# A Direct Proof of a Theorem concerning Treed Overrings

Authors: Ahmed Ayache
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

---

## Abstract

We improve and provide a direct proof of a theorem that answered a question by Dobbs about treed overrings. More precisely, we show that ifR is an integrally closed local domain and each of its proper overrings is treed, then R is a valuation domain or R is a PVD with associated valuation overring V such that tr.deg[V/M:R/M]=1.


---

## Body

## 1. Introduction

All rings considered are assumed to be commutative integral domains with identity. Throughout this paper,R is an integral domain with quotient field qf(R). We denote by Spec(R) its set of prime ideals. As usual, an overring T of R is a ring contained between R and qf(R), and if, in addition, T≠R, we say that T is a proper overring of R. For an element u∈qf(R), the conductor of u to R is an ideal of R defined by (R:u)={r∈R:ru∈R}. If dimR denotes the Krull dimension of R, then the valuative dimension dimvR of R is defined to be the supermum (possibly ∞) of dimT as T ranges over the set of overrings of R. For a ring extension R⊂S of integral domains, we denote by tr.deg[S:R] the transcendence degree of the quotient field of S over that of R.Let us recall some basic definitions:(i)
Ris said to be agoing-down domain if the extension R⊂T is going down for every domain T containing R [1–4]. The most natural examples of going-down domains are arbitrary Prüfer domains and domains of (Krull) dimension 1.(ii)
R is called atreed domain if its spectrum Spec(R), endowed with the natural partial ordering, is a tree. It was shown in [2, Theorem 2.2] that each going-down domain must be treed. The converse, however, is false (cf. [3, Example 2.2]).(iii)
Following Hedstrom and Houston [5, 6], R is called apseudovaluation domain (or, in short, PVD) if each prime ideal P of R is strongly prime; that is, whenever x∈qf(R) and y∈qf(R) are such that xy∈P, then either x∈P or y∈P. Recall from [5, Theorem 2.7] that R is a PVD if and only if there is a (uniquely determined) valuation overring V of R such that Spec(R)=Spec(V) as sets; in this case, V is called the canonically associated valuation overring of R. It is shown in [7, Proposition 2.6] that the PVDs may be characterized as the Cartesian products V×Fk arising from a valuation domain (V,M) and a field k⊆F=V/M.In general, we have the following diagram of (irreversible) implications:R is a PVD ⇒R is going-down ⇒R is treed.This paper is primary concerned with treed domains. In 1987, Dobbs has constructed a nonintegrally closed, two-dimensionalR which is not going down, although each of its overrings is treed [1, Example 2.3]. It has since been an open question whether an integrally closed, local treed domain (R,M) of valuative dimension two such that R/M is an algebraically closed field and each overring of R is treed is necessarily going down. In 2008, Ayache and Jarboui have provided an affirmative answer by showing that an integrally closed domain of valuative dimension two such that each of its overrings is treed is necessarily going down [8, Corollary 1]. In 2014, the author answered the question of Dobbs in the general context by proving that an integrally closed local domain R such that each of its overrings is treed is a PVD [9, Theorem 10]. The proof was too long and took many steps. In this paper, we present a direct result of the same proof with a slight improvement. Indeed, we do prove that if R is an integrally closed local domain and each of itsproper overrings is treed, then R is a valuation domain or R is a PVD with associated valuation overring V such that tr.deg[V/M:R/M]=1.


## 2. Main Theorem

The following result, which we label here as Proposition1 for the sake of reference, collects some needed properties concerning treed domains [10, Corollary 2.9 & Lemma 2.10].Proposition 1.
LetR⊂S be an extension of integral domains and P a prime ideal of R. If each intermediate ring between R and S is treed, then(i)
there are no prime idealsQo⊂Q1⊂Q2 of  S lying over P;(ii)
for every prime idealQ of S lying over P one has tr.deg[S/Q:R/P]≤1. In particular, if P is nonmaximal, then R/P⊂S/Q is algebraic.Theorem 2.
Let(R,M) be an integrally closed local domain. If each proper overring of R is treed, then R is a valuation domain or R is a PVD with associated valuation overring V such that tr.deg[V/M:R/M]=1.Proof.
Assume thatR is not a valuation domain. According to [11, Theorem 19.15], there is a valuation overring (V,Qo) of R centred on M such that tr.deg[V/Qo:R/M]>0. Let K be the residue field of V and let T:=R+Qo be the pullback ring V×KR/M. Then T is a PVD with associated valuation overring V [7, Proposition 2.6]. Furthermore, as every intermediate ring between T and V is treed, then tr.deg[V/Qo:T/Qo]=tr.deg[V/Qo:R/M]≤1 (Proposition 1(ii)); that is, tr.deg[V/Qo:R/M]=1. Our task is to prove that R=T by showing that Qo=M. Assume, by way of contradiction, that M⊂Qo and let u∈Qo∖M. Then u-1∉V, so u∉R and u-1∉R. Therefore, (R:u)⊆M and (R:u-1)⊆M. By the u,u-1-lemma, MR[u] is a prime ideal of R[u] lying over M [4, Remark, page 168].
(1) IdentifyR[u]:=R[X]/P∗, where P∗ is a nonzero prime ideal of the polynomial ring R[X] such that P∗∩R=(0). Then MR[u] can be written in the form MR[u]=M[X]/P∗. It follows that R[u]/MR[u]≅R[X]/M[X]≅(R/M)[X] is a principal ideal domain and tr.deg[R[u]/MR[u]:R/M]=1.
(2) SetN:=Qo∩R[u]. We necessarily have MR[u]⊂N, such for if N=MR[u], then u∈Qo∩R[u]=MR[u], so u-1 is a root of a polynomial over R and some coefficient of which is a unit. By the u,u-1-lemma, either u∈R or u-1∈R, a contradiction. In fact, we can say that MR[u]⊂N are consecutive since N/MR[u] is a nonzero prime of R[u]/MR[u], and N is a maximal ideal of R[u].
(3) LetPo be the union of the nonmaximal prime ideals of V. If Po=Qo, then there is a prime ideal Q′ of V such that Q′⊂Qo and u∈Q′. Set P′:=Q′∩R[u]. If P′=N, then Q′⊂Qo are necessarily consecutive (Proposition 1(i)), and a fortiori Po=Q′, a contradiction. Thus, P′⊂N. As R[u] is treed, P′ and MR[u] are comparable. But since MR[u]⊂N are consecutive, we have P′⊆MR[u], so u∈Q′∩R[u]⊆MR[u]. Again, the u,u-1-lemma leads to a contradiction. Hence, Po⊂Qo are consecutive.
(4) Note thatPo is a common prime ideal of T and V [5, Theorem 2.7] and T/Po is the pullback ring V/Po×KR/M. Then T/Po is a PVD with associated valuation overring V/Po [7, Proposition 2.6]. Furthermore, we have dimvT/Po=2 [12, Théorème 1(ii)]. Therefore, there is a rank two-valuation overring (W′,M′) of T/Po. According to [13, Proposition 1.3(a)], W′ is comparable to V/Po under containment. In fact, we necessarily have T/Po⊂W′⊂V/Po since dimW′=2, while dimV/Po=1. There is a valuation ring (W1,M1) such that T⊂W1⊂V and W′=W1/Po. Thus, Po⊂Qo⊂M1 is a maximal chain of prime ideals of W1.
(5) We claim that there is no prime idealQ of V contained in Qo and lying over MR[u]. Indeed, assume that Q∩R[u]=MR[u]. Because every intermediate ring between R[u] and V is treed and MR[u] is not a maximal ideal of R[u], tr.deg[V/Q:R[u]/MR[u]]=0 (Proposition 1(ii)). By application of [14, Theorem 1.5], the algebraic extension R[u]/MR[u]⊂V/Q satisfies the inequality formula:(1)htQoQ+tr.deg⁡VQo:RuN≤htNMRu+tr.deg⁡VQ:RuMRu.Since tr.deg[V/Q:R[u]/MR[u]]=0 and ht(N/MR[u])=1, it follows that∗tr.degVQo:RuN=0.
On the other hand, by application of [14, Corollary 1.4] to the extension R/M⊂R[u]/MR[u], we get(2)htNMRu+tr.deg⁡RuN:RM≤tr.deg⁡RuMRu:RM=1.Hence, ∗∗tr.degRuN:RM=0.
From∗ and ∗∗, we obtain the contradiction(3)tr.deg⁡VQo:RM=tr.deg⁡VQo:RuN+tr.deg⁡RuN:RM=0.
(6) SetP1=Po∩R[u]. Then P1 and MR[u] are both contained in N. As R[u] is treed, then P1 and MR[u] are comparable. Since every intermediate ring between R[u] and V is treed and Qo∩R[u]=M1∩R[u]=N, then Po does not lie over N (Proposition 1(i)). Thus P1⊆MR[u]. In view of the point (5), we must have P1⊂MR[u].
(7) By virtue of [11, Theorem 19.7], there is a valuation overring (W2,M2) of R[u] centred on N and having a chain of prime ideals Po⊂Q1⊂M2 lying over the chain P1⊂MR[u]⊂N. Note that Qo and Q1 are incomparable. Indeed, the inclusion Q1⊂Qo does not hold by the point (5), and the inclusion Qo⊆Q1 is impossible since Qo∩R[u]=N⊈Q1∩R[u]=MR[u]. It follows that W1 and W2 are incomparable. Set S:=W1∩W2; then S is a semilocal Prüfer domain with m1=M1∩S and m2=M2∩S as maximal ideals. Let I:=m1∩m2 and H:=R+I. Then H is a proper overring of R (R[u]⊂H) (cf. [15, Theorem 1.4]) and H is a local ring with maximal ideal I. Moreover, there is a bijective correspondence preserving inclusion from the set of prime ideals of S which do not contain I and the set of prime ideals of H which do not contain I. It follows that Qo∩H and Q1∩H are incomparable and are strictly contained in the maximal ideal I of H, a contradiction, since H is treed. Hence, Qo=M and R=T, as desired.



---

*Source: 938648-2015-07-08.xml*