WHEN IS EACH PROPER OVERRING OF R AN S(EIDENBERG)-DOMAIN?

A domain R is called a maximal “non-S” subring of a field L if R ⊂ L, R is not an S-domain and each domain T such that R ⊂ T ⊆ L is an S-domain. We show that maximal “non-S”subrings R of a field L are the integrally closed pseudo-valuation domains satisfying dim(R) = 1, dimv(R) = 2 and L = qf(R).


Introduction
Throughout this paper, R → S denotes an extension of commutative integral domains, qf(R) the quotient field of an integral domain R and tr.deg[S : R] the transcendence degree of qf(S) over qf(R).If tr.deg[S : R] = 0, we say that S is algebraic over R. We recall that a ring R of finite Krull dimension n is a Jaffard ring if its valuative dimension (the limit of the sequence (dim(R[X 1 , . . ., X n ]) − n, n ∈ N)) dim v (R), is also n.Prüfer domains and Noetherian domains are Jaffard domains.Recall that a domain R is 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 R where R is a non Jaffard domain and each ring T , R ⊂ T ⊆ L is 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 R such that each domain contained between R and 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 R is not an S-domain and each ring T , R ⊂ T ⊆ L is an S-domain.First we show that if R is a maximal 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 T be a ring, I an ideal of T , D be a subring of T/I and let R be the subring of T defined by the following pullback construction: Following [4], we say that R is 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 T and shares the ideal I with 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 p of R is strongly prime, in the sense that whenever x, y ∈ qf(R) satisfy xy ∈ p, then either x ∈ p or y ∈ p, equivalently, in case R has a (uniquely determined) valuation overring V such that Spec(R) = Spec(V ) as sets, equivalently (by [ 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.

Main results
Let R be a domain contained in a field L. We say that R is a maximal "non-S"subring of L if R is not an S-domain and each ring T such that R ⊂ T ⊆ L is an S-domain.
First of all, we establish the following: Proposition 2.1.Let R be a domain and L a field containing R. If R is a maximal "non-S"subring of L, then L = qf(R).
Proof: First notice that L is algebraic over R. Indeed, if not then there exists an element t of L transcendental 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] R is a field which contradicts the fact that R is 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 α is algebraic over R.
Thus there exists an element r ∈ R such that rα is integral over 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 L can be reduced to the case where L = qf(R).Now notice that if R is a maximal "non-S"subring of qf(R), then R is 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 R be a domain.Then the following statements are equivalent: (i) R is a maximal "non-S" subring of qf(R); (ii) R is an integrally closed PVD with dim(R) = 1 and dim v (R) = 2.
Proof: (i) ⇒ (ii).We have already noticed that R is integrally closed.On the other hand since R is not an S-domain, then there is a height 1 prime ideal p of R such that ht(p[X]) = 2. Then there is a nonzero prime ideal where u is an algebraic element over R. By [8, Corollary 19.7], there is a valuation overring W of R 1 containing a prime ideal P of height 1 such that Assume that R = (V q , qV q , R p /pR p ), then the domain (V q , qV q , R p /pR p ) is a proper overring of R and it should be an S-domain and by [11,Proposition 1.4], we get tr.deg[V q /qV q : R p /pR p ] = 0 which is impossible.Therefore R := (V q , qV q , R p /pR p ). Hence R is a PVD (cf.[2]).
Our task now is to show that tr.deg[V q /qV q : R p /pR p ] = 1.The extension R p /pR p ⊂ V q /qV q can not be algebraic since R is not an S-domain [11,Proposition 1.4].Assume that tr.deg[V q /qV q : R p /pR p ] ≥ 2, and let X, Y be two transcendental algebraically independent elements of V q /qV q over R p /pR p .Then the domain T := (V q , qV q , (R p /pR p )[X]) is a proper overring of R, thus T is an S-domain.Hence by [11, Proposition 1.4], we get tr.deg[V q /qV q : (R p /pR p )[X]] = 0, which is impossible.Hence tr.deg[V q /qV q : R p /pR p ] = 1.Therefore by [1, Proposition 2.5], dim(R) = 1 and dim v (R) = 2.
(ii) ⇒ (i).Since R is a PVD, then R := (V, M, k), where V is a valuation domain with maximal ideal M and k is a field.It is clear that R is not an S-domain because tr.deg[V/M : R/M ] = 1.Now, let T be a domain such that R ⊂ T ⊆ qf(R).Then by [3, Lemma 1.3], either T is an overring of V , so it is an S-domain, or T is an intermediate domain between R and V , so Hence R is a maximal "non-S" subring of qf(R).Now we determine when a pullback R is a maximal "non-S" subring of its quotient field.We recall some notation for conductors.If R is a domain and I, J are R-submodules of qf(R), then (I : If R is a PVD with associated valuation domain V and maximal ideal M , assume that R = V , then M is not a principal ideal of R and 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.(ii) ⇒ (i).Since D ⊂ K is not an algebraic extension, then R is not an S-domain [11,Proposition 1.4].The ring T is a PVD, so there is a valuation domain W with maximal ideal M such that T := (W, M, K).

Theorem 2 . 3 .Case 1 : 1 . 2 :
Let T be a domain, M a maximal ideal of T and D a subring of the field K = T/M.Let R := (T, M, D).Then the following statements are equivalent: (i) R is a maximal "non-S" subring of qf(R); (ii) D is a field algebraically closed in (M : M )/M , with tr.deg[K : D] = 1 and T is a one-dimensional Jaffard PVD.Proof: (i) ⇒ (ii).By Theorem 2.2, R is a PVD.Hence there exists a valuation domain V with m as a maximal ideal such thatR := (V, m, k),where k is a field.Since T is an overring of R, then by [3, Lemma1.3],either R ⊂ T ⊆ V or V ⊆ T .If R ⊂ T ⊆ V ,then T shares the ideal m with R and V , so T := (V, m, T /m).But we have M ⊆ m (since R is local with maximal ideal m).Thus M = m because M is a maximal ideal of T .Hence T := (V, M, K), D = R/M = R/m = k, so D is a field.On the other hand R is integrally closed (Theorem 2.2), thus D is algebraically closed in V /M = (M : M )/M .We have dim(T ) = dim(V ) = dim(R) = 1, and since T is an S-domain, then dim(T ) = dim v (T ) = 1.Now tr.deg[K : D] = dim v (R) − dim v (T ) = Case If T is an overring of V , then T = V since V is a one-dimensional valuation domain.Thus m = M .This yields D = R/M = R/m = k and it is obvious that D is algebraically closed in V/M = (M : M )/M .On the other hand tr.deg[K : D] = dim v (R) − dim v (T ) = 1.
But R := (T, M, D).Hence R is a PVD with associated valuation domain W = (M : M ).Furthermore, dim(R) = dim(T ) = 1 and dim v (R) = dim v (T ) + dim v (D) + tr.deg[K : D] = 2. Since D is algebraically closed in W/M ,then R is integrally closed.Thus by Theorem 2.2, R is a maximal "non-S" subring of qf(R).