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Page 1: Some questions concerning proper subrings

Ricerche mat.DOI 10.1007/s11587-014-0188-6

Some questions concerning proper subrings

Noômen Jarboui · Manar El Islam Toumi ·Salma Trabelsi

Received: 11 February 2014 / Revised: 1 May 2014© Università degli Studi di Napoli "Federico II" 2014

Abstract Given a ring extension R ⊂ S of integral domains, it is shown that if eachproper subring of S containing R is integrally closed, then S is integrally closed. Asan application, we show that if each proper subring of S containing R is a valuation(resp., Prüfer, resp. Principal ideal) domain, then so is S.

Keywords Integral domain · Intermediate ring · Overring · Ring extension · Integralextension · Integrally closed · Prüfer domain · Dedekind domain · Valuation domain ·Noetherian ring

Mathematics Subject Classification (2010) 13B02 · 13A18 · 13B22 · 13B25 ·13C13

1 Introduction

All rings considered in this paper are assumed to be commutative, integral domainsand to contain a unity element. If R is a subring of S, we assume that the unity of Sis contained in R, and hence is the unity of R. If R is a ring, then q f (R) denotes thequotient field of R.

If R ⊂ S is a ring extension, we use the term S-overring of R to mean a subring ofS containing R, a proper S-overring of R is an S-overring of R distinct from S. By anoverring of R, we mean a q f (R)-overring of R. Several papers (for example [1,14])

Communicated by Marco Fontana.

N. Jarboui (B) · M. E. I. Toumi · S. TrabelsiDepartment of Mathematics, Faculty of Sciences of Sfax, University of Sfax, P.O. Box: 1171,3000 Sfax, Tunisiae-mail: [email protected]

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have dealt with what are called Noetherian pairs of rings, PID pairs of rings. In a generalcontext, the definitions are as follows. If P is a ring-theoretic property and if R is asubring of S, then (R, S) is said to be a P-pair if each S-overring of R has property P .In considering a pair (R, S) of rings and a class C of intermediate rings, one questionthat naturally arises asks whether (R, S) is a P-pair if each element of C has propertyP . Several papers are devoted to study this question when the class C consists of allS-overrings of R distinct from R and R does not satisfy P (see for example [2,4–6,11–13]). In this case R is said to be a maximal non-P subring of S (more precisely, R doesnot satisfy P while every subring of S properly containing R satisfies P). These papersconsidered the properties valuation, Prüfer, Noetherian, Jaffard, and ACCP subringsand determined several conditions for an integral domain to be a maximal non-Psubring. In this paper we consider the particular case of this problem where C is theclass of all proper S-overrings of R and P is the property integrally closed. Our studyis motivated both by the significance of integrally closed domains (in particular Prüferand valuation domains) for commutative algebra and several neighboring fields suchas algebraic geometry, algebraic number theory, homological algebra, multiplicativeideal theory and analytic function theory on one hand, and on the other hand by therecent increasing interest in the study of pairs of rings as mentioned above. Our maincontribution in this paper is Theorem 1 which states that if each proper S-overring ofR is integrally closed, then S is integrally closed and if moreover R is not a field, thenR and S have the same quotient field. As a consequence, Corollary 3 states that givena ring theoretic property P satisfying the following three conditions:

(i) Each field satisfies P .(ii) R has property P ⇒ R is integrally closed.

(iii) R satisfies P ⇒ each overring of R satisfies P .

Then if each proper S-overring of R has property P , then so does S.As examples of such property P , one can take P := valuation, Prüfer, PID,

Dedekind, etc...Throughout the paper, we use “⊂” to denote proper containment and “⊆” to denote

containment. The maximal spectrum of a ring R is denoted Max(R). Any unexplainedterminology is standard as in [10].

2 Results

We begin with property concerning integrally closed domains, mentioned above. Butfirst recall that a domain R is integrally closed or normal if every element of thequotient field of R that is integral over R is in R. Recall also from [7] that a pair ofrings (R, S) is called a normal pair, if T is integrally closed in S for each intermediatering T (i.e. for each ring R ⊆ T ⊆ S). The most natural example of a normal pair(R, S) arises when R is an arbitrary Prüfer domain and S is its quotient field (cf. [[10]Theorems 23.4(1) and 26.1(1)]). In case R ⊂ S are rings sharing the same nonzeroideal I , then (R, S) is a normal pair if and only if (R/I, S/I ) is a normal pair. Indeed,the rings contained between R/I and S/I are the rings of the type A/I where A isan intermediate ring between R and S. As A = S×S/I A/I , a well known fact aboutpullbacks (see [9, Corollary 1.5 (5)]) shows that A/I is integrally closed in S/I if and

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Some questions concerning proper subrings

only if A is integrally closed in S. Recall from [8] that the pair of domains (R, S) isintegrally closed if R is a subring of S and all intermediate rings (including R andS) are integrally closed. It was proved in [8, Lemma 1] that if (R, S) is an integrallyclosed pair and R is not a field, then S is contained in the quotient field of R. Now, ifR is a field and S is a ring containing R, then one can check easily that (R, S) is anintegrally closed pair if and only if S is a field algebraic over R. It is worth noticing thatthe concepts: “integrally closed pair” and “normal pair” are different. More preciselyan integrally closed pair is a normal pair, but the converse is false in general. One canconsider the pair (R, S) where R = Z2 + XQ(

√2)[X ] and S = Q + XQ(

√2)[X ]

(Z2 is the ring of fractions of the integers Z with respect to the element 2). The ringsR and S share the same ideal I = XQ(

√2)[X ] and as (R/I, S/I ) = (Z2, Q) is a

normal pair (because Z2 is a valuation, hence a Prüfer domain, with quotient field Q),then it follows from the above comments that (R, S) is also a normal pair. However Sis not integrally closed. Thus (R, S) is not an integrally closed pair.

Theorem 1 Suppose R is a proper subring of a ring S and that each proper S-overringof R is integrally closed, then S is integrally closed. If moreover R is not a field, thenS is an overring of R.

Proof We claim that S is algebraic over R. Indeed, if there exists a transcendentalelement z ∈ S over the ring R then the ring T = R[z2, z3] is a proper S-overring ofR which is not integrally closed (since z ∈ q f (T ) and z is integral over T but z �∈ T ),a contradiction. Now, we will discuss the following two cases.

Case 1. R is a field. In this case S is also a field since S is algebraic over R. Hence Sis integrally closed.

Case 2. R is not a field. Let K denote the quotient field of R. In this case we demonstratefirst that S ⊆ K . Indeed, assume by way of contradiction that S �⊆ K and picks ∈ S \ K . As s is algebraic over R, then there exists a nonzero element r ∈ R suchthat rs is integral over R. Let x denote the element rs. It is clear that x ∈ S \ K . Letf (X) = Xn+∑

0≤i≤n−1 ai Xi be the monic minimal polynomial of x with coefficientsin R. Since R is not a field, then pick y a nonzero element of R such that y is notinvertible in R and consider the S-overring T of R where T = R[yx]. One can checkeasily that q f (T ) = K (x) = K (yx) and that each element z of T can be representedas a polynomial function of yx , of degree <n. Indeed, let z ∈ T then there exists apolynomial g of R[X ] such that z = g(yx). Set h(X) = Xn +∑

0≤i≤n−1 ai yn−i X i ∈R[X ]. The division algorithm implies that g(X) = q(X)h(X) + r(X), where q, r ∈R[X ] and either r = 0 or the degree of r is <n, and this representation is unique.Thus z = g(yx) = q(yx)h(yx) + r(yx) = q(yx)yn f (x) + r(yx) = r(yx), asdesired. We claim that x �∈ T . Indeed, assume that x ∈ T then by the above commentthere exist r0, r1, ..., rn−1 ∈ R such that x = r0 + r1(yx) + · · · + rn−1(yx)n−1.As {1, x, ..., xn−1} is a basis of K (x) as a K -vector space then by identification ofthe coefficients of the latter linear combination we get 1 = r1 y. Hence y would beinvertible in R, absurd. On the other hand, it is obvious that x ∈ q f (T ) and x isintegral over T since it is integral over R. Thus, T is not integrally closed. HenceT = S, another contradiction since x ∈ S but x �∈ T . Now as each proper S-overringT of R is integrally closed and S ⊆ K , then it follows that T is integrally closed in

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S. Thus (R, S) is a normal pair. Set Max(R) = {Mi | i ∈ I } the set of maximalideals of R. For i in I , set Pi the prime ideal of R such that SMi = RPi . Accordingto [3, Lemma 3.1], S = ⋂

i∈I SMi = ⋂i∈I RPi . Therefore S is integrally closed as an

intersection of integrally closed domains with the same quotient field.

For any ring extension R ⊂ S, we let [R, S] denote the set of intermediate rings,i.e. rings T such that R ⊆ T ⊆ S.

Corollary 1 Suppose R is a proper subring of a ring S and let C be a finite class ofproper S-overrings of R such that R �∈ C. If each element in [R, S] \ C is integrallyclosed, then (R, S) is an integrally closed pair. If moreover R is not a field, then S isan overring of R.

Proof Let R1 be a minimal element in C. Since each proper R1-overring of R isintegrally closed, then according to Theorem 1, R1 is integrally closed. Now, let R2 bea minimal element of C\{R1}. Since each proper R2-overring of R is integrally closed,then according to Theorem 1, R2 is integrally closed. Using this procedure, we showthat each element of C is integrally closed, which completes the proof.

Taking R to be a prime subring of S in Theorem 1 yields the following absolutecase of this result.

Corollary 2 If the ring S admits a proper subring, and if each proper subring of S isintegrally closed, then S is integrally closed.

Let P be a ring theoretic property satisfying the following three conditions:

(i) Each field satisfies P .(ii) R satisfies P ⇒ R is integrally closed.

(iii) R satisfies P ⇒ each overring of R satisfies P .

As examples of such property P , one can take P := valuation, Prüfer, PID, Dedekind,etc...

As an application of the proof of Theorem 1, we get the following result.

Corollary 3 Let P be a ring theoretic property satisfying the above three conditions(i) − (i i i), then if each proper S-overring of R has property P , then so does S.

Proof Since each proper S-overring of R has property P , then each proper S-overringof R is integrally closed. Argue as in the proof of Theorem 1, if R is a field then S isalso a field. Hence S satisfies P . Now, if R is not a field, then S is an overring of R.Thus S satisfies P . The desired conclusion. Corollary 4 Let P be a ring theoretic property satisfying the above three conditions(i) − (i i i). If the ring S admits a proper subring, and if each proper subring of S hasproperty P , then so does S.

As a consequence of Corollary 1, we have the following.

Corollary 5 Let P be a ring theoretic property satisfying the above three conditions(i) − (i i i). Let C be a finite class of proper S-overrings of R such that R �∈ C. If eachelement in [R, S] \ C has property P , then (R, S) is P-pair.

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Acknowledgments The authors extend their thanks to the referee for his/her valuable suggestions.

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