# Characterizations of Pairs of Rings with Few Non-Dedekind Intermediary Rings

Authors: Naseam Al-Kuleab; Noômen Jarboui
Journal: Journal of Mathematics
 (2022)

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## Abstract

We determine the Dedekind domain pairs of rings; that is, pairs of ringsR⊂S such that each intermediary ring in between R and S is a Dedekind domain. We also establish that if R⊂S is an extension of rings having only one non-Dedekind intermediary ring, then necessarily R is not Dedekind and so R is a maximal non-Dedekind domain subring of S. Maximal non-Dedekind domain subrings R of S are identified in the following cases: (1) R is not integrally closed, (2) R is integrally closed and either SuppS/R<∞ or MaxR<∞, (3) S is a field, (4) R is a valuation domain, and (5) R⊂S is an integral extension. We also provide some classifications of pairs of rings having exactly two non-Dedekind domain intermediary rings.


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## Body

## 1. Introduction

LetP be any ring-theoretic property. Let B be a ring containing A as a proper subring. The pair of rings A,B is called a P-pair if every subring of B containing A has property P. If A is a proper subring of a ring B, then we say that A is a maximal non-P subring of B if A does not have P; however any subring of B containing A properly has P. The structure of P pairs has already been studied for several choices of the property P, so there is a well-developed theory about the topic. On the contrary, there is a long standing line of research which investigates the structure of maximal non-P subrings (see, for instance, [1–12]). Motivated by some works in group theory (cf. [13, 14]), the second named author and Al Subaiei [15] have started a new research program by asking the following more general question: “Given a ring-theoretic property P, a pair of rings R⊂S and a nonnegative integer n. Provide necessary and/or sufficient conditions so that there exist n intermediary rings in between R and S which do not satisfy P.” We retain the same notation and terminology as in [15]. More precisely, let R,S denote the set of intermediary rings in between R and S, and let R,SP denote the set of rings T∈R,S satisfying P. Set R,Snon−P the complement of R,SP in R,S. Using this terminology, it follows that R,S is a P-pair if R,Snon−P is the empty set and R is a maximal non-P subring of S if R,Snon−P is exactly the singleton set R. It is worth noting that among the difficult questions in ring theory is to get information about the structures of both R and S by examining the properties of R,SP or R,Snon−P. In [15], the above question was discussed for the property “Prüfer” and 0≤n≤2. In [16], the same question was investigated for the property “Artinian” and 0≤n≤2. The aim of this paper is to continue our recent program by considering the special case of this question when the property we are going to study is P:= “Dedekind domain.” A ring R is a Dedekind domain if and only if R is Prüfer and Noetherian. Dedekind domains have been thoroughly studied in the literature, so many of their properties are well-known (cf. [17–19]). The motivations for our work are numerous, but we will limit ourselves to only two. We mention first the importance of Dedekind domains in commutative algebra and many nearby fields such as algebraic geometry and algebraic number theory and secondly the growing interest in the study of pairs of rings as we explained at the beginning of the introduction. In this work, and as the abstract indicated, we will focus our attention on the cases where 0≤n≤2. The present research article is structured as follows: Section 2 is devoted to answer the above question in the case n=0. More precisely, Theorem 1 states that if R is a subring of S, then R,S is a Dedekind domain pair if and only if either R⊆S is an algebraic field extension or R is a Dedekind domain but not a field and R⊆S⊆qfR, where qfR denotes the quotient field of R. As a consequence, we classify in Corollary 1, rings whose all subrings are Dedekind domains. Section 3 is addressed to the study of pairs of rings R⊂S such that R,Snon−P=1 (as usual E denotes the cardinality of the set E). We demonstrate in Lemma 1, that in this case R is a fortiori non-Dedekind and so R should be maximal non-Dedekind domain subring of S. Theorem 2 is concerned with maximal non-Dedekind domain subrings when the bottom ring is not Prüfer. As a consequence, we derive in Corollary 2 that if R is a maximal non-Dedekind domain subring of S, then S and R have the same quotient field, R⊂S is a P-extension and R is integrally closed (if and) only if R is Prüfer. Corollary 3 treats the particular case where S is integral over R. Theorems 3 and 5 identify such domains when the bottom ring is a Prüfer domain with some additional conditions on its spectrum. A complete characterization of maximal non-Dedekind subrings of a field is established in Theorem 4. In Section 4, we deal with pairs of rings R⊂S such that R,Snon−P=2. We provide characterizations of these pairs R,S in Theorem 6 in case R⊂R′⊂S and R′ is a semilocal domain. An identification of such pairs in case R≠R′ and S is a field or in case S is integral over R is provided in Corollaries 6 and 7, respectively.Among the essential tools used in this work is the concept of minimal extensions. A ring extensionR⊂S is called minimal if R is a proper subring of S and R,S=R,S. A key result proved by Sato et al. ([20], page 1738, lines 8–13), stated that if R is a domain which is not a field, then any domain which is a minimal ring extension of R must be an overring of R. According to [21], Théorème 2.2(i) and Lemme 1.3, if R⊂S is a minimal extension and R is not a field, then there is a unique maximal ideal M of R called the crucial maximal ideal of R⊂S such that the canonical injective ring homomorphism RM⟶SM can be viewed as a minimal ring extension, while the canonical ring homomorphism RP⟶SP is an isomorphism for all prime ideals P of R, except M (cf. [21, 22] and [23], p. 37). If moreover S is integral over R, then M=R:S:=x∈R|xS⊆R (cf. [21], Théorème 2.2(ii)). For more information about minimal extensions, we invite the interested reader to consult [21–23].All the rings considered throughout this paper are commutative, are assumed to contain an identity element and all ring homomorphisms are unital. We denote bySpecR (respectively, MaxR) the set of prime ideals (respectively, the set of maximal ideals) of R. For any P∈SpecR, the height of P, denoted htRP, is the supremum of lengths m of chains P0⊂P1⊂…⊂Pm=P in SpecR. If R⊂S are rings and P∈SpecR, then SP:=SR∖P, R¯S:=x∈S|xis integral overR is the integral closure of R in S and R′≔R¯qfR is the integral closure of R. An overring of an integral domain R is an intermediary ring in between R and its quotient field. Such an overring is termed proper if it contains properly R. As usual, the symbol “⊆” is used for containment and “⊂” for proper containment. Any notation or terminology is standard as in the books [18, 24].


## 2. Dedekind Pairs of Rings

We begin by recalling the following definition.Definition 1.
IfR is a subring of S, we say that R,S is a Dedekind domain pair if every intermediary ring in between R and S is a Dedekind domain.Now we are going to characterize Dedekind domain pairs of rings.Theorem 1.
LetR⊆S be integral domains. Then the following conditions are equivalent:(1)
The pair of ringsR,S is Dedekind.(2)
EitherR is a Dedekind domain but not a field and S is an overring of R or R⊆S is an algebraic extension of fields.Proof.
The implication (1)⇒(2) follows readily from ([25], Theorem 3.2) since any Dedekind domain is Prüfer. Let us prove that (2) implies (1). If R⊆S is an algebraic extension of fields, then any intermediary ring is a field, so a Dedekind domain. Assume now that R is a Dedekind domain which is not a field and R⊆S⊆qfR. Then any intermediary ring in between R and S would be an overring of R, and hence a Dedekind domain.
We derive the following consequence where we characterize domainsR whose each subring is a Dedekind domain.Corollary 1.
Every subring of an integral domainR is a Dedekind domain if and only if either ℤ⊆R⊆ℚ or ℤp⊆R is an algebraic field extension, where p is a prime number.Proof.
Since the prime subringπ of R is a Dedekind domain, it follows that each subring of R is a Dedekind domain if and only if π,R is a Dedekind domain pair. If charR=0, then π=ℤ. So, by using Theorem 1, ℤ,R is a Dedekind domain pair if and only if R⊆qfℤ=ℚ. Now, if charR=p is prime, then π=ℤp and again by virtue of Theorem 1, ℤp,R is a Dedekind domain pair if and only if R is a field algebraic over ℤp. The proof is complete.


## 3. When isR,Snon−P = 1?

In this section, we shall study pairs of ringsR⊂S such that R,Snon−P=1. We start our investigation with the following useful lemma.Lemma 1.
LetR⊂S be integral domains. Then the following hold true:(1)
IfR,Snon−P=n∈ℕ, then R is never a Dedekind domain.(2)
R,Snon−P=1if and only if R is a maximal non-Dedekind domain subring of S.Proof.
(1)
Assume, by way of contradiction, thatR is a Dedekind domain and pick T a minimal element of R,Snon−P. Obviously, each proper subring of T containing R is a Dedekind domain. Hence ([10], Corollary 3), asserts that T should be a Dedekind domain, the desired contradiction.(2)
Follows immediately from the first assertion.In what follows we identify integral domainsR⊂S such that R is not a Prüfer domain and R,Snon−P=1. But, first recall [3] that if R⊂S are rings, then we say that R has a minimal overring A in S, if A≠R and every subring B of S properly containing R must contain A. It is worth mentioning that if R⊂S are rings such that S is an integrally closed overring of R, then R′=R¯S. This fact will be freely used in the remainder of this paper.Theorem 2.
LetR⊂S be integral domains such that R is not Prüfer. Then the following conditions are equivalent:(1)
R is a maximal non-Dedekind domain subring of S.(2)
R′ is a Dedekind domain, S is an overring of R′ and R′ is the unique minimal overring of R in S.(3)
R′ is a Dedekind domain, S is an overring of R′, R⊂R′ is a minimal extension and R=S∩RR:R′.Proof.
(1)⇒ (2) Since R is not Prüfer and each subring of S properly containing R is Prüfer (as it is Dedekind), then R is a maximal non-Prüfer domain subring of S. It follows from [25], Theorem 2.5, that R′ is the unique minimal overring of R in S and S is an overring of R′. As R≠R′ and R is a maximal non-Dedekind domain subring of S, then R′ is a Dedekind domain.
(2)⇔ (3) argue as in [25], Theorem 2.5.
(2)⇒ (1) since R≠R′, then R is not integrally closed, and a fortiori R is not a Dedekind domain. Now, let T∈R,S∖R. Then T⊇R′ since R′ is the unique minimal overring of R in S. Thus, T is a Dedekind domain as an overring of the Dedekind domain R′.
Recall [26] that if R is a subring of S, then we say that an element s in S is primitive over R if s is a root of a polynomial fX∈RX with unit content. We say that R⊆S is a primitive extension (in short, P-extension) in case any element of S is primitive over R. Clearly, each integral extension is necessarily a P-extension. Moreover, for any integral domain R, the ring extension R⊆qfR is a P-extension if and only if R′ is Prüfer (cf. [26], Theorem 5). Note that if R⊆S⊆T is a tower of rings and R⊆T is a P-extension, then so is R⊆S. Recall also that R,S is called a normal pair if R⊆S and T¯S=T for any T∈R,S (cf. [27]). Ben Nasr and Jarboui have established a relationship between normal pairs and P-extensions (for not necessarily integral domains). In fact, they have shown ([28], Theorem 1) that R,S is a normal pair of rings if and only if R is integrally closed in S and R⊆S is a P-extension.
The following result shows (among other facts) thatS must be an overring of R if R is a maximal non-Dedekind domain subring of S.Corollary 2.
LetR⊂S be integral domains. Assume that R is a maximal non-Dedekind domain subring of S. Then the following conditions hold:(1)
R and S have the same quotient field.(2)
R⊂S is a P-extension.(3)
R is integrally closed if and only if R is Prüfer.Proof.
(1)
IfR is not a Prüfer domain, then Theorem 2 permits one to conclude that qfS=qfR. Now, we suppose that R is a Prüfer domain. As each subring of S properly containing R is Prüfer (as it is Dedekind), it follows that R,S is a Prüfer domain pair. Moreover, R cannot be a field since it is not Dedekind. Thus [25], Theorem 3.2, guarantees that qfS=qfR.(2)
It was proven in [28], Theorem 2, that if both A⊂B and B⊂C are P-extensions, then A⊂C is also a P-extension. Our task is to show that if R is a maximal non-Dedekind domain subring of S, then R⊂S is a P-extension. If R is not a Prüfer domain, then Theorem 2 implies that R′⊂S is a P-extension since R′ is Prüfer and S is an overring of R′ (cf. [26], Theorem 5). But, R⊂R′ is a P-extension since it is integral. Therefore, R⊂S is also a P-extension, as desired. Assume now that R is Prüfer. Then, by using [26], Theorem 2, we derive that R⊂S is a P-extension, since S is an overring of R.(3)
The “If  ” part is immediate. For the “Only if” part, assume thatR is integrally closed. Then R is integrally closed in S since qfR=qfS. As R⊂S is a P-extensions, it follows that R,S is a normal pair. Now, since S is a Prüfer domain, it follows from [25], Lemma 2.2, that R is also a Prüfer domain. This completes the proof.
We derive from Theorem2 the following result which treats the case where S is integral over R.Corollary 3.
Assume thatR is a subring of the integral domain S and S is integral over R. Then the following conditions are equivalent:(1)
R is a maximal non-Dedekind subring of S.(2)
R⊂S is minimal, R is not a field and S is a Dedekind domain.Proof.
(1)⇒ (2) The ring R cannot be a field since it is not Dedekind. On the other hand, Corollary 2 ensures that qfR=qfS. As R⊂S is an integral extension, then S⊆R′. We deduce that R cannot be integrally closed and so R cannot be a Prüfer domain. On the other hand, as S is an integrally closed domain, then R′⊆S. Hence, S=R′. Now, Theorem 2 asserts that R⊂S=R′ is a minimal extension.
(2)⇒ (1) Since R⊂S is a minimal extension and R is not a field, then [20], page 1738, lines 8–13, guarantees that qfR=qfS. As S is integral over R, then a fortiori S⊆R′. Therefore, R≠R′. This proves that R is not a Dedekind domain since it is not integrally closed. The proof is complete.
For our convenience, we quote here some tools and some well-known facts which will be used repeatedly in the sequel. One of the most important results related to normal pairs is the result of [27], Theorem 1, which states that if R is a local domain, then R,S is a normal pair if and only if there is a divided prime ideal P of R (that is, PRP=P) such that S=RP and R/P is a valuation domain. This result has been generalized in [29], Theorem 2.5, where the authors have proved that if R,S is a normal pair and M is a maximal ideal of R, then SM=RQ, where QM is a divided prime ideal of RM. In these conditions, R/QM/Q is a valuation domain and the set of prime ideals of R in between Q and M is totally ordered.
Next, we characterize pairs of ringsR⊂S satisfying R,Snon−P=1 in case R is a semilocal Prüfer domain.Theorem 3.
Assume thatR⊂S are integral domains, where R is a semilocal Prüfer domain. Then the following conditions are equivalent:(1)
R is a maximal non-Dedekind domain subring of S.(2)
R is not Noetherian, R has a Dedekind minimal overring Ω in S and S is an overring of R.Proof.
(1)⇒ (2) it follows from Corollary 2 that S is an overring of R. As R is a Prüfer domain, then R,S is a normal pair. Notice that R is not a Noetherian domain since it is a Prüfer domain but not a Dedekind domain. Set MaxR:=M1,…,Mr. As R is a maximal non-Noetherian subring of S, it follows from [3], Proposition 3.6, that there is a unique maximal ideal Mj of R such that RMj is not Noetherian. Let Qj be the prime ideal of R such that SMj=RQj. We have SMj≠RMj since SMj is Noetherian, whereas RMj is not. Set Ω:=∩RMi|i≠j∩RPj, where Qj⊆Pj⊂Mj satisfying htMj/Pj=1. Clearly, Ω∈R,S. Moreover, Ω≠R since MaxΩ≠MaxR (indeed, the maximal ideals of Ω are Mi, i≠j (and Pj if Pj is not comparable to Mi, i≠j)). Hence, Ω is a Dedekind domain. For each i≠j, we have ΩMi=RMi. For i=j, we have RMj⊆ΩMj⊆RPj. Thus ΩMj=RMj or ΩMj=RPj. But if ΩMj=RMj, then Ω=∩1≤i≤rΩMi=∩1≤i≤rRMi=R, a contradiction. Therefore, ΩMj=RPj. Now, let T∈R,S∖R; then T=∩1≤i≤rRHi, where Hi is the prime ideal of R such that TMi=RHi. Since R⊂T⊆S, then Qi⊆Hi⊆Mi for each i and TMj≠RMj since TMj is Noetherian. Hence, Ω⊆T. This proves that Ω is a Dedekind minimal overring of R in S.
(2)⇒ (1) let Ω be the unique (Dedekind) minimal overring of R in S. Then for each T∈R,S∖R, T is a Dedekind domain as an overring of the Dedekind domain Ω. As R is not a Dedekind domain (as it is not Noetherian), then R is a maximal non-Dedekind domain subring of S. The proof is complete.Lemma 2.
Assume thatR is maximal non-Dedekind domain subring of S. Then the following hold true:(1)
IfR is a valuation domain, then dimR≤2.(2)
IfR is a Prüfer domain and S is a field, then R is a valuation domain.Proof.
(1)
By using Theorem3, R has a Dedekind minimal overring Ω in S. Thus, R⊂Ω is a minimal extension. An application of [29], Corollary 4.5, yields that dimR≤dimΩ+1≤2.(2)
AsS is a field and R is a maximal non-Dedekind domain subring of S, it follows from Corollary 2 that S=qfR. In order to prove that R is a valuation domain, it suffices to show that R is a local domain. For, let M1 and M2 be two (distinct) maximal ideals of R. Then there exist a∈M1 and b∈M2 such that a+b=1. Notice that R≠R1/a and R≠R1/b but R=R1/a∩R1/b. Since R1/a and R1/b are Dedekind domains, then R would be a Krull domain. But R is also Prüfer, thus R is a Dedekind domain, the desired contradiction. Consequently, R is local.
Our next result provides a classification of maximal non-Dedekind domain subrings of a field.Theorem 4.
LetS be a field and let R be a subring of S. Then the following conditions are equivalent:(1)
R is a maximal non-Dedekind domain subring of S.(2)
Exactly one of the following conditions holds true:(a)
R is a local domain with quotient field S, R′ is a Dedekind domain and the ring extension R⊂R′ is minimal.(b)
S is the quotient field of R and either R is a rank 2 valuation domain with a discrete rank 1 valuation overring or R is a non-Noetherian rank 1 valuation domain.Proof.
(1)⇒ (2) if R is not Prüfer, then condition (a) is satisfied according to Theorem 2. Suppose now that R is Prüfer. Then, Lemma 2 yields that R is a valuation domain and dimR≤2. If dimR=1, then clearly R is a rank 1 nondiscrete valuation domain. If dimR=2, then R is a rank 2 valuation domain with a discrete rank 1 valuation overring.
(2)⇒ (1) this is trivial.Corollary 4.
LetR⊂S be integral domains such that R is a valuation domain. Then the following conditions are equivalent:(1)
R is a maximal non-Dedekind domain subring of S.(2)
Exactly one of the following conditions holds true:(a)
dimR=2 and S is a rank 1 discrete valuation overring of R.(b)
S is the quotient field of R and either R is a rank 2 valuation domain with a discrete rank 1 valuation overring or R is a non-Noetherian rank 1 valuation domain.Proof.
(1)⇒ (2) if S is a field, then Theorem 4 applies easily and so condition (b) is satisfied. Assume now that S is not a field. As R is a valuation domain, then S is a (discrete) rank 1 valuation overring of R (since it is a Noetherian). The fact that dimR=2 follows immediately from Lemma 2.
(2)⇒ (1) this is trivial.Corollary 5.
IfR is a maximal non-Dedekind domain subring of S, then dimR≤2.Proof.
It follows from Theorem2 (and Corollary 2 (3)) that if R is not integrally closed, then R′ is a Dedekind domain. As any integral extension preserves Krull dimensions, we get dimR=1. Now, if R is integrally closed (or equivalently Prüfer), then for any maximal ideal M of R, the localization RM is a valuation domain. So we have two cases to consider:(i)
IfRM is a Dedekind domain, then RM is a rank 1 discrete valuation domain.(ii)
IfRM is not a Dedekind domain, then RM is a maximal non-Dedekind domain subring of SM. Thus, Lemma 2 ensures that dimRM≤2.
In all cases, we havedimR=supdimRM|M∈MaxR≤2.
We close this section with the following result. We will make use of the following two facts:(i)
IfR⊂S are rings, then(1)SuppSR≔SuppRSR≔P∈SpecR|RP≠SP≔SR∖P.(ii)
IfI is a nonzero ideal of a ring R, then(2)ΩRI≔x∈qfR|∀y∈I,xyn∈Rfor some integern≥1,is an overring ofR called the Kaplansky transform of R with respect to I, and can be expressed as ΩRI=∩RP|P∈SpecR,I⊈P (cf. [30, 31]).Theorem 5.
LetR⊂S be an extension of integral domains such that R is a Prüfer domain and SuppS/R<∞, then the following conditions are equivalent:(1)
R is a maximal non-Dedekind domain subring of S.(2)
R is not Noetherian and R has a Dedekind minimal overring in S.In this caseR,S=2 or 3.Proof.
(1)⇒ (2) first of all, note that R is not Noetherian since it is Prüfer but not Dedekind. Moreover, R,S is a normal pair because S is an overring of the Prüfer domain R (see Corollary 2). Since SuppS/R<∞, it follows that R⊂ΩRM is a minimal extension, where M represents any maximal ideal of SuppS/R ([32], Corollary 2.6). Now, if SuppS/R contains two maximal ideals M and M′, then R⊂ΩRM and R⊂ΩRM′ are minimal extensions. As ΩRM and ΩRM′ are Krull domains, it results that R=ΩRM∩ΩRM′ is a Krull domain, so a Dedekind domain since it is already a Prüfer domain, a contradiction. Thus SuppS/R contains a unique maximal ideal M. Using again ([32], Corollary 2.6), ΩRM is the unique ring T in R,S such that R⊂T is a minimal extension. It follows from [32], Corollary 2.7, that R,S,⊆ is totally ordered and so ΩRM is the unique minimal overring of R in S and R,S=1+SuppS/R. But dimR≤2 (cf. Corollary 5) and SuppS/R,⊆ is totally ordered (cf. [32], Corollary 2.7). Thus, SuppS/R contains at most 2 elements M and an eventual prime ideal Q contained in M. Therefore, R,S=2 or 3.
(2)⇒ (1) this is clear.


## 4. When isR,Snon−P = 2?

In this section, we start our investigation with the following result which generalizes Corollary2. But, we point out that any minimal ring extension R⊂S is a P-extension. Indeed, if R⊂S is integral, then obviously R⊂S is a P-extension. If R⊂S is (integrally) closed, then R,S is a normal pair and so R⊂S is a P-extension. We will make use of this fact in what follows.Lemma 3.
LetR⊂S be integral domains such that R,Snon−P=n≥1. Then R and S have the same quotient field and R⊂S is a P-extension. Moreover, if R is integrally closed, then R is a Prüfer domain.Proof.
We argue by induction onn. The case n=1 follows readily from Corollary 2. Assume that for any ring extension A⊂B such that A,Bnon−P=k≤n, we have qfA=qfB and A⊂B is a P-extension. Let R⊂S be a ring extension such that R,Snon−P=n+1. By Lemma 1, R is not a Dedekind domain. Thus, let T be a minimal element in the set R,Snon−P∖R. A new, Lemma 1 guarantees that R⊂T is a minimal extension and so qfR=qfT by [20], page 1738, lines 8–13, and R⊂T is a P-extension. On the contrary, as |T,Snon−P|≤n, then the induction hypothesis ensures that qfT=qfS and T⊂S is a P-extension. Therefore, we get qfR=qfS on one hand and on the other hand, R⊂S is a P-extension, by using the transitivity of the “P-extension” property as explained in the proof of Corollary 2. The “moreover” assertion can be proved along the same lines as in the proof of Corollary 2. The proof is complete.
Given a ring extensionR⊆S, we will use the following notation: MSuppRS/R≔MaxR∩SuppRS/R. Recall that a ring extension is said to satisfy FCP if, when R,S is viewed as a poset under inclusion, each chain in R,S is finite. In [33], Knebusch and Zhang introduced the following concept, which can be considered as a relativization of the concept of a Prüfer ring. A ring extension R⊆S is called a Prüfer extension if R⊆T is a flat epimorphism (in the category of commutative rings) for each T∈R,S. Any Prüfer extension is integrally closed. It is noteworthy that a ring extension R⊆S is a P extension if and only if R¯S⊆S is a Prüfer extension: cf. [28], Corollary 1. (This is a far-reaching generalization of the above-mentioned result of Gilmer and Hoffmann ([26], Theorem 5.) For an arbitrary ring extension R⊆S, it was proved in [33], Theorem 5.2, page 47, that R⊆S is a Prüfer extension if and only if R,S is a normal pair. Recall also from [33] that any ring extension R⊆S has a maximum Prüfer subextension R⊆R˜S, called the Prüfer hull of R in S. (Note that R˜S was denoted by PR,S in [33].) We say that a ring extension R⊆S is Prüfer-closed (or that R is Prüfer-closed in S) if R=R˜S.
The next theorem identifies pairs of ringsR⊂S such that R,Snon−P=2 under some additional conditions.Theorem 6.
LetR⊂S be integral domains such that R⊂R′⊂S and R′ is semilocal. Then the following conditions are equivalent:(1)
R,Snon−P=2.(2)
Exactly one of the following conditions holds:(a)
S is not a Dedekind domain and R⊂S is a minimal extension.(b)
R′ is a Dedekind domain, S is an overring of R′, R,R′ is a chain of length 2, and R,S=R,R′∪R′,S.(c)
R′ is a Dedekind domain, S is an overring of R′, and the ring extensions R⊂R′ and R⊂S∩RR:R′ are minimal.(d)
R⊂R′ is a minimal extension, R′ is not Noetherian, R′ has a Dedekind minimal overring Ω in S, S⊆qfR and R:R′⊆N for each N∈MSuppS/R′.Proof.
(1)⇒ (2) if R⊂S is a minimal extension, then we are done. Next, we will assume that R⊂S is not a minimal extension. Hypothetically, there are exactly two non-Dedekind domains in the set R,S. One of these two domains is R by Lemma 1 and the other, say T, is such that R⊂T is a minimal extension accordingly to Lemma 1. Therefore, T is a maximal non-Dedekind domain subring of S. Now, we will distinguish the following cases:Case 1.
T is not a Prüfer domain.
In this case,T is not integrally closed accordingly to Theorem 2. In particular, T≠R′. As R≠R′, it follows that R′ is a Dedekind domain. Next, we discuss the following subcases.Subcase 1.
R¯T=T.
LetB∈R,R′∖R,T. Then B is a Dedekind domain. Thus R′⊆B and so B=R′. This proves that R,R′=R⊂T⊂R′ is a chain of length 2. On the other hand, let B∈R,S∖R,R′. Then, B is a Dedekind domain since B≠R and B≠T. As R¯B⊆B inherits the “P-extension” property from R⊂S, we conclude then that R¯B,B is a normal pair. Applying ([25], Lemma 2.2), it results that R¯B is a Prüfer domain. But, R¯B lies in between R and R′, then a fortiori R¯B=R′. Hence R′⊆B, and so B∈R′,S. We have proved that R,S=R,R′∪R′,S.Subcase 2.
R¯T=R.
We claim that the ring extensionR⊂R′ is minimal. Indeed, if there exists a ring B such that R⊂B⊆R′, then B≠T because R is integrally closed in T. Thus B is a Dedekind domain. In particular, B is integrally closed. This implies that B=R′. This proves our claim. Set M=R:R′. Since R⊂T is a (closed) minimal extension, then either RM⊂TM is a (closed) minimal extension or TM=RM. The former case is impossible by virtue of [34], Lemma 3. Hence, TM=RM. It follows that T⊆S∩RM. If T is properly contained in S∩RM, then S∩RM would be a Dedekind domain and so an integrally closed domain. On the other hand, for any maximal ideal m of R distinct from M, Rm is integrally closed since Rm=Rm′. It follows that R=∩m∈MaxRRm=∩m∈MaxR∖MRm∩RM=∩m∈MaxR∖MRm∩S∩RM is integrally closed as an intersection of integrally closed domains with the same quotient field, which is a contradiction. Thus T=S∩RM.Case 2.
T is a Prüfer domain.
In this case,R would be a maximal non-Prüfer subring of T. Using [25], Theorem 2.5, we derive that R′=T. It follows that R′ is a (semilocal) Prüfer domain and a maximal non-Dedekind subring of S. Thus, Theorem 3 implies that R′ is not Noetherian, R′ has a Dedekind minimal overring in S and S⊆qfR. Finally, we will show that R:R′⊆N for every N∈MSuppS/R′. As R′ is a semilocal Prüfer domain and R′ is a maximal non-Noetherian subring of S, it follows from [3], Proposition 3.9, that R′⊂S is an FCP extension. Since R⊂R′ is an FCP extension, it results that R⊂S is also an FCP extension (cf. [35], Theorem 3.13). Thus, in order to show that R:R′⊆N for each N∈MSuppS/R′, we will demonstrate that R⊂S is a Prüfer-closed extension (cf. [36], Theorem 2.7). To this end, assume by way of contradiction that R≠R˜S. As R′ and R˜S are incomparable under the usual set inclusion, we deduce that each subring of R˜S properly containing R is a Dedekind domain. Hence, R would be a maximal non-Dedekind subring of R˜S, which contradicts Theorem 2 since R′ must be an intermediary ring in between R and R˜S.
(2)⇒ (1) Suppose condition (a) satisfied. By assumption, R is not a Dedekind domain since R≠R′. Moreover, S is not Dedekind. Therefore, R,Snon−P=2. Suppose now condition (b) verified. As R,R′=R⊂T⊂R′, it results that R and T are not Dedekind domains since they are not integrally closed. Let B∈R,S such that B≠R and B≠T. Since B is integrally closed, B∈R′,S. Now, as R′ is a Dedekind domain, then so is B. Assume condition (c) fulfilled, then obviously the domains R and S∩RM (where M=R:R′) are not Dedekind domains since they are not integrally closed. Let B∈R,S and assume that B≠R and B≠S∩RM. As R⊂R′ is a minimal extension and R¯B∈R,R′, it follows that either R=R¯B or R′=R¯B. If R=R¯B, then R would be integrally closed in B. On the other hand, R⊂R′ is an FCP extension (since it is minimal) and R′⊂S is also an FCP extension since R′ is a semilocal Dedekind domain and S is an overring of R′ (see [37], Theorem 1.5). Hence, R⊂S is an FCP extension. Thus, R⊂S is a P-extension by virtue of [36], Proposition 2.4. As B∈R,S, then R⊂B is also a P-extension. We deduce that the pair R,B is normal. According to [34], Lemma 3, BM=RM. So B⊆S∩RM. But R⊂S∩RM is a minimal extension, thus B=S∩RM, a contradiction. Now, if R′=R¯B, then B would be a Dedekind domain as an overring of the Dedekind domain R′. If condition (d) is satisfied, then it is obvious that R′ and R are not Dedekind domains. Now combining theorems in [36], Theorem 2.5 and Theorem 2.7, we deduce that R,S=R,R′∪R′,S. As each ring T in R′,S∖R′ must contain Ω, then T would be a Dedekind domain as an overring of the Dedekind domain Ω. This completes the proof of Theorem 6.Corollary 6.
LetR⊂S be integral domains such that R≠R′ and S is a field. Then the following conditions are equivalent:(1)
R,Snon−P=2.(2)
Exactly one of the following conditions hold:(a)
R′ is a Dedekind domain with quotient field S, R,R′ is a chain of length 2, and R,S=R,R′∪R′,S.(b)
R′ is a Dedekind domain with quotient field S, R⊂R′ and R⊂RR:R′ are minimal extensions.(c)
S=qfR, R⊂R′ is a minimal extension and either R′ is a rank 2 valuation domain with a discrete rank 1 valuation overring or R′ is a non-Noetherian rank 1 valuation domain.Corollary 7.
Assume thatR⊂S is an integral extension. Then the following conditions are equivalent:(1)
R,Snon−P=2.(2)
Exactly one of the following conditions hold:(a)
S is not a Dedekind domain and R⊂S is a minimal extension.(b)
S is a Dedekind domain which is not a field and R,S=3.



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*Source: 4642508-2022-04-28.xml*