International Journal of Mathematics and Mathematical Sciences

Volume 2015, Article ID 938648, 3 pages

http://dx.doi.org/10.1155/2015/938648

## A Direct Proof of a Theorem concerning Treed Overrings

Faculty of Science, Sana’a University, P.O. Box 12460, Sana’a, Yemen

Received 15 April 2015; Accepted 23 June 2015

Academic Editor: David E. Dobbs

Copyright © 2015 Ahmed Ayache. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### 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 if is an integrally closed local domain and each of its *proper* overrings is treed, then is a valuation domain or is a PVD with associated valuation overring such that .

#### 1. Introduction

All rings considered are assumed to be commutative integral domains with identity. Throughout this paper, is an integral domain with quotient field . We denote by its set of prime ideals. As usual, an overring of is a ring contained between and , and if, in addition, , we say that is a proper overring of . For an element , the conductor of to is an ideal of defined by . If denotes the Krull dimension of , then the valuative dimension of is defined to be the supermum (possibly ) of as ranges over the set of overrings of . For a ring extension of integral domains, we denote by the transcendence degree of the quotient field of over that of .

Let us recall some basic definitions:(i)is said to be a* going-down domain* if the extension is going down for every domain containing [1–4]. The most natural examples of going-down domains are arbitrary Prüfer domains and domains of (Krull) dimension 1.(ii) is called a* treed domain* if its spectrum , endowed with the natural partial ordering, is a tree. It was shown in [2, Theorem ] that each going-down domain must be treed. The converse, however, is false (cf. [3, Example ]).(iii)Following Hedstrom and Houston [5, 6], is called a* pseudovaluation domain* (or, in short, PVD) if each prime ideal of is strongly prime; that is, whenever and are such that , then either or . Recall from [5, Theorem ] that is a PVD if and only if there is a (uniquely determined) valuation overring of such that as sets; in this case, is called the canonically associated valuation overring of . It is shown in [7, Proposition ] that the PVDs may be characterized as the Cartesian products arising from a valuation domain and a field .

In general, we have the following diagram of (irreversible) implications: is a PVD is going-down is treed.

This paper is primary concerned with treed domains. In 1987, Dobbs has constructed a nonintegrally closed, two-dimensional which is not going down, although each of its overrings is treed [1, Example ]. It has since been an open question whether an integrally closed, local treed domain of valuative dimension two such that is an algebraically closed field and each overring of 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 ]. In 2014, the author answered the question of Dobbs in the general context by proving that an integrally closed local domain such that each of its overrings is treed is a PVD [9, Theorem ]. 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 is an integrally closed local domain and each of its* proper* overrings is treed, then is a valuation domain or is a PVD with associated valuation overring such that .

#### 2. Main Theorem

The following result, which we label here as Proposition 1 for the sake of reference, collects some needed properties concerning treed domains [10, Corollary & Lemma ].

Proposition 1. *Let be an extension of integral domains and a prime ideal of . If each intermediate ring between and is treed, then*(i)*there are no prime ideals of lying over ;*(ii)*for every prime ideal of lying over one has . In particular, if is nonmaximal, then is algebraic.*

*Theorem 2. Let be an integrally closed local domain. If each proper overring of is treed, then is a valuation domain or is a PVD with associated valuation overring such that .*

*Proof. *Assume that is not a valuation domain. According to [11, Theorem ], there is a valuation overring of centred on such that . Let be the residue field of and let be the pullback ring . Then is a PVD with associated valuation overring [7, Proposition ]. Furthermore, as every intermediate ring between and is treed, then (Proposition 1(ii)); that is, . Our task is to prove that by showing that . Assume, by way of contradiction, that and let . Then , so and . Therefore, and . By the -lemma, is a prime ideal of lying over [4, Remark, page 168].

(1) Identify , where is a nonzero prime ideal of the polynomial ring such that . Then can be written in the form . It follows that is a principal ideal domain and .

(2) Set . We necessarily have , such for if , then , so is a root of a polynomial over and some coefficient of which is a unit. By the -lemma, either or , a contradiction. In fact, we can say that are consecutive since is a nonzero prime of , and is a maximal ideal of .

(3) Let be the union of the nonmaximal prime ideals of . If , then there is a prime ideal of such that and . Set . If , then are necessarily consecutive (Proposition 1(i)), and a fortiori , a contradiction. Thus, . As is treed, and are comparable. But since are consecutive, we have , so . Again, the -lemma leads to a contradiction. Hence, are consecutive.

(4) Note that is a common prime ideal of and [5, Theorem ] and is the pullback ring . Then is a PVD with associated valuation overring [7, Proposition ]. Furthermore, we have [12, Théorème (ii)]. Therefore, there is a rank two-valuation overring of . According to [13, Proposition (a)], is comparable to under containment. In fact, we necessarily have since , while . There is a valuation ring such that and . Thus, is a maximal chain of prime ideals of .

(5) We claim that there is no prime ideal of contained in and lying over . Indeed, assume that . Because every intermediate ring between and is treed and is not a maximal ideal of , (Proposition 1(ii)). By application of [14, Theorem ], the algebraic extension satisfies the inequality formula:Since and , it follows thatOn the other hand, by application of [14, Corollary ] to the extension , we getHence, From and , we obtain the contradiction(6) Set . Then and are both contained in . As is treed, then and are comparable. Since every intermediate ring between and is treed and , then does not lie over (Proposition 1(i)). Thus . In view of the point (5), we must have .

(7) By virtue of [11, Theorem ], there is a valuation overring of centred on and having a chain of prime ideals lying over the chain . Note that and are incomparable. Indeed, the inclusion does not hold by the point (5), and the inclusion is impossible since . It follows that and are incomparable. Set ; then is a semilocal Prüfer domain with and as maximal ideals. Let and . Then is a proper overring of () (cf. [15, Theorem ]) and is a local ring with maximal ideal . Moreover, there is a bijective correspondence preserving inclusion from the set of prime ideals of which do not contain and the set of prime ideals of which do not contain . It follows that and are incomparable and are strictly contained in the maximal ideal of , a contradiction, since is treed. Hence, and , as desired.

*Conflict of Interests*

*The author declares that there is no conflict of interests regarding the publication of this paper.*

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