Abstract

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

1. Introduction

Let be any ring-theoretic property. Let be a ring containing as a proper subring. The pair of rings is called a -pair if every subring of containing has property . If is a proper subring of a ring , then we say that is a maximal non- subring of if does not have ; however any subring of containing properly has . The structure of pairs has already been studied for several choices of the property , 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- subrings (see, for instance, [112]). 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 , a pair of rings and a nonnegative integer . Provide necessary and/or sufficient conditions so that there exist intermediary rings in between and which do not satisfy .” We retain the same notation and terminology as in [15]. More precisely, let denote the set of intermediary rings in between and , and let denote the set of rings satisfying . Set the complement of in . Using this terminology, it follows that is a -pair if is the empty set and is a maximal non- subring of if is exactly the singleton set . It is worth noting that among the difficult questions in ring theory is to get information about the structures of both and by examining the properties of or . In [15], the above question was discussed for the property “Prüfer” and . In [16], the same question was investigated for the property “Artinian” and . 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 “Dedekind domain.” A ring is a Dedekind domain if and only if is Prüfer and Noetherian. Dedekind domains have been thoroughly studied in the literature, so many of their properties are well-known (cf. [1719]). 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 . The present research article is structured as follows: Section 2 is devoted to answer the above question in the case . More precisely, Theorem 1 states that if is a subring of , then is a Dedekind domain pair if and only if either is an algebraic field extension or is a Dedekind domain but not a field and , where denotes the quotient field of . 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 such that (as usual denotes the cardinality of the set ). We demonstrate in Lemma 1, that in this case is a fortiori non-Dedekind and so should be maximal non-Dedekind domain subring of . 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 is a maximal non-Dedekind domain subring of , then and have the same quotient field, is a -extension and is integrally closed (if and) only if is Prüfer. Corollary 3 treats the particular case where is integral over . 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 such that . We provide characterizations of these pairs in Theorem 6 in case and is a semilocal domain. An identification of such pairs in case and is a field or in case is integral over is provided in Corollaries 6 and 7, respectively.

Among the essential tools used in this work is the concept of minimal extensions. A ring extension is called minimal if is a proper subring of and . A key result proved by Sato et al. ([20], page 1738, lines 8–13), stated that if is a domain which is not a field, then any domain which is a minimal ring extension of must be an overring of . According to [21], Théorème 2.2(i) and Lemme 1.3, if is a minimal extension and is not a field, then there is a unique maximal ideal of called the crucial maximal ideal of such that the canonical injective ring homomorphism can be viewed as a minimal ring extension, while the canonical ring homomorphism is an isomorphism for all prime ideals of , except (cf. [21, 22] and [23], p. 37). If moreover is integral over , then (cf. [21], Théorème 2.2(ii)). For more information about minimal extensions, we invite the interested reader to consult [2123].

All the rings considered throughout this paper are commutative, are assumed to contain an identity element and all ring homomorphisms are unital. We denote by (respectively, ) the set of prime ideals (respectively, the set of maximal ideals) of . For any , the height of , denoted , is the supremum of lengths of chains in . If are rings and , then , is the integral closure of in and is the integral closure of . An overring of an integral domain is an intermediary ring in between and its quotient field. Such an overring is termed proper if it contains properly . 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. If is a subring of , we say that is a Dedekind domain pair if every intermediary ring in between and is a Dedekind domain.

Now we are going to characterize Dedekind domain pairs of rings.

Theorem 1. Let be integral domains. Then the following conditions are equivalent:(1)The pair of rings is Dedekind.(2)Either is a Dedekind domain but not a field and is an overring of or 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 is an algebraic extension of fields, then any intermediary ring is a field, so a Dedekind domain. Assume now that is a Dedekind domain which is not a field and . Then any intermediary ring in between and would be an overring of , and hence a Dedekind domain.
We derive the following consequence where we characterize domains whose each subring is a Dedekind domain.

Corollary 1. Every subring of an integral domain is a Dedekind domain if and only if either or is an algebraic field extension, where is a prime number.

Proof. Since the prime subring of is a Dedekind domain, it follows that each subring of is a Dedekind domain if and only if is a Dedekind domain pair. If , then . So, by using Theorem 1, is a Dedekind domain pair if and only if . Now, if is prime, then and again by virtue of Theorem 1, is a Dedekind domain pair if and only if is a field algebraic over . The proof is complete.

3. When is  = 1?

In this section, we shall study pairs of rings such that . We start our investigation with the following useful lemma.

Lemma 1. Let be integral domains. Then the following hold true:(1)If , then is never a Dedekind domain.(2) only if is a maximal non-Dedekind domain subring of .

Proof. (1)Assume, by way of contradiction, that is a Dedekind domain and pick a minimal element of . Obviously, each proper subring of containing is a Dedekind domain. Hence ([10], Corollary 3), asserts that should be a Dedekind domain, the desired contradiction.(2)Follows immediately from the first assertion.In what follows we identify integral domains such that is not a Prüfer domain and . But, first recall [3] that if are rings, then we say that has a minimal overring in , if and every subring of properly containing must contain . It is worth mentioning that if are rings such that is an integrally closed overring of , then . This fact will be freely used in the remainder of this paper.

Theorem 2. Let be integral domains such that is not Prüfer. Then the following conditions are equivalent:(1) is a maximal non-Dedekind domain subring of .(2) is a Dedekind domain, is an overring of and is the unique minimal overring of in .(3) is a Dedekind domain, is an overring of , is a minimal extension and .

Proof. (1)  (2) Since is not Prüfer and each subring of properly containing is Prüfer (as it is Dedekind), then is a maximal non-Prüfer domain subring of . It follows from [25], Theorem 2.5, that is the unique minimal overring of in and is an overring of . As and is a maximal non-Dedekind domain subring of , then is a Dedekind domain.
(2)  (3) argue as in [25], Theorem 2.5.
(2)  (1) since , then is not integrally closed, and a fortiori is not a Dedekind domain. Now, let . Then since is the unique minimal overring of in . Thus, is a Dedekind domain as an overring of the Dedekind domain .
Recall [26] that if is a subring of , then we say that an element in is primitive over if is a root of a polynomial with unit content. We say that is a primitive extension (in short, -extension) in case any element of is primitive over . Clearly, each integral extension is necessarily a -extension. Moreover, for any integral domain , the ring extension is a -extension if and only if is Prüfer (cf. [26], Theorem 5). Note that if is a tower of rings and is a -extension, then so is . Recall also that is called a normal pair if and for any (cf. [27]). Ben Nasr and Jarboui have established a relationship between normal pairs and -extensions (for not necessarily integral domains). In fact, they have shown ([28], Theorem 1) that is a normal pair of rings if and only if is integrally closed in and is a -extension.
The following result shows (among other facts) that must be an overring of if is a maximal non-Dedekind domain subring of .

Corollary 2. Let be integral domains. Assume that is a maximal non-Dedekind domain subring of . Then the following conditions hold:(1) and have the same quotient field.(2) is a -extension.(3) is integrally closed only if is Prüfer.

Proof. (1)If is not a Prüfer domain, then Theorem 2 permits one to conclude that . Now, we suppose that is a Prüfer domain. As each subring of properly containing is Prüfer (as it is Dedekind), it follows that is a Prüfer domain pair. Moreover, cannot be a field since it is not Dedekind. Thus [25], Theorem 3.2, guarantees that .(2)It was proven in [28], Theorem 2, that if both and are -extensions, then is also a -extension. Our task is to show that if is a maximal non-Dedekind domain subring of , then is a -extension. If is not a Prüfer domain, then Theorem 2 implies that is a -extension since is Prüfer and is an overring of (cf. [26], Theorem 5). But, is a -extension since it is integral. Therefore, is also a -extension, as desired. Assume now that is Prüfer. Then, by using [26], Theorem 2, we derive that is a -extension, since is an overring of .(3)The “If  ” part is immediate. For the “Only if” part, assume that is integrally closed. Then is integrally closed in since . As is a -extensions, it follows that is a normal pair. Now, since is a Prüfer domain, it follows from [25], Lemma 2.2, that is also a Prüfer domain. This completes the proof.We derive from Theorem 2 the following result which treats the case where is integral over .

Corollary 3. Assume that is a subring of the integral domain and is integral over . Then the following conditions are equivalent:(1) is a maximal non-Dedekind subring of .(2) is minimal, is not a field and is a Dedekind domain.

Proof. (1) (2) The ring cannot be a field since it is not Dedekind. On the other hand, Corollary 2 ensures that . As is an integral extension, then . We deduce that cannot be integrally closed and so cannot be a Prüfer domain. On the other hand, as is an integrally closed domain, then . Hence, . Now, Theorem 2 asserts that is a minimal extension.
(2) (1) Since is a minimal extension and is not a field, then [20], page 1738, lines 8–13, guarantees that . As is integral over , then a fortiori . Therefore, . This proves that 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 is a local domain, then is a normal pair if and only if there is a divided prime ideal of (that is, ) such that and is a valuation domain. This result has been generalized in [29], Theorem 2.5, where the authors have proved that if is a normal pair and is a maximal ideal of , then , where is a divided prime ideal of . In these conditions, is a valuation domain and the set of prime ideals of in between and is totally ordered.
Next, we characterize pairs of rings satisfying in case is a semilocal Prüfer domain.

Theorem 3. Assume that are integral domains, where is a semilocal Prüfer domain. Then the following conditions are equivalent:(1) is a maximal non-Dedekind domain subring of .(2) is not Noetherian, has a Dedekind minimal overring in and is an overring of .

Proof. (1)  (2) it follows from Corollary 2 that is an overring of . As is a Prüfer domain, then is a normal pair. Notice that is not a Noetherian domain since it is a Prüfer domain but not a Dedekind domain. Set . As is a maximal non-Noetherian subring of , it follows from [3], Proposition 3.6, that there is a unique maximal ideal of such that is not Noetherian. Let be the prime ideal of such that . We have since is Noetherian, whereas is not. Set , where satisfying . Clearly, . Moreover, since (indeed, the maximal ideals of are , (and if is not comparable to , )). Hence, is a Dedekind domain. For each , we have . For , we have . Thus or . But if , then , a contradiction. Therefore, . Now, let ; then , where is the prime ideal of such that . Since , then for each and since is Noetherian. Hence, . This proves that is a Dedekind minimal overring of in .
(2)  (1) let be the unique (Dedekind) minimal overring of in . Then for each , is a Dedekind domain as an overring of the Dedekind domain . As is not a Dedekind domain (as it is not Noetherian), then is a maximal non-Dedekind domain subring of . The proof is complete.

Lemma 2. Assume that is maximal non-Dedekind domain subring of . Then the following hold true:(1)If is a valuation domain, then .(2)If is a Prüfer domain and is a field, then is a valuation domain.

Proof. (1)By using Theorem 3, has a Dedekind minimal overring in . Thus, is a minimal extension. An application of [29], Corollary 4.5, yields that .(2)As is a field and is a maximal non-Dedekind domain subring of , it follows from Corollary 2 that . In order to prove that is a valuation domain, it suffices to show that is a local domain. For, let and be two (distinct) maximal ideals of . Then there exist and such that . Notice that and but . Since and are Dedekind domains, then would be a Krull domain. But is also Prüfer, thus is a Dedekind domain, the desired contradiction. Consequently, is local.Our next result provides a classification of maximal non-Dedekind domain subrings of a field.

Theorem 4. Let be a field and let be a subring of . Then the following conditions are equivalent:(1) is a maximal non-Dedekind domain subring of .(2) one of the following conditions holds true:(a) is a local domain with quotient field , is a Dedekind domain and the ring extension is minimal.(b) is the quotient field of and either is a rank 2 valuation domain with a discrete rank 1 valuation overring or is a non-Noetherian rank 1 valuation domain.

Proof. (1) (2) if is not Prüfer, then condition (a) is satisfied according to Theorem 2. Suppose now that is Prüfer. Then, Lemma 2 yields that is a valuation domain and . If , then clearly is a rank 1 nondiscrete valuation domain. If , then is a rank 2 valuation domain with a discrete rank 1 valuation overring.
(2) (1) this is trivial.

Corollary 4. Let be integral domains such that is a valuation domain. Then the following conditions are equivalent:(1) is a maximal non-Dedekind domain subring of .(2) one of the following conditions holds true:(a) and is a rank 1 discrete valuation overring of .(b) is the quotient field of and either is a rank 2 valuation domain with a discrete rank 1 valuation overring or is a non-Noetherian rank 1 valuation domain.

Proof. (1) (2) if is a field, then Theorem 4 applies easily and so condition (b) is satisfied. Assume now that is not a field. As is a valuation domain, then is a (discrete) rank 1 valuation overring of (since it is a Noetherian). The fact that follows immediately from Lemma 2.
(2) (1) this is trivial.

Corollary 5. If is a maximal non-Dedekind domain subring of , then .

Proof. It follows from Theorem 2 (and Corollary 2 (3)) that if is not integrally closed, then is a Dedekind domain. As any integral extension preserves Krull dimensions, we get . Now, if is integrally closed (or equivalently Prüfer), then for any maximal ideal of , the localization is a valuation domain. So we have two cases to consider:(i)If is a Dedekind domain, then is a rank 1 discrete valuation domain.(ii)If is not a Dedekind domain, then is a maximal non-Dedekind domain subring of . Thus, Lemma 2 ensures that .In all cases, we have .
We close this section with the following result. We will make use of the following two facts:(i)If are rings, then(ii)If is a nonzero ideal of a ring , thenis an overring of called the Kaplansky transform of with respect to , and can be expressed as (cf. [30, 31]).

Theorem 5. Let be an extension of integral domains such that R is a Prüfer domain and , then the following conditions are equivalent:(1) is a maximal non-Dedekind domain subring of .(2) is not Noetherian and has a Dedekind minimal overring in .

In this case or 3.

Proof. (1)  (2) first of all, note that is not Noetherian since it is Prüfer but not Dedekind. Moreover, is a normal pair because is an overring of the Prüfer domain (see Corollary 2). Since , it follows that is a minimal extension, where represents any maximal ideal of ([32], Corollary 2.6). Now, if contains two maximal ideals and , then and are minimal extensions. As and are Krull domains, it results that is a Krull domain, so a Dedekind domain since it is already a Prüfer domain, a contradiction. Thus contains a unique maximal ideal . Using again ([32], Corollary 2.6), is the unique ring in such that is a minimal extension. It follows from [32], Corollary 2.7, that is totally ordered and so is the unique minimal overring of in and . But (cf. Corollary 5) and is totally ordered (cf. [32], Corollary 2.7). Thus, contains at most 2 elements and an eventual prime ideal contained in . Therefore, or 3.
(2)  (1) this is clear.

4. When is  = 2?

In this section, we start our investigation with the following result which generalizes Corollary 2. But, we point out that any minimal ring extension is a -extension. Indeed, if is integral, then obviously is a -extension. If is (integrally) closed, then is a normal pair and so is a -extension. We will make use of this fact in what follows.

Lemma 3. Let be integral domains such that . Then and have the same quotient field and is a -extension. Moreover, if is integrally closed, then is a Prüfer domain.

Proof. We argue by induction on . The case follows readily from Corollary 2. Assume that for any ring extension such that , we have and is a -extension. Let be a ring extension such that . By Lemma 1, is not a Dedekind domain. Thus, let be a minimal element in the set . A new, Lemma 1 guarantees that is a minimal extension and so by [20], page 1738, lines 8–13, and is a -extension. On the contrary, as , then the induction hypothesis ensures that and is a -extension. Therefore, we get on one hand and on the other hand, is a -extension, by using the transitivity of the “-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 extension , we will use the following notation: . Recall that a ring extension is said to satisfy FCP if, when is viewed as a poset under inclusion, each chain in 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 is called a Prüfer extension if is a flat epimorphism (in the category of commutative rings) for each . Any Prüfer extension is integrally closed. It is noteworthy that a ring extension is a P extension if and only if 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 , it was proved in [33], Theorem 5.2, page 47, that is a Prüfer extension if and only if is a normal pair. Recall also from [33] that any ring extension has a maximum Prüfer subextension , called the Prüfer hull of in . (Note that was denoted by in [33].) We say that a ring extension is Prüfer-closed (or that is Prüfer-closed in ) if .
The next theorem identifies pairs of rings such that under some additional conditions.

Theorem 6. Let be integral domains such that and is semilocal. Then the following conditions are equivalent:(1).(2) one of the following conditions holds:(a) is not a Dedekind domain and is a minimal extension.(b) is a Dedekind domain, is an overring of , is a chain of length 2, and .(c) is a Dedekind domain, is an overring of , and the ring extensions and are minimal.(d) is a minimal extension, is not Noetherian, has a Dedekind minimal overring in , and for each .

Proof. (1) (2) if is a minimal extension, then we are done. Next, we will assume that is not a minimal extension. Hypothetically, there are exactly two non-Dedekind domains in the set . One of these two domains is by Lemma 1 and the other, say , is such that is a minimal extension accordingly to Lemma 1. Therefore, is a maximal non-Dedekind domain subring of . Now, we will distinguish the following cases:

Case 1. is not a Prüfer domain.
In this case, is not integrally closed accordingly to Theorem 2. In particular, . As , it follows that is a Dedekind domain. Next, we discuss the following subcases.

Subcase 1. .
Let . Then is a Dedekind domain. Thus and so . This proves that is a chain of length 2. On the other hand, let . Then, is a Dedekind domain since and . As inherits the “-extension” property from , we conclude then that is a normal pair. Applying ([25], Lemma 2.2), it results that is a Prüfer domain. But, lies in between and , then a fortiori . Hence , and so . We have proved that .

Subcase 2. .
We claim that the ring extension is minimal. Indeed, if there exists a ring such that , then because is integrally closed in . Thus is a Dedekind domain. In particular, is integrally closed. This implies that . This proves our claim. Set . Since is a (closed) minimal extension, then either is a (closed) minimal extension or . The former case is impossible by virtue of [34], Lemma 3. Hence, . It follows that . If is properly contained in , then would be a Dedekind domain and so an integrally closed domain. On the other hand, for any maximal ideal of distinct from , is integrally closed since . It follows that is integrally closed as an intersection of integrally closed domains with the same quotient field, which is a contradiction. Thus .

Case 2. is a Prüfer domain.
In this case, would be a maximal non-Prüfer subring of . Using [25], Theorem 2.5, we derive that . It follows that is a (semilocal) Prüfer domain and a maximal non-Dedekind subring of . Thus, Theorem 3 implies that is not Noetherian, has a Dedekind minimal overring in and . Finally, we will show that for every . As is a semilocal Prüfer domain and is a maximal non-Noetherian subring of , it follows from [3], Proposition 3.9, that is an FCP extension. Since is an FCP extension, it results that is also an FCP extension (cf. [35], Theorem 3.13). Thus, in order to show that for each , we will demonstrate that is a Prüfer-closed extension (cf. [36], Theorem 2.7). To this end, assume by way of contradiction that . As and are incomparable under the usual set inclusion, we deduce that each subring of properly containing is a Dedekind domain. Hence, would be a maximal non-Dedekind subring of , which contradicts Theorem 2 since must be an intermediary ring in between and .
(2) (1) Suppose condition (a) satisfied. By assumption, is not a Dedekind domain since . Moreover, is not Dedekind. Therefore, . Suppose now condition (b) verified. As , it results that and are not Dedekind domains since they are not integrally closed. Let such that and . Since is integrally closed, . Now, as is a Dedekind domain, then so is . Assume condition (c) fulfilled, then obviously the domains and (where ) are not Dedekind domains since they are not integrally closed. Let and assume that and . As is a minimal extension and , it follows that either or . If , then would be integrally closed in . On the other hand, is an FCP extension (since it is minimal) and is also an FCP extension since is a semilocal Dedekind domain and is an overring of (see [37], Theorem 1.5). Hence, is an FCP extension. Thus, is a -extension by virtue of [36], Proposition 2.4. As , then is also a -extension. We deduce that the pair is normal. According to [34], Lemma 3, . So . But is a minimal extension, thus , a contradiction. Now, if , then would be a Dedekind domain as an overring of the Dedekind domain . If condition (d) is satisfied, then it is obvious that and are not Dedekind domains. Now combining theorems in [36], Theorem 2.5 and Theorem 2.7, we deduce that . As each ring in must contain , then would be a Dedekind domain as an overring of the Dedekind domain . This completes the proof of Theorem 6.

Corollary 6. Let be integral domains such that and is a field. Then the following conditions are equivalent:(1).(2) one of the following conditions hold:(a) is a Dedekind domain with quotient field , is a chain of length 2, and .(b) is a Dedekind domain with quotient field , and are minimal extensions.(c), is a minimal extension and either is a rank 2 valuation domain with a discrete rank 1 valuation overring or is a non-Noetherian rank 1 valuation domain.

Corollary 7. Assume that is an integral extension. Then the following conditions are equivalent:(1).(2) one of the following conditions hold:(a) is not a Dedekind domain and is a minimal extension.(b) is a Dedekind domain which is not a field and .

Data Availability

All data required for this paper are included within this paper.

Conflicts of Interest

The authors declare no conflicts of interest.

Acknowledgments

This work was supported through the Annual Funding track by the Deanship of Scientific Research, Vice Presidency for Graduate Studies and Scientific Research, King Faisal University, Saudi Arabia (Project no. GRANT274).