- Email: [email protected]
Lifting Chains of Primes to Integral Extensions
JOURNAL OF ALGEBRA ARTICLE NO.
185, 162]174 Ž1996.
0318
Lifting Chains of Primes to Integral Extensions Stephen McAdam Department of Mathematics, Uni¨ ersity of Texas, Austin, Texas 78712
and Chandni Shah* Department of Mathematics, Uni¨ ersity of California, Ri¨ erside, California 92521 Communicated by Richard G. Swan Received August 31, 1995
1. INTRODUCTION Let R be a Noetherian domain; we consider SpecŽ R . as a partially ordered set under inclusion. The goal of this paper is to raise Žand answer a special case of. the following question: Ž1.1. Question. Let V be a finite subset of SpecŽ R .. Which partially ordered sets arise as the sets of all primes lying over V in some finitely generated integral extension domain T of R? That is, for which partially ordered sets W does there exist an integral extension domain T of R such that W(V
T
s Q g Spec Ž T . N Q l R g V 4 ?
The following unpublished result of Christel Rotthaus and Sylvia Wiegand answers a special case of Ž1.1.. Let R s K w X 1 , . . . , X n x with K a field and the X i indeterminates. Let Pi s Ž X 1 , . . . , X i . R, 0 F i F n, and let V be the partially ordered set 0 s P0 ; P1 ; ??? ; Pn4 . Their result states that a finite partially ordered set W is order isomorphic to V T for some finitely generated integral extension domain T of R if and only if Ž1. W * Current address: Department of Mathematics, Colgate University, Hamilton, NY 13346. E-mail: [email protected] 162 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
LIFTING CHAINS OF PRIMES
163
contains a unique minimal element and Ž2. every maximal saturated chain of elements of W has length n. Note that in the above result, the partially ordered set V is a chain, RrPi is integrally closed for each i, and PirPiy1 is not contained in the Jacobson radical of RrPiy1 , for each i G 1. We show that the Rotthaus]Wiegand hypothesis can be weakened to just those assumptions, leaving their conclusion unchanged. That is, we prove the following. Ž1.2. THEOREM. Let V be a chain 0 s P0 ; P1 ; ??? ; Pn4 of prime ideals in R. For each i, assume that RrPi is integrally closed, and for each i G 1, assume that PirPiy1 is not contained in the Jacobson radical of RrPiy1. Then a finite partially ordered set W is order isomorphic to V T for some finitely generated integral extension domain T of R if and only if Ž1. W contains a unique minimal element and Ž2. e¨ ery maximal saturated chain of elements in W has length n. Furthermore, if such a T exists, then it can be chosen to be a simple integral extension of R. In fact, the result we eventually prove in Ž3.1. is somewhat stronger than Ž1.2.. However, before dealing with it, we must introduce integrally discerning prime ideals.
2. INTEGRALLY DISCERNING PRIMES Ž2.1. DEFINITION. If P is a prime ideal in R and if p is a prime ideal in the polynomial ring Rw X x such that p l R s P but p / PRw X x, then we say that p is an upper to P. If, furthermore, p contains a monic polynomial, then we say p is an integral upper to P. ŽSee wK, Section 1-5x for elementary facts about uppers.. Ž2.2. DEFINITION. Let P, Q be prime ideals in R and let q be an upper to Q in Rw X x. We use UŽ P, q. to denote the set p N p is an upper to P and q ; p4 . Ž2.3. DEFINITION. A nonzero prime ideal P in R is said to be integrally discerning if for every finite nonempty set U of integral uppers to P, there is an integral upper k to 0 such that UŽ P, k. s U. Our first lemma strengthens the definition just given. It will be needed in the next section. Ž2.4. LEMMA. Let P be an integrally discerning prime ideal in R, and let U be a finite nonempty set of integral uppers to P. Then there are < R < integral uppers k to 0 such that UŽ P, k. s U. Ž Here, < R < denotes the cardinality of R, which is infinite since R is not a field..
164
MCADAM AND SHAH
Proof. Since P is integrally discerning, there is an integral upper k to 0 such that UŽ P, k. s U. Now Rw X xrk is a simple integral extension of R, and so it can be written in the form Rw u x for some u integral over R. Here k is the kernel of Rw X x ª Rw u x. For every b g P, let k b be the kernel of Rw X x ª Rw u q b x. Note that k b s g Ž X y b . N g Ž X . g k4 . Also note that g Ž X . and g Ž X y b . differ by a polynomial in PRw X x. It easily follows that UŽ P, k b . s UŽ P, k. s U. Since Ž u q b . is a root of every polynomial of k b , it follows that for at most finitely many c g P, we have k c s k b . On the other hand, < P < is easily seen to equal < R <, and so there are < R < distinct uppers k b to 0, b g P, such that UŽ P, k. s U. In order to understand integrally discerning primes, we need two more definitions. We follow each of them with some relevant facts. Ž2.5. DEFINITION. A nonzero prime ideal P in R is called strongly comaximizable if for every m G 2, there is a finitely generated integral extension domain T of R in which there are exactly m primes lying over P, and those m primes are pairwise comaximal. Ž2.6. THEOREM wM3, Ž1.4.x. If P is a prime ideal in R and if P is not contained in the Jacobson radical of R, then P is strongly comaximizable. Ž2.7. THEOREM wM3, Ž1.5. Ži. m Žii.x. Let P be a prime ideal in R, and let n G 2 be a fixed integer. Then P is strongly comaximizable if and only if there is a finitely generated integral extension domain T of R in which there are exactly n primes lying o¨ er P, and those n primes are pairwise comaximal. Ž2.8. DEFINITION. The domain R is said to be pseudo-integrally closed if for every integral upper k to 0 in Rw X x, there is a monic polynomial f Ž X . such that k s Rad f Ž X . Rw X x. We recall a well-known fact. Ž2.9. LEMMA wM3, Ž2.4.x. If R is an integrally closed domain, and if k is an integral upper to 0 in Rw X x, then k is generated by a monic polynomial. Ž2.10.. Remarks. Ža. If R is integrally closed, then it is pseudointegrally closed. This is immediate from Ž2.9.. Žb. If P is a prime ideal in R, then RrP is pseudo-integrally closed if and only if for every integral upper p to P, there is a monic polynomial f Ž X . g Rw X x such that p s RadŽ P, f Ž X .. Rw X x. This follows easily from the fact that if p is a prime ideal of Rw X x with PRw X x : p, and if bars denote reduction modulo P, then p is an integral upper to P in Rw X x if and only if p is an integral upper to 0 in Rw X x. Žc. In Ž2.16., we show that if R contains a field of characteristic 0, then R is integrally closed if and only if it is pseudo-integrally closed.
LIFTING CHAINS OF PRIMES
165
Žd. In Ž2.22., we give an example of a domain which is pseudointegrally closed, but not integrally closed. Že. In Ž2.21., we show that if R is pseudo-integrally closed, then it satisfies the famous Cohen]Seidenberg going down theorem Žwhich is known to hold for integrally closed domains.. Our next theorem shows the relation between the various definitions we have made. It forms the basis of our understanding of integrally discerning primes. Ž2.11. THEOREM. statements: Ža. Žb. Žc.
Let P be a prime ideal in R. Consider the following
P is strongly comaximizable and RrP is pseudo-integrally closed. P is integrally discerning. P is strongly comaximizable.
Then Ža. « Žb. « Žc.. Proof. Ža. « Žb. Assume Ža. holds. Let p 1 , . . . , p m Ž m ) 0. be integral uppers to P. We must show that there exists an integral upper k to 0 such that UŽ P, k. s p 1 , . . . , p m 4 . Since P is strongly comaximizable, there is an integral extension domain T of R in which there are exactly m primes, P1 , . . . , Pm , lying over P, and these m primes are pairwise comaximal. ŽIf m s 1, take T s R.. Since RrP is pseudo-integrally closed, Ž2.10.Žb. shows that for each i, there is a monic polynomial f i Ž X . g Rw X x with p i s RadŽ P, f i Ž X .. Rw X x. For each i, let a i be a root of f i Ž X ., and in D s T w a1 , . . . , a m x, let Pi1 , . . . , Pi g i be all of the primes of D which lie over Pi in T. Note that if Ji s Pi1 l ??? l Pi g i , then the ideals J1 , . . . , Jm are pairwise comaximal in D Žsince P1 , . . . , Pm are pairwise comaximal in T .. Pick b g D with b ' a i mod Ji , 1 F i F m, and let k9 s Ž X y b . Dw X x. Now k9 is an integral upper to 0 in Dw X x, and it is not hard to see that k s k9 l Rw X x is an integral upper to 0 in Rw X x. We claim k is the integral upper we seek. First, we show that k ; p i for every i. Since b ' a i mod Ji , and since Ji : Pi1 , we have k9 ; Ž Pi1 , X y a i . Dw X x. We easily see that Ž Pi1 , X y a i . Dw X x l Rw X x is an integral upper to P. Since f i Ž X . g Ž X y a i . Dw X x, f i Ž X . g Ž Pi1 , X y a i . Dw X x l Rw X x, and since the only upper to P which contains f i Ž X . is p i , it follows that Ž Pi1 , X y a i . Dw X x l Rw X x s p i . Since k9 ; Ž Pi1 , X y a i . Dw X x, we see that k s k9 l Rw X x ; p i . Next, suppose that k ; p, for some upper p to P in Rw X x. We must show that p is one of p 1 , . . . , p m . By going up, there is some prime q of Dw X x with k9 ; q and q l Rw X x s p. An easy exercise shows that q is an integral upper to some prime of D lying over P. However, the only primes
166
MCADAM AND SHAH
of D lying over P are the various Pi j , 1 F i F m, 1 F j F g i . Thus, for some such i and j, q is an integral upper to Pi j . Since k9 ; q, we see that X y b g q. As b ' a i mod Pi j , we see that X y a i g q. As Ž Pi j , X y a i . Dw X x is an upper to Pi j and is contained in q, we see that q s Ž Pi j , X y a i . Dw X x. However, as argued in the preceding paragraph, Ž Pi j , X y a i . Dw X x l Rw X x s p i . Since q l Rw X x s p, we have p s p i , as desired. This completes the proof of Ža. « Žb.. Žb. « Žc. Assume Žb. holds. Then there is an integral upper k to 0 such that UŽ P, k. s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 . Let T s Rw X xrk. Then T is a simple integral extension domain of R, and in T there are exactly two primes lying over P Žnamely Ž P, X . Rw X xrk and Ž P, X q 1. Rw X xrk. and those two primes are Žobviously. pairwise comaximal. By Ž2.7. Žwith n s 2., P is strongly comaximizable. The next corollary shows that integrally discerning primes are quite common. Ž2.12. COROLLARY. If P is a prime ideal in R such that RrP is integrally closed and P is not contained in the Jacobson radical of R, then P is integrally discerning. Proof. Since P is not contained in the Jacobson radical of R, Ž2.6. shows that P is strongly comaximizable. Since RrP is integrally closed, Ž2.10.Ža. shows that RrP is pseudo-integrally closed. Thus Ž2.11.Ža. « Žb. shows that P is integrally discerning. For the sake of reference, we state the following. Ž2.13. LEMMA wM1, Lemma 3x. Let f Ž X . be a monic polynomial in Rw X x. Let k be a prime ideal of Rw X x minimal o¨ er f Ž X . Rw X x. Then k is an upper to 0. Ž2.14. LEMMA. statements:
Let P be a prime ideal in R. Consider the following
Ža. RrP is pseudo-integrally closed. Žb. For e¨ ery integral upper p to P, there is an integral upper k to 0 with UŽ P, k. s p4 . Then Ža. implies Žb., and if R is pseudo-integrally closed, Ža. and Žb. are equi¨ alent. Proof. Suppose Ža. holds, and let p be an integral upper to P. By Ž2.10.Žb., there is a monic f Ž X . g Rw X x with p s RadŽ P, f Ž X .. Rw X x. Let k be a prime of Rw X x with k : p and with k minimal over f Ž X . Rw X x. By Ž2.13., k is an Žobviously integral. upper to 0. We already have p g UŽ P, k.. Let p9 g U Ž P, k.. Then f Ž X . g k : p9, and P : p9, so p s
LIFTING CHAINS OF PRIMES
167
RadŽ P, f Ž X .. Rw X x : p9. As p and p9 are both uppers to P, p9 s p. Thus UŽ P, k. s p4 , so that Žb. holds. Now suppose that R is pseudo-integrally closed and that Žb. holds. Let p be an integral upper to P. By Žb., there is an integral upper k to 0 such that UŽ P, k. s p4 . Since R is pseudo-integrally closed, there is a monic polynomial f Ž X . g Rw X x such that k s Rad f Ž X . Rw X x. We claim that p s RadŽ P, f Ž X .. Rw X x. Let p9 be a prime minimal over Ž P, f Ž X .. Rw X x. By Ž2.13. Žused modulo P ., p9 is an upper to P. Since k s Rad f Ž X . Rw X x : p9, we have p9 g UŽ P, k. s p4 . Thus p9 s p, which proves the claim. Now Ž2.10.Žb. shows that Ža. holds. When R is pseudo-integrally closed, we can improve Ž2.11.. Ž2.15. THEOREM. Let R be pseudo-integrally closed and P a prime ideal in R. Then P is integrally discerning if and only if P is strongly comaximizable and RrP is pseudo-integrally closed. Proof. Suppose P is strongly comaximizable and RrP is pseudointegrally closed. Then Ž2.11.Ža. « Žb. shows that P is integrally discerning. Conversely, suppose P is integrally discerning. By Ž2.11.Žb. « Žc., P is strongly comaximizable. Now let p be an integral upper to P. Since P is integrally discerning, there is an integral upper k to 0 such that UŽ P, k. s p4 . Since R is pseudo-integrally closed, Ž2.14. shows that RrP is pseudointegrally closed. Ž2.16. LEMMA. If R contains a field of characteristic 0, then R is integrally closed if and only if it is pseudo-integrally closed. Proof. If R is integrally closed, then it is pseudo-integrally closed by Ž2.10.Ža.. Thus, assume that R is not integrally closed. Let R9 be the integral closure of R. Pick a g R9 y R. Let k be the kernel of Rw X x ª Rw a x, so that k is an integral upper to 0. We will show that R is not pseudo-integrally closed, by showing that there is no monic polynomial f Ž X . with k s Rad f Ž X . Rw X x. Suppose to the contrary that such an f Ž X . exists. Since a f R, there is a maximal ideal M of R such that a f R S , where S s R y M. Thus a g RXS y R S . A well-known and easy exercise shows that ay1 is not in R S . Now the kernel of R S w X x ª R S w a x is k S . Since we are assuming k s Rad f Ž X . R w X x, we have k S s Rad f Ž X . R S w X x. Summarizing, we are in the local domain Ž R S , MS ., we know that R S contains a field of characteristic 0, we have an element a in the quotient field of R S such that a f R S and ay1 f R S , and the kernel of R S w X x ª R S w a x equals Rad f Ž X . R S w X x. By wMS, Ž3.3.x, we must have f Ž X . g MS R S w X x. This contradicts that f Ž X . is monic.
168
MCADAM AND SHAH
Ž2.17. COROLLARY. Let R be integrally closed, and assume R contains a field of characteristic 0. Let P be a prime ideal in R. Then P is integrally discerning if and only if P is strongly comaximizable and RrP is integrally closed. Proof. Since R contains a field of characteristic 0, so does RrP. Therefore this corollary follows from Ž2.15. and Ž2.16.. Ž2.18. COROLLARY. Let R be an integrally closed Noetherian Hilbert domain containing a field of characteristic 0. Let P be a nonzero prime ideal of R. Then P is integrally discerning if and only if RrP is integrally closed. Proof. Since the Jacobson radical of R is 0, Ž2.6. shows that P is strongly comaximizable. Thus the corollary follows from Ž2.17.. In the definition of integrally discerning prime Ž2.3., we require the existence of an integral upper to 0 for every finite nonempty set U of integral uppers to P. The following result shows that it is enough to check only certain Žsmall. sets U of uppers if R is pseudo-integrally closed. Ž2.19. PROPOSITION. Let R be pseudo-integrally closed and let P be a nonzero prime ideal in R. Then the following are equi¨ alent: Ža. P is integrally discerning. Žb. If U is a set of integral uppers to P such that either U has size 1, or U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 , then there is an integral upper k to 0 such that UŽ P, k. s U. Proof. Ža. « Žb. is trivial. Thus suppose that Žb. holds. Letting U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 , the argument in Ž2.11.Žb. « Žc. shows that P is strongly comaximizable. Therefore, by Ž2.15., it will suffice to show that RrP is pseudo-integrally closed. However, that follows from Žb. and Ž2.14.. We will use Ž2.20. and Ž2.21. in the next section. Ž2.20. LEMMA. Let K ; Q ; P be prime ideals in R, and suppose that RrK is pseudo-integrally closed. Let k and p be uppers to K and P respecti¨ ely, with k an integral upper, and with k ; p. Then there is an upper q to Q with k ; q ; p. Proof. Working modulo K, we will assume K s 0 and R is pseudointegrally closed. Thus there is a monic polynomial f Ž X . g Rw X x with k s Rad f Ž X . Rw X x. Note that Ž Q, f Ž X .. Rw X x : p. Let q be a prime ideal in Rw X x with q : p, and with q minimal over Ž Q, f Ž X .. Rw X x. By Ž2.13. Žapplied modulo Q ., q is an upper to Q. Obviously k s Rad f Ž X . Rw X x ; q ; p.
LIFTING CHAINS OF PRIMES
169
Ž2.21. THEOREM. Let R be pseudo-integrally closed, and let T be an integral extension domain of R. Then R : T satisfies going down. Proof. Let Q ; P be prime ideals of R, and let P9 be a prime ideal of T with P9 l R s P. We must show that there is a prime ideal Q9 of T with Q9 ; P9 and with Q9 l R s Q. By wM2, Proposition 2x, we may assume that T s Rw a x for some a g T. Let k be the kernel of the natural map s : Rw X x ª Rw a x s T. Then k is an integral upper to 0, and we may replace T by Rw X xrk. Now the prime P9 is of the form prk for some upper p to P with k ; p. By Ž2.20. Žapplied with K s 0., there is an upper q to Q with k ; q ; p. Thus qrk ; prk. As it is easily seen that Žqrk. l R s Q, we take Q9 to be qrk. We give an example of a Noetherian domain which is pseudo-integrally closed, but not integrally closed. Ž2.22. EXAMPLE. Let Z 2 be the integers modulo 2, and let Y be an indeterminate. Let T s Z 2 ww Y xx, and let R be the subdomain of T consisting of those power series whose Y term has coefficient 0. Clearly R is not integrally closed Žas its integral closure is T .. Let k be an integral upper to 0. To show that R is pseudo-integrally closed, we must show that k s Rad f Ž X . Rw X x for some monic polynomial f Ž X . g Rw X x. Now there is an integral upper k9 to 0 in T w X x such that k s k9 l Rw X x. As T is integrally closed, Ž2.9. shows that there is a monic g Ž X . g T w X x such that k9 s Ž g Ž X ..T w X x. We leave to the reader the exercise of showing that if f Ž X . s g 2 Ž X ., then f Ž X . g Rw X x, and k s Rad f Ž X . Rw X x.
3. LIFTING CHAINS OF PRIMES Recalling the introduction, we saw that Ž1.2. generalizes the Rotthaus]Wiegand result mentioned there. We are now ready to prove a stronger result than Ž1.2.. ŽSee the remark following the statement.. Ž3.1. THEOREM. Let V be a chain 0 s P0 ; P1 ; ??? ; Pn4 of prime ideals in R. Suppose that R is pseudo-integrally closed. Also suppose that for each i G 1, PirPiy1 is integrally discerning. Then a finite partially ordered set W is order isomorphic to V T for some finitely generated integral extension domain T of R if and only if Ž1. W contains a unique minimal element, and Ž2. e¨ ery maximal saturated chain of elements in W has length n. Furthermore, if such a T exists, then it can be chosen to be a simple integral extension of R. Remark. In Ž1.2., we assume that RrP0 s R is integrally closed. By Ž2.10.Ža., that implies R is pseudo-integrally closed. Also in Ž1.2., we
170
MCADAM AND SHAH
assume that for each i G 1, PirPiy1 is not contained in the Jacobson radical of RrPiy1 , and Ž RrPiy1 .rŽ PirPiy1 . ( RrPi is integrally closed. By Ž2.12., that implies that PirPiy1 is integrally discerning. Thus, the hypothesis of Ž1.2. implies the hypothesis of Ž3.1., so that Ž3.1. implies Ž1.2.. We first need a lemma. Ž3.2. LEMMA. Let T be an integral extension of R. Let K ; Q ; P be prime ideals of R. Let K 9 ; P9 be prime ideals of T, lying o¨ er K and P, respecti¨ ely. If RrK is pseudo-integrally closed, then there is a prime ideal Q9 of T, lying o¨ er Q, such that K 9 ; Q9 ; P9. Proof. By Ž2.21., RrK : TrK 9 satisfies going down. Now P9rK 9 is a prime in TrK 9 and Ž P9rK 9. l Ž RrK . s PrK. Since QrK ; PrK, there is a prime Q9rK 9 in TrK 9 lying over QrK, with Q9rK 9 ; P9rK 9. Clearly Q9 satisfies the lemma. Proof of Ž3.1.. First we claim that for each i, RrPi is pseudo-integrally closed. For i s 0, this is given by the hypothesis that R is pseudointegrally closed. Now inductively assume that it holds for RrPiy1. Since PirPiy 1 is integrally discerning, Ž 2.15 . shows that RrPi ( Ž RrPiy1 .rŽ PirPiy1 . is pseudo-integrally closed. Let T be a finitely generated integral extension domain of R. Clearly V T has a unique minimal element, namely 0 g SpecŽT .. Therefore, we must show that if C is a maximal saturated chain of primes in V T, then C has length n. By maximality of C , we see that the minimal element in C must be 0. Also by going up and the maximality of C , we see that the maximal element of C must lie over Pn . We now claim that for each i, C contains a prime lying over Pi . The cases i s 0 and i s n have already been done. For each i such that 0 - i - n, the claim now follows easily from Ž3.2., the first paragraph of this proof, and the fact that C is saturated. The claim shows that the length of C is at least n. On the other hand, incomparability shows that distinct primes in C contract to distinct primes in R, and so the length of C is at most n. Therefore that length equals n. ŽNote that this direction of the proof only used that RrPi is pseudo-integrally closed for each i.. Conversely, let W be a finite partially ordered set having a unique minimal element, such that every maximal saturated chain of primes in W has length n. We must construct a simple integral extension domain T of R such that V T is order isomorphic to W . For each i, let W i s ¨ g W N height ¨ s i4 . Using reverse induction, for each i, we construct a set Pi s p ¨ N ¨ g W i 4 of integral uppers to Pi in a bijective correspondence to W i , in such a way that if i - j, then UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4 . Let Pn be any collection of distinct integral uppers to Pn in a bijective correspondence to Wn . Now suppose that Pi has already been constructed.
LIFTING CHAINS OF PRIMES
171
We must construct Piy1. Thus, for every ¨ g W iy1 , we must find a corresponding integral upper p ¨ to Piy1 , such that for each j ) i y 1, UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4 . We start by considering j s i. Let bars denote reduction modulo Piy1. Since Pi is integrally discerning, Ž2.4. shows that in Rw X x there are infinitely many integral uppers p ¨ to 0, such that UŽ Pi , p ¨ . s p u N u g W i and ¨ - u4 . ŽThe infinitude of choices is important, since if ¨ and ¨ 9 are distinct in W iy1 , and if u N u g W i and ¨ - u4 s u N u g W i and ¨ 9 - u4 , we can pick distinct p ¨ and p ¨ 9 corresponding to ¨ and ¨ 9, respectively, and so preserve the bijectiveness of the correspondence we are constructing.. Pick one such p ¨ and let p ¨ be its inverse image under the map Rw X x ª Rw X x. It is easily seen that p ¨ is an integral upper to Piy1 , and UŽ Pi , p ¨ . s p u N u g W i and ¨ - u4 . Now we show that for every j such that i - j F n, UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4. First suppose that p g UŽ Pj , p ¨ .. Then p is an upper to Pj and p ¨ ; p. We have Piy1 ; Pi ; Pj , and we have uppers p ¨ ; p to Piy1 and Pj , respectively, with p ¨ an integral upper. By the opening paragraph of this proof, RrPiy1 is pseudo-integrally closed. Thus Ž2.20. shows there is an upper q to Pi with p ¨ ; q ; p. As q g UŽ Pi , p ¨ ., and as we already know what UŽ Pi , p ¨ . equals, we see that q s p u for some u g W i with ¨ - u. Since p is an upper to Pj and p u s q ; p, we see that p g UŽ Pj , p u .. However, since p u g Pi Žwhich we are inductively assuming has already been constructed., we know that UŽ Pj , p u . s p w N w g W j and u - w4 . Thus p s p w for some w g W j with u - w. As ¨ - u - w, we see that p g p w N w g W j and ¨ - w4 . Therefore, UŽ Pj , p ¨ . : p w N w g W j and ¨ - w4. To show the other inclusion, pick p w for some w g W j with ¨ - w. Since p w is Žby induction. an upper to Pj , in order to show that p w g UŽ Pj , p ¨ . we need only show that p ¨ ; p w . Since ¨ g W iy1 , height ¨ s i y 1. Similarly height w s j ) i. Now ¨ - w forms a chain of elements in W . That short chain can be embedded in a maximal saturated chain, which by assumption must have length n. Therefore, it is easily seen that there exists an element u g W i with ¨ - u - w. We already know Žby construction of p ¨ . UŽ Pi , p ¨ . s p u N u g W i and ¨ - u4 , and Žby induction. UŽ Pj , p u . s p w N w g W j and u - w4 . Since ¨ - u - w, we have p u g UŽ Pi , p ¨ . and p w g UŽ Pj , p u .. Thus p ¨ ; p u , and p u ; p w , so that p ¨ ; p w . This completes the proof that UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4 . Having now inductively established the existence of a bijection as described above, we consider the minimal element of W Žwhich, having height 0, is in W 0 . and its image under our correspondence, which we call p 0 . Thus p 0 is an integral upper to P0 s 0, and by the above construction, for each i G 1, UŽ Pi , p 0 . s Pi . Now let T s Rw X xrp 0 , which is a simple integral extension domain of R. It is easily seen that for each i, the primes
172
MCADAM AND SHAH
in T which lie over Pi are p ¨ rp 0 N p ¨ g Pi 4 s p ¨ rp 0 N ¨ g W i 4 . This gives a bijection from W to V T. Also, if i - j, and if ¨ g W i and w g W j , our construction shows that p ¨ rp 0 ; p w rp 0 if and only if p ¨ ; p w if and only if ¨ - w, and so our bijection is an order isomorphism between W and V T.
4. CONCERNING GENERALIZATIONS It is difficult to assess the potential for strengthening Ž3.1.. We suspect that the assumption PirPiy1 is integrally discerning for each i G 1 can be somewhat weakened, but we do not know by how much. The Žhidden. assumption of strong comaximizability in Ž3.1. Žwhich by Ž2.11.Žb. « Žc. is implied by the assumption of integral discernibility . certainly plays an important role, at least if maximal ideals are involved. We illustrate this in our next example. Ž4.1. EXAMPLE. Let Ž R, M . be a local ŽNoetherian. domain, and let V be the chain of prime ideals 0 ; P ; M. Assume P is not strongly comaximizable. Then there does not exist an integral extension domain T of R such that V T is the partially ordered set
nT M1
M2
P1
P2 0
ŽNote that W s V T does satisfy the hypothesis of Ž3.1.. See wM3, Ž3.6.x for an example of a prime not strongly comaximizable.. Proof. Suppose there exists a finitely generated integral extension domain T of R such that V T is as above. Then in T, P1 and P2 are comaximal. By Ž2.7. Žwith n s 2., P is strongly comaximizable. This is a contradiction of the hypothesis. Finally, we consider the question of extending Ž3.1. to partially ordered sets V which are not chains. That is not difficult to do, but only in an awkward and unsatisfying way. To explain this, we need a definition. Ž4.2. DEFINITION. Let P1 , . . . , Pn be nonzero prime ideals in R. We say that the set P1 , . . . , Pn4 is integrally discerning if for every choice U1 , . . . , Un ,
LIFTING CHAINS OF PRIMES
173
with each Ui a finite nonempty set of integral uppers to Pi , there is an integral upper k to 0 with UŽ Pi , k. s Ui , 1 F i F n. There is a generalization of Ž3.1. in which V satisfies the following condition: if P g V and if the set of primes which are directly above P in the partially ordered set V is Q1 , . . . , Q m 4 , then Q1rP, . . . , Q m rP 4 is an integrally discerning set of primes contained in Spec Ž RrP .. However, integrally discerning sets of primes of size bigger than one are rare. In particular, if R is pseudo integrally closed, the primes involved must be pairwise comaximal, as Ž4.3. shows. Therefore, the generalization of Ž3.1. does not consider ‘‘typical’’ partially ordered sets V , and so we choose not to pursue it. Ž4.3. THEOREM. Let n G 2. If P1 , . . . , Pn are pairwise comaximal, and if RrPi is pseudo-integrally closed for each i, then P1 , . . . , Pn4 is integrally discerning. Furthermore, if R is pseudo-integrally closed, then the con¨ erse also holds. Proof. The proof of the initial segment is very similar to the proof of Ž2.11.Ža. « Žb.. It begins by letting Ui be a finite nonempty set of integral uppers to Pi for each i. Now if the size of Ui is m i , then there is a finitely generated integral extension domain T of R such that for each i, there are exactly m i primes of T lying over Pi , and those m i primes are pairwise comaximal wM3, Ž1.4.x. The rest of the argument is a straightforward variation of the proof of Ž2.11.Ža. « Žb., and we leave the details to the reader. Next, assume that R is pseudo-integrally closed, and that P1 , . . . , Pn4 is integrally discerning. It is easily seen that each Pi is integrally discerning, and so by Ž2.15., RrPi is pseudo-integrally closed. Now if P and Q are distinct primes in P1 , . . . , Pn4 , then we must show that P and Q are comaximal. Suppose to the contrary that M is a maximal ideal of R containing both P and Q. As it is easily seen that P, Q4 is integrally discerning, there is an integral upper k to 0 in Rw X x such that UŽ P, k. s Ž P, X . Rw X x4 and UŽ Q, k. s Ž Q, X q 1. Rw X x4 . There is an element u, integral over R, such that Rw X xrk ( Rw u x. Under this isomorphism, the image of Ž P, X . Rw X xrk is Ž P, u. Rw u x, and the image of Ž Q, X q 1. Rw X xrk is Ž Q, u q 1. Rw u x. Since Ž Q, u q 1. Rw u x lies over Q, and Q : M, by going up there is a maximal ideal N of Rw u x which lies over M, with Ž Q, u q 1. Rw u x : N. Since P : M and N lies over M, by Ž2.21. there is a prime P9 of Rw u x lying over P, with P9 : N. However, since UŽ P, k. s Ž P, X . Rw X x4 , we see that Ž P, u. Rw u x is the only prime ideal in Rw u x which lies over P. Thus P9 s Ž P, u. Rw u x. It follows that N contains both Ž P, u. Rw u x and Ž Q, u q 1. Rw u x. This is clearly impossible.
174
MCADAM AND SHAH
We close with a corollary which is not really relevant, but which is pleasant enough to deserve mention. Ž4.4. COROLLARY. Suppose that P and Q are distinct prime ideals in R such that RrP and RrQ are both pseudo-integrally closed. Then the following are equi¨ alent: Ža. P, Q4 is integrally discerning. Žb. There is an integral upper k to 0 in Rw X x such that UŽ P, k. s Ž P, X . Rw X x4 and UŽ Q, k. s Ž Q, X q 1. Rw X x4 . Proof. It is obvious that Ža. implies Žb.. Thus, suppose that Žb. holds. Let u be as in the second half of the proof of Ž4.3.. Thus Ž P, u. Rw u x is the unique prime of Rw u x lying over P, and similarly, Ž Q, u q 1. Rw u x is the unique prime of Rw u x lying over Q. Obviously Ž P, u. Rw u x and Ž Q, u q 1. Rw u x are pairwise comaximal. Now Rw u xrŽ Q, u q 1. Rw u x ( Ž R w X xrk .r ŽŽ Q, X q 1. R w X xrk . ( R w X xr Ž Q, X q 1. R w X x ( RrQ. Therefore, Rw u xrŽ Q, u q 1. Rw u x is pseudo-integrally closed, and similarly, the same is true of Rw u xrŽ P, u. Rw u x. By Ž4.3., Ž P, u. Rw u x, Ž Q, u q 1. Rw u x4 is integrally discerning in Rw u x. Since Ž P, u. Rw u x and Ž Q, u q 1. Rw u x are the unique primes of Rw u x lying over P and Q, respectively, it is not difficult to see that P, Q4 is integrally discerning in R.
ACKNOWLEDGMENT The authors wish to thank the referee for helpful suggestions concerning presentation.
REFERENCES wKx wM1x wM2x wM3x wMSx
I. Kaplansky, ‘‘Commutative Rings,’’ Univ. of Chicago Press, Chicago, 1974. S. McAdam, Going down, Duke Math. J. 39 Ž1972., 633]636. S. McAdam, Going down and open extensions, Canad. J. Math. 27 Ž1975., 111]114. S. McAdam, Strongly comaximizable primes, J. Algebra 170 Ž1994.. S. McAdam and C. Shah, Substructures of Spec Rw X x, J. Algebra, to appear.
185, 162]174 Ž1996.
0318
Lifting Chains of Primes to Integral Extensions Stephen McAdam Department of Mathematics, Uni¨ ersity of Texas, Austin, Texas 78712
and Chandni Shah* Department of Mathematics, Uni¨ ersity of California, Ri¨ erside, California 92521 Communicated by Richard G. Swan Received August 31, 1995
1. INTRODUCTION Let R be a Noetherian domain; we consider SpecŽ R . as a partially ordered set under inclusion. The goal of this paper is to raise Žand answer a special case of. the following question: Ž1.1. Question. Let V be a finite subset of SpecŽ R .. Which partially ordered sets arise as the sets of all primes lying over V in some finitely generated integral extension domain T of R? That is, for which partially ordered sets W does there exist an integral extension domain T of R such that W(V
T
s Q g Spec Ž T . N Q l R g V 4 ?
The following unpublished result of Christel Rotthaus and Sylvia Wiegand answers a special case of Ž1.1.. Let R s K w X 1 , . . . , X n x with K a field and the X i indeterminates. Let Pi s Ž X 1 , . . . , X i . R, 0 F i F n, and let V be the partially ordered set 0 s P0 ; P1 ; ??? ; Pn4 . Their result states that a finite partially ordered set W is order isomorphic to V T for some finitely generated integral extension domain T of R if and only if Ž1. W * Current address: Department of Mathematics, Colgate University, Hamilton, NY 13346. E-mail: [email protected] 162 0021-8693r96 $18.00 Copyright Q 1996 by Academic Press, Inc. All rights of reproduction in any form reserved.
LIFTING CHAINS OF PRIMES
163
contains a unique minimal element and Ž2. every maximal saturated chain of elements of W has length n. Note that in the above result, the partially ordered set V is a chain, RrPi is integrally closed for each i, and PirPiy1 is not contained in the Jacobson radical of RrPiy1 , for each i G 1. We show that the Rotthaus]Wiegand hypothesis can be weakened to just those assumptions, leaving their conclusion unchanged. That is, we prove the following. Ž1.2. THEOREM. Let V be a chain 0 s P0 ; P1 ; ??? ; Pn4 of prime ideals in R. For each i, assume that RrPi is integrally closed, and for each i G 1, assume that PirPiy1 is not contained in the Jacobson radical of RrPiy1. Then a finite partially ordered set W is order isomorphic to V T for some finitely generated integral extension domain T of R if and only if Ž1. W contains a unique minimal element and Ž2. e¨ ery maximal saturated chain of elements in W has length n. Furthermore, if such a T exists, then it can be chosen to be a simple integral extension of R. In fact, the result we eventually prove in Ž3.1. is somewhat stronger than Ž1.2.. However, before dealing with it, we must introduce integrally discerning prime ideals.
2. INTEGRALLY DISCERNING PRIMES Ž2.1. DEFINITION. If P is a prime ideal in R and if p is a prime ideal in the polynomial ring Rw X x such that p l R s P but p / PRw X x, then we say that p is an upper to P. If, furthermore, p contains a monic polynomial, then we say p is an integral upper to P. ŽSee wK, Section 1-5x for elementary facts about uppers.. Ž2.2. DEFINITION. Let P, Q be prime ideals in R and let q be an upper to Q in Rw X x. We use UŽ P, q. to denote the set p N p is an upper to P and q ; p4 . Ž2.3. DEFINITION. A nonzero prime ideal P in R is said to be integrally discerning if for every finite nonempty set U of integral uppers to P, there is an integral upper k to 0 such that UŽ P, k. s U. Our first lemma strengthens the definition just given. It will be needed in the next section. Ž2.4. LEMMA. Let P be an integrally discerning prime ideal in R, and let U be a finite nonempty set of integral uppers to P. Then there are < R < integral uppers k to 0 such that UŽ P, k. s U. Ž Here, < R < denotes the cardinality of R, which is infinite since R is not a field..
164
MCADAM AND SHAH
Proof. Since P is integrally discerning, there is an integral upper k to 0 such that UŽ P, k. s U. Now Rw X xrk is a simple integral extension of R, and so it can be written in the form Rw u x for some u integral over R. Here k is the kernel of Rw X x ª Rw u x. For every b g P, let k b be the kernel of Rw X x ª Rw u q b x. Note that k b s g Ž X y b . N g Ž X . g k4 . Also note that g Ž X . and g Ž X y b . differ by a polynomial in PRw X x. It easily follows that UŽ P, k b . s UŽ P, k. s U. Since Ž u q b . is a root of every polynomial of k b , it follows that for at most finitely many c g P, we have k c s k b . On the other hand, < P < is easily seen to equal < R <, and so there are < R < distinct uppers k b to 0, b g P, such that UŽ P, k. s U. In order to understand integrally discerning primes, we need two more definitions. We follow each of them with some relevant facts. Ž2.5. DEFINITION. A nonzero prime ideal P in R is called strongly comaximizable if for every m G 2, there is a finitely generated integral extension domain T of R in which there are exactly m primes lying over P, and those m primes are pairwise comaximal. Ž2.6. THEOREM wM3, Ž1.4.x. If P is a prime ideal in R and if P is not contained in the Jacobson radical of R, then P is strongly comaximizable. Ž2.7. THEOREM wM3, Ž1.5. Ži. m Žii.x. Let P be a prime ideal in R, and let n G 2 be a fixed integer. Then P is strongly comaximizable if and only if there is a finitely generated integral extension domain T of R in which there are exactly n primes lying o¨ er P, and those n primes are pairwise comaximal. Ž2.8. DEFINITION. The domain R is said to be pseudo-integrally closed if for every integral upper k to 0 in Rw X x, there is a monic polynomial f Ž X . such that k s Rad f Ž X . Rw X x. We recall a well-known fact. Ž2.9. LEMMA wM3, Ž2.4.x. If R is an integrally closed domain, and if k is an integral upper to 0 in Rw X x, then k is generated by a monic polynomial. Ž2.10.. Remarks. Ža. If R is integrally closed, then it is pseudointegrally closed. This is immediate from Ž2.9.. Žb. If P is a prime ideal in R, then RrP is pseudo-integrally closed if and only if for every integral upper p to P, there is a monic polynomial f Ž X . g Rw X x such that p s RadŽ P, f Ž X .. Rw X x. This follows easily from the fact that if p is a prime ideal of Rw X x with PRw X x : p, and if bars denote reduction modulo P, then p is an integral upper to P in Rw X x if and only if p is an integral upper to 0 in Rw X x. Žc. In Ž2.16., we show that if R contains a field of characteristic 0, then R is integrally closed if and only if it is pseudo-integrally closed.
LIFTING CHAINS OF PRIMES
165
Žd. In Ž2.22., we give an example of a domain which is pseudointegrally closed, but not integrally closed. Že. In Ž2.21., we show that if R is pseudo-integrally closed, then it satisfies the famous Cohen]Seidenberg going down theorem Žwhich is known to hold for integrally closed domains.. Our next theorem shows the relation between the various definitions we have made. It forms the basis of our understanding of integrally discerning primes. Ž2.11. THEOREM. statements: Ža. Žb. Žc.
Let P be a prime ideal in R. Consider the following
P is strongly comaximizable and RrP is pseudo-integrally closed. P is integrally discerning. P is strongly comaximizable.
Then Ža. « Žb. « Žc.. Proof. Ža. « Žb. Assume Ža. holds. Let p 1 , . . . , p m Ž m ) 0. be integral uppers to P. We must show that there exists an integral upper k to 0 such that UŽ P, k. s p 1 , . . . , p m 4 . Since P is strongly comaximizable, there is an integral extension domain T of R in which there are exactly m primes, P1 , . . . , Pm , lying over P, and these m primes are pairwise comaximal. ŽIf m s 1, take T s R.. Since RrP is pseudo-integrally closed, Ž2.10.Žb. shows that for each i, there is a monic polynomial f i Ž X . g Rw X x with p i s RadŽ P, f i Ž X .. Rw X x. For each i, let a i be a root of f i Ž X ., and in D s T w a1 , . . . , a m x, let Pi1 , . . . , Pi g i be all of the primes of D which lie over Pi in T. Note that if Ji s Pi1 l ??? l Pi g i , then the ideals J1 , . . . , Jm are pairwise comaximal in D Žsince P1 , . . . , Pm are pairwise comaximal in T .. Pick b g D with b ' a i mod Ji , 1 F i F m, and let k9 s Ž X y b . Dw X x. Now k9 is an integral upper to 0 in Dw X x, and it is not hard to see that k s k9 l Rw X x is an integral upper to 0 in Rw X x. We claim k is the integral upper we seek. First, we show that k ; p i for every i. Since b ' a i mod Ji , and since Ji : Pi1 , we have k9 ; Ž Pi1 , X y a i . Dw X x. We easily see that Ž Pi1 , X y a i . Dw X x l Rw X x is an integral upper to P. Since f i Ž X . g Ž X y a i . Dw X x, f i Ž X . g Ž Pi1 , X y a i . Dw X x l Rw X x, and since the only upper to P which contains f i Ž X . is p i , it follows that Ž Pi1 , X y a i . Dw X x l Rw X x s p i . Since k9 ; Ž Pi1 , X y a i . Dw X x, we see that k s k9 l Rw X x ; p i . Next, suppose that k ; p, for some upper p to P in Rw X x. We must show that p is one of p 1 , . . . , p m . By going up, there is some prime q of Dw X x with k9 ; q and q l Rw X x s p. An easy exercise shows that q is an integral upper to some prime of D lying over P. However, the only primes
166
MCADAM AND SHAH
of D lying over P are the various Pi j , 1 F i F m, 1 F j F g i . Thus, for some such i and j, q is an integral upper to Pi j . Since k9 ; q, we see that X y b g q. As b ' a i mod Pi j , we see that X y a i g q. As Ž Pi j , X y a i . Dw X x is an upper to Pi j and is contained in q, we see that q s Ž Pi j , X y a i . Dw X x. However, as argued in the preceding paragraph, Ž Pi j , X y a i . Dw X x l Rw X x s p i . Since q l Rw X x s p, we have p s p i , as desired. This completes the proof of Ža. « Žb.. Žb. « Žc. Assume Žb. holds. Then there is an integral upper k to 0 such that UŽ P, k. s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 . Let T s Rw X xrk. Then T is a simple integral extension domain of R, and in T there are exactly two primes lying over P Žnamely Ž P, X . Rw X xrk and Ž P, X q 1. Rw X xrk. and those two primes are Žobviously. pairwise comaximal. By Ž2.7. Žwith n s 2., P is strongly comaximizable. The next corollary shows that integrally discerning primes are quite common. Ž2.12. COROLLARY. If P is a prime ideal in R such that RrP is integrally closed and P is not contained in the Jacobson radical of R, then P is integrally discerning. Proof. Since P is not contained in the Jacobson radical of R, Ž2.6. shows that P is strongly comaximizable. Since RrP is integrally closed, Ž2.10.Ža. shows that RrP is pseudo-integrally closed. Thus Ž2.11.Ža. « Žb. shows that P is integrally discerning. For the sake of reference, we state the following. Ž2.13. LEMMA wM1, Lemma 3x. Let f Ž X . be a monic polynomial in Rw X x. Let k be a prime ideal of Rw X x minimal o¨ er f Ž X . Rw X x. Then k is an upper to 0. Ž2.14. LEMMA. statements:
Let P be a prime ideal in R. Consider the following
Ža. RrP is pseudo-integrally closed. Žb. For e¨ ery integral upper p to P, there is an integral upper k to 0 with UŽ P, k. s p4 . Then Ža. implies Žb., and if R is pseudo-integrally closed, Ža. and Žb. are equi¨ alent. Proof. Suppose Ža. holds, and let p be an integral upper to P. By Ž2.10.Žb., there is a monic f Ž X . g Rw X x with p s RadŽ P, f Ž X .. Rw X x. Let k be a prime of Rw X x with k : p and with k minimal over f Ž X . Rw X x. By Ž2.13., k is an Žobviously integral. upper to 0. We already have p g UŽ P, k.. Let p9 g U Ž P, k.. Then f Ž X . g k : p9, and P : p9, so p s
LIFTING CHAINS OF PRIMES
167
RadŽ P, f Ž X .. Rw X x : p9. As p and p9 are both uppers to P, p9 s p. Thus UŽ P, k. s p4 , so that Žb. holds. Now suppose that R is pseudo-integrally closed and that Žb. holds. Let p be an integral upper to P. By Žb., there is an integral upper k to 0 such that UŽ P, k. s p4 . Since R is pseudo-integrally closed, there is a monic polynomial f Ž X . g Rw X x such that k s Rad f Ž X . Rw X x. We claim that p s RadŽ P, f Ž X .. Rw X x. Let p9 be a prime minimal over Ž P, f Ž X .. Rw X x. By Ž2.13. Žused modulo P ., p9 is an upper to P. Since k s Rad f Ž X . Rw X x : p9, we have p9 g UŽ P, k. s p4 . Thus p9 s p, which proves the claim. Now Ž2.10.Žb. shows that Ža. holds. When R is pseudo-integrally closed, we can improve Ž2.11.. Ž2.15. THEOREM. Let R be pseudo-integrally closed and P a prime ideal in R. Then P is integrally discerning if and only if P is strongly comaximizable and RrP is pseudo-integrally closed. Proof. Suppose P is strongly comaximizable and RrP is pseudointegrally closed. Then Ž2.11.Ža. « Žb. shows that P is integrally discerning. Conversely, suppose P is integrally discerning. By Ž2.11.Žb. « Žc., P is strongly comaximizable. Now let p be an integral upper to P. Since P is integrally discerning, there is an integral upper k to 0 such that UŽ P, k. s p4 . Since R is pseudo-integrally closed, Ž2.14. shows that RrP is pseudointegrally closed. Ž2.16. LEMMA. If R contains a field of characteristic 0, then R is integrally closed if and only if it is pseudo-integrally closed. Proof. If R is integrally closed, then it is pseudo-integrally closed by Ž2.10.Ža.. Thus, assume that R is not integrally closed. Let R9 be the integral closure of R. Pick a g R9 y R. Let k be the kernel of Rw X x ª Rw a x, so that k is an integral upper to 0. We will show that R is not pseudo-integrally closed, by showing that there is no monic polynomial f Ž X . with k s Rad f Ž X . Rw X x. Suppose to the contrary that such an f Ž X . exists. Since a f R, there is a maximal ideal M of R such that a f R S , where S s R y M. Thus a g RXS y R S . A well-known and easy exercise shows that ay1 is not in R S . Now the kernel of R S w X x ª R S w a x is k S . Since we are assuming k s Rad f Ž X . R w X x, we have k S s Rad f Ž X . R S w X x. Summarizing, we are in the local domain Ž R S , MS ., we know that R S contains a field of characteristic 0, we have an element a in the quotient field of R S such that a f R S and ay1 f R S , and the kernel of R S w X x ª R S w a x equals Rad f Ž X . R S w X x. By wMS, Ž3.3.x, we must have f Ž X . g MS R S w X x. This contradicts that f Ž X . is monic.
168
MCADAM AND SHAH
Ž2.17. COROLLARY. Let R be integrally closed, and assume R contains a field of characteristic 0. Let P be a prime ideal in R. Then P is integrally discerning if and only if P is strongly comaximizable and RrP is integrally closed. Proof. Since R contains a field of characteristic 0, so does RrP. Therefore this corollary follows from Ž2.15. and Ž2.16.. Ž2.18. COROLLARY. Let R be an integrally closed Noetherian Hilbert domain containing a field of characteristic 0. Let P be a nonzero prime ideal of R. Then P is integrally discerning if and only if RrP is integrally closed. Proof. Since the Jacobson radical of R is 0, Ž2.6. shows that P is strongly comaximizable. Thus the corollary follows from Ž2.17.. In the definition of integrally discerning prime Ž2.3., we require the existence of an integral upper to 0 for every finite nonempty set U of integral uppers to P. The following result shows that it is enough to check only certain Žsmall. sets U of uppers if R is pseudo-integrally closed. Ž2.19. PROPOSITION. Let R be pseudo-integrally closed and let P be a nonzero prime ideal in R. Then the following are equi¨ alent: Ža. P is integrally discerning. Žb. If U is a set of integral uppers to P such that either U has size 1, or U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 , then there is an integral upper k to 0 such that UŽ P, k. s U. Proof. Ža. « Žb. is trivial. Thus suppose that Žb. holds. Letting U s Ž P, X . Rw X x, Ž P, X q 1. Rw X x4 , the argument in Ž2.11.Žb. « Žc. shows that P is strongly comaximizable. Therefore, by Ž2.15., it will suffice to show that RrP is pseudo-integrally closed. However, that follows from Žb. and Ž2.14.. We will use Ž2.20. and Ž2.21. in the next section. Ž2.20. LEMMA. Let K ; Q ; P be prime ideals in R, and suppose that RrK is pseudo-integrally closed. Let k and p be uppers to K and P respecti¨ ely, with k an integral upper, and with k ; p. Then there is an upper q to Q with k ; q ; p. Proof. Working modulo K, we will assume K s 0 and R is pseudointegrally closed. Thus there is a monic polynomial f Ž X . g Rw X x with k s Rad f Ž X . Rw X x. Note that Ž Q, f Ž X .. Rw X x : p. Let q be a prime ideal in Rw X x with q : p, and with q minimal over Ž Q, f Ž X .. Rw X x. By Ž2.13. Žapplied modulo Q ., q is an upper to Q. Obviously k s Rad f Ž X . Rw X x ; q ; p.
LIFTING CHAINS OF PRIMES
169
Ž2.21. THEOREM. Let R be pseudo-integrally closed, and let T be an integral extension domain of R. Then R : T satisfies going down. Proof. Let Q ; P be prime ideals of R, and let P9 be a prime ideal of T with P9 l R s P. We must show that there is a prime ideal Q9 of T with Q9 ; P9 and with Q9 l R s Q. By wM2, Proposition 2x, we may assume that T s Rw a x for some a g T. Let k be the kernel of the natural map s : Rw X x ª Rw a x s T. Then k is an integral upper to 0, and we may replace T by Rw X xrk. Now the prime P9 is of the form prk for some upper p to P with k ; p. By Ž2.20. Žapplied with K s 0., there is an upper q to Q with k ; q ; p. Thus qrk ; prk. As it is easily seen that Žqrk. l R s Q, we take Q9 to be qrk. We give an example of a Noetherian domain which is pseudo-integrally closed, but not integrally closed. Ž2.22. EXAMPLE. Let Z 2 be the integers modulo 2, and let Y be an indeterminate. Let T s Z 2 ww Y xx, and let R be the subdomain of T consisting of those power series whose Y term has coefficient 0. Clearly R is not integrally closed Žas its integral closure is T .. Let k be an integral upper to 0. To show that R is pseudo-integrally closed, we must show that k s Rad f Ž X . Rw X x for some monic polynomial f Ž X . g Rw X x. Now there is an integral upper k9 to 0 in T w X x such that k s k9 l Rw X x. As T is integrally closed, Ž2.9. shows that there is a monic g Ž X . g T w X x such that k9 s Ž g Ž X ..T w X x. We leave to the reader the exercise of showing that if f Ž X . s g 2 Ž X ., then f Ž X . g Rw X x, and k s Rad f Ž X . Rw X x.
3. LIFTING CHAINS OF PRIMES Recalling the introduction, we saw that Ž1.2. generalizes the Rotthaus]Wiegand result mentioned there. We are now ready to prove a stronger result than Ž1.2.. ŽSee the remark following the statement.. Ž3.1. THEOREM. Let V be a chain 0 s P0 ; P1 ; ??? ; Pn4 of prime ideals in R. Suppose that R is pseudo-integrally closed. Also suppose that for each i G 1, PirPiy1 is integrally discerning. Then a finite partially ordered set W is order isomorphic to V T for some finitely generated integral extension domain T of R if and only if Ž1. W contains a unique minimal element, and Ž2. e¨ ery maximal saturated chain of elements in W has length n. Furthermore, if such a T exists, then it can be chosen to be a simple integral extension of R. Remark. In Ž1.2., we assume that RrP0 s R is integrally closed. By Ž2.10.Ža., that implies R is pseudo-integrally closed. Also in Ž1.2., we
170
MCADAM AND SHAH
assume that for each i G 1, PirPiy1 is not contained in the Jacobson radical of RrPiy1 , and Ž RrPiy1 .rŽ PirPiy1 . ( RrPi is integrally closed. By Ž2.12., that implies that PirPiy1 is integrally discerning. Thus, the hypothesis of Ž1.2. implies the hypothesis of Ž3.1., so that Ž3.1. implies Ž1.2.. We first need a lemma. Ž3.2. LEMMA. Let T be an integral extension of R. Let K ; Q ; P be prime ideals of R. Let K 9 ; P9 be prime ideals of T, lying o¨ er K and P, respecti¨ ely. If RrK is pseudo-integrally closed, then there is a prime ideal Q9 of T, lying o¨ er Q, such that K 9 ; Q9 ; P9. Proof. By Ž2.21., RrK : TrK 9 satisfies going down. Now P9rK 9 is a prime in TrK 9 and Ž P9rK 9. l Ž RrK . s PrK. Since QrK ; PrK, there is a prime Q9rK 9 in TrK 9 lying over QrK, with Q9rK 9 ; P9rK 9. Clearly Q9 satisfies the lemma. Proof of Ž3.1.. First we claim that for each i, RrPi is pseudo-integrally closed. For i s 0, this is given by the hypothesis that R is pseudointegrally closed. Now inductively assume that it holds for RrPiy1. Since PirPiy 1 is integrally discerning, Ž 2.15 . shows that RrPi ( Ž RrPiy1 .rŽ PirPiy1 . is pseudo-integrally closed. Let T be a finitely generated integral extension domain of R. Clearly V T has a unique minimal element, namely 0 g SpecŽT .. Therefore, we must show that if C is a maximal saturated chain of primes in V T, then C has length n. By maximality of C , we see that the minimal element in C must be 0. Also by going up and the maximality of C , we see that the maximal element of C must lie over Pn . We now claim that for each i, C contains a prime lying over Pi . The cases i s 0 and i s n have already been done. For each i such that 0 - i - n, the claim now follows easily from Ž3.2., the first paragraph of this proof, and the fact that C is saturated. The claim shows that the length of C is at least n. On the other hand, incomparability shows that distinct primes in C contract to distinct primes in R, and so the length of C is at most n. Therefore that length equals n. ŽNote that this direction of the proof only used that RrPi is pseudo-integrally closed for each i.. Conversely, let W be a finite partially ordered set having a unique minimal element, such that every maximal saturated chain of primes in W has length n. We must construct a simple integral extension domain T of R such that V T is order isomorphic to W . For each i, let W i s ¨ g W N height ¨ s i4 . Using reverse induction, for each i, we construct a set Pi s p ¨ N ¨ g W i 4 of integral uppers to Pi in a bijective correspondence to W i , in such a way that if i - j, then UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4 . Let Pn be any collection of distinct integral uppers to Pn in a bijective correspondence to Wn . Now suppose that Pi has already been constructed.
LIFTING CHAINS OF PRIMES
171
We must construct Piy1. Thus, for every ¨ g W iy1 , we must find a corresponding integral upper p ¨ to Piy1 , such that for each j ) i y 1, UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4 . We start by considering j s i. Let bars denote reduction modulo Piy1. Since Pi is integrally discerning, Ž2.4. shows that in Rw X x there are infinitely many integral uppers p ¨ to 0, such that UŽ Pi , p ¨ . s p u N u g W i and ¨ - u4 . ŽThe infinitude of choices is important, since if ¨ and ¨ 9 are distinct in W iy1 , and if u N u g W i and ¨ - u4 s u N u g W i and ¨ 9 - u4 , we can pick distinct p ¨ and p ¨ 9 corresponding to ¨ and ¨ 9, respectively, and so preserve the bijectiveness of the correspondence we are constructing.. Pick one such p ¨ and let p ¨ be its inverse image under the map Rw X x ª Rw X x. It is easily seen that p ¨ is an integral upper to Piy1 , and UŽ Pi , p ¨ . s p u N u g W i and ¨ - u4 . Now we show that for every j such that i - j F n, UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4. First suppose that p g UŽ Pj , p ¨ .. Then p is an upper to Pj and p ¨ ; p. We have Piy1 ; Pi ; Pj , and we have uppers p ¨ ; p to Piy1 and Pj , respectively, with p ¨ an integral upper. By the opening paragraph of this proof, RrPiy1 is pseudo-integrally closed. Thus Ž2.20. shows there is an upper q to Pi with p ¨ ; q ; p. As q g UŽ Pi , p ¨ ., and as we already know what UŽ Pi , p ¨ . equals, we see that q s p u for some u g W i with ¨ - u. Since p is an upper to Pj and p u s q ; p, we see that p g UŽ Pj , p u .. However, since p u g Pi Žwhich we are inductively assuming has already been constructed., we know that UŽ Pj , p u . s p w N w g W j and u - w4 . Thus p s p w for some w g W j with u - w. As ¨ - u - w, we see that p g p w N w g W j and ¨ - w4 . Therefore, UŽ Pj , p ¨ . : p w N w g W j and ¨ - w4. To show the other inclusion, pick p w for some w g W j with ¨ - w. Since p w is Žby induction. an upper to Pj , in order to show that p w g UŽ Pj , p ¨ . we need only show that p ¨ ; p w . Since ¨ g W iy1 , height ¨ s i y 1. Similarly height w s j ) i. Now ¨ - w forms a chain of elements in W . That short chain can be embedded in a maximal saturated chain, which by assumption must have length n. Therefore, it is easily seen that there exists an element u g W i with ¨ - u - w. We already know Žby construction of p ¨ . UŽ Pi , p ¨ . s p u N u g W i and ¨ - u4 , and Žby induction. UŽ Pj , p u . s p w N w g W j and u - w4 . Since ¨ - u - w, we have p u g UŽ Pi , p ¨ . and p w g UŽ Pj , p u .. Thus p ¨ ; p u , and p u ; p w , so that p ¨ ; p w . This completes the proof that UŽ Pj , p ¨ . s p w N w g W j and ¨ - w4 . Having now inductively established the existence of a bijection as described above, we consider the minimal element of W Žwhich, having height 0, is in W 0 . and its image under our correspondence, which we call p 0 . Thus p 0 is an integral upper to P0 s 0, and by the above construction, for each i G 1, UŽ Pi , p 0 . s Pi . Now let T s Rw X xrp 0 , which is a simple integral extension domain of R. It is easily seen that for each i, the primes
172
MCADAM AND SHAH
in T which lie over Pi are p ¨ rp 0 N p ¨ g Pi 4 s p ¨ rp 0 N ¨ g W i 4 . This gives a bijection from W to V T. Also, if i - j, and if ¨ g W i and w g W j , our construction shows that p ¨ rp 0 ; p w rp 0 if and only if p ¨ ; p w if and only if ¨ - w, and so our bijection is an order isomorphism between W and V T.
4. CONCERNING GENERALIZATIONS It is difficult to assess the potential for strengthening Ž3.1.. We suspect that the assumption PirPiy1 is integrally discerning for each i G 1 can be somewhat weakened, but we do not know by how much. The Žhidden. assumption of strong comaximizability in Ž3.1. Žwhich by Ž2.11.Žb. « Žc. is implied by the assumption of integral discernibility . certainly plays an important role, at least if maximal ideals are involved. We illustrate this in our next example. Ž4.1. EXAMPLE. Let Ž R, M . be a local ŽNoetherian. domain, and let V be the chain of prime ideals 0 ; P ; M. Assume P is not strongly comaximizable. Then there does not exist an integral extension domain T of R such that V T is the partially ordered set
nT M1
M2
P1
P2 0
ŽNote that W s V T does satisfy the hypothesis of Ž3.1.. See wM3, Ž3.6.x for an example of a prime not strongly comaximizable.. Proof. Suppose there exists a finitely generated integral extension domain T of R such that V T is as above. Then in T, P1 and P2 are comaximal. By Ž2.7. Žwith n s 2., P is strongly comaximizable. This is a contradiction of the hypothesis. Finally, we consider the question of extending Ž3.1. to partially ordered sets V which are not chains. That is not difficult to do, but only in an awkward and unsatisfying way. To explain this, we need a definition. Ž4.2. DEFINITION. Let P1 , . . . , Pn be nonzero prime ideals in R. We say that the set P1 , . . . , Pn4 is integrally discerning if for every choice U1 , . . . , Un ,
LIFTING CHAINS OF PRIMES
173
with each Ui a finite nonempty set of integral uppers to Pi , there is an integral upper k to 0 with UŽ Pi , k. s Ui , 1 F i F n. There is a generalization of Ž3.1. in which V satisfies the following condition: if P g V and if the set of primes which are directly above P in the partially ordered set V is Q1 , . . . , Q m 4 , then Q1rP, . . . , Q m rP 4 is an integrally discerning set of primes contained in Spec Ž RrP .. However, integrally discerning sets of primes of size bigger than one are rare. In particular, if R is pseudo integrally closed, the primes involved must be pairwise comaximal, as Ž4.3. shows. Therefore, the generalization of Ž3.1. does not consider ‘‘typical’’ partially ordered sets V , and so we choose not to pursue it. Ž4.3. THEOREM. Let n G 2. If P1 , . . . , Pn are pairwise comaximal, and if RrPi is pseudo-integrally closed for each i, then P1 , . . . , Pn4 is integrally discerning. Furthermore, if R is pseudo-integrally closed, then the con¨ erse also holds. Proof. The proof of the initial segment is very similar to the proof of Ž2.11.Ža. « Žb.. It begins by letting Ui be a finite nonempty set of integral uppers to Pi for each i. Now if the size of Ui is m i , then there is a finitely generated integral extension domain T of R such that for each i, there are exactly m i primes of T lying over Pi , and those m i primes are pairwise comaximal wM3, Ž1.4.x. The rest of the argument is a straightforward variation of the proof of Ž2.11.Ža. « Žb., and we leave the details to the reader. Next, assume that R is pseudo-integrally closed, and that P1 , . . . , Pn4 is integrally discerning. It is easily seen that each Pi is integrally discerning, and so by Ž2.15., RrPi is pseudo-integrally closed. Now if P and Q are distinct primes in P1 , . . . , Pn4 , then we must show that P and Q are comaximal. Suppose to the contrary that M is a maximal ideal of R containing both P and Q. As it is easily seen that P, Q4 is integrally discerning, there is an integral upper k to 0 in Rw X x such that UŽ P, k. s Ž P, X . Rw X x4 and UŽ Q, k. s Ž Q, X q 1. Rw X x4 . There is an element u, integral over R, such that Rw X xrk ( Rw u x. Under this isomorphism, the image of Ž P, X . Rw X xrk is Ž P, u. Rw u x, and the image of Ž Q, X q 1. Rw X xrk is Ž Q, u q 1. Rw u x. Since Ž Q, u q 1. Rw u x lies over Q, and Q : M, by going up there is a maximal ideal N of Rw u x which lies over M, with Ž Q, u q 1. Rw u x : N. Since P : M and N lies over M, by Ž2.21. there is a prime P9 of Rw u x lying over P, with P9 : N. However, since UŽ P, k. s Ž P, X . Rw X x4 , we see that Ž P, u. Rw u x is the only prime ideal in Rw u x which lies over P. Thus P9 s Ž P, u. Rw u x. It follows that N contains both Ž P, u. Rw u x and Ž Q, u q 1. Rw u x. This is clearly impossible.
174
MCADAM AND SHAH
We close with a corollary which is not really relevant, but which is pleasant enough to deserve mention. Ž4.4. COROLLARY. Suppose that P and Q are distinct prime ideals in R such that RrP and RrQ are both pseudo-integrally closed. Then the following are equi¨ alent: Ža. P, Q4 is integrally discerning. Žb. There is an integral upper k to 0 in Rw X x such that UŽ P, k. s Ž P, X . Rw X x4 and UŽ Q, k. s Ž Q, X q 1. Rw X x4 . Proof. It is obvious that Ža. implies Žb.. Thus, suppose that Žb. holds. Let u be as in the second half of the proof of Ž4.3.. Thus Ž P, u. Rw u x is the unique prime of Rw u x lying over P, and similarly, Ž Q, u q 1. Rw u x is the unique prime of Rw u x lying over Q. Obviously Ž P, u. Rw u x and Ž Q, u q 1. Rw u x are pairwise comaximal. Now Rw u xrŽ Q, u q 1. Rw u x ( Ž R w X xrk .r ŽŽ Q, X q 1. R w X xrk . ( R w X xr Ž Q, X q 1. R w X x ( RrQ. Therefore, Rw u xrŽ Q, u q 1. Rw u x is pseudo-integrally closed, and similarly, the same is true of Rw u xrŽ P, u. Rw u x. By Ž4.3., Ž P, u. Rw u x, Ž Q, u q 1. Rw u x4 is integrally discerning in Rw u x. Since Ž P, u. Rw u x and Ž Q, u q 1. Rw u x are the unique primes of Rw u x lying over P and Q, respectively, it is not difficult to see that P, Q4 is integrally discerning in R.
ACKNOWLEDGMENT The authors wish to thank the referee for helpful suggestions concerning presentation.
REFERENCES wKx wM1x wM2x wM3x wMSx
I. Kaplansky, ‘‘Commutative Rings,’’ Univ. of Chicago Press, Chicago, 1974. S. McAdam, Going down, Duke Math. J. 39 Ž1972., 633]636. S. McAdam, Going down and open extensions, Canad. J. Math. 27 Ž1975., 111]114. S. McAdam, Strongly comaximizable primes, J. Algebra 170 Ž1994.. S. McAdam and C. Shah, Substructures of Spec Rw X x, J. Algebra, to appear.