When is each proper overring of R an S(Eidenberg)-domain?

Publ. Mat. 46 (2002), 435–440 WHEN IS EACH PROPER OVERRING OF RAN S(EIDENBERG)-DOMAIN? Noˆ omen Jarboui Abstract A domain Ris called a maximal “non-S” subring of a field Lif R⊂L,Ris not an S-domain and each domain Tsuch that R⊂ T⊆Lis an S-domain. We show that maximal “non-S”subrings R of a field Lare the integrally closed pseudo-valuation domains satisfying dim(R)=1,dim v(R) = 2 and L=qf(R). 1. Introduction Throughout this paper, R→Sdenotes an extension of commutative integral domains, qf(R) the quotient field of an integral domain Rand tr.deg[S:R] the transcendence degree of qf(S) over qf(R). If tr.deg[S: R] = 0, we say that Sis algebraic over R. We recall that a ring Rof finite Krull dimension nis a Jaffard ring if its valuative dimension (the limit of the sequence (dim(R[X1,...,X n]) −n,n∈N)) dimv(R), is also n. Pr¨ufer domains and Noetherian domains are Jaffard domains. Recall that a domain Ris an S-domain [12] if for each height 1 prime ideal p of R, the extended prime p[X] in one indeterminate is also height 1 in R[X]. We assume familiarity with these concepts as in [1] and [12]. In [3], the author and M. Ben Nasr considered maximal non-Jaffard subrings of a field L, that is, the domains Rwhere Ris a non Jaffard domain and each ring T,R⊂T⊆Lis Jaffard. They characterized these domains in terms of pseudo-valuation domains. On the other hand the author and I. Yengui in [11] studied the domains Rsuch that each domain contained between Rand its quotient field is an S-domain. They are said to be absolutely S-domains. To complete this circle of ideas and to honor Seidenberg we deal with maximal “non-S” subring(s) of a field; that is, the domains R, where Ris not an S-domain and each ring T, R⊂T⊆Lis an S-domain. First we show that if Ris a maximal 2000 Mathematics Subject Classification. Primary: 13B02; Secondary: 13C15, 13A17, 13A18, 13B25, 13E05. Key words. Jaffard domain, S-domain, valuation domain, Krull dimension, pullback. 436 N. Jarboui “non-S” subring of a field L, then L= qf(R). Hence, we may restrict ourselves to the case where L= qf(R). Let us recall some terminology: Let Tbe a ring, Ian ideal of T,Dbe a subring of T/I and let Rbe the subring of Tdefined by the following pullback construction: R−−−−→D       T−−−−→T/I Following [4], we say that Ris the ring of the (T,I,D)construction and we set R:= (T,I,D). Note that R:= (T,I,D) if and only it is contained in Tand shares the ideal Iwith the ring T. The (T,I,D) constructions were considered for the first time in [7], in the contest of general pullback construction. Particularly the last construction to be noted here concerns the notion of a pseudo-valuation domain (for short, a PVD), which was introduced by J. R. Hedstrom and E. G. Houston [9] and has been studied subsequently in [2], [5], [6] and [10]. A domain R is said to be a PVD in case each prime ideal pof Ris strongly prime, in the sense that whenever x, y ∈qf(R) satisfy xy ∈p, then either x∈por y∈p, equivalently, in case Rhas a (uniquely determined) valuation overring Vsuch that Spec(R) = Spec(V) as sets, equivalently (by [2, Proposition 2.6]) in case Ris a pullback of the form V×Kk, where Vis a valuation domain with residue field Kand kis a subfield of K. As the terminology suggests, any valuation domain is a PVD [9, Proposition 1.1]. Although the converse is false [9, Example 2.1], any PVD must, at least, be local [9, Corollary 1.3]. The main result of this paper is Theorem 2.2, which states that Ris a maximal “non-S” subring of qf(R) if and only if Ris an integrally closed pseudo-valuation domain with dim(R) = 1 and dimv(R) = 2. As an application of Theorem 2.2, we give necessary and sufficient conditions for certain pullbacks to be maximal “non-S” subrings of their quotient fields. 2. Main results Let Rbe a domain contained in a field L. We say that Ris a maximal “non-S”subring of Lif Ris not an S-domain and each ring Tsuch that R⊂T⊆Lis an S-domain. First of all, we establish the following: Proposition 2.1. Let Rbe a domain and La field containing R.IfR is a maximal “non-S”subring of L, then L= qf(R). Proper Overrings 437 Proof: First notice that Lis algebraic over R. Indeed, if not then there exists an element tof Ltranscendental over R. Hence each overring of R[t] should be an S-domain that is R[t] is an absolutely S-domain. Hence by [11, Proposition 1.14] Ris a field which contradicts the fact that Ris not an S-domain. Now our task is to show that L= qf(R). Assume that qf(R)⊂L, and let α∈L\qf(R). Then αis algebraic over R. Thus there exists an element r∈Rsuch that rα is integral over R.Thus R⊂R[rα] is an integral extension. But R[rα] is an S-domain. Hence R is an S-domain, the desired contradiction to complete the proof. As a direct consequence of Proposition 2.1, the study of maximal “non-S” subring(s) of a field Lcan be reduced to the case where L= qf(R). Now notice that if Ris a maximal “non-S”subring of qf(R), then Ris integrally closed. Indeed, if R=R, then Ris an S-domain, and hence so is R(since R⊂Ris an integral extension), which is impossible. Our main result is the following: Theorem 2.2. Let Rbe a domain. Then the following statements are equivalent: (i) Ris a maximal “non-S” subring of qf(R); (ii) Ris an integrally closed PVD with dim(R)=1and dimv(R)=2. Proof: (i) ⇒(ii). We have already noticed that Ris integrally closed. On the other hand since Ris not an S-domain, then there is a height 1 prime ideal pof Rsuch that ht(p[X]) = 2. Then there is a nonzero prime ideal Pof R[X] contained in p[X] such that P∩R= (0). Thus Ris a subring of R1=R[X]/P which is isomorphic to R[u], where uis an algebraic element over R.By[8, Corollary 19.7], there is a valuation overring Wof R1containing a prime ideal Pof height 1 such that P∩R1=p[X]/P. Denoting V=W∩qf(R), Vis a valuation overring of Rcontaining a height 1 prime ideal q=P∩qf(R)[8, Theorem 19.16] such that q∩R=p.Now,tr.deg[W/P:V/q]=0[8, Theorem 19.16]. Hence tr.deg[V/q :R/p]=tr.deg[W/P:R/p] ≥tr.deg[R1/(p[X]/P ):R/p] = tr.deg[(R[X]/P )/(p[X]/P ):R/p] = tr.deg[(R[X]/p[X]) : R/p]=1. 438 N. Jarboui Assume that R=(Vq,qV q,R p/pRp), then the domain (Vq,qV q,R p/pRp) is a proper overring of Rand it should be an S-domain and by [11, Proposition 1.4], we get tr.deg[Vq/qVq:Rp/pRp] = 0 which is impossible. Therefore R:= (Vq,qV q,R p/pRp). Hence Ris a PVD (cf. [2]). Our task now is to show that tr.deg[Vq/qVq:Rp/pRp] = 1. The extension Rp/pRp⊂Vq/qVqcan not be algebraic since Ris not an S-domain [11, Proposition 1.4]. Assume that tr.deg[Vq/qVq:Rp/pRp]≥2, and let X,Ybe two transcendental algebraically independent elements of Vq/qVqover Rp/pRp. Then the domain T:= (Vq,qV q,(Rp/pRp)[X]) is a proper overring of R,thusTis an S-domain. Hence by [11, Proposition 1.4], we get tr.deg[Vq/qVq:(Rp/pRp)[X]] = 0, which is impossible. Hence tr.deg[Vq/qVq:Rp/pRp] = 1. Therefore by [1, Proposition 2.5], dim(R) = 1 and dimv(R)=2. (ii) ⇒(i). Since Ris a PVD, then R:= (V,M,k), where Vis a valuation domain with maximal ideal Mand kis a field. It is clear that Ris not an S-domain because tr.deg[V/M :R/M] = 1. Now, let Tbe a domain such that R⊂T⊆qf(R). Then by [3, Lemma 1.3], either Tis an overring of V, so it is an S-domain, or Tis an intermediate domain between Rand V,soT:= (V,M,D), where R/M ⊂D⊆V/M. Since R is integrally closed, then tr.deg[V/M :D] = 0. Thus Tis an S-domain. Hence Ris a maximal “non-S” subring of qf(R). Now we determine when a pullback Ris a maximal “non-S” subring of its quotient field. We recall some notation for conductors. If Ris a domain and I,Jare R-submodules of qf(R), then (I:J)={x∈ qf(R)|xJ ⊂I}.IfRis a PVD with associated valuation domain Vand maximal ideal M, assume that R=V, then Mis not a principal ideal of Rand V=(M:M)[2, Proposition 2.3], and by [2, Lemma 2.4], we get V=(R:M)=(M:M). We establish the following theorem. Theorem 2.3. Let Tbe a domain, Ma maximal ideal of Tand Da subring of the field K=T/M.LetR:= (T,M,D). Then the following statements are equivalent: (i) Ris a maximal “non-S” subring of qf(R); (ii) Dis a field algebraically closed in (M:M)/M , with tr.deg[K: D]=1and Tis a one-dimensional Jaffard PVD. Proof: (i) ⇒(ii). By Theorem 2.2, Ris a PVD. Hence there exists a valuation domain Vwith mas a maximal ideal such that R:= (V, m, k), where kis a field. Since Tis an overring of R, then by [3, Lemma 1.3], either R⊂T⊆Vor V⊆T. Proper Overrings 439 Case 1: If R⊂T⊆V, then Tshares the ideal mwith Rand V,so T:= (V, m, T/m). But we have M⊆m(since Ris local with maximal ideal m). Thus M=mbecause Mis a maximal ideal of T. Hence T:= (V,M,K), D=R/M =R/m =k,soDis a field. On the other hand Ris integrally closed (Theorem 2.2), thus Dis algebraically closed in V/M =(M:M)/M . We have dim(T) = dim(V) = dim(R) = 1, and since Tis an S-domain, then dim(T) = dimv(T) = 1. Now tr.deg[K: D] = dimv(R)−dimv(T)=1. Case 2: If Tis an overring of V, then T=Vsince Vis a one-dimensional valuation domain. Thus m=M. This yields D=R/M =R/m =k and it is obvious that Dis algebraically closed in V/M =(M:M)/M . On the other hand tr.deg[K:D] = dimv(R)−dimv(T)=1. (ii) ⇒(i). Since D⊂Kis not an algebraic extension, then Ris not an S-domain [11, Proposition 1.4]. The ring Tis a PVD, so there is a valuation domain Wwith maximal ideal Msuch that T:= (W, M, K). But R:= (T,M,D). Hence Ris a PVD with associated valuation domain W=(M:M). Furthermore, dim(R) = dim(T) = 1 and dimv(R) = dimv(T) + dimv(D)+tr.deg[K:D] = 2. Since Dis algebraically closed in W/M, then Ris integrally closed. Thus by Theorem 2.2, Ris a maximal “non-S” subring of qf(R). Acknowledgement. The author express thanks to the referees for valuable suggestions. References [1] D. F. Anderson, A. Bouvier, D. E. Dobbs, M. Fontana and S. Kabbaj, On Jaffard domains, Exposition. Math. 6(2) (1988), 145–175. [2] D. F. Anderson and D. E. Dobbs, Pairs of rings with the same prime ideals, Canad. J. Math. 32(2) (1980), 362–384. [3] M. Ben Nasr and N. Jarboui, Maximal non-Jaffard subrings of a field, Publ. Mat. 44(1) (2000), 157–175. [4] P.-J. Cahen, Couples d’anneaux partageant un id´eal, Arch. Math. (Basel) 51(6) (1988), 505–514. [5] D. E. Dobbs, Coherence, ascent of going-down, and pseudovaluation domains, Houston J. Math. 4(4) (1978), 551–567. [6] D. E. Dobbs, On the weak global dimension of pseudo-valuation domains, Canad. Math. Bull. 21(2) (1978), 159–164. [7] M. Fontana, Topologically defined classes of commutative rings, Ann. Mat. Pura Appl. (4) 123 (1980), 331–355. 440 N. Jarboui [8] R. Gilmer,“Multiplicative ideal theory”, Pure and Applied Mathematics 12, Marcel Dekker, Inc., New York, 1972. [9] J. R. Hedstrom and E. G. Houston, Pseudo-valuation domains, Pacific J. Math. 75(1) (1978), 137–147. [10] J. R. Hedstrom and E. G. Houston, Pseudo-valuation domains. II, Houston J. Math. 4(2) (1978), 199–207. [11] N. Jarboui and I. Yengui, Absolutely S-domains and pseudo polynomial rings, Colloq. Math. (to appear). [12] I. Kaplansky,“Commutative rings”, revised edition, The University of Chicago Press, Chicago, Ill.-London, 1974. Department of Mathematics Faculty of Sciences of Sfax 3018 Sfax, BP 802 Tunisia E-mail address:[email protected] Primera versi´o rebuda el 8 de novembre de 2001, darrera versi´o rebuda el 16 de maig de 2002.