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One Counterexample for Two Open Questions about the Rings R(X) and R〈X〉
Journal of Algebra 222, 424᎐427 Ž1999. doi:10.1006rjabr.1999.8009, available online at http:rrwww.idealibrary.com on
One Counterexample for Two Open Questions about the Rings RŽ X . and R ² X : Ihsen Yengui Department of Mathematics, Faculty of Sciences, Sfax, Tunisia Communicated by Walter Feit Received February 16, 1999
We propose to give an answer to two open questions: is R ² X : strong S Žresp., catenarian . when RŽ X . is? We construct a ring R such that RŽ X . is both catenarian and strong S whereas R ² X : is neither catenarian nor strong S. 䊚 1999 Academic Press
INTRODUCTION Let R be a domain and denote by Rw n x the ring of polynomials in n indeterminates with coefficients in R, Žfor n s 1, Rw1x s Rw X x is the ring of polynomials in one indeterminate.. Then RŽ X . s Sy1 Rw X x is the Nagata ring Žin one indeterminate., where S is the multiplicative subset of Rw X x formed by polynomials whose coefficients generate R, and R ² X : s Uy1 Rw X x the Serre conjecture ring, where U is the multiplicative subset formed by the monic polynomials. By induction, one finally defines the Nagata and Serre conjecture rings in n indeterminates as R Ž n . s R Ž n y 1. Ž X .
and
R ² n: s R ² n y 1:² X : .
Clearly, RŽ n. is a localization of R ² n:; hence if the latter ring is a strong S-ring or if it is catenarian, so is the former one. In this paper, we answer the open question about the converse w3, Question 2.8x. Indeed, we construct a counterexample showing that the converse does not hold. Recall that a domain R is strong S Žresp., catenarian . if, for each consecutive pair p ; q of primes in R, the extended primes pw X x ; qw X x are consecutive in Rw X x Žresp., ht Ž q . s ht Ž p . q 1.. A stably strong S-domain Žresp., a universally catenarian domain. R is a domain R such that Rw n x is strong S Žresp., catenarian . for any n. 424 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
QUESTIONS ABOUT RINGS R Ž X . AND R ² X :
425
We use ‘‘; ’’ to denote proper containment. If K : L are two fields we denote t.d.w L : K x the transcendence degree of L over K.
THE EXAMPLE Let K be a field, and T a semi-local universally catenarian K-algebra with two maximal ideals M and N, respectively, of heights 1 and 2 such that t.d.w TrM : K x s 1, and t.d.w TrN : K x s 0, that is, TrN is algebraic over K Žone can proceed as in w2, Exemple Ax.. Let I s M l N, D a one-dimensional valuation domain with quotient field K, and R s ŽT, I, D . s y1 Ž D ., where denotes the natural homomorphism from T onto TrI. The spectrum of R is formed by Ž0., I, primes of height 1 contained in I, and a unique maximal m Žcorresponding to the maximal ideal of D .. Moreover, by w2, Theoreme ´ ` 1x, dim ¨ R s 3. Thus, for all n, ht Ž mw n x. s 3 s ht Ž I w n x. q 1 and dim RŽ n. s dim R ² n: s dim Rw n x y n s ht Ž mw n x. s 3. First claim: for all n, the Nagata ring RŽ n. is catenarian. It is enough to show that for two consecutive primes P ; Q of Rw n x, such that Q : mw n x, one has ht Ž Q . s ht Ž P . q 1. We consider two cases: ᎏP does not contain I; then the chain lifts in T w1, Proposition 4x. In this case, Q cannot be above m since no prime of T is above m. Therefore Q ; mw n x, and thus, ht Ž Q . F 2. Clearly, the claim follows. ᎏP contains I; then I w n x : P ; Q : mw n x. Since we have ht Ž I w n x. s 2, and ht Ž mw n x. s 3, it follows necessarily that P s I w n x and Q s mw n x. Then, for all n, RŽ n. is also strong S. In particular, RŽ X . is catenarian and strong S. Second claim: R ² X : is not catenarian. Let t g N, t⬘ g M be such that t q t⬘ s 1 in the ring T, and x g m but x f I, in the ring R. Denote by L the quotient field of T Žand thus also of R .. Define the prime ideals Q⬘ and P⬘ of T w X x simply as Q⬘ s f g T w X x such that f Ž trx . g M 4 , P⬘ s f g T w X x such that f Ž trx . s 0 4 s Ž xX y t . L w X x l T w X x . Then P⬘ is an upper to Ž0., Q⬘ is an upper to M, and P⬘ ; Q⬘. Then let Q s Q⬘ l R w X x s f g R w X x such that f Ž trx . g M 4 , P s P⬘ l R w X x s f g R w X x such that f Ž trx . s 0 4 s Ž xX y t . L w X x l R w X x .
426
IHSEN YENGUI
P and Q are two prime ideals of Rw X x such that P ; Q. It is easy to see that P is an upper to Ž0., and Q is an upper to I. For this, let us consider the polynomial g s Ž xX y t . Ž xX y t⬘ . s x 2 X 2 y x Ž t q t⬘ . X q tt⬘ s x 2 X 2 y xX q tt⬘, where tt⬘ g NM s I, but x 2 f I. Hence g g P Ža fortiori g g Q . but g f I w X x. In particular, we have ht Ž P . s 1, and ht Ž Q . s 3. We show that P and Q are consecutive in Rw X x. By way of contradiction, we suppose that we have P ; J ; Q, and consider two cases: ᎏJ contains I. Since Q is an upper to I, it follows that J s I w X x. However, the polynomial g above is such that g g P and g f I w X x. This provides a contradiction. ᎏJ does not contain I. The chain P ; J ; Q lifts in T w X x as P⬘ ; J⬘ ; Q⬙ w1, Proposition 4x. Thus ht Ž Q⬙ . G 3, and necessarily Q⬙ is above N Žthe only prime of height 2 in T .. However, we have Ž xX y t . g P⬘, and t g N, a fortiori Ž xX y t . g Q⬙ and t g Q⬙. Thus xX g Q s Q⬙ l Rw X x. By definition, we would then have t s x Ž trx . g M, a contradiction. To prove our claim Ž R ² X : is not catenarian ., it is enough to show that Q lifts in R ² X :. By way of contradiction suppose that a unitary polynomial f is in Q. Write f s X n q a ny 1 X ny 1 q ⭈⭈⭈ qa0 , with a k g R, 0 F k F n y 1. By definition, we have n
f Ž trx . s Ž trx . q a ny 1 Ž trx .
ny1
q ⭈⭈⭈ qa0 s ␣ ,
where ␣ g M. Then t n q a ny 1 xt ny 1 q ⭈⭈⭈ qa0 x n s x n␣ . Since t ' 1 Žmod M ., we have 1 q x Ž a ny 1 q ⭈⭈⭈ qa0 x ny 1 . g M l R s I, a contradiction. Last claim: R ² X : is not a strong S-ring. The chain P ; Q in Rw X x is such that Q is minimal containing P and I w X x. This chain lifts as P⬘ ; Q⬘
QUESTIONS ABOUT RINGS R Ž X . AND R ² X :
427
in T w X x w1, Proposition 4x. Since Q⬘ is an upper to M and Q is an upper to I we have t.d. T w X x rQ⬘ : R w X x rQ s t.d. w TrM : RrI x s t.d. w TrM : K x s 1. From w1, Lemme 6x, we conclude that ht Rw2xŽ Qw1xrP w1x. G 2 and hence R ² X : is not a strong S-ring.
REFERENCES 1. P.-J. Cahen, Couple d’anneaux partageant un ideal, Arch. Math. Ž Basel . 51 Ž1988., 505᎐514. 2. P.-J. Cahen, Construction B, I, D et anneaux localement ou residuellement de Jaffard, ´ Arch. Math. Ž Basel . 54 Ž1990., 125᎐141. 3. P.-J. Cahen, S. Kabbaj, and Z. E. Khayyari, Krull and valuative dimension of the Serre conjecture ring R ² n:, in Lecture Notes in Pure and Applied Mathematics, Vol. 185, pp. 173᎐185, Dekker, New York, 1997.
One Counterexample for Two Open Questions about the Rings RŽ X . and R ² X : Ihsen Yengui Department of Mathematics, Faculty of Sciences, Sfax, Tunisia Communicated by Walter Feit Received February 16, 1999
We propose to give an answer to two open questions: is R ² X : strong S Žresp., catenarian . when RŽ X . is? We construct a ring R such that RŽ X . is both catenarian and strong S whereas R ² X : is neither catenarian nor strong S. 䊚 1999 Academic Press
INTRODUCTION Let R be a domain and denote by Rw n x the ring of polynomials in n indeterminates with coefficients in R, Žfor n s 1, Rw1x s Rw X x is the ring of polynomials in one indeterminate.. Then RŽ X . s Sy1 Rw X x is the Nagata ring Žin one indeterminate., where S is the multiplicative subset of Rw X x formed by polynomials whose coefficients generate R, and R ² X : s Uy1 Rw X x the Serre conjecture ring, where U is the multiplicative subset formed by the monic polynomials. By induction, one finally defines the Nagata and Serre conjecture rings in n indeterminates as R Ž n . s R Ž n y 1. Ž X .
and
R ² n: s R ² n y 1:² X : .
Clearly, RŽ n. is a localization of R ² n:; hence if the latter ring is a strong S-ring or if it is catenarian, so is the former one. In this paper, we answer the open question about the converse w3, Question 2.8x. Indeed, we construct a counterexample showing that the converse does not hold. Recall that a domain R is strong S Žresp., catenarian . if, for each consecutive pair p ; q of primes in R, the extended primes pw X x ; qw X x are consecutive in Rw X x Žresp., ht Ž q . s ht Ž p . q 1.. A stably strong S-domain Žresp., a universally catenarian domain. R is a domain R such that Rw n x is strong S Žresp., catenarian . for any n. 424 0021-8693r00 $35.00 Copyright 䊚 2000 by Academic Press All rights of reproduction in any form reserved.
QUESTIONS ABOUT RINGS R Ž X . AND R ² X :
425
We use ‘‘; ’’ to denote proper containment. If K : L are two fields we denote t.d.w L : K x the transcendence degree of L over K.
THE EXAMPLE Let K be a field, and T a semi-local universally catenarian K-algebra with two maximal ideals M and N, respectively, of heights 1 and 2 such that t.d.w TrM : K x s 1, and t.d.w TrN : K x s 0, that is, TrN is algebraic over K Žone can proceed as in w2, Exemple Ax.. Let I s M l N, D a one-dimensional valuation domain with quotient field K, and R s ŽT, I, D . s y1 Ž D ., where denotes the natural homomorphism from T onto TrI. The spectrum of R is formed by Ž0., I, primes of height 1 contained in I, and a unique maximal m Žcorresponding to the maximal ideal of D .. Moreover, by w2, Theoreme ´ ` 1x, dim ¨ R s 3. Thus, for all n, ht Ž mw n x. s 3 s ht Ž I w n x. q 1 and dim RŽ n. s dim R ² n: s dim Rw n x y n s ht Ž mw n x. s 3. First claim: for all n, the Nagata ring RŽ n. is catenarian. It is enough to show that for two consecutive primes P ; Q of Rw n x, such that Q : mw n x, one has ht Ž Q . s ht Ž P . q 1. We consider two cases: ᎏP does not contain I; then the chain lifts in T w1, Proposition 4x. In this case, Q cannot be above m since no prime of T is above m. Therefore Q ; mw n x, and thus, ht Ž Q . F 2. Clearly, the claim follows. ᎏP contains I; then I w n x : P ; Q : mw n x. Since we have ht Ž I w n x. s 2, and ht Ž mw n x. s 3, it follows necessarily that P s I w n x and Q s mw n x. Then, for all n, RŽ n. is also strong S. In particular, RŽ X . is catenarian and strong S. Second claim: R ² X : is not catenarian. Let t g N, t⬘ g M be such that t q t⬘ s 1 in the ring T, and x g m but x f I, in the ring R. Denote by L the quotient field of T Žand thus also of R .. Define the prime ideals Q⬘ and P⬘ of T w X x simply as Q⬘ s f g T w X x such that f Ž trx . g M 4 , P⬘ s f g T w X x such that f Ž trx . s 0 4 s Ž xX y t . L w X x l T w X x . Then P⬘ is an upper to Ž0., Q⬘ is an upper to M, and P⬘ ; Q⬘. Then let Q s Q⬘ l R w X x s f g R w X x such that f Ž trx . g M 4 , P s P⬘ l R w X x s f g R w X x such that f Ž trx . s 0 4 s Ž xX y t . L w X x l R w X x .
426
IHSEN YENGUI
P and Q are two prime ideals of Rw X x such that P ; Q. It is easy to see that P is an upper to Ž0., and Q is an upper to I. For this, let us consider the polynomial g s Ž xX y t . Ž xX y t⬘ . s x 2 X 2 y x Ž t q t⬘ . X q tt⬘ s x 2 X 2 y xX q tt⬘, where tt⬘ g NM s I, but x 2 f I. Hence g g P Ža fortiori g g Q . but g f I w X x. In particular, we have ht Ž P . s 1, and ht Ž Q . s 3. We show that P and Q are consecutive in Rw X x. By way of contradiction, we suppose that we have P ; J ; Q, and consider two cases: ᎏJ contains I. Since Q is an upper to I, it follows that J s I w X x. However, the polynomial g above is such that g g P and g f I w X x. This provides a contradiction. ᎏJ does not contain I. The chain P ; J ; Q lifts in T w X x as P⬘ ; J⬘ ; Q⬙ w1, Proposition 4x. Thus ht Ž Q⬙ . G 3, and necessarily Q⬙ is above N Žthe only prime of height 2 in T .. However, we have Ž xX y t . g P⬘, and t g N, a fortiori Ž xX y t . g Q⬙ and t g Q⬙. Thus xX g Q s Q⬙ l Rw X x. By definition, we would then have t s x Ž trx . g M, a contradiction. To prove our claim Ž R ² X : is not catenarian ., it is enough to show that Q lifts in R ² X :. By way of contradiction suppose that a unitary polynomial f is in Q. Write f s X n q a ny 1 X ny 1 q ⭈⭈⭈ qa0 , with a k g R, 0 F k F n y 1. By definition, we have n
f Ž trx . s Ž trx . q a ny 1 Ž trx .
ny1
q ⭈⭈⭈ qa0 s ␣ ,
where ␣ g M. Then t n q a ny 1 xt ny 1 q ⭈⭈⭈ qa0 x n s x n␣ . Since t ' 1 Žmod M ., we have 1 q x Ž a ny 1 q ⭈⭈⭈ qa0 x ny 1 . g M l R s I, a contradiction. Last claim: R ² X : is not a strong S-ring. The chain P ; Q in Rw X x is such that Q is minimal containing P and I w X x. This chain lifts as P⬘ ; Q⬘
QUESTIONS ABOUT RINGS R Ž X . AND R ² X :
427
in T w X x w1, Proposition 4x. Since Q⬘ is an upper to M and Q is an upper to I we have t.d. T w X x rQ⬘ : R w X x rQ s t.d. w TrM : RrI x s t.d. w TrM : K x s 1. From w1, Lemme 6x, we conclude that ht Rw2xŽ Qw1xrP w1x. G 2 and hence R ² X : is not a strong S-ring.
REFERENCES 1. P.-J. Cahen, Couple d’anneaux partageant un ideal, Arch. Math. Ž Basel . 51 Ž1988., 505᎐514. 2. P.-J. Cahen, Construction B, I, D et anneaux localement ou residuellement de Jaffard, ´ Arch. Math. Ž Basel . 54 Ž1990., 125᎐141. 3. P.-J. Cahen, S. Kabbaj, and Z. E. Khayyari, Krull and valuative dimension of the Serre conjecture ring R ² n:, in Lecture Notes in Pure and Applied Mathematics, Vol. 185, pp. 173᎐185, Dekker, New York, 1997.