . g pi j : qi j and since f p Ž X . factors completely in R**w X x, we see that there is a linear polynomial X y a i j g qi j . Clearly qi j s Ž Q i j , X y a i j . R**w X x. ŽNote that for each pi j g U1 j ??? j Un , we now have selected a single qi j in R**w X x lying over pi j .. We claim that we can find < R** < s < R < uppers L to 0 in R**w X x such that each such L is contained in each qi j for 1 F i F n and 1 F j F m i , but L is not contained in any other uppers to any Q i j for 1 F i F n, 1 F j F ri except the qi j just mentioned. Suppose that our claim is true. Then for any such L, let K s L l Rw X x. Note that K is an upper to 0 in Rw X x Žand note that < R < such K arise in this way.. We will now show that any such K satisfies our theorem. That is, for 1 F i F n, we will show that UŽ Pi , K . s Ui . First, pick p g Ui and recall that p s pi j for some j Ž1 F j F m i .. Since L ; qi j and qi j l Rw X x s pi j , we see that K ; pi j s p, as desired. Conversely, suppose that p is an upper to Pi and K ; p. Since L lies over K, by going up, there is a prime q in R**w X x lying over p, with L ; q. Of course, q is an upper to some prime in R** lying over Pi . That is, q is an upper to some prime in R** lying over Pi . That is, q is an upper to one of Q i j for 1 F j F ri . By our Žas yet to be proved. claim about L, q must be one of qi j for some j Ž1 F j F m i .. Thus p s q l Rw X x s qi j l Rw X x s pi j g Ui . We now see that to complete the proof of our theorem, we need only prove the claim made at the start of this paragraph. Before proceeding, we will simplify notation. At present, we have a Noetherian domain R** and a finite set of maximal ideals Q i j for 1 F i F n and 1 F j F ri . For some of the Q i j Žnamely, when 1 F j F m i . we have an upper qi j s Ž Q i j , x y a i j . R**w X x, while for other Q i j Žnamely, when m i - j F ri . we do not have any upper to Q i j . We seek < R** < uppers L to 0 such that the set q ¬ q is an upper to any Q i j Ž1 F i F n, 1 F j F ri . and L ; q4 s qi j ¬ 1 F i F n, 1 F j F m i 4 . Changing notation, we will replace R** by T and Q i j Ž1 F i F n, 1 F j F ri . with maximal ideals M1 , . . . , Mh and N1 , . . . , Nk . For 1 F i F h, we consider an upper m i s Ž Mi , X y a i .T w X x Žand we do not consider any upper to any of N1 , . . . , Nk .. We seek < T < uppers L to 0 such that the set q ¬ q is an upper to one of
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M1 , . . . , Mh , N1 , . . . , Nk , and L ; q4 s m1 , . . . , m h 4 . ŽWe will first construct one such L and show how to vary the construction to find < T < of them.. By the Chinese remainder theorem, pick 0 / b g T with b ' 1 mod Mi for 1 F i F h and b ' 0 mod Nj for 1 F j F k, and c g T with c ' a i mod Mi for 1 F i F h and c ' 1 mod Nj for 1 F j F k. Now let Q1 , . . . , Q w , Q wq1 , . . . , Q z be all of the height 1 primes of T which contain b, ordered so that the first w of these also contains c, while the remaining ones do not contain c. For 1 F s F w, we claim that Q s does not contain any of M1 , . . . , Mh , N1 , . . . , Nk . Otherwise, since Q s has height 1, we would have Q s equals some Mi or some Nj . This cannot be, since each of Q1 , . . . , Q w contains both b and c, while no Mi contains b and no Nj contains c. This proves the claim. It now follows from the prime avoidance lemma that there is an element 0 / g g M1 l ??? l Mh l N1 l ??? l Nk l Q wq 1 l ??? l Q z with g f Q1 j ??? j Q w . Note we now have that no height 1 prime of T contains both b and c q g. Note also that c q g is congruent to c modulo each of M1 , . . . , Mh , N1 , . . . , Nk . Let L g be the kernel of the map T w X x ª T wŽ c q g .rb x. Clearly L g is an upper to 0 in T w X x, and we claim it has the property we desire. To see this, we first show that L g is the radical of Ž bX y Ž c q g ..T w X x. For this, consider a height 1 prime L9 of T w X x which contains bX y Ž c q g .. If L9 l T is not 0, then it has height 1 and L9 s Ž L9 l T .T w X x. This implies that b and c q g are both contained in the height 1 prime L9 l T. As we know that cannot be, we see that L9 l T s 0. Thus L9 is an upper to 0 containing bX y Ž c q g .. However, the only upper to 0 containing bX y Ž c q g . is easily seen to be L g , so that L9 s L g . Therefore L g is the radical of Ž bX y Ž c q g ..T w X x. We now show that L g behaves as desired. First, consider one of our uppers m i s Ž Mi , X y a i .T w X x. Since b ' 1 mod Mi and c q g ' c ' a i mod Mi , we see that bX y Ž c q g . g m i . As L g is the radical of Ž bX y Ž c q g ..T w X x, we see that L g ; m i , as desired. Furthermore, since b ' 1 mod Mi , when taken modulo Mi , bX y Ž c q g . is not zero, but does have degree 1. It follows immediately that bX y Ž c q g . is contained in at most one upper to Mi . Therefore, q ¬ q is an upper to any of M1 , . . . , Mh and L g ; q4 s m1 , . . . , m h 4 . In order to complete the proof that L g behaves as desired, we must now show that q ¬ q is an upper to any of N1 , . . . , Nk and L g ; q4 is empty. Suppose to the contrary that for some j with 1 F j F k, there is an upper q to Nj with L g ; q. Thus bX y Ž c q g . g q. However, the construction of b gives b g Nj , so that bX g Nj T w X x ; q. Thus, we have c q g g q l T s Nj . This contradicts that c q g ' c ' 1 mod Nj . We have now found an upper L s L g to 0 of the desired sort. It remains only to show that there are < T < such L. Since every ideal in T w X x is finitely generated, the number of such L cannot exceed < T <, and so we must show
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there exist at least < T < such L. L g was defined in terms of an element g in Ž M1 l ??? l Mh l N1 l ??? l Nk l Q wq1 l ??? l Q z . y Ž Q1 j ??? j Q w .. A new choice of g gives rise to a new L g . Pick 0 / e g Q1 l ??? l Q w . We see that as d varies through T, Ž1 q ed . g gives rise to < T < different possible replacements for g, and we are done. The proof of Lemma 2.2 requires the next lemma. Ž4.3. LEMMA. Let Ž R, M . be a quasilocal domain. Let K be an upper to 0 in Rw X x with K MRw X x. Let Rw X xrK , Rw u x Ž for some u algebraic o¨ er R . and let T be the integral closure of R in Rw u x. Let H be a prime ideal in Rw u x with H l R s M. If N s H l T, then TN s Ž Rw u x.H . Proof. Since H in Rw u x lies over M in R, N in T lies over M in R. Let L be the kernel of the map T w X x ª T w u x s Rw u x. Obviously L l Rw X x s K, and so since K MRw X x, clearly L NT w X x. Let S s T y N, so that TN s TS . Of course LS NS TS w X x. ŽNote also that u is in the quotient field of T wK, Sect. 1-6, Exercise 35x.. Since TS is integrally closed in Ž Rw u x.S s TS w u x and since TS has NS as its unique maximal ideal, the u, uy1 lemma wK, Sect. 1-6, Exercise 31x shows that either u g TS or uy1 g TS or LS : NS TS w X x. However, we already know that the last case does not hold. Suppose that u f TS . ŽWe will get a contradiction.. Then we see that uy1 is a nonunit in TS , and so uy1 g NS . Therefore, no prime in TS w u x can lie over NS Žsince uy1 is a unit in TS w u x.. However, since H is a prime in Rw u x s T w u x and H l T s N, we see that HS is a prime in ŽT w u x.S s TS w u x lying over NS . This contradiction shows that u g TS , and so TS s TS w u x s Ž Rw u x.S . As we already know HS is a prime in Ž Rw u x.S lying over NS , we see that HS s NS . It follows that H is the unique ideal of Rw u x maximal with respect to being disjoint from S. This implies that Ž Rw u x.S s Ž Rw u x.H . Thus TN s TS s Ž Rw u x.H , as desired. We give the proof of Lemma 2.2. Proof of Lemma 2.2. Suppose that Rw X xrL , Rw u x Žfor u an element algebraic over R .. Let H be the image under this isomorphism of q2rL, so that H is a prime of Rw u x lying over Q2 . Let T be the integral closure of R in Rw u x and let N s H l T. By Lemma 4.3, TN s Ž Rw u x.H . Since H lies over Q2 , N lies over Q2 in the integral extension domain T of the integrally closed domain R. Since Q1 ; Q2 , by the well known going down theorem, there is a prime P of T with P ; N and P l R s Q1. Since TN s Ž Rw u x.H , PN s Q H for some prime Q of Rw u x with Q ; H and Q l R s Q1. If q1rL is the inverse image of Q under Rw X xrL , Rw u x, then q1 l R s Q1 and L ; q1 ; q2 Žthis last since q1rL ; q2rL.. It only remains to show that q1 is an upper
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to Q1 in Rw X x. If not, then we must have q1 s Q1 Rw X x, in which case L ; q1 s Q1 Rw X x ; Q2 Rw X x. This contradicts the hypothesis and completes the proof. We give the proof of Theorem 2.6. Proof of Theorem 2.6. In this and the next paragraph, we lift the problem to R9w X x s D9w X x. Let Q1 , . . . , Q s , Q sq1 , . . . , Q m be all of the primes of D9 which lie over any of P1 , . . . , Pn , ordered so that Q1 , . . . , QS are the minimal members of the set Q1 , . . . , Q m 4 . We note that D9 is not Henselian. We now construct finite sets Vj of uppers to Q j Ž1 F j F m., in such a way that GU and minimality hold. First, we construct Vj for 1 F j F s Žcorresponding to the minimal members of Q1 , . . . , Q m 4.. Suppose 1 F j F s, and Q j l R s Pi . Let Vj s q ¬ q is an upper to Q j in D9w X x and q l Rw X x g Ui 4 . We now augment V1 , . . . , Vs to V1 , . . . , Vs , Vsq1 , . . . , Vm . We do this as follows. For s q 1 F j F m, if Q j l R s Pi , then let Vj s q ¬ q is an upper to Q j with q l Rw X x g Ui , and there is some k Ž1 F k F s . and some q9 g Vk such that q9 ; q4 . We leave the verification that GU and minimality hold to the reader. By Theorem 2.1 applied to the pairwise incomparable primes Q1 , . . . , Q s , there are < D9 < s < R < uppers L to 0 in D9w X x, such that for 1 F j F s, UŽ Q j , L. s Vj . By Theorem 2.5, we also know that for s q 1 F j F m, either UŽ Q j , L. s Vj or L ; Q j 9w X x. For each L as before, let K s L l Rw X x. We claim that these < R < K satisfy the conclusion of our theorem. We first show that such a K is contained in every member of U1 j ??? j Un by showing that it is contained in every minimal member of that set. Thus, let p g Ug be minimal in U1 j ??? j n . Since U1 , . . . , Un satisfy minimality, there is a prime Q j in Spec R9 s Spec D9 lying over Pg with Q j minimal in the set Q1 , . . . , Q m . By our ordering, we must have 1 F j F s. Let q be an upper to Q j with q l Rw X x s p. By definition of Vj Žsince 1 F j F s ., q g Vj . By choice of L Žrecalling K s L l Rw X x., L ; q. Contracting to Rw X x, we see that K ; p, as desired. This shows that every prime in U1 j ??? j Un contains K. We now claim that for 1 F i F n, either UŽ Pi , K . s Ui or K : Pi Rw X x. ŽLater, we will eliminate the second possibility for 1 F i F r.. Suppose K Pi Rw X x. We must show that UŽ Pi , K . s Ui . The previous paragraph shows one containment between these two sets. For the other, let p be an upper to Pi with K ; p. ŽWe must show p g Ui .. As L lies over K, by going up there is a prime q of D9w X x lying over p with L ; q. Of course q is an upper to some Q j lying over Pi . Since K Pi Rw X x, clearly L Q j D9w X x. Therefore, by what Theorem 2.5 already told us about L, we
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know that UŽ Q j , L. s Vj . Since L ; q, q g Vj . It follows that p s q l Rw X x g Ui , as desired. This proves our claim. In order to complete the proof of the theorem, in view of the claim just proved, it will suffice to show that if 1 F i F r Ži.e., if Pi is minimal in P1 , . . . , Pn4., then K Pi Rw X x. Suppose K : Pi Rw X x. By going up, there is a prime of D9w X x which contains L and lies over Pi Rw X x. We easily see that prime has the form Q t D9w X x, where Q t is minimal among Q1 , . . . , Q m . Thus 1 F t F s. However, for such a t, we know that L is only contained in finitely many uppers to Q t Žnamely, those uppers in Vt . and so L Q t D9w X x. This contradiction completes the proof. ACKNOWLEDGMENT The authors wish to thank Roger Wiegand for helpful suggestions concerning presentation.
REFERENCES wHMx R. Heitmann and S. McAdam, Comaximizable primes, Proc. Amer. Math. Soc. 112 Ž1991., 661]669. wHWx W. Heinzer and S. Wiegand, Prime ideals in two-dimensional polynomial rings, Proc. Amer. Math. Soc. 107 Ž1989., 577]586. wKx I. Kaplansky, ‘‘Commutative Rings,’’ University of Chicago Press, Chicago, 1974. wM1x S. McAdam, Uppers in Rw X x, Comm. Algebra 22 Ž1994., 499]528. wM2x S. McAdam, Strongly comaximizable primes, J. Algebra 170 Ž1994., 206]228. wM3x S. McAdam, Finite posets in Uspec Rw X x, J. Algebra 172 Ž1995., 167]183. wMSx S. McAdam and C. Shah, Lifting chains of primes to integral extensions, J. Algebra 185 Ž1996. 162]174. wNx M. Nagata, ‘‘Local Rings,’’ Interscience, New York, 1962. wSex F. Serio, Uppers to zero in intersections of prime ideals, Comm. Algebra 22 Ž1994., 1349]1362. wSh1x C. Shah, Prime ideals lying over zero in polynomial rings, J. Algebra 175 Ž1995., 188]198. wSh2x C. Shah, Affine and projective lines over 1-dimensional semi-local domains, Proc. Amer. Math. Soc., 124 Ž1996. 697]705. wTx H. T. Tang, Gauss’ lemma, Proc. Amer. Math. Soc. 35 Ž1972., 372]376. wWx R. Wiegand, The prime spectrum of a two-dimensional affine domain, J. Pure Appl. Algebra 40 Ž1986., 209]214.
JOURNAL OF ALGEBRA ARTICLE NO.
Substructures of Spec Rw X x Stephen McAdam Department of Mathematics, Colgate Uni¨ ersity, Hamilton, New York 13346
and Chandni Shah Department of Mathematics, Uni¨ ersity of California, Ri¨ erside, California 92521 Communicated by Mel¨ in Hochster Received August 2, 1994
1. INTRODUCTION This work is part of a loosely related series of papers attempting to understand the poset structure of Spec Rw X x, with R a Noetherian ring. An underlying difficulty in this endeavor is that the nature of R itself has very subtle influences on the poset structure of Spec Rw X x. That fact is illustrated in wWx Žin which Wiegand characterizes Spec Zw X x, with Z the integers., and also plays a large role in the present work. Because of this, it appears to be too difficult to understand all of Spec Rw X x except in cases in which Spec R is very simple Žsee wHWx and wSh2x., and Wiegand’s work in wWx seems to be as complicated a case in which understanding all of Spec Rw X x is feasible. Therefore this paper, like wM1x and wM3x before it, only attempts to understand a finite chunk of Spec Rw X x. We begin with three definitions. DEFINITION. If P is a prime ideal in the ring R and if p is a prime ideal in the polynomial ring Rw X x, then we call p an upper to P in Rw X x if p l R s P, but p / PRw X x. ŽSee wK, Sect. 1-5x for basic facts about uppers.. DEFINITION. Let R be a domain, let P be a nonzero prime ideal in R, and let K be an upper to 0 in Rw X x. We use UŽ P, K . to denote the set p ¬ p is an upper to P in Rw X x, and K ; p4 . 184 0021-8693r97 $25.00 Copyright Q 1997 by Academic Press All rights of reproduction in any form reserved.
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DEFINITION. Let 0 ; P ; M be prime ideals in the domain R. Let U be a finite set of uppers to P in Rw X x, and assume that for each p g U, p MRw X x. We will say that M overwhelms U if for every upper K to 0 in Rw X x with UŽ P, K . s U, we have K ; MRw X x. The question raised in this paper is Žfor R a Noetherian but not a local Henselian domain. what conditions on R, P, M, and U are required to assure that U is not overwhelmed by M ? Let us briefly outline what considerations lead us to that question. We need a fourth definition. DEFINITION. Let P1 , . . . , Pn be distinct nonzero prime ideals in the domain R and, for 1 F i F n, let Ui be a finite, possibly empty, set of uppers to Pi in Rw X x. We will say that U1 , . . . , Un are algebraically discernible if there is an upper K to 0 in Rw X x such that for 1 F i F n, UŽ Pi , K . s Ui . The question we originally asked was what sets of uppers U1 , . . . , Un are algebraically discernible? We felt that if a satisfactory answer to that question was available, then we could strengthen the results reported in wSex, wSh1x, wM1x, and wM3x. Exploring this original question led us down some unexpected paths. Let us describe that journey in the easiest possible setting Žwhich nonetheless encompasses all the main ideas of the situation.. Let Ž R, M . be an integrally closed, local non-Henselian domain and let P be a prime ideal of R with 0 / P / M. Let U and V be finite sets of uppers to P and M, respectively, such that for each p g U, p MRw X x. We hoped to show that U, V are algebraically discernible. We first proved the following theorem. THEOREM A. There are infinitely many uppers K to 0 in Rw X x such that UŽ P, K . s U. That was a good start. We only needed to find one such K which also satisfied UŽ M, K . s V. However, we then showed that this was impossible for all but one choice of V. THEOREM B. With notation as before, if U, V are algebraically discernible, then V s q ¬ q is an upper to M in Rw X x and there is a p g U with p ; q4 . Remark. Since for p g U we are assuming p MRw X x, it follows that p can be contained in at most finitely many uppers to M. Thus, since U is finite, the set V of Theorem B is also finite Žwhether or not U, V are algebraically discernible .. Somewhat naively, we then hoped to show for the special choice of V in Theorem B, that U, V are algebraically discernible. Instead, we discovered that the best we could do was the following theorem.
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THEOREM C. With notation as before, if V s q ¬ q is an upper to M in Rw X x and there is a p g U with p ; q4 , then U, V are algebraically discernible if and only if M does not o¨ erwhelm U. Theorem C is interesting. It says that if K is an upper to 0 with UŽ P, K . s U, then either UŽ M, K . s V Žas originally hoped. or K : MRw X x. ŽIn the latter case UŽ M, K . consists of all uppers to M, and so is as far from equaling V as possible.. We set about looking for an example in which M overwhelmed U. Some examples appear in Section 3. The easiest to describe Žbut the hardest to prove. is as follows. With ŽT, N . a non-Henselian Discrete Valuation Ring ŽDVR., let R s T ww Y xx, P s Ž Y . R, and M s Ž N, Y . R. If U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 , then M overwhelms U. ŽSee Example 3.12 and wM2, Ž3.6.x.. Thus, our original desire to understand when U, V are algebraically discernible was frustrated by the fact that M might overwhelm U, and so our goal became Žand still is. to determine under what circumstances M overwhelms U. ŽThe examples in Section 3 show that the question of when M overwhelms U is rather subtle. Those examples do not hint at any clear pattern.. In Section 2, we discuss Theorems A, B, and C, though in more generality than stated here. ŽHowever, we defer some of the more technical arguments until Section 4.. In Section 3 we give examples of situations in which M does and does not overwhelm U. 2. ALGEBRAICALLY DISCERNIBLE UPPERS Notation. Throughout this section, R will be a domain and P1 , . . . , Pn will be distinct nonzero prime ideals of R. For 1 F i F n, Ui will be a finite set of uppers to Pi in Rw X x. While we are primarily interested in the case that R is Noetherian, the ideas are easily extended to the case that R lies between some Noetherian domain D and the integral closure of D. R will never be a field. Also, we will assume that R is not a quasilocal Henselian domain. Our first theorem shows that when P1 , . . . , Pn are pairwise incomparable, U1 , . . . , Un are always algebraically discernible. ŽAs the proof of Theorem 2.1 is highly technical, and contributes little to the understanding of the main ideas of this paper, we defer the proof until Section 4.. Ž2.1. THEOREM. Suppose there is a Noetherian domain D with integral closure D9 such that D : R : D9. Suppose R is not Henselian. If P1 , . . . , Pn are pairwise incomparable, then U1 , . . . , Un are algebraically discernible. In fact, there are < R < uppers K to 0 in Rw X x such that for 1 F i F n, UŽ Pi , K . s Ui . Ž Here, < R < denotes the cardinality of R..
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Remarks. Ža. Theorem A of the Introduction is a special case of Theorem 2.1. Žb. In the case that R s Z, the integers, Theorem 2.1 also follows from Wiegand’s characterization of Zw X x in wWx. The question of which U1 , . . . , Un are algebraically discernible, is similar to the question of which rings satisfy Wiegand’s final axiom characterizing Spec Zw X x. We now turn to the general case. The next lemma is crucial. Ž2.2. LEMMA. Let 0 ; Q1 ; Q2 be prime ideals in an integrally closed domain R. Suppose that L ; q2 are uppers in Rw X x to 0 and Q2 , respecti¨ ely. Suppose that L Q2 Rw X x. Then there is an upper q1 to Q1 with L ; q1 ; q2 . We defer the proof of Lemma 2.2 until Section 4. Ž2.3. PROPOSITION. In the Ž arbitrary. domain R, if U1 , . . . , Un are algebraically discernible, then the following two conditions must hold. GU: If p ; q, where p g U1 j ??? j Un and q is an upper to Pi Žsome i, 1 F i F n., then q g Ui . Minimality: Suppose that p g Ui Žsome i, 1 F i F n. and p is minimal in the set U1 j ??? j Un . Then Žwith R9 the integral closure of R . there is a Q g Spec R9 with Q l R s Pi such that Q is minimal in the set Q9 g Spec R9 ¬ Q9 l R g P1 , . . . , Pn44 . Proof. To prove GU, suppose that p ; q with p g Uj Žsome j, 1 F j F . n and with q an upper to Pi Žsome i, 1 F i F n.. As we are assuming that U1 , . . . , Un are algebraically discernible, there is an upper K to 0 in Rw x x with UŽ Ph , K . s Uh for 1 F h F n. Since p g Uj , K ; p ; q. Thus q g UŽ Pi , K . s Ui . This proves GU. As for minimality, let p g Ui and suppose that p is minimal in the set U1 j ??? j Un . Since p g Ui s UŽ Pi , K ., K ; p. By lying over, there is a prime L in R9w X x with L l Rw X x s K Žso that L is an upper to 0 in R9w X x., and by going up, there is a prime q of R9w X x with L ; q and with q l Rw X x s p. Let q l R9 s Q. Thus Q l R s Pi and q is an upper to Q in R9w X x. We will show that Q is minimal in the set Q9 g Spec R9 ¬ Q9 l R g P1 , . . . , Pn44 . If not, then for some 1 F j F n there is some Q9 g Spec R9 with Q9 ; Q and Q9 l R s Pj . Since K is contained in only finitely many uppers to Pi Žnamely, those uppers in Ui ., clearly K Pi Rw X x. It follows that L QR9w X x. We have 0 ; Q9 ; Q, L ; q, and L QR9w X x. By Lemma 2.2, there is an upper q9 to Q9 in R9w X x with L ; q9 ; q. Intersecting with Rw X x Žand letting p9 s q9 l Rw X x., we have K ; p9 ; p. Since p9 is an upper to Pj , we have p9 g UŽ Pj , K . s Uj . This contradicts that p is minimal in U1 j ??? j Un , and so completes the proof.
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Ž2.4. LEMMA. Let R be integrally closed, and suppose that both GU and minimality hold. Also suppose that P1 , . . . , Pr are the minimal members of P1 , . . . , Pn4 . Then for r q 1 F i F n, Ui s q ¬ q is an upper to Pi and there exists a p g U1 j ??? j Ur with p ; q4 . Proof. Suppose that q is an upper to Pi Ž r q 1 F i F n. and there exists a p g U1 j ??? j Ur with p ; q. By GU, q g Ui . Conversely, suppose that q g Ui . Surely there is a p which is minimal in U1 j ??? j Un with p : q. Suppose p g Uj , so that p is an upper to Pj . Since minimality holds and since R is integrally closed, we see that Pj is minimal in P1 , . . . , Pn4 . Thus 1 F j F r. This shows that p g U1 j ??? j Ur , as desired. Since i ) r G j, it also shows that p / q, so that p ; q, as desired. Remark. Theorem B of the Introduction is a special case of Lemma 2.4 Žcombined with Proposition 2.3.. The converse of Proposition 2.3 does not hold, as examples in the next section show. The best we can get is a partial converse. We consider that first in the integrally closed case. Ž2.5. THEOREM. Let R be integrally closed and suppose that U1 , . . . , Un satisfy GU and minimality. Suppose that P1 , . . . , Pr are the minimal members of P1 , . . . , Pn4 . Also, suppose that K is an upper to 0 in Rw X x such that for 1 F i F r, UŽ Pi , K . s Ui . Ž By Theorem 2.1, if R is Noetherian and nonHenselian, then < R < such K exist.. Then for r q 1 F j F n, either UŽ Pj , K . s Uj or K ; Pj Rw X x. Proof. Suppose K Pj Rw X x Žfor some j, r q 1 F j F n.. Let UŽ Pj , K . s Vj . ŽWe will show that Vj s Uj .. Since the intersection of any infinite set of uppers to Pj is Pj Rw X x Žwhich does not contain K ., we see that Vj is finite. Consider the primes P1 , . . . , Pr , Pj and the corresponding finite sets of uppers U1 , . . . , Ur , Vj . Since UŽ Pi , K . s Ui for 1 F i F r and UŽ Pj , K . s Vj , we see that U1 , . . . , Ur , Vj are algebraically discernible. By Proposition 2.3, U1 , . . . , Ur , Vj satisfy GU and minimality. By Lemma 2.4, Vj s q ¬ q is an upper to Pj and there exists a p g U1 j ??? j Ur with p ; q4 . On the other hand, U1 , . . . , Un also satisfy GU and minimality Žby hypothesis., and so Lemma 2.4 shows that Uj equals that same set. Thus UŽ Pj , K . s Vj s Uj . Remark. Theorem C of the Introduction follows easily from Theorem 2.5. We now drop the assumption that R is integrally closed. The next theorem is proved in Section 4. Ž2.6. THEOREM. Let D be a Noetherian domain with integral closure D9 and suppose that D : R : D9. Suppose R is not Henselian. Let P1 , . . . , Pr , Prq1 , . . . , Pn be distinct nonzero prime ideals in R, ordered such that P1 , . . . , Pr are the minimal members of the set P1 , . . . , Pn4 . Suppose that
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GU and minimality hold. Then there exist < R < uppers K to 0 in Rw X x such that for 1 F i F r, UŽ Pi , K . s Ui , while for r q 1 F i F n, either UŽ Pi , K . s Ui or K ; Pi Rw X x.
3. EXAMPLES Notation. In this section, we restrict our attention to the special case that 0 ; P ; M are prime ideals in the Noetherian domain R, and U is a finite set of uppers to P such that for all p g U, p MRw X x. Suppose Žmomentarily. that R is integrally closed. Also suppose that V is a finite set of uppers to M such that U, V satisfy GU and minimality. Then Theorem 2.5 shows that either U, V are algebraically discernible or M overwhelms U. ŽSee the Introduction.. Therefore, in this section we explore through examples when M does or does not overwhelm U. ŽThus, we do not consider V, just U. Also, not all the examples have R integrally closed.. We can detect no pattern in our examples, and so consider the question of when M overwhelms U a subtle one. Let us begin by considering the case that U consists of a single upper to P. Thus, U s p4 , where p is an upper to P and p MRw X x. Our first three examples discuss this case. Ž3.1. EXAMPLE. With notation as before, suppose that RrP is integrally closed. Also suppose U s p4 , with p an upper to P such that p MRw X x. Then M does not overwhelm U. Proof. Let overbars denote reduction modulo P. By wT, Theorem Bx, p is generated by the set of polynomials of minimal degree which are contained in p. Note that each such polynomial is in no other upper to 0 in Rw X x except p. ŽTo see this, note that the analogous statement is easily seen to be true in F w X x, with F the quotient field of R, and so our statement follows from the fact that F w X x is a localization of Rw X x.. Therefore, since p MRw X x, there is a nonconstant polynomial b Ž X . g p y MRw X x with b Ž X . not contained in any upper to 0 in Rw X x except p. Pick f Ž X . g Rw X x such that f Ž X . s b Ž X .. Thus f Ž X . is contained in p y MRw X x, and f Ž X . is not contained in any other upper to P except p. Let K be a prime in Rw X x with f Ž X . g K : p and with K minimal over f Ž X . Rw X x. By the principal ideal theorem, height K s 1. Let K l R s Q. We must have either Q s 0 and K is an upper to 0, or height Q s 1 and K s QRw X x. However, the latter case is impossible, since if it held, we would have f Ž X . g K s QRw X x s Ž K l R . Rw X x : Ž p l R . Rw X x s PRw X x : MRw X x, which is a contradiction. Thus K is an upper to 0. Obviously K is contained in p by construction and is not contained in any
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other upper to P wsince that is true of f Ž X . g K x. Therefore, UŽ P, K . s p4 s U. Now since f Ž X . f MRw X x, we see that K MRw X x. Thus M does not overwhelm U. Ž3.2. Question. With notation as at the start of this section, is it true that R M rPM is integrally closed if and only if for each upper p to P with p MRw X x, M does not overwhelm U s p4 ? Concerning the preceding question, if R M rPM is integrally closed, then M does not overwhelm U s p4 for any upper p to P with p MRw X x. This follows easily from Example 3.1. The next two examples lend credence to the converse, since in the next two examples, R M rPM is not integrally closed and M does overwhelm U s p4 , for carefully chosen p. ŽNonetheless, we suspect that the converse is false and that a counterexample can be constructed which avoids the assumptions of either of the next two examples.. Ž3.3. LEMMA. Let Ž R, M . be a quasilocal domain containing a field of characteristic 0 and let F be the quotient field of R. Suppose that a g F and that a f R and ay1 f R. Let p be the kernel of the map Rw X x ª Rw a x. If f Ž X . is a polynomial contained in p, but not contained in any other upper to 0 in Rw X x except p, then f Ž X . g MRw X x. Proof. Obviously p s Ž X y a . F w X x l Rw X x. Since contraction gives a one-to-one map from the uppers to 0 in F w X x to the uppers to 0 in Rw X x, we see that f Ž X . is not contained in any upper to 0 in F w X x except Ž X y a . F w X x. Thus, in the Unique Factorization Domain ŽUFD. F w X x, the only prime polynomial dividing f Ž X . is X y a , so that f Ž X . s aŽ X y a . k , for some a g F and some integer k G 1. For some h with 0 F h F k, consider the degree k y h coefficient of this polynomial. That coefficient is in R Žsince f Ž X . g Rw X x. and has the form ma a h for some nonzero integer m. We will show that each a a h Ž0 F h F k . is in M. This will show that f Ž X . g MRw X x as desired. Since R contains a field of characteristic 0, we see that m represents a unit in R. Thus a a h is in R for 0 F h F k. Letting h s 0, we see that a g R. Letting h s 1 and recalling that a f R, we see that a is not a unit in R, so that a g M. Now consider an h with 0 - h F k. If a a h f M, then Žsince a a h g R . we have that a a h is a unit of R, so that ay1 s a a hy 1ra a h g R. This contradiction shows that a a h g M for all 0 F h F k, and completes the argument that f Ž X . g MRw X x. Ž3.4. EXAMPLE. Let P be a nonzero prime ideal in the local UFD Ž R, M .. Suppose that RrP is not integrally closed, but does contain a field of characteristic 0. Then there is an upper p to P in Rw X x with p MRw X x, such that M overwhelms U s p4 .
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Proof. Let R s RrP and let R9 be the integral closure of R. Pick any a g R9 y R. ŽA well known and easy exercise shows that ay1 f R.. Let p be the kernel of the map Rw X x ª Rw a x. Since a is integral over R, p contains a monic polynomial. Thus p MRw X x. Let p be the inverse image of p under the map Rw X x ª Rw X x. Now p is easily seen to be an upper to P in Rw X x and p MRw X x. We will show that M overwhelms p4 . Assume that K is an upper to 0 in Rw X x and that UŽ P, K . s p4 . We must show that K ; MRw X x. Since R and hence Rw X x is a UFD, we see that K is principal. Let K s f Ž X . Rw X x. Thus f Ž X . is contained in p, but is not contained in any other upper to P except p Žsince the same is true of K .. Since p is the image of p in Rw X x, we have that f Ž X . is in p, but is not in any other upper to 0 in Rw X x except p. By Lemma 3.3, f Ž X . g MRw X x. Thus f Ž X . g MRw X x, so that K s f Ž X . Rw; MRw X x. Ž3.5. LEMMA. Let P ; Q ; M be prime ideals in the integrally closed Ž not necessarily Noetherian. domain R and let U be a finite set of uppers to P such that for each p g U, p MRw X x. Let W s q ¬ q is an upper to Q and there is a p g U with p ; q4 , V1 s m ¬ m is an upper to M and there is a p g U with p ; m4 , and V2 s m ¬ m is an upper to M and there is a q g W with q ; m4 . If M does not o¨ erwhelm U, then V1 s V2 . Proof. Suppose M does not overwhelm U. Then there is an upper K to 0 in Rw X x with UŽ P, K . s U such that K MRw X x. Since no p g U is contained in MRw X x, V1 is finite. We easily see that U, V1 satisfy GU and minimality. By Theorem 2.5, we see that UŽ M, K . s V1. Obviously K QRw X x. Thus W is finite and we easily see that U, W satisfy GU and minimality. By Theorem 2.5, UŽ Q, K . s W. Since no prime in U is contained in MRw X x, no prime in W is contained in MRw X x. Thus, V2 is finite. As W, V2 satisfy GU and minimality, Theorem 2.5 shows that UŽ M, K . s V2 . Thus V1 s UŽ M, K . s V2 . Ž3.6. EXAMPLE. Let P be a nonzero prime ideal in the integrally closed local domain Ž R, M .. Let T s RrP and N s MrP. Assume that height N G 2. Also assume that in the integral closure of T, N1X and N2X are distinct primes lying over N, with height N1X G 2. ŽThus, T s RrP is not integrally closed.. Pick a g N1X y N2X and let p be the kernel of the map Rw X x ª T w X x ª T w a x. Then p is an upper to P in Rw X x, p MRw X x, and M overwhelms U s p4 . Proof. Since T w a x is a simple integral extension of T s RrP, clearly p is an upper to P in Rw X x, and p contains a monic polynomial, so that p MRw X x. In order to show that M overwhelms U s p4 , we will use Lemma 3.5. Thus we will now find a special prime Q with P ; Q ; M.
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Let N1 s N1X l T w a x and N2 s N2X l T w a x. Note that N1 / N2 and height N1 G 2. Let S be the set of height 1 primes of T w a x which are contained in N1 and let SX s Q9 g S ¬ QX N2 4 . The principal ideal theorem shows that N1 s D Q9 g S4 . Thus N1 : D Q9 g S94 j N2 . If S9 is finite, the prime avoidance lemma shows that N1 is contained in either N2 or in some Žheight 1. prime in S9. As neither of these is true, we see that S9 is infinite. Since T w a x is a finite T-module Žand contained in the quotient field of . T , ŽT :T w a x. s b g T ¬ bT w a x : T 4 / 0. Thus, only finitely many height 1 primes of T w a x can contain ŽT :T w a x.. Therefore, we may select Q9 g S9 such that ŽT :T w a x. Q9. Since Q9 g S9, we have that Q9 ; N1 and Q9 N2 . Since ŽT :T w a x. Q9, an easy exercise shows that Q9 is the unique prime of T w a x lying over Q9 l T. Note that since Q9 ; N1 , Q9 l T ; N. Letting Q be the inverse image in R of Q9 l T, we see that P ; Q ; M. Let W, V1 , and V2 be as in Lemma 3.5. We will show that V1 / V2 . Since the prime N2 of T w a x lies over N in T, under the isomorphism w R X xrp , T w a x, N2 corresponds to a prime mrp, where m is an upper to M in Rw X x. By definition, m g V1. We need only show that m f V2 . Suppose m g V2 . Then there is a q g W with q ; m. Since q g W, we have p ; q. Since qrp ; mrp, under the isomorphism Rw X xrp , T w a x, qrp corresponds to a prime q9 of T w a x contained in N2 . Since Žbeing in W . q is an upper to Q Žwhich taken modulo P is Q9 l T ., we see that q9 in T w a x lies over Q9 l T in T. Since Q9 is the only prime of T w a x lying over Q9 l T, we see that q9 s Q9. This contradicts that Q9 is not contained in N2 . Therefore, V1 / V2 , and so by Lemma 3.5, M overwhelms p4 . This completes our study of the case that U consists of a single upper to P. We now consider more general U. Ž3.7. LEMMA. Let P be a prime ideal in the Noetherian domain R, with P not contained in the Jacobson radical of R. Let g Ž X . be a nonconstant polynomial in Rw X x. Then there is a polynomial hŽ X . g Rw X x such that hŽ X . ' g Ž X . modulo P and hŽ X . cannot be factored into the product of two nonconstant polynomials. Proof. Let N be a maximal ideal of R not containing P, so that P and N 2 are comaximal. By wK, Theorem 77x, N 2 / N. Suppose that g Ž X . s a d X d q ??? qa0 . Pick bd f N, b 0 g N y N 2 , and, for 0 - i - d, bi g N. Now for 0 F i F d, use the Chinese remainder theorem to pick c i ' a i mod P and c i ' bi mod N 2 . Let hŽ X . s c d X d q ??? qc0 . Clearly hŽ X . ' g Ž X . mod P. Also, we clearly have c d f N, c 0 g N y N 2 , and, for 0 - i d, c i g N. The argument used to prove Eisenstein’s criterion shows that hŽ X . cannot be factored into the product of two nonconstant polynomials.
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The next example strengthens both the assumption and the conclusion of Example 3.1. Ž3.8. EXAMPLE. Let 0 ; P ; M be prime ideals in the Noetherian UFD R and suppose that RrP is integrally closed. Also suppose that P is not contained in the Jacobson radical of R. If U is any finite set of uppers to P in Rw X x such that p MRw X x for all p g U, then M does not overwhelm U. Proof. Let overbars denote reduction modulo P. For each p g U, p is an upper to 0 in the integrally closed domain Rw X x, and so Žas in the proof of Example 3.1., wT, Theorem Bx shows there is a nonconstant polynomial b p Ž X . g p y MRw X x with b p Ž X . not contained in any upper to 0 in Rw X x except p. Let b Ž X . s P b p Ž X . over all p g U. Clearly b Ž X . f MRw X x, and the only uppers to 0 in Rw X x which contain b Ž X . are the p for p g U. Pick g Ž X . g Rw X x with g Ž X . s b Ž X .. By Lemma 3.7, pick hŽ X . g Rw X x such that hw X x s g w X x s b Ž X . and hŽ X . cannot be factored into the product of two nonconstant polynomials. Obviously hŽ X . f MRw X x, and the only uppers to P in Rw X x which contain hŽ X . are the uppers p g U Žsince these statements are true modulo P .. Since R is a UFD, we can write hŽ X . s cf Ž X . with c g R and with the greatest common divisor of the coefficients of f Ž X . equaling 1. Thus, the only constant factors if f Ž X . are units in R. Since also cf Ž X . s hŽ X . cannot be factored into the product of two nonconstant polynomials, the same is true of f Ž X .. Therefore, f Ž X . is irreducible in the UFD Rw X x. Since cf Ž X . s hŽ X . f MRw X x, we have f Ž X . f MRw X x and c f M. Thus c f P, and so c is not contained in any upper to P. Since we already know that the set of uppers to P in Rw X x which contain cf Ž X . s hŽ X . is exactly the set U, we see that the set of uppers to P in Rw X x which contain f Ž X . is also the set U. Let K s f Ž X . Rw X x. Since f Ž X . is irreducible Žand nonconstant. in Rw X x, K is an upper to 0 in Rw X x. Since f Ž X . f MRw X x, K MRw X x. Since the set of uppers to P which contain f Ž X . is exactly U, UŽ P, K . s U. This shows that M does not overwhelm U. Before presenting the next example Žour most interesting one., we must introduce a concept. DEFINITION. The nonzero prime ideal P in the Noetherian domain R is called strongly comaximizable if for every positive integer m, there is a finitely generated integral extension domain T of R such that there are exactly m primes in T which lie over P, and those m primes in T are pairwise comaximal.
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Strongly comaximizable primes are studied in wM2x. In particular, wM2, Ž1.4.x says that if P is a prime not contained in the Jacobson radical of the Noetherian domain R, then P is strongly comaximizable. Thus strongly comaximizable primes are plentiful. ŽAlso, since strong comaximizability is easily seen to be preserved under localization, it is easy to construct strongly comaximizable primes in local domains.. While it is easy to find strongly comaximizable primes, finding primes which are not strongly comaximizable is significantly harder. Here is an example of one. Let T be an integrally closed Noetherian domain, let R s T ww Y xx, and let P s Ž Y . R. P is not strongly comaximizable wM2, Ž3.6.x. We need the following results from wM2x. Ž3.9. LEMMA. Suppose that the prime P is not a strongly comaximizable prime ideal in the Noetherian domain R. Then the following holds. If N is a maximal ideal in R9, the integral closure of R, and if P : N l R, then there is a prime ideal Q of R9 with Q : N and Q l R s P. Proof. This follows from wM2, Ž1.12.Ža.x. ŽWe leave it to the reader to add the little argument necessary to compensate for the fact that R9 might not be finitely generated over R, the key ingredient of that argument being the fact that only finitely many primes of R9 lie over P wN, Ž33.10.x.. Ž3.10. LEMMA. Let P be a prime ideal in the Noetherian domain R. The following are equi¨ alent. Ži. P is strongly comaximizable. Žii. There is a finitely generated integral extension domain T of R in which all of the primes which lie o¨ er P can be ordered P1 , . . . , Pr , Prq1 , . . . , Pn , with 0 - r - n, in such a way that P1 l ??? l Pr and Prq1 l ??? l Pn are comaximal. Proof. This is wM2, Ž1.5.Ži. m Žiii.x. The next lemma is similar to Lemma 2.4, but instead of assuming that R is integrally closed, we assume that P is not strongly comaximizable. Ž3.11. LEMMA. Suppose that P is not a strongly comaximizable prime in the local domain Ž R, M ., with P / M. Suppose that U and V are a finite sets of uppers to P and M, respecti¨ ely, such that U, V satisfy GU and minimality. Then V s q ¬ q is an upper to M, and there is a p g U with p ; q4 . Proof. If p g U and if p ; q with q an upper to M, then q g V by GU. Conversely, suppose that q g V. We must show there is a p g U with p ; q. If not, then q is minimal in U j V, and so by minimality, in the integral closure R9 of R, there is a prime N lying over M such that N does not contain any prime of R9 lying over P. This contradicts Lemma 3.9.
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Ž3.12. EXAMPLE. With notation as at the start of this section, and with Ž R, M . local, suppose that P is a prime of R which is not strongly comaximizable. Let U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 . Then M overwhelms U. Proof. Let K be an upper to 0 in Rw X x such that UŽ P, K . s U. In order to show that M overwhelms U, we must show that K ; MRw X x. Suppose that K MRw X x Žand we will derive a contradiction.. Let UŽ M, K . s V. ŽSince the intersection of any infinite set of uppers to M is MRw X x, we see that V is finite.. By definition, we have that U, V are algebraically discernible. By Proposition 2.3, U, V satisfy GU and minimality. Using Lemma 3.11, it is now trivial to see that we must have UŽ M, K . s V s Ž M, X . Rw X x, Ž M, X q 1. Rw X x4 . Now Rw X xrK , Rw u x for some u algebraic over R. Since we know exactly which two uppers to M contain K, we see that in Rw u x there are exactly two primes lying over M, namely, Ž M, u. Rw u x and Ž M, u q 1. Rw u x. Similarly, the only two primes lying over P are Ž P, u. Rw u x and Ž P, u q 1. Rw u x. Let T be the integral closure of R in Rw u x and let L be the kernel of w T X x ª T w u x s Rw u x. Let Q s Ž P, u. Rw u x l T and let qrL be the preimage of Ž P, u. Rw u x under the isomorphism T w X xrL , Rw u x. Now q is an upper to Q in T w X x, and since u g Ž P, u. Rw u x, we see that X g q. Thus q s Ž Q, X .T w X x. Similarly, if N s Ž M, u. Rw u x l T, then the pre-image of Ž M, u. Rw u x in T w X xrL is Ž N, X .T w X xrL. Obviously Q : N. We now claim that N is the only maximal ideal of T which contains Q. Suppose that N9 / N is another maximal ideal of T with Q : N9. Then since L ; Ž Q, X .T w X x : Ž N 9, X .T w X x, under T w X xrL , R w u x, Ž N9, X .T w X xrL would correspond to a prime in Rw u x which lies over M in R and which contains u. This is impossible since we already know that the only prime in Rw u x which lies over M and contains u is Ž M, u. Rw u x Žwhich corresponds to Ž N, X .T w X xrL.. This proves the claim that N is the only maximal ideal of T containing Q. We next claim that Q is the only prime of T which lies over P and is contained in N. Note that Lemma 4.3 shows TN s Ž Rw u x.Ž M , u. Rw ux. Since Ž P, u. Rw u x is the only prime in Rw u x which is contained in Ž M, u. Rw u x and lies over P, we see that there is only one prime in T which is contained in N and lies over P. Clearly Q is that prime. We claim that Ž P, u q 1. Rw u x l T / Q. If equality held, then we would have Q ; Ž M, u q 1. Rw u x l T. Since we already know that N is the only maximal ideal of T containing Q, we have Ž M, u q 1. Rw u x l T s N. By Lemma 4.3 Žused twice. we get that Ž Rw u x.Ž M, u. Rw ux s Ž Rw u x.Ž M , uq1. Rw ux. The unique maximal ideal of this ring is seen to contain both u and u q 1, a contradiction.
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We wish to use Lemma 3.10, which refers to a finitely generated extension of R. Since T might not be finitely generated over R, in this paragraph and the next, we will cut T down a bit. For now, we claim that only finitely many primes in T lie over P. Since u is in the quotient field of T, write u s arb with a and b in T. Then T lies between Rw a, b x and its integral closure. Since Rw a, b x is Noetherian and integral over R, only finitely many primes of Rw a, b x can lie over P, and so the same is true of T, by wN, Ž33.10.x. Let Q s Q1 , Q2 , . . . , Q h be all of the primes of T lying over P. Since we have already seen that Ž P, u q 1. Rw u x l T / Q s Q1 , we see that h G 2. We have also previously seen that none of Q2 , . . . , Q h is contained in N. Pick an element b g Ž Q2 l ??? l Q h . y N. The primes of Rw b x which lie over P are Q1 l Rw b x, Q2 l Rw b x, . . . , Q h l Rw b x. The first of these is contained in N l Rw b x and is not contained in any other maximal ideal of Rw b x Žsince Q1 is contained in N but not in any other maximal ideal of T .. On the other hand, none of Q2 l Rw b x, . . . , Q h l Rw b x is contained in N l Rw b x Žsince b f N .. Therefore, we see that the two ideals Q1 l Rw b x and F Q i l Rw b x ¬ 2 F i F h4 are comaximal. Now Lemma 3.10 shows that P is strongly comaximizable. This contradicts the hypothesis and completes the proof. Remark. Let T be a non-Henselain DVR, let R s T ww Y xx, and let P s Ž Y . R. By wM2, Ž3.6.x, P is not strongly comaximizable. However, RrP , T is integrally closed. Thus Example 3.12 shows that the conclusion of Example 3.1 cannot be extended to a set U consisting of two uppers to P. Furthermore, since T is a DVR, R is a UFD. Thus Example 3.12 shows that in Example 3.8, the assumption that P is not contained in the Jacobson radical is important Žalthough we wonder if it can be weakened to assuming that P is strongly comaximizable.. Before giving our final example, we need another concept. DEFINITION. The nonzero prime ideal P in the Noetherian domain R is called integrally discerning if for every finite set U of uppers to P such that every p g U contains a monic polynomial, there is an upper K to 0 in Rw X x such that K contains a monic polynomial and UŽ P, K . s U. Integrally discerning primes are studied in wMSx, which shows, among other things, that if P is not contained in the Jacobson radical of R and if RrP is integrally closed, then P is integrally discerning. Thus integrally discerning primes are plentiful. Ž3.13. EXAMPLE. With notation as at the start of this section, suppose that P is integrally discerning. If every p g U contains a monic polynomial, then M does not overwhelm U.
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Proof. Since P is integrally discerning, there is an upper K to 0 in Rw X x such that UŽ P, K . s U and K contains a monic polynomial. Obviously K MRw X x, and so M does not overwhelm U.
4. TECHNICAL ARGUMENTS In order to prove Theorem 2.1, we need the following easy consequence of wHW, Ž1.1.x. ŽFor a strengthening of this, see wHMx.. Ž4.1. LEMMA. Let P be a nonzero prime ideal in the Noetherian domain R. Furthermore, suppose that either R is not Henselian or R is Henselian but P is not the maximal ideal of R. Then for any positi¨ e integer m, there is a finitely generated integral extension domain of R in which at least m primes lie o¨ er P. The next corollary follows pleasantly from Lemma 4.1. Ž4.2. COROLLARY. Let R be a Noetherian domain and let R S be a localization of R. If R S is a local Henselian domain, but not the quotient field of R, then R S s R Ž and R is a local Henselian domain.. Proof of Theorem 2.1. We first reduce to the case that R is Noetherian. Since Dw X x : Rw X x : D9w X x and D is Noetherian, any prime in Dw X x has at most finitely many primes of Rw X x lying over it by wN, Ž33.10.x. It is therefore an easy exercise to find a ring T, finitely generated over D Žand hence Noetherian., with D : T : R, such that for 1 F i F n, Pi is the only prime of R lying over Pi l T, and for each p g U1 j ??? j Un , p is the only prime of Rw X x lying over p l T w X x. Let Wi s p l T w X x ¬ p g Ui 4 . Suppose that we can find < T < s < R < uppers L to 0 in T w x x such that for each i, UŽ Pi l T, L. s Wi . Then it is easily seen that for any such L, if K is a prime in Rw X x lying over L, then K is an upper to 0 and UŽ Pi , K . s Ui . ŽSince D : T : R : D9 and R is not Henselian, T is not Henselian.. Therefore, we may replace R by T, Pi by the Pi l T, and Ui by Wi . In other words, we may simply assume that R is Noetherian. Let S s R y Ž P1 j ??? j Pn .. We leave to the reader the easy verification of the fact that we can replace R by R S , Pi by the Ž Pi .S , and Ui by pS ¬ p g Ui 4 Žhere simply using that uppers to 0 in R S w X x have the form K S , where K is an upper to 0 in Rw X x.. Since P1 , . . . , Pn are pairwise incomparable, we see that the primes Ž Pi .S Ž1 F i F n. are all maximal in R S . Also, by Corollary 4.2, R S is not Henselian. The upshot is that we may now assume that R is a Noetherian, non-Henselian domain and that P1 , . . . , Pn are all maximal ideals. For 1 F i F n, suppose Ui s pi1 , . . . , pi m i 4 . By repeated applications of Lemma 4.1, we can find a finite integral extension domain R* of R such that there are at least m i distinct primes of R* lying over Pi . Now pick any
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p g U1 j ??? j Un . Say p g Ui . Since p is an upper to the maximal ideal Pi , p has the form p s Ž Pi , f p Ž X .. Rw X x, where f p Ž X . is a monic polynomial. Suppose to R* we adjoin all of the roots of f p Ž X . for all p g U1 j ??? j Un , and call the resulting ring R** Žan integral extension domain of R .. Note that in R**, there are at least m i primes lying over Pi Ž1 F i F n. and, for any p g U1 j ??? j Un , the polynomial f p Ž X . factors into a product of linear polynomials in R**w X x. For i s 1, 2, . . . , n, suppose that Q i1 , . . . , Q i r i are all of the distinct Žmaximal. prime ideals of R** lying over Pi , and note that we already have ri G m i . Pick any p g Ui , so that according to our earlier notation, p s pi j for some j Ž1 F j F m i .. By going up, there is an upper qi j to Q i j in R**w X x with qi j l Rw X x s pi j . Since f p Ž X . g pi j : qi j and since f p Ž X . factors completely in R**w X x, we see that there is a linear polynomial X y a i j g qi j . Clearly qi j s Ž Q i j , X y a i j . R**w X x. ŽNote that for each pi j g U1 j ??? j Un , we now have selected a single qi j in R**w X x lying over pi j .. We claim that we can find < R** < s < R < uppers L to 0 in R**w X x such that each such L is contained in each qi j for 1 F i F n and 1 F j F m i , but L is not contained in any other uppers to any Q i j for 1 F i F n, 1 F j F ri except the qi j just mentioned. Suppose that our claim is true. Then for any such L, let K s L l Rw X x. Note that K is an upper to 0 in Rw X x Žand note that < R < such K arise in this way.. We will now show that any such K satisfies our theorem. That is, for 1 F i F n, we will show that UŽ Pi , K . s Ui . First, pick p g Ui and recall that p s pi j for some j Ž1 F j F m i .. Since L ; qi j and qi j l Rw X x s pi j , we see that K ; pi j s p, as desired. Conversely, suppose that p is an upper to Pi and K ; p. Since L lies over K, by going up, there is a prime q in R**w X x lying over p, with L ; q. Of course, q is an upper to some prime in R** lying over Pi . That is, q is an upper to some prime in R** lying over Pi . That is, q is an upper to one of Q i j for 1 F j F ri . By our Žas yet to be proved. claim about L, q must be one of qi j for some j Ž1 F j F m i .. Thus p s q l Rw X x s qi j l Rw X x s pi j g Ui . We now see that to complete the proof of our theorem, we need only prove the claim made at the start of this paragraph. Before proceeding, we will simplify notation. At present, we have a Noetherian domain R** and a finite set of maximal ideals Q i j for 1 F i F n and 1 F j F ri . For some of the Q i j Žnamely, when 1 F j F m i . we have an upper qi j s Ž Q i j , x y a i j . R**w X x, while for other Q i j Žnamely, when m i - j F ri . we do not have any upper to Q i j . We seek < R** < uppers L to 0 such that the set q ¬ q is an upper to any Q i j Ž1 F i F n, 1 F j F ri . and L ; q4 s qi j ¬ 1 F i F n, 1 F j F m i 4 . Changing notation, we will replace R** by T and Q i j Ž1 F i F n, 1 F j F ri . with maximal ideals M1 , . . . , Mh and N1 , . . . , Nk . For 1 F i F h, we consider an upper m i s Ž Mi , X y a i .T w X x Žand we do not consider any upper to any of N1 , . . . , Nk .. We seek < T < uppers L to 0 such that the set q ¬ q is an upper to one of
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M1 , . . . , Mh , N1 , . . . , Nk , and L ; q4 s m1 , . . . , m h 4 . ŽWe will first construct one such L and show how to vary the construction to find < T < of them.. By the Chinese remainder theorem, pick 0 / b g T with b ' 1 mod Mi for 1 F i F h and b ' 0 mod Nj for 1 F j F k, and c g T with c ' a i mod Mi for 1 F i F h and c ' 1 mod Nj for 1 F j F k. Now let Q1 , . . . , Q w , Q wq1 , . . . , Q z be all of the height 1 primes of T which contain b, ordered so that the first w of these also contains c, while the remaining ones do not contain c. For 1 F s F w, we claim that Q s does not contain any of M1 , . . . , Mh , N1 , . . . , Nk . Otherwise, since Q s has height 1, we would have Q s equals some Mi or some Nj . This cannot be, since each of Q1 , . . . , Q w contains both b and c, while no Mi contains b and no Nj contains c. This proves the claim. It now follows from the prime avoidance lemma that there is an element 0 / g g M1 l ??? l Mh l N1 l ??? l Nk l Q wq 1 l ??? l Q z with g f Q1 j ??? j Q w . Note we now have that no height 1 prime of T contains both b and c q g. Note also that c q g is congruent to c modulo each of M1 , . . . , Mh , N1 , . . . , Nk . Let L g be the kernel of the map T w X x ª T wŽ c q g .rb x. Clearly L g is an upper to 0 in T w X x, and we claim it has the property we desire. To see this, we first show that L g is the radical of Ž bX y Ž c q g ..T w X x. For this, consider a height 1 prime L9 of T w X x which contains bX y Ž c q g .. If L9 l T is not 0, then it has height 1 and L9 s Ž L9 l T .T w X x. This implies that b and c q g are both contained in the height 1 prime L9 l T. As we know that cannot be, we see that L9 l T s 0. Thus L9 is an upper to 0 containing bX y Ž c q g .. However, the only upper to 0 containing bX y Ž c q g . is easily seen to be L g , so that L9 s L g . Therefore L g is the radical of Ž bX y Ž c q g ..T w X x. We now show that L g behaves as desired. First, consider one of our uppers m i s Ž Mi , X y a i .T w X x. Since b ' 1 mod Mi and c q g ' c ' a i mod Mi , we see that bX y Ž c q g . g m i . As L g is the radical of Ž bX y Ž c q g ..T w X x, we see that L g ; m i , as desired. Furthermore, since b ' 1 mod Mi , when taken modulo Mi , bX y Ž c q g . is not zero, but does have degree 1. It follows immediately that bX y Ž c q g . is contained in at most one upper to Mi . Therefore, q ¬ q is an upper to any of M1 , . . . , Mh and L g ; q4 s m1 , . . . , m h 4 . In order to complete the proof that L g behaves as desired, we must now show that q ¬ q is an upper to any of N1 , . . . , Nk and L g ; q4 is empty. Suppose to the contrary that for some j with 1 F j F k, there is an upper q to Nj with L g ; q. Thus bX y Ž c q g . g q. However, the construction of b gives b g Nj , so that bX g Nj T w X x ; q. Thus, we have c q g g q l T s Nj . This contradicts that c q g ' c ' 1 mod Nj . We have now found an upper L s L g to 0 of the desired sort. It remains only to show that there are < T < such L. Since every ideal in T w X x is finitely generated, the number of such L cannot exceed < T <, and so we must show
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there exist at least < T < such L. L g was defined in terms of an element g in Ž M1 l ??? l Mh l N1 l ??? l Nk l Q wq1 l ??? l Q z . y Ž Q1 j ??? j Q w .. A new choice of g gives rise to a new L g . Pick 0 / e g Q1 l ??? l Q w . We see that as d varies through T, Ž1 q ed . g gives rise to < T < different possible replacements for g, and we are done. The proof of Lemma 2.2 requires the next lemma. Ž4.3. LEMMA. Let Ž R, M . be a quasilocal domain. Let K be an upper to 0 in Rw X x with K MRw X x. Let Rw X xrK , Rw u x Ž for some u algebraic o¨ er R . and let T be the integral closure of R in Rw u x. Let H be a prime ideal in Rw u x with H l R s M. If N s H l T, then TN s Ž Rw u x.H . Proof. Since H in Rw u x lies over M in R, N in T lies over M in R. Let L be the kernel of the map T w X x ª T w u x s Rw u x. Obviously L l Rw X x s K, and so since K MRw X x, clearly L NT w X x. Let S s T y N, so that TN s TS . Of course LS NS TS w X x. ŽNote also that u is in the quotient field of T wK, Sect. 1-6, Exercise 35x.. Since TS is integrally closed in Ž Rw u x.S s TS w u x and since TS has NS as its unique maximal ideal, the u, uy1 lemma wK, Sect. 1-6, Exercise 31x shows that either u g TS or uy1 g TS or LS : NS TS w X x. However, we already know that the last case does not hold. Suppose that u f TS . ŽWe will get a contradiction.. Then we see that uy1 is a nonunit in TS , and so uy1 g NS . Therefore, no prime in TS w u x can lie over NS Žsince uy1 is a unit in TS w u x.. However, since H is a prime in Rw u x s T w u x and H l T s N, we see that HS is a prime in ŽT w u x.S s TS w u x lying over NS . This contradiction shows that u g TS , and so TS s TS w u x s Ž Rw u x.S . As we already know HS is a prime in Ž Rw u x.S lying over NS , we see that HS s NS . It follows that H is the unique ideal of Rw u x maximal with respect to being disjoint from S. This implies that Ž Rw u x.S s Ž Rw u x.H . Thus TN s TS s Ž Rw u x.H , as desired. We give the proof of Lemma 2.2. Proof of Lemma 2.2. Suppose that Rw X xrL , Rw u x Žfor u an element algebraic over R .. Let H be the image under this isomorphism of q2rL, so that H is a prime of Rw u x lying over Q2 . Let T be the integral closure of R in Rw u x and let N s H l T. By Lemma 4.3, TN s Ž Rw u x.H . Since H lies over Q2 , N lies over Q2 in the integral extension domain T of the integrally closed domain R. Since Q1 ; Q2 , by the well known going down theorem, there is a prime P of T with P ; N and P l R s Q1. Since TN s Ž Rw u x.H , PN s Q H for some prime Q of Rw u x with Q ; H and Q l R s Q1. If q1rL is the inverse image of Q under Rw X xrL , Rw u x, then q1 l R s Q1 and L ; q1 ; q2 Žthis last since q1rL ; q2rL.. It only remains to show that q1 is an upper
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to Q1 in Rw X x. If not, then we must have q1 s Q1 Rw X x, in which case L ; q1 s Q1 Rw X x ; Q2 Rw X x. This contradicts the hypothesis and completes the proof. We give the proof of Theorem 2.6. Proof of Theorem 2.6. In this and the next paragraph, we lift the problem to R9w X x s D9w X x. Let Q1 , . . . , Q s , Q sq1 , . . . , Q m be all of the primes of D9 which lie over any of P1 , . . . , Pn , ordered so that Q1 , . . . , QS are the minimal members of the set Q1 , . . . , Q m 4 . We note that D9 is not Henselian. We now construct finite sets Vj of uppers to Q j Ž1 F j F m., in such a way that GU and minimality hold. First, we construct Vj for 1 F j F s Žcorresponding to the minimal members of Q1 , . . . , Q m 4.. Suppose 1 F j F s, and Q j l R s Pi . Let Vj s q ¬ q is an upper to Q j in D9w X x and q l Rw X x g Ui 4 . We now augment V1 , . . . , Vs to V1 , . . . , Vs , Vsq1 , . . . , Vm . We do this as follows. For s q 1 F j F m, if Q j l R s Pi , then let Vj s q ¬ q is an upper to Q j with q l Rw X x g Ui , and there is some k Ž1 F k F s . and some q9 g Vk such that q9 ; q4 . We leave the verification that GU and minimality hold to the reader. By Theorem 2.1 applied to the pairwise incomparable primes Q1 , . . . , Q s , there are < D9 < s < R < uppers L to 0 in D9w X x, such that for 1 F j F s, UŽ Q j , L. s Vj . By Theorem 2.5, we also know that for s q 1 F j F m, either UŽ Q j , L. s Vj or L ; Q j 9w X x. For each L as before, let K s L l Rw X x. We claim that these < R < K satisfy the conclusion of our theorem. We first show that such a K is contained in every member of U1 j ??? j Un by showing that it is contained in every minimal member of that set. Thus, let p g Ug be minimal in U1 j ??? j n . Since U1 , . . . , Un satisfy minimality, there is a prime Q j in Spec R9 s Spec D9 lying over Pg with Q j minimal in the set Q1 , . . . , Q m . By our ordering, we must have 1 F j F s. Let q be an upper to Q j with q l Rw X x s p. By definition of Vj Žsince 1 F j F s ., q g Vj . By choice of L Žrecalling K s L l Rw X x., L ; q. Contracting to Rw X x, we see that K ; p, as desired. This shows that every prime in U1 j ??? j Un contains K. We now claim that for 1 F i F n, either UŽ Pi , K . s Ui or K : Pi Rw X x. ŽLater, we will eliminate the second possibility for 1 F i F r.. Suppose K Pi Rw X x. We must show that UŽ Pi , K . s Ui . The previous paragraph shows one containment between these two sets. For the other, let p be an upper to Pi with K ; p. ŽWe must show p g Ui .. As L lies over K, by going up there is a prime q of D9w X x lying over p with L ; q. Of course q is an upper to some Q j lying over Pi . Since K Pi Rw X x, clearly L Q j D9w X x. Therefore, by what Theorem 2.5 already told us about L, we
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know that UŽ Q j , L. s Vj . Since L ; q, q g Vj . It follows that p s q l Rw X x g Ui , as desired. This proves our claim. In order to complete the proof of the theorem, in view of the claim just proved, it will suffice to show that if 1 F i F r Ži.e., if Pi is minimal in P1 , . . . , Pn4., then K Pi Rw X x. Suppose K : Pi Rw X x. By going up, there is a prime of D9w X x which contains L and lies over Pi Rw X x. We easily see that prime has the form Q t D9w X x, where Q t is minimal among Q1 , . . . , Q m . Thus 1 F t F s. However, for such a t, we know that L is only contained in finitely many uppers to Q t Žnamely, those uppers in Vt . and so L Q t D9w X x. This contradiction completes the proof. ACKNOWLEDGMENT The authors wish to thank Roger Wiegand for helpful suggestions concerning presentation.
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