n-absorbing ideal factorization in commutative rings

Abstract

In this article, we show that Mori domains, pseudo-valuation domains, and n-absorbing ideals, the three seemingly unrelated notions in commutative ring theory, are interconnected. In particular, we prove that an integral domain R is a Mori locally pseudo-valuation domain if and only if each proper ideal of R is a finite product of 2-absorbing ideals of R. Moreover, every ideal of a Mori locally almost pseudo-valuation domain can be written as a finite product of 3-absorbing ideals. To provide concrete examples of such rings, we study rings of the form $A+\mathit{\text{XB}}[X]$ where A is a subring of a commutative ring B and X is indeterminate, which is of independent interest, and along with several characterization theorems, we prove that in such a ring, each proper ideal is a finite product of n-absorbing ideals for some $n\ge 2$ if and only if $A\subseteq B$ is essentially a finite product of field extensions. A complete description of when an order of a quadratic number field is a locally pseudo valuation domain, a locally almost pseudo valuation domain or a locally conducive domain is given.

KEYWORDS:

• 2-absorbing ideals
• almost pseudo-valuation domains
• Mori domains
• pseudo-valuation domains
• strongly Laskerian rings

2020 MATHEMATICS SUBJECT CLASSIFICATION:

• Primary: 13A15
• Secondary: 13B30
• 13F05
• 13G05

1 Introduction

Pseudo-valuation domains (PVDs) were defined by Hedstrom and Houston [44], as a generalization of valuation domains. Due to its several interesting properties, pseudo-valuation domains have been extensively studied [32, 45], and their generalizations were also considered. Among them are locally pseudo-valuation domains (LPVDs) considered by Dobbs and Fontana [28], and almost pseudo-valuation domains (APVDs) introduced by Badawi and Houston [14].

This paper aims to investigate the ring-theoretic properties of a certain class of commutative rings including that of pseudo-valuation domains, almost pseudo-valuation domains and locally pseudo-valuation domains. In Section 2, we gather well-known properties of these classes of integral domains, and prove the APVD-counterparts of known theorems concerning PVDs. Several results proved in this section are used throughout the paper. In Section 3, we study Mori LPVDs. It is shown that a locally pseudo-valuation domain R is a Mori domain if and only if the complete integral closure $R^{*}$ of R is a Dedekind domain such that the contraction map $\text{Spec}(R^{*})\to \text{Spec}(R)$ is bijective (cf. Corollary 21). We also show that if R is a Mori LPVD, then the Nagata ring of R is a Mori LPVD if and only if the integral closure of R and the complete integral closure of R coincide.

We also relate LPVDs with n-absorbing ideals of a commutative ring introduced by Anderson and Badawi in 2011 [6]. Using this concept, the AF-dimension of a ring R can be defined as the smallest n such that each proper ideal of R is a finite product of n-absorbing ideals of R. Then a Dedekind domain is exactly an integral domain whose AF-dimension is one. We focus on the fact that since every prime ideal is a 2-absorbing ideal, some of the properties of Dedekind domains can be inherited by integral domains with AF-dimension at most two. Motivated by the fact that Dedekind domains possess several interesting properties that connects numerous classes of integral domains, we show that domains with AF-dimension at most two have similar properties. In particular, we show that an integral domain R is a Mori LPVD if and only if the AF-dimension of R is at most two. Using this, we show that an LPVD is strongly Laskerian if and only if it is a Mori domain (Lemma 27), extending [15, Corollary 3.7]. Motivated by the globalized pseudo-valuation domains (GPVDs) introduced by Dobbs and Fontana [28], we provide pullback descriptions of Mori GPVDs.

In Section 4, we give a structure theorem of TAF-rings corresponding to [37, Theorem 39.2], and show that certain Noetherian domains with finite AF-dimension can be constructed from pullbacks. This result is used to classify orders of quadratic number fields in terms of AF-dimensions and LPVDs, LAPVDs and locally conducive domains, extending [29, Theorem 2.5].

In Section 5, we classify reduced rings of the form $A+\mathit{\text{XB}}[X]$ whose Krull dimension is 1, where $A\subseteq B$ is an extension of commutative unital rings and X is an indeterminate, in order to provide concrete examples of TAF-rings. In particular, we show that $A+\mathit{\text{XB}}[X]$ is a TAF-ring exactly when the contraction map $\text{Spec}(B)\to \text{Spec}(A)$ is a bijection and both A and B are direct product of finitely many fields.

2 Ring-theoretic properties of almost pseudo-valuation domains

Throughout this paper, every ring is assumed to be nonzero, commutative and unital. A quasilocal ring (respectively, semi-quasilocal ring) is a ring with only one maximal ideal (respectively, finitely many maximal ideals). When R is an integral domain, K will denote its quotient field, and the integral closure (respectively, complete integral closure) of R in K will be denoted by $R\prime$ (respectively, $R^{*})$. $\mathbb{N},\mathbb{Z},\mathbb{Q}$ and $\mathbb{R}$ denote the set of natural numbers, the ring of integers, the field of rational numbers and that of real numbers, respectively. The nilradical (respectively, the set of prime ideals) of a ring R will be denoted by Nil(R) (respectively, $\text{Spec}(R)$).

Definition 1.

Let R be an integral domain with quotient field K.

1. An ideal I of R is strongly prime (respectively, strongly primary) if given $x,y\in K$ with $\mathit{\text{xy}}\in I$, either $x\in I$ or $y\in I$ (respectively, either $x\in I$ or $y^n\in I$ for some $n\in \mathbb{N}$).

2. R is said to be a valuation domain if the set of ideals of R is totally ordered under set inclusion.

3. A DVR is a Noetherian valuation domain that is not a field.

4. R is a Prüfer domain if R_M is a valuation domain for each maximal ideal M of R.

5. R is a pseudo-valuation domain, or PVD in short, if every prime ideal of R is strongly prime.

6. R is an almost pseudo-valuation domain, or APVD in short, if every prime ideal of R is strongly primary.

7. An overring of R is a ring T such that $R\subseteq T\subseteq K$. An overring T of R is proper if $T\ne R$.

8. By a valuation overring of R we mean an overring of R that is also a valuation domain.

9. Given two R-submodules I, J of K, the set $\{x\in \mathit{\text{K⏧xJ}}\subseteq I\}$ will be denoted by $I:J$.

10. A nonzero R-submodule I of K is said to be a fractional ideal if $R:I$ contains a nonzero element.

11. A fractional ideal I of R is invertible if $I(R:I)=R$.

12. R is conducive if each overring of R other than K is a fractional ideal of R.

13. R is locally conducive if R_M is conducive for each maximal ideal M of R.

14. R is seminormal if for any $x\in K$ such that $x^2\in R$ and $x^3\in R$, we have $x\in R$.

Any unexplained terminology is standard, as in [10, 37], or [49].

We first collect known results concerning PVDs, APVDs and conducive domains.

Theorem 2.

Every valuation domain is a PVD, and every PVD is an APVD.

Proof.

The first statement follows from [44, Proposition 1.1], while the second follows from the definition. □

Theorem 3.

Let M be a maximal ideal of a domain R. Then the following are equivalent.

1. R is a PVD (respectively, an APVD).

2. M is strongly prime (respectively, strongly primary).

3. M is a prime ideal (respectively, a primary ideal) of a valuation overring of R.

4. $M:M$ is a valuation domain, and M (respectively, the radical ideal of M in $M:M$) is the maximal ideal of $M:M$.

Proof.

Note that if $M:M$ is a valuation domain, then R is quasilocal [5, Corollary 3.4].

$(1)\iff (2)$: Follows from [5, Proposition 3.1 and Corollary 3.6] and [14, Theorem 3.4].

$(1)\iff (3)$: Follows from [5, Proposition 3.11] and [14, Theorem 3.4].

$(1)\iff (4)$: Follows from [5, Theorem 2.1] and [14, Theorem 3.4]. □

Theorem 4.

An integral domain R is conducive if and only if $R:V\ne (0)$ for some valuation overring V of R. In particular, every valuation domain is conducive.

Proof.

This is just [27, Theorem 3.2]. □

An ideal I of a ring R is divided if $I⊊\mathit{\text{aR}}$ for each $a\in R\setminus I$. A ring R is divided if each prime ideal of R is divided. If R_M is a divided ring for each maximal ideal M of R, then we say that R is locally divided. It is easy to see that zero-dimensional rings and one-dimensional domains are locally divided. We also have the following result due to [14, Proposition 2.13.(1) and Theorem 4.1].

Theorem 5.

Every APVD is divided and conducive.

Given an integral domain R with quotient field K, F(R) will denote the set of fractional ideals of R. Recall that the map $v:F(R)\to F(R)$ that sends I to $R:(R:I)$ for each $I\in F(R)$, defines the famous v-operation on an integral domain R [37, Chapter 34]. For each $I\in F(R)$, the image of I under v is denoted by I^v. This is the most famous example of so-called star operation that proved to be extremely useful in terms of classification of integral domains, and have become a main topic of multiplicative theory of ideals since Gilmer’s modern treatment of star operations [37, Chapter 32], which is based upon Krull’s work [51].

If $I^v=I$ for a fractional ideal I of R, then we say that I is a divisorial ideal of R. An ideal of R that is also a divisorial ideal of R is said to be a divisorial integral ideal of R. A domain R is said to be a divisorial domain if every nonzero ideal of R is a divisorial integral ideal. It is well-known that a valuation domain is a divisorial domain if and only if its maximal ideal is a divisorial integral ideal. In fact, we have the following.

Theorem 6.

[37, Exercise 34.12] Let R be a valuation domain with maximal ideal N. Then the following are equivalent.

1. $N\ne N^2$.

2. N is a principal ideal of R.

3. N is a divisorial integral ideal of R.

4. R is a divisorial domain.

Moreover, if $N=N^2$, then $\{\mathit{\text{aN⏧a}}\in K\setminus \{0\left\}\right\}$ is the set of nondivisorial integral ideals of R.

Recall that when R is a PVD, there exists a unique valuation overring V of R such that $\text{Spec}(R)=\text{Spec}(V)$ as sets [44, Theorem 2.7]. Such V is called the associated valuation overring of R. Note that $V=M:M$ where M is the maximal ideal of R (cf. Theorem 3).

Lemma 7.

Let R be a domain. Then the following are equivalent.

1. R is an APVD.

2. $R\prime$ is a PVD with associated valuation overring $M:M$ for some maximal ideal M of R.

In particular, $\text{dim}(R)=\text{dim}(M:M)$ if R is an APVD with maximal ideal M.

Proof.

One can easily see that any integral domain satisfying either (1) or (2) must be quasilocal. Hence, we may assume that R is quasilcoal with maximal ideal M.

(1) $\Rightarrow$ (2): Let R be an APVD. By [14, Proposition 3.7], $R\prime$ is a PVD, M is an ideal of $R\prime$ (so $R\prime \subseteq M:M$), and the maximal ideal N of $R\prime$ is the radical of M in $R\prime$. By Theorem 3 the maximal ideal N₀ of $M:M$ is the radical of M in $M:M$. Then $N_0\cap R\prime =\sqrt{M}\cap R\prime =\sqrt{M}=N$, where the first radical is over $M:M$ and the second over $R\prime$. It follows that N = N₀. Hence, $M:M$ is the associated valuation overring of $R\prime$ [44, Theorem 2.7].

(2) $\Rightarrow$ (1): Suppose that (2) holds. Then M is an ideal of $R\prime$. Let P be the radical of M in $R\prime$. Since $R\prime$ is a PVD, the set of prime ideals of $R\prime$ is totally ordered under inclusion [44, Corollary 1.3], and P is a prime ideal of $R\prime$. Thus, $P\cap R=M$, and P must be the maximal ideal of $R\prime$. Since $M:M$ is the associated valuation overring of $R\prime$, P is the maximal ideal of $M:M$ by the comment preceding this lemma. Hence, the radical of M in $M:M$ is P, and R must be an APVD by Theorem 3.

The remaining assertion now follows from the fact that $\text{dim}(R)=\text{dim}(R\prime )$. □

From now on, if R is an APVD, then M (respectively, N) will denote the maximal ideal of R (respectively, $R\prime$), and V the associated valuation overring of $R\prime$. In this case, we will also call V the associated valuation overring of R.

Corollary 8.

Let R be an APVD.

1. R is a PVD if and only if M = N.

2. R is a valuation domain if and only if R = V.

3. Either R = V, or $M=R:V$ is a divisorial integral ideal of R.

Proof.

(1): If M = N, then M is strongly prime by Lemma 7, and R is a PVD. Conversely, if R is a PVD, then M is strongly prime. Since M is a proper ideal of $R\prime$, it must be a prime ideal of $R\prime$. Since M is an N-primary ideal of $R\prime$, we must have M = N.

(2): If R is a valuation domain, then R is a PVD, so M = N by (1). Since N is the maximal ideal of V, we have R = V [37, Theorem 17.6]. The converse is obvious.

(3): We have $M\subseteq R:V$ since M is an ideal of both R and V. Suppose that $R\ne V$. Then $R:V$ is a proper ideal of R, so $R:V\subseteq M$ and $M=R:V$. On the other hand, R is a divisorial integral ideal of R. Therefore, $M=R:V$ is a divisorial integral ideal of R [37, Theorem 34.1.(3)]. □

Given rings A, B, C with unital ring homomorphisms $u:A\to C$ and $v:B\to C$, where v is surjective, the pullback of u with respect to v, denoted by $A{\times }_{C}B$, is the ring $\{(a,b)\in A\times B⏧u(a)=v(b)\}$. In this paper, we are interested in the case when B is an integral domain, $C=B/I$ for some ideal I of B, A is a subring of C, and u (respectively, v) is the canonical inclusion (respectively, canonical projection). In this case, $D=A{\times }_{C}B$ is an integral domain, B is an overring of D, and $B^{*}=D^{*}$ [34, Lemma 1.1.4.(10)]. Note that such D can be identified to the ring $v^{-1}(A)$ [34, Lemma 1.1.4.(11)]. The notion of a pullback of a domain was proved to be immensely useful, in terms of producing examples and proving theorems. For instance, it is well-known that every PVD arises from a pullback of valuation domains. Precisely, an integral domain R is a PVD if and only if $R=L{\times }_{V/M}V$ for some valuation domain V with maximal ideal M and a subfield L of V/M [7, Proposition 2.6]. Motivated by this result, we present a pullback characterization of APVDs in the following proposition.

Proposition 9.

Let R be an integral domain. Then R is an APVD if and only if there exists a valuation domain V with maximal ideal N, an N-primary ideal M and a field $L\subseteq V/M$ such that R is the pullback $L{\times }_{V/M}V$. In this case, M is the maximal ideal of R, $V=M:M$ and L is the residue field of R.

Proof.

Suppose that R is an APVD, and let $L=R/M$. By the comment preceding this proposition, we can assume that R is not a PVD. Then $R:V=M$ by Corollary 8, so R is the pullback of the given form (cf. [19, Theorem 1]).

Conversely, assume that R is a pullback of the given form, and let $\varphi \prime :R\to L$ the ring homomorphism induced by the canonical projection map $\varphi :V\to V/M$. Then $M=\mathit{\text{ker}}(\varphi \prime )$ is an ideal of R. In fact, M is the maximal ideal of R since $L\cong R/\mathit{\text{ker}}(\varphi \prime )$. It follows that V is an overring of R, and R is an APVD by Theorem 3. Note that $M:M=V$ [34, Theorem 4.2.6]. □

Recall that an integral domain is said to be a Mori domain if it satisfies the ascending chain condition on divisorial integral ideals. Our next goal is to classify Mori APVDs. We first need the following lemma.

Lemma 10.

Let R be an APVD such that N is a divisorial ideal of R. Then we have the following.

1. Each nonzero ideal of V is a divisorial ideal of R.

2. $J^v=\mathit{\text{JV}}$ for each nonprincipal ideal J of R.

Proof.

If I is an ideal of V, then $I^v\subseteq N:(N:I)$ [57, Proposition 1.20]. On the other hand, $N:(N:I)=I$ [21, Proposition 4.1]. Thus, (1) follows. For (2), we adapt the proof of [44, Proposition 2.14]. Choose a nonprincipal ideal J of R. Notice that if $x\in R:J$, then $\mathit{\text{xJ}}\subseteq R$, and $\mathit{\text{xJ}}\ne R$ since J cannot be a principal ideal of R. Hence, $\mathit{\text{xJ}}\subseteq M$, and thereby $\mathit{\text{xJV}}\subseteq \mathit{\text{MV}}=M\subseteq R$ implies $x\in R:\mathit{\text{JV}}$. Thus, $R:J\subseteq R:\mathit{\text{JV}}$. Since JV is a divisorial ideal of R by (1), we then have $\mathit{\text{JV}}={(\mathit{\text{JV}})}^v\subseteq J^v$. Since $J^v\subseteq {(\mathit{\text{JV}})}^v=\mathit{\text{JV}}$, (2) follows. □

Corollary 11.

Let R be an integral domain that is not a field. Then the following are equivalent.

1. R is a Mori APVD.

2. $M:M$ is a DVR for some maximal ideal M of R.

Proof.

(1) $\Rightarrow$ (2): Suppose that R is a Mori APVD with maximal ideal M. Then $M:M$ is a valuation domain by Proposition 9. Since a Mori valuation domain is a DVR, we only need to show that $M:M$ is a Mori domain. If $M:M=R$, then we have nothing to prove. Suppose that $M:M\ne R$. Then M is a divisorial integral ideal of R by Corollary 8.(3). Hence, $M:M=R:M$ is a Mori domain [16, Corollary 11].

(2) $\Rightarrow$ (1): Suppose that (2) holds, and let N be the maximal ideal of $V=M:M$. Then the radical of M in V is N, so R must be an APVD with maximal ideal M by Theorem 3. It also follows that N = aV and $M=a^{n}V$ for some $a\in N$ and $n\in \mathbb{N}$. If R = V, then R is clearly a Mori domain. Assume that $R\ne V$. Then by Corollary 8.(3), M is a divisorial integral ideal of R, and V is a divisorial ideal of R. Hence, N is a divisorial ideal of R. Now, let ${\left\{I_{\alpha }\right\}}_{\alpha \in \mathcal{A}}$ be an ascending chain of divisorial integral ideals of R. By Lemma 10, $I_{\alpha }$ is either a principal ideal of R or an ideal of V for each $\alpha \in \mathcal{A}$. Note that R must satisfy ACCP since every nonunit of R is a nonunit of V, while R has the ascending chain condition on nonprincipal divisorial integral ideals since V is a Noetherian ring. Hence, the chain ${\left\{I_{\alpha }\right\}}_{\alpha \in \mathcal{A}}$ must be stationary, and R is Mori. □

Lemma 12.

Let R be an APVD that is not a field. Then the following hold.

1. $R^{*}$ is a one-dimensional valuation domain.

2. $R^{*}$ is the associated valuation overring of R if and only if R is one-dimensional.

3. R is Mori if and only if the associated valuation overring of R is a DVR. In this case $R^{*}$ is the associated valuation overring of R.

Proof.

Let M be the maximal ideal of R, $L=R/M$ and $V=M:M$ the associated valuation overring of R. Then R is a pullback of $L{\times }_{V/M}V$ by Proposition 9. In particular, R and V have the same complete integral closure [34, Lemma 1.1.4.(10)]. Since R is conducive by Theorem 5, $R^{*}$ is completely integrally closed [42, Corollary 6]. Moreover, $R^{*}$ is a valuation domain since it is an overring of a valuation domain V. Now (1) and (2) both follow from [37, Theorem 17.5.(3)]. The first assertion of (3) is an immediate consequence of Corollary 11, while the second one then follows from (2) and Lemma 7. □

Following [4, Definition 2.4], a nonzero nonunit element a of a ring R is irreducible if for each $b,c\in R$ such that a = bc, either aR = bR or aR = cR. A ring R is atomic if each nonzero nonunit element of R can be written in at least one way as a finite product of irreducible elements of R. If a domain R has the ascending chain condition on its principal ideals, then we say that R satisfies ACCP. A domain that satisfies ACCP is atomic, but the converse fails in general [43]. Our next result shows that in an APVD, these two properties actually coincide.

Corollary 13.

The following are equivalent for an integral domain R.

1. R is a Mori PVD.

2. R is a PVD that satisfies ACCP.

3. R is an atomic PVD.

4. There exists a maximal ideal M of R such that $M:M$ is a DVR whose maximal ideal is M.

Proof.

We only need to show (3) $\Rightarrow$ (4) $\Rightarrow$ (1), which follows from [3, Theorem 5.1 and Corollary 5.2] and Corollary 11. □

Proposition 14.

Let R be an integral domain that is not a PVD. Then the following are equivalent.

1. R is an APVD that satisfies ACCP.

2. R is an atomic APVD.

3. $M:M$ is a one-dimensional valuation domain for some maximal ideal M of R.

Proof.

(1) $\Rightarrow$ (2) is well-known as mentioned in the comment preceding Corollary 13. Since every APVD is divided by Theorem 5, (2) $\Rightarrow$ (3) follows from Lemma 7 and [11, Proposition 19]. Suppose that R satisfies (3). Then M is a nonzero ideal of a one-dimensional valuation domain $M:M$, so R is an APVD by Theorem 3. By Corollary 8 and our assumption that R is not a PVD, $M\ne N$. Thus, there exists $a\in N\setminus M$, and we have $M\subseteq \mathit{\text{aV}}$. Suppose that there exists a strictly ascending chain $(f_1)⊊(f_2)⊊\cdots$ of nonzero principal ideals of R. Then $f_i\in f_{i+1}M\subseteq a{f}_{i+1}V$ for each $i\in \mathbb{N}$. Since V is one-dimensional, $f_1\in \underset{i\in \mathbb{N}}{\cap }{a}^{i}V=(0)$ [37, Theorem 17.1.(3)], a contradiction. Hence, R must satisfy ACCP. □

Proposition 15.

Let R be a PVD with maximal ideal M. Then the following are equivalent.

1. Every overring of R is Mori.

2. Every overring of R satisfies ACCP.

3. Every overring of R is atomic.

4. $R\prime$ is a DVR.

5. R is Mori and every overring of R is a PVD.

Proof.

(1) $\Rightarrow$ (2) $\Rightarrow$ (3): Trivial.

$(3)\iff (4)$: Follows from [30, Theorem 12] and Lemma 7.

$(4)\Rightarrow (5)$: If (4) holds, then $R\prime =M:M$. By [9, Corollary 2.2] and Corollary 13, (5) follows.

$(5)\Rightarrow (4)$: Assume (5), and we have $R=M:M$ by [9, Corollary 2.2]. Hence, (4) follows from Corollary 13.

$(3)\Rightarrow (1)$: Suppose that (3) is true. Given an overring T of R, T is an atomic PVD since (3) implies (5). Hence, T is Mori by Corollary 13, which yields (1). □

3 TAF-domains, Mori domains and locally pseudo-valuation domains

In 2007, Badawi [12] generalized the notion of a prime ideal as follows. Let I be an ideal of a ring R. We say that I is a 2-absorbing ideal of R if for any $a_1,a_2,a_3\in R$ such that $a_1{a}_2{a}_3\in I$, either $a_1{a}_2\in I, a_1{a}_3\in I$ or $a_2{a}_3\in I$. It is easy to see that every prime ideal is 2-absorbing, but the converse fails, for the product of two distinct maximal ideals of a ring is 2-absorbing [6, Theorem 2.6], but not prime. In [6], Anderson and Badawi generalized this concept further by defining an ideal I of a ring R to be an n-absorbing ideal of R if for any $a_1,\dots ,a_{n+1}\in R$ such that $a_1\cdots a_{n+1}\in I$, there exists $i\in \{1,\dots ,n+1\}$ such that $\prod _{j\in \{1,\dots ,n+1\}\setminus \left\{i\right\}}{a}_j\in I$. Given a proper ideal I of a ring R, the minimal $n\in \mathbb{N}$ such that I is n-absorbing is denoted by ${\omega }_R(I)$, and we set ${\omega }_R(I)=\infty$ when I is not n-absorbing for any $n\in \mathbb{N}$. From now on, we will call a ring R finite-absorbing if ${\omega }_R(I)\in \mathbb{N}$ for every proper ideal I of R.

A ring in which every proper ideal is a finite product of prime ideals is said to be a general ZPI-ring [37, Chapter 39]. On the other hand, Mukhtar et.al. [54] considered rings in which each proper ideal is a finite product of 2-absorbing ideals, and called such rings TAF-rings. It follows that every general ZPI-ring is a TAF-ring. The main portion of this section consists of various ring-theoretic properties of TAF-domains, i.e., TAF-rings that are also integral domains. For instance, it is well-known that an integral domain is a general ZPI-ring (in fact, a Dedekind domain) if and only if it is a Noetherian Prüfer domain, and in Proposition 19 we show that a similar criterion holds for TAF-domains. Using this criterion, we prove the structure theorem for TAF-rings analogous to [37, Theorem 39.2] in section 4.

Recall that an integral domain R is a locally pseudo-valuation domain, or an LPVD in short, if R_M is a PVD for each maximal ideal M of R [28, Proposition 2.2]. An integral domain in which each nonzero nonunit is contained in only finitely many maximal ideals is of finite character. We begin with a collection of useful facts concerning Mori domains.

Theorem 16.

Let R be an integral domain.

1. R is a Mori domain if and only if for each nonzero ideal I of R, there exists a finitely generated ideal J of R such that $J\subseteq I$ and $R:I=R:J$.

2. If R is a one-dimensional Mori domain, then R is of finite character.

3. If R is Mori, then R_S is Mori for each multiplicatively closed subset S of R.

4. Suppose that R is of finite character. Then R is Mori if and only if it is locally Mori.

Proof.

(1): This is well-known. For instance, see [31, Proposition 2.6.11].

(2): This is exactly [35, Lemma 3.11].

(3): [31, Proposition 3.3.25].

(4): By (3), every Mori domain is locally Mori. The converse follows from [53, Theorem 4.7] and [62, Corollary 5]. □

Theorem 17.

Let R be an integral domain.

1. If R is a Mori domain and $R:R^{*}\ne (0)$, then $R^{*}$ is a Krull domain. Moreover, R has Krull dimension 1 if and only if $R^{*}$ is a Dedekind domain.

2. If R is a seminormal Mori domain that has Krull dimension 1, then $R^{*}$ is a Dedekind domain.

3. If $R^{*}$ is a Krull domain, then ${(R_S)}^{*}={(R^{*})}_S$ for each multiplicatively closed subset S of R.

Proof.

[16, Corollary 18], [20, Corollary 3.4], [17, Theorem 2.9], and [20, Lemma 3.1]. □

The following characterization is well-known.

Theorem 18.

Let R be an integral domain that is not a field. Then the following are equivalent.

1. R is a Dedekind domain.

2. R is of finite character, and R_M is a DVR for each maximal ideal M of R.

3. R is a Noetherian Prüfer domain.

4. R is an integrally closed Noetherian locally conducive domain.

Proof.

(1) $\iff$ (2): [37, Theorem 37.2].

(1) $\iff$ (3): [37, Theorem 37.1].

(3) $\Rightarrow$ (4): Let R be a Noetherian Prüfer domain. Then R is integrally closed [37, Theorem 26.2], and R_M is a DVR for each maximal ideal M of R by the equivalence of (2) and (3). Since every valuation domain is conducive by Theorem 4, R is locally conducive.

(4) $\Rightarrow$ (2): Suppose that (4) holds. Let M be a maximal ideal of R. Then R_M is a conducive Krull domain [53, Theorems 12.1, 12.4], which is a DVR [27, Corollary 2.5]. Consequently, R is a one-dimensional Noetherian domain, and must be of finite character. □

We have an analogous characterization for TAF-domains.

Proposition 19.

Let R be an integral domain. Then the following are equivalent.

1. R is a TAF-domain.

2. R is of finite character, and R_M is a Mori PVD for each maximal ideal M of R.

3. R is a Mori LPVD.

4. R is a seminormal Mori locally conducive domain.

In particular, R has Krull dimension at most 1.

Proof.

We may assume that R is not a field.

(1) $\Rightarrow$ (2): Suppose that R is a TAF-domain. Then R is of finite character and R_M is a TAF-domain for each maximal ideal M of R [54, Theorem 4.4]. Hence, R_M is an atomic PVD for each maximal ideal M of R by [54, Theorem 4.3], and Corollary 13 then yields (2) (cf. [50, Theorem 4.1]).

(2) $\Rightarrow$ (3): Follows from Theorem 16.

(3) $\Rightarrow$ (4): Assume (3). Then R is seminormal [28, Remarks 2.4], and locally conducive [27, Proposition 2.1].

(4) $\Rightarrow$ (1): Suppose (4) holds, and let M be a maximal ideal of R. Then R_M is a seminormal Mori conducive domain. Hence, ${(R_M)}^{*}$ is a conducive Krull domain by Theorem 17, which is a DVR [27, Corollary 2.5]. Hence, R_M has Krull dimension 1 by Theorem 17. It follows that R_M is a PVD [27, Corollary 2.6], and R is of finite character by Theorem 16. Now R is a TAF-domain by [54, Theorem 4.4]. □

Proposition 19 tells us that every TAF-domain is a Mori domain. In fact, we can say something stronger. Note that an integral domain R is Mori whenever the polynomial ring $R[X]$ is Mori, but the converse fails in general, as Roitman proved [61, Theorem 8.4].

Proposition 20.

If R is a TAF-domain, then $R[X]$ is a Mori domain.

Proof.

Let R be a TAF-domain. Then R is a Mori LPVD, R has finite character and R_M is a Mori domain for each maximal ideal M of R by Proposition 19, so by [59, Proposition 3.14] we may assume that R is quasilocal with maximal ideal M. Then R is a Mori PVD, and V is a DVR by Corollary 11. Hence, $V[X]$ is Mori (in fact, Noetherian). Let $F=R/M, L=V/M$, and T be the quotient field of $F[X]$. Then $F[X]=T\cap L[X]$, so $R[X]=R{[X]}_{M[X]}\cap V[X]$ and $R[X]$ is Mori [60, Corollary 4.16]. □

We next “globalize” Lemma 12.(3). A ring extension $R\subseteq T$ is said to be unibranched if, for each prime ideal P of R, there exists exactly one prime ideal Q of T such that $Q\cap R=P$. By a locally almost pseudo-valuation domain, or an LAPVD in short, we mean an integral domain R such that R_M is an APVD for each maximal ideal M.

Corollary 21.

Let R be an LAPVD. Then R is a Mori domain if and only if the following two conditions are satisfied.

1. $R^{*}$ is a Dedekind domain.

2. $R\subseteq R^{*}$ is a unibranched extension.

Proof.

Assume that R is a Mori LAPVD, and choose a maximal ideal M of R. Then R_M is a Mori APVD by Theorem 16. We claim that ${(R^{*})}_{R\setminus M}$ is a DVR. To avoid triviality, R_M is assumed to be an integral domain that is not a field. We consider two cases:

Case 1: M is an invertible ideal of R.

If so, then MR_M is a principal ideal of R_M, and ${(R_M)}^{*}=M{R}_M:M{R}_M=R_M$ is a DVR by Lemma 12. Since $R_M\subseteq {(R^{*})}_{R\setminus M}\subseteq {(R_M)}^{*}$, it follows that ${(R^{*})}_{R\setminus M}$ is a DVR.

Case 2: M is not an invertible ideal of R.

In this case, we have $R:M=M:M$. Moreover, since R is Mori, there exists a finitely generated ideal J of R such that $J\subseteq M$ and $R:M=R:J$ by Theorem 16.(1). Thus, ${(M:M)}_{R\setminus M}\subseteq M{R}_M:M{R}_M\subseteq R_M:M{R}_M\subseteq R_M:J{R}_M={(R:J)}_{R\setminus M}={(R:M)}_{R\setminus M}={(M:M)}_{R\setminus M}$. Hence, ${(M:M)}_{R\setminus M}=M{R}_M:M{R}_M$ is a DVR by Corollary 11. On the other hand, R_M is a one-dimensional quasilocal domain by Lemma 12, so ${(R_M)}^{*}⊊K$ [27, Proposition 4.3.(ii)]. Since ${(M:M)}_{R\setminus M}\subseteq {(R^{*})}_{R\setminus M}\subseteq {(R_M)}^{*}⊊K$, we deduce that ${(R^{*})}_{R\setminus M}={(R_M)}^{*}$ is a DVR.

Now the claim is proved, and we can see that there exists exactly one prime ideal N of $R^{*}$ such that $N\cap R=M$. Such N is a maximal ideal of $R^{*}$, so $R\subseteq R^{*}$ is a unibranched extension and ${(R^{*})}_{N\prime }$ is DVR for each maximal ideal $N\prime$ of $R^{*}$. Since R is a one-dimensional Mori domain, R is of finite character (Theorem 16.(2)) and so is $R^{*}$. Thus, $R^{*}$ is Dedekind by Theorem 18.

Conversely, suppose that R satisfies (1) and (2), and let M be a maximal ideal of R. Then there exists unique maximal ideal N of $R^{*}$ that contracts to M, and M has height 1 by (2). Thus, R_M is a one-dimensional APVD, and by Theorem 17 and Lemma 12 ${(R_M)}^{*}={(R^{*})}_{R\setminus M}={(R^{*})}_N$ is its associated valuation overring, which is a DVR by (1). Hence, R_M is a Mori domain. Moreover, since $R^{*}$ is of finite character, so is R by (2). Therefore, R is a Mori domain by [62, Corollary 5]. □

Remark 22.

It is well-known that an integral domain is Dedekind if and only if it is an integrally closed Noetherian ring with Krull dimension at most 1 [49, Theorem 96]. This equivalence cannot be generalized to TAF-domains the way Proposition 19 does. In other words, a TAF-domain is a seminormal Mori domain with Krull dimension at most 1, but the converse fails in general. For instance, let $R=\mathbb{Q}+(X^2-1)\mathbb{R}[X]$ where X an indeterminate. Then R is the pullback $\mathbb{Q}{\times }_{T/M}T$ where $T=\mathbb{R}[X]$ and $M=(X^2-1)T=(X-1)T\cap (X+1)T$. Hence, R is a seminormal Mori one-dimensional domain and M is a maximal ideal of R [20, Proposition 4.1 and Theorem 4.3.(3)]. However, $R\subseteq R^{*}=T$ is not a unibranched extension since the prime ideals $(X-1)T$ and $(X+1)T$ of T both lie over M. Hence, by Proposition 19 and Corollary 21, R cannot be a TAF-domain.

Unlike Dedekind domains, a TAF-domain may not be integrally closed. For instance, let X be an indeterminate and F a field that is not algebraically closed. If L is an algebraic closure of F, then $R=F+\mathit{\text{XL}}[X]$ is a TAF-domain [54, Corollary 4.8], but R is not integrally closed. On the other hand, we have the following.

Proposition 23.

An integral overring of a TAF-domain is a TAF-domain.

Proof.

Let R be a TAF-domain and T an integral overring of R. If N is a maximal ideal of T and $M=N\cap R$, then $T_{R\setminus M}$ must be a PVD whose maximal ideal is the maximal ideal of R_M [44, Theorem 1.7], because it is an integral overring of R_M. Therefore, $T_N=T_{R\setminus M}$, and $R\subseteq T$ is unibranched. It follows that T is of finite character, since R is of finite character (Proposition 19). It also follows that the associated valuation overring of T_N equals that of R_M, so T_N is Mori by Lemma 12. □

In the next corollary, we present a relation between various conditions on Mori LPVDs (equivalently, TAF-domains), and partially extend Proposition 15. Recall from [55] that an integral domain R is an i-domain if the contraction map $\text{Spec}(T)\to \text{Spec}(R)$ is injective for each overring T of R. On the other hand, as defined in [49, p.26], an integral domain R is an S-domain if $\mathit{\text{PR}}[X]$ is a height 1 prime ideal of $R[X]$ whenever P is a height 1 prime ideal of R. A ring R is a strong S-ring if R/N is an S-domain for every prime ideal N of R.

Corollary 24.

Let R be a Mori LPVD (equivalently, a TAF-domain). Then the following are equivalent.

1. Each overring of R is a Mori LPVD.

2. Each overring of R is a Mori domain.

3. Each overring of R is an LPVD.

4. $R\prime =R^{*}$.

5. (5) $R\prime$ is a Dedekind domain.

6. (6) $R\prime$ is a Prüfer domain.

7. (7) $R\prime$ is a Krull domain.

8. R is an i-domain.

9. R is a strong S-ring.

10. (10) $\text{dim}(R[X])\le 2$.

Proof.

Note first that (1) $\Rightarrow$ (2) and (1) $\Rightarrow$ (3) are trivial, while (1) $\Rightarrow$ (4) and (2) $\Rightarrow$ (5) can be derived from Proposition 19 and [18, Proposition 3.3, Theorem 3.4] ((2) $\Rightarrow$ (5) also follows from [30, Theorem 12]). On the other hand, the equivalence of (3), (6) and (8) follows from [28, Theorem 2.9].

(4) $\Rightarrow$ (5): Follows from Corollary 21.

(5) $\iff$ (7): Follows from Proposition 19, [54, Theorem 3.1] and the well-known fact that an integral domain is Dedekind if and only if it is Krull domain with Krull dimension at most 1 [53, Theorem 12.5].

(5) $\Rightarrow$ (8): Follows from [28, Theorem 2.9].

(8) $\Rightarrow$ (1): Suppose that (8) holds, and let T be an overring of R. Note first that by Proposition 19 R is an i-domain of finite character, and so is T. Let N be a maximal ideal of T. Then by Proposition 19, it suffices to show that T_N is a Mori PVD. We may assume that $T_N\ne K$. Let $M=N\cap R$. Then M is a maximal ideal of R since R is one-dimensional. Now R_M is a Mori PVD i-domain. Hence, $(R_M)\prime$ is the associated valuation overring of R_M [28, Corollary 2.10], which is a DVR by Corollary 13. Thus, T_N is an integral overring of R_M [36, Proposition 1.16.(3)], and MR_M is the maximal ideal of T_N. It follows that T_N is a Mori PVD by Corollary 13.

(5) $\Rightarrow$ (9): If $R\prime$ is Dedekind, then it is a strong S-ring [52, Proposition 2.5], so R is a strong S-ring [52, Corollary 4.7].

(9) $\Rightarrow$ (10): Since R has Krull dimension at most 1 by Proposition 19, the conclusion follows from [49, Theorem 39].

(10) $\Rightarrow$ (4): Since R has Krull dimension at most 1, R is a strong S-ring if and only if R is an S-domain. If $\text{dim}(R[X])\le 2$, then for each maximal ideal M of R, $\text{dim}(R_M[X])=\text{dim}(R{[X]}_{R\setminus M})\le 2$ and R_M is a strong S-ring by [45, Theorem 2.5]. Now $(R_M)\prime$ is the associated valuation overring of R_M [45, Remark 2.6]. Hence, ${(R\prime )}_{R\setminus M}=(R_M)\prime ={(R_M)}^{*}={(R^{*})}_{R\setminus M}$. Since this equality holds for arbitrary maximal ideal M of R, we must have $R\prime =R^{*}$. □

Recall that an integral domain R is said to be a globalized pseudo-valuation domain, or a GPVD in short, if there exists a Prüfer overring T of R satisfying the following two conditions.

1. $R\subseteq T$ is a unibranched extension.

2. There exists a nonzero radical ideal A common to T and R such that each prime ideal of T (respectively, R) which contains A is a maximal ideal of T (respectively, R).

As mentioned in [28, pp.155–156], the class of GPVDs is a stronger globalization of that of PVDs than that of LPVDs, in the sense that every GPVD is an LPVD, and given a maximal ideal M of R, there exists unique maximal ideal N of T such that T_N is the associated valuation overring of R_M. In this case, T is uniquely determined by the above conditions, and is called the Prüfer domain associated to R. From now on, when R is a GPVD, T will denote the Prüfer domain associated to R. However, note that even a Noetherian LPVD may not be a GPVD [28, Example 3.4].

Proposition 25.

Let R be a GPVD. Then R is Mori if and only if T is a Dedekind domain.

Proof.

Suppose that R is Mori. Since every GPVD is an LPVD, the conclusion follows from Corollary 24. Conversely, suppose that T is Dedekind. Then given a maximal ideal M of R, there exists unique maximal ideal N of T such that T_N is the associated valuation overring of R_M . T_N is a DVR since T is Dedekind, and R_M must be Mori by Corollary 13. On the other hand, T is of finite character, and so is R since $R\subseteq T$ is unibranched. Therefore, R is a Mori domain by Theorem 16.(4). □

Lemma 26.

Let R be a Mori domain. Then R is a GPVD if and only if $R:R^{*}\ne (0)$ and R is an LPVD. In this case, $T=R^{*}$ and $A=R:R^{*}$, where T and A are as mentioned in the definition of a GPVD.

Proof.

Suppose that R is a GPVD. Given a maximal ideal M of R, R_M is a Mori PVD, and there exists the unique prime ideal N of T such that $N\cap R=M$, as mentioned in the paragraph preceding Proposition 25. Notice that ${(R_M)}^{*}$ is the associated valuation overring of R_M by Lemma 12. Then by [17, Lemma 3.1] and Corollary 21 we have ${(R^{*})}_{R\setminus M}={(R_M)}^{*}=T_N=T_{R\setminus M}$, and $T=R^{*}$ by globalization. Moreover, $R:R^{*}\ne (0)$ since it contains A.

Conversely, suppose that $R:R^{*}\ne (0)$ and R is an LPVD, and let $I=R:R^{*}$. Then R is a TAF-ring by Proposition 19, so $R^{*}$ is a Dedekind domain and $R\subseteq R^{*}$ is a unibranched extension by Corollary 21. If I = R, then $R=R^{*}$ is a GPVD. Suppose that I is a proper ideal of R. Since $R^{*}$ is a fractional ideal of R, $R_S:{(R^{*})}_S=I_S$ for each multiplicatively closed subset S of R by Theorem 16.(1). On the other hand, since $R^{*}$ is a Dedekind domain, there exist $a_1,\dots ,a_n\in \mathbb{N}$ and maximal ideals $N_1,\dots ,N_n$ of $R^{*}$ such that $I=N_1^{a_1}\cdots N_n^{a_n}$. Let $M_i=N_i\cap R$ for each $i\in \{1,\dots ,n\}$. Then $\{M_1,\dots ,M_n\}$ is the set of maximal ideals of R containing I, and $I{R}_{M_i}=R_{M_i}:{(R^{*})}_{R\setminus M_i}=R_{M_i}:{(R_{M_i})}^{*}=M_i{R}_{M_i}$ by Theorem 17.(3) and Lemma 12.(3). Thus, a_i = 1 for each i, and I is an intersection of maximal ideals of $R^{*}$. Thus, I is a common radical ideal of R and $R^{*}$. Moreover, $I=M_1\cap \cdots \cap M_n$, so each prime ideal of $R^{*}$ (respectively, R) which contains I is a maximal ideal of $R^{*}$ (respectively, R). We conclude that R is a GPVD with $R^{*}$ its associated Prüfer domain. □

Recall that a ring in which each proper ideal is a finite intersection of primary ideals is said to be Laskerian. A Laskerian ring in which each primary ideal contains a power of its radical is said to be strongly Laskerian. In the next lemma, we record that an LPVD is Mori if and only if it is strongly Laskerian, extending a result of Barucci [15, Corollary 3.7].

Lemma 27.

The following are equivalent when R is an LPVD.

1. R is strongly Laskerian.

2. R is finite-absorbing.

3. R is Mori.

Proof.

(1) $\Rightarrow$ (2): This is [25, Lemma 19].

(2) $\Rightarrow$ (3): Note that by [25, Lemma 30] and Proposition 19, we may assume that R is a PVD with maximal ideal M. But then the conclusion follows from [25, Corollary 54] and Corollary 13.

(3) $\Rightarrow$ (2): Follows from Proposition 19. □

As we have seen from Proposition 19, TAF-domains have some interesting ring-theoretic properties. On the other hand, in [54], the authors focus on Noetherian TAF-domains. In particular, they characterized when a Noetherian domain R with $R:R\prime \ne (0)$ is a TAF-domain [54, Corollary 4.10]. Note that if R is a Noetherian domain, then R is Mori and $R\prime =R^{*}$. The next theorem is motivated by this observation. Recall that a ring is reduced if its zero ideal is a radical ideal.

Theorem 28.

(cf. [54, Corollary 4.10]) Let R be an integral domain that is not a field. Then the following are equivalent.

1. R is a TAF-domain such that $R:R^{*}\ne (0)$.

2. R is a Mori GPVD.

3. There exists a Dedekind domain T, distinct maximal ideals $N_1,\dots ,N_n$ of T and field extensions $K_i\subseteq T/N_i$ such that R is a pullback domain ${\pi }^{-1}({\prod }_{i=1}^n{K}_i)$ where $\pi :T\to {\prod }_{i=1}^n(T/N_i)$ is the canonical map.

4. $R^{*}$ is a Dedekind domain and $R/(R:R^{*})\subseteq R^{*}/(R:R^{*})$ is a unibranched extension of Artinian reduced rings.

Proof.

(1) $\iff$ (2): Follows from Proposition 19 and Lemma 26.

(2) $\Rightarrow$ (3): If (2) holds, then R is a Mori GPVD with $R^{*}$ its associated Prüfer domain by Lemma 26. Let $T=R^{*}$ and let A be the ideal mentioned in the definition of a GPVD. Since $T=R^{*}$ is a Dedekind domain by Corollary 21, A is a finite product of maximal ideals of T, say, $N_1,\dots ,N_n$. Note that since A is a radical ideal of a Dedekind domain, $N_1,\dots ,N_n$ are distinct maximal ideals of T. If we let $M_i=N_i\cap R$ and $K_i=R/M_i$ for each $i\in \{1,\dots ,n\}$, then $A={\cap }_{i=1}^n{N}_i={\cap }_{i=1}^n{M}_i$, and R is the pullback of the form stated in (2) (cf. [28, Theorem 3.1]).

(3) $\Rightarrow$ (2) and (3) $\Rightarrow$ (4): Suppose that (3) holds. Then T is an overring of R, and $R^{*}=T^{*}=T$ by [34, Lemma 1.1.4.(10)] and the fact that Dedekind domains are completely integrally closed. By Chinese remainder theorem it follows that $R:T=N_1\cap \cdots \cap N_n$. Let M be a maximal ideal of R. If $R:T\subseteq M$, then $M=N_i\cap R$ for some $i\in \{1,\dots ,n\}$ [34, Lemma 1.1.4.(6)], and R_M is the pullback ${\pi }_i^{-1}(K_i)$ where ${\pi }_i:T_{N_i}\to T_{N_i}/N_i{T}_{N_i}$ is the canonical map (cf. [34, Lemma 1.1.6]). Therefore, R_M is a PVD by [7, Proposition 2.6]. On the other hand, if $R:T\cancel{\subseteq }M$, then there exists unique prime ideal N of T such that $M=N\cap R$, and R_M is isomorphic to T_N [34, Lemma 1.1.4.(3)], which is a DVR. It also follows that R is an LPVD and $R\subseteq T$ is a unibranched extension. Hence, by Corollary 21, R is Mori. Now by Lemma 26, R is a GPVD and (2) follows. Since $R\subseteq T$ is unibranched, so is $R/R:T\to T/R:T$. Since $R:T$ is an intersection of n maximal ideals (as an ideal of both R and T), $R/R:T$ and $T/R:T$ are both isomorphic to a product of n fields by Chinese remainder theorem, so they are Artinian reduced rings, and (4) follows.

(4) $\Rightarrow$ (3): Assume (4). Since $R/(R:R^{*})$ is Artinian, if $R:R^{*}$ is the zero ideal, then R is a field, which is a contradiction. Therefore, $R:R^{*}\ne (0)$. Since $R^{*}$ is a Dedekind domain, we have $R:R^{*}=N_1\cdots N_n$ for some distinct maximal ideals of $R^{*}$. Letting $T=R^{*}$, we have (3). □

Let R be an integral domain. Given $f\in R[X]$, let c(f) be the ideal of R generated by the coefficients of f. Then $N=\{f\in R[X]⏧c(f)=R\}$ and $N_v=\{f\in R[X]⏧{(c(f))}^v=R\}$ are multiplicatively closed subsets of $R[X]$ [48, Proposition 2.1]. $R{[X]}_N$ is usually denoted by R(X), and is called the Nagata ring of R.

Proposition 29.

The following are equivalent for an integral domain R.

1. R(X) is a TAF-domain.

2. Every overring of R(X) is a TAF-domain.

3. Every overring of R is a TAF-domain.

Proof.

(1) $\Rightarrow$ (3): Assume that R(X) is a TAF-domain. Then R is an LPVD, $R\prime$ is a Prüfer domain and every overring of R(X) is an LPVD [24, Corollary 3.9]. Then $R{[X]}_{N_v}$, being an overring of R(X), is a Mori LPVD by Corollary 24. Since a strictly ascending chain of divisorial integral ideals ${\left\{I_i\right\}}_{i\in \mathbb{N}}$ of R induces a strictly ascending chain of divisorial integral ideals ${\left\{I_i{[X]}_{N_v}\right\}}_{i\in \mathbb{N}}$ of $R{[X]}_{N_v}$ by [48, Proposition 2.8], R must be a Mori domain. It follows that each overring of R is a TAF-domain by Corollary 24.

(3) $\Rightarrow$ (2): Suppose that every overring of R is a TAF-domain. Then by Proposition 20, $R[X]$ is a Mori domain, and so is R(X), being a localization of $R[X]$ (Theorem 16.(3)). On the other hand, R is an LPVD and $R\prime$ is a Dedekind domain by Corollary 24, so each overring of R(X) is an LPVD by [24, Corollary 3.9]. Therefore, by Proposition 19 and Corollary 24, each overring of R(X) is a TAF-domain.

(2) $\Rightarrow$ (1): Trivial. □

4 TAF-rings and FAF-domains

The main theorem of this section is Theorem 32 which generalizes Proposition 19 to commutative rings with zero divisors. The key part of its proof is taken from [3, Theorem 5.1]. A prime ideal P of a commutative ring R is said to be strongly prime if aP and bR are comparable for any two elements a, b of R. Note that for an integral domain R, this definition coincides with the notion of strongly prime ideal in Definition 1 (cf. [5, Proposition 3.1]). Similarly, a ring R is said to be a pseudo-valuation ring or a PVR if some maximal ideal of R is strongly prime [8, Lemma 1 and Theorem 2]. A ring R is said to be a locally pseudo-valuation ring or an LPVR if R_M is a PVR for each maximal ideal M of R. Recall that, as we defined it in the paragraph preceding Corollary 13, a nonzero nonunit element a of a ring R is irreducible if for each $b,c\in R$ such that a = bc, either aR = bR or aR = cR.

Lemma 30.

Let R be a quasilocal ring with maximal ideal M.

1. For any $r,s\in R$ such that r = rs, either r = 0 or s is a unit of R.

2. Let a be a nonzero irreducible element of R. If a = bc for some $b,c\in R$, then one of b and c is a unit of R.

3. R is a PVR if and only if for any two ideals I, J of R, I and JM are comparable.

4. Let R be a PVR. Then every ideal of R is comparable to M², and $M^2=\mathit{\text{aM}}$ for each nonzero irreducible element a of R.

Proof.

(1): If r = rs, then $r(1-s)=0$. If s is not a unit of R, then $1-s$ is a unit of R since R is quasilocal. Therefore, r = 0.

(2): If a = bc, then without loss of generality we have aR = bR, and b = ar for some $r\in R$. Now a = arc, so c is a unit of R by (1).

(3) This follows from [8, Theorem 5].

(4) The first assertion follows from (3). For the second assertion, let a be a nonzero irreducible element of R. Then for each $b\in M, a\notin \mathit{\text{bM}}$ by (2). Since R is a PVR, we must have $\mathit{\text{bM}}\subseteq \mathit{\text{aR}}$, from which it follows that $M^2\subseteq \mathit{\text{aR}}$. Then $M^2=\mathit{\text{aI}}$ for some ideal I of R. Since $M^2\ne \mathit{\text{aR}}, I\subseteq M$ and $\mathit{\text{aI}}\subseteq \mathit{\text{aM}}\subseteq M^2$. Therefore, $M^2=\mathit{\text{aM}}$. □

In the next lemma, we extend Corollary 13 and part of [54, Theorem 4.3] to commutative rings with zero divisors.

Lemma 31.

Let R be a quasilocal ring with maximal ideal M. Then the following are equivalent.

1. R is a TAF-ring.

2. R is strongly Laskerian and every ideal of R is comparable to M².

3. R is a strongly Laskerian PVR.

4. R is a PVR that satisfies ACCP.

5. R is an atomic PVR.

6. R is an atomic ring, and for each nonzero proper ideal I of R there exists $n\in \mathbb{N}$ such that $M^n\subseteq I⊊M^{n-1}$.

Proof.

Note that strongly Laskerian rings satisfy ACCP [46, Corollary 3.6.(b)], and every ring that satisfies ACCP is atomic [4, Theorem 3.2]. Therefore, we have $(3)\Rightarrow (4)\Rightarrow (5)$.

$(5)\Rightarrow (6)$: Let R be an atomic PVR and I a nonzero proper ideal of R. Choose a nonzero $a\in I$. Since R is atomic, $a=a_1\cdots a_n$ for some (nonzero) irreducible elements $a_1,\dots ,a_m$ of R. Then $M^{m+1}=\mathit{\text{aM}}\subseteq I$, where the first equality follows from Lemma 30.(3). Thus, there exists the smallest $n\in \mathbb{N}$ such that $M^n\subseteq I$. If n = 1, then we are done. If $n\ge 2$, then $M^{n-2}M=M^{n-1}\cancel{\subseteq }I$, so $I⊊M^{n-2}M=M^{n-1}$ by Lemma 30.(3).

$(6)\Rightarrow (5)$: Assume (6), and let $a,b\in R$. Then we only need to show that aM and bR are comparable. We may assume that a, b are nonzero nonunits of R. Then $M^n\subseteq \mathit{\text{aR}}⊊M^{n-1}$ and $M^m\subseteq \mathit{\text{bR}}⊊M^{m-1}$ for some $n,m\in \mathbb{N}$. If n < m, then $\mathit{\text{bR}}⊊M^{m-1}\subseteq M^n\subseteq \mathit{\text{aR}}$. It follows that bR = aI for some proper ideal I of R, so $\mathit{\text{bR}}\subseteq \mathit{\text{aM}}$. On the other hand, if $n\ge m$, then $\mathit{\text{aM}}\subseteq M^n\subseteq M^m\subseteq \mathit{\text{bR}}$.

$(5)\Rightarrow (1)$: Suppose that (5) holds. If 0 is irreducible in R, then R is an integral domain and the conclusion follows from Corollary 13 and Proposition 19. Assume that 0 is not irreducible in R. Since R is atomic, $0=a_1\cdots a_n$ for some (nonzero) irreducible elements $a_1,\dots ,a_n$ of R. Then $M^{n+1}=a_1\cdots a_{n}M=(0)$, where the first equality follows from Lemma 30.(3). Thus, M is nilpotent, and R is finite-absorbing [25, Theorem 27]. Since every ideal of R is comparable to M² by Lemma 30.(3), R is a TAF-ring [25, Proposition 33].

$(1)\Rightarrow (2)$: [25, Proposition 33].

$(2)\Rightarrow (3)$. Suppose that (2) holds. If $M=M^2$, then $M=\{x\in R⏧x\in \mathit{\text{xM}}\}$ [22, Exercise 29.(d), Chapter IV, Section 2], so $M=(0)$ and R is a field. Hence, we may assume that $M\ne M^2$. Notice that by Lemma 30.(3), given irreducible elements $a_1,\dots ,a_n$ of R, $a_1\cdots a_{n}M=M^{n+1}$. Now choose $a,b\in R$. We need to show that aM and bR are comparable. We may assume that $a,b\in M\setminus \left\{0\right\}$. Note that R is atomic as mentioned in the beginning of this proof, so there exist irreducible elements $a_1,\dots ,a_n,b_1,\dots ,b_m$ of R such that $a=a_1\cdots a_n$ and $b=b_1\cdots b_m$. Suppose that n < m. Then $\mathit{\text{bR}}=b_1\cdots b_{m}R\subseteq M^m\subseteq M^{n+1}=a_1\cdots a_{n}M=\mathit{\text{aM}}$. On the other hand, if $n\ge m$, then $\mathit{\text{aM}}=a_1\cdots a_{n}M=M^{n+1}\subseteq M^{m+1}=b_1\cdots b_{m}M\subseteq b_1\cdots b_{m}R=\mathit{\text{bR}}$. Hence, M is strongly prime, and R is a PVR [8, Theorem 2]. □

One can notice from the proof that a quasilocal TAF-ring that is not an integral domain is exactly a ring R with nilpotent maximal ideal M such that every ideal of R is comparable to $M^2.$ We will call such a ring a $\#$-ring, an ad-hoc notation that will be used in Theorem 32 only. Now we can derive the promised result.

Theorem 32.

(cf. [37, Theorem 39.2])Let R be a ring. Then the following are equivalent.

1. R is a TAF-ring.

2. R is strongly Laskerian, and for each maximal ideal M of R, every M-primary ideal of R is comparable to M².

3. R is a strongly Laskerian LPVR.

4. R is a finite-absorbing LPVR.

5. $R=R_1\times \cdots \times R_r$, where R_i is either a $\#$-ring or a Mori LPVD (equivalently, a TAF-domain) for each $i\in \{1,\dots ,r\}$.

Proof.

We may assume that R is not a field.

(1) $\Rightarrow$ (2): Follows from [25, Proposition 33].

(2) $\Rightarrow$ (3): We mimic the proof of [3, Theorem 5.1]. Note first that if a ring R satisfies (2), then so is R_M for each maximal ideal M of R. Hence, we may assume that R is a quasilocal ring that satisfies (2), and the conclusion follows from Lemma 31.

(3) $\iff$ (4): Note that every LPVR is locally divided [8, Lemma 1.(a)]. Hence, (3) $\iff$ (4) follows from [25, Lemma 20.(2)].

(4) $\Rightarrow$ (1): By [6, Theorem 2.5] and [25, Corollary 32], we may assume that R is quasilocal, so Lemma 31 yields the conclusion.

(1) $\iff$ (5): Follows from [54, Proposition 2.4, Theorem 3.3], Proposition 19 and Lemma 31. □

In [1], the authors introduced the AF-dimension of a ring R, denoted by AF-dim(R), which is the minimum positive integer n such that every proper ideal of R can be written as a finite product of n-absorbing ideals of R (if such n does not exist, set AF-dim $(R)=\infty$). We call R an FAF-ring (finite absorbing factorization ring) if AF-dim(R) is finite. An integral domain that is also an FAF-ring will be called an FAF-domain. The authors of [1] themselves presented several examples of rings and computed their AF-dimensions. All of the rings considered in such examples, however, were Noetherian. In the remainder of this section, motivated by the result that an integral domain R is a Mori LPVD if and only if $\text{AF‐dim}(R)\le 2$ (Proposition 19), we show that $\text{AF‐dim}(R)\le 3$ whenever R is a Mori LAPVD, and construct a non-Noetherian example that attains the equality. We also prove that an integral domain R may have AF-dimension 3 without being Mori.

The following result enables us to compute the AF-dimension of a domain locally.

Lemma 33.

(cf. [25, Corollary 31], [1, Theorem 4.3]) Let R be an integral domain and $n\in \mathbb{N}$. Then the following are equivalent.

1. $\text{AF‐dim}(R)\le n$.

2. R is of finite character and $\text{AF‐dim}(R_M)\le n$ for each maximal ideal M of R.

Theorem 34.

(cf. [1, Theorem 5.4]) Let R be a Mori domain such that $R:R^{*}$ is nonzero. Then the following are equivalent.

1. R is an FAF-domain.

2. R_M is an FAF-domain for each maximal ideal M of R.

3. $R\subseteq R^{*}$ is a unibranched extension of one-dimensional domains.

4. R is locally conducive.

Proof.

We may again assume that R is not a field. Note that if R satisfies one of $(1),(2),(3)$, and (4), then R has Krull dimension 1 by [1, Theorem 4.1], [18, Theorem 2.2] and Theorem 16.(3), so $R^{*}$ is Dedekind and ${(R^{*})}_{R\setminus M}={(R_M)}^{*}$ for each maximal ideal M of R by Theorem 17.

(1) $\Rightarrow$ (2): Follows from [1, Proposition 3.5].

(2) $\Rightarrow$ (3): Suppose that R satisfies (2), and fix a maximal ideal M of R. We have to show that the contraction map $f:\text{Spec}(R^{*})\to \text{Spec}(R)$ is bijective. Since R has Krull dimension 1, the surjectivity of f follows from [20, Proposition 1.1]. On the other hand, by Theorem 16.(3) R_M is a Mori FAF-domain such that $R_M:{(R_M)}^{*}\supseteq {(R:R^{*})}_{R\setminus M}\ne (0)$. Hence, by [1, Lemma 5.2], there exists only one maximal ideal of ${(R^{*})}_{R\setminus M}$ that contracts to MR_M. Thus, f is injective.

(3) $\Rightarrow$ (4) and (1): Suppose that $R\subseteq R^{*}$ is a unibranched extension of one-dimensional domains. Since $R^{*}$ is of finite character, so is R. Thus, there are only finitely many maximal ideals of R that contains $R:R^{*}$. Consider a maximal ideal M of R, and let N be the maximal ideal of $R^{*}$ that contracts to M. If M does not contain $R:R^{*}$, then $(R:R^{*})R_M=R_M$, and ${(R^{*})}_{R\setminus M}={(R^{*})}_{R\setminus M}{R}_M={(R^{*})}_{R\setminus M}(R:R^{*})R_M\subseteq R_M$. Hence, $R_M={(R^{*})}_{R\setminus M}={(R^{*})}_N$ is a DVR, which is conducive by Theorem 4. If $R:R^{*}\subseteq M$, then $R_M:{(R^{*})}_N\supseteq {(R:R^{*})}_M\ne (0)$ and ${(R^{*})}_N={(R_M)}^{*}$ is a DVR that is also an overring of R_M. Therefore, R_M is a conducive domain by Theorem 4, and (4) follows. Moreover, R_M is an FAF-domain since $N{(R^{*})}_N\cap R_M=M{R}_M$ [1, Lemma 5.1]. By Lemma 33, (1) follows.

$(4)\Rightarrow (3)$: Let R be a locally conducive domain and M a maximal ideal of R. Then R_M is a Mori conducive domain, so ${(R_M)}^{*}$ has only two overrings: ${(R_M)}^{*}$ and K [27, Proposition 4.3]. Hence, ${(R_M)}^{*}$ is a one-dimensional valuation domain by [37, Theorem 19.6], and $R_M\subseteq {(R_M)}^{*}={(R^{*})}_{R\setminus M}$ is a unibranched extension of one-dimensional domains. Hence, there exists unique maximal ideal N of $R^{*}$ that contracts to M, and (3) follows. □

The proof of [1, Lemma 5.1] gives a useful upper bound of the AF-dimension of a quasilocal conducive domain with a discrete one-dimensional valuation overring. In the next lemma, we present a result that works for a different class of integral domains.

Lemma 35.

Let R be a finite-absorbing quasilocal one-dimensional domain with maximal ideal M.

1. If I is an ideal of R and $n\in \mathbb{N}$, then ${\omega }_R(I)\le n$ if and only if $M^n\subseteq I$.

2. Suppose that $M^2=\mathit{\text{aM}}$ for some $a\in M$, and $n=\mathrm{\text{max}}\{{\omega }_R(I)⏧I\text{is an ideal of}R\text{such that}I\cancel{\subseteq }{M}^2\}$ for some $n\in \mathbb{N}$. Then $\text{AF‐dim}(R)=n$.

Proof.

(1): The statement follows from [25, Lemma 4].

(2): Assume first that n = 1. If I is an ideal of R such that $M^2⊊I\subseteq M$, then $\omega (I)=1$ by the definition of n, so I is a prime ideal. Since I is a nonzero ideal by construction, we must have I = M. Therefore, $M^2=\mathit{\text{aM}}⊊\mathit{\text{aR}}\subseteq M$ implies that aR = M. In other words, R is Noetherian by Cohen’s theorem, and must be a DVR [37, Theorem 38.1]. Thus, $\text{AF‐dim}(R)=1$, and we are done.

Suppose that $n\ge 2$. Since there exists an ideal I of R such that $I\cancel{\subseteq }{M}^2$ and ${\omega }_R(I)=n$ by our assumption, it follows that $\text{AF‐dim}(R)\ge n$. Now choose an ideal I₀ of R contained in M². Since $M^2=\mathit{\text{aM}}$, we have $I_0\subseteq \mathit{\text{aM}}$ and $I_0=a{I}_1$ for some proper ideal I₁ of R. We also have ${\omega }_R(I_0)>{\omega }_R(I_1)$. Indeed, if ${\omega }_R(I_0)=m$ for some $m\in \mathbb{N}$ (such m exists since R is finite-absorbing), then $M^m=a{M}^{m-1}$, so $M^{m-1}\subseteq I_1$ and ${\omega }_R(I_1)\le m-1$ by (1). Now, either $I_1\cancel{\subseteq }{M}^2$ and I₁ is an n-absorbing ideal of R, or $I_1=a{I}_2$ for some proper ideal I₂ of R with ${\omega }_R(I_1)>{\omega }_R(I_2)$. Iterating this process, we deduce that $I_0=a^r{I}_r$ for some $r\in \mathbb{N}$ and an ideal I_r of R that is n-absorbing. Since $M^2=\mathit{\text{aM}}\subseteq \mathit{\text{aR}}$, we have ${\omega }_R(\mathit{\text{aR}})\le 2\le n$ by (1). Therefore I₀ is a finite product of n-absorbing ideals of R, and $\text{AF‐dim}(R)\le n$. □

Theorem 36.

1. Let R be a one-dimensional APVD such that M is a principal ideal of V. Then $\text{AF‐dim}(R)\le 3$.

2. Every Mori LAPVD has AF-dimension at most 3.

3. Let R be a Mori domain. Then we have the following.

   $\text{AF‐dim}(R)=\{\begin{array}{cc}1\hfill & \mathit{\text{if and only if}}R\mathit{\text{ is }}a\mathit{\text{ Pr}}ü\mathit{\text{fer domain}},\hfill \\ 2\hfill & \mathit{\text{if and only if}}R\mathit{\text{ is an LPVD that is not }}a\mathit{\text{ Pr}}ü\mathit{\text{fer domain}},\hfill \\ 3\hfill & \mathit{\text{if}}R\text{ is an LAPVD that is not an LPVD}.\hfill \end{array}$

Proof.

(1): Let M = aV for some $a\in V$. Note that V is one-dimensional by Lemma 7 and $a\in M$. We also have ${\cap }_{i=1}^{\infty }{M}^i=(0)$ [37, Theorem 17.1]. Hence, given a proper ideal I of R, either $\mathit{\text{IV}}=(0)$ or $M^i\subseteq \mathit{\text{IV}}$ for some i. The former yields that $I=(0)$, which is a prime ideal. The latter gives $M^{i+1}\subseteq \mathit{\text{IVM}}=\mathit{\text{IM}}\subseteq I$. Note that for each nonzero proper ideal J of R, $\omega (J)\le n$ if and only if $M^n\subseteq J$ [25, Lemma 4]. Hence, we must have ${\omega }_R(I)\le i+1$, and R is finite-absorbing. If $I\cancel{\subseteq }{M}^2$, then $\mathit{\text{IV}}\cancel{\subseteq }{M}^2$ and $M^2⊊\mathit{\text{IV}}$, so $M^3\subseteq \mathit{\text{IM}}\subseteq I$ and ${\omega }_R(I)\le 3$. Therefore, $\text{AF‐dim}(R)\le 3$ by Lemma 35.

(2): Let R be a Mori LAPVD. We may assume that R is not a field. Since every Mori APVD has Krull dimension at most 1 by Corollary 11, R is one-dimensional. Then by Theorem 16.(4), R is of finite character. Hence, by Lemma 33, we may assume that R is a Mori APVD. Then $V=M:M$ is a DVR by Corollary 11, so R is one-dimensional and M is a principal ideal of V, and $\text{AF‐dim}(R)\le 3$ by (1).

(3): We may assume that R is not a field. Since R is a Mori domain, R is Prüfer if and only if R is Dedekind [31, Corollary 2.6.21], from which the first case follows. The second and third assertions then follow from (2) and Proposition 19. □

Note that given $n\in \mathbb{N}$, one can construct a commutative ring with zero divisors whose AF-dimension equals n [1, Proposition 3.8]. However, every FAF-domain we mentioned so far, including the examples in [1], is an LAPVD that has AF-dimension at most 3. Hence, one may ask whether we can construct an integral domain that has AF-dimension n, where $n\ge 4$ is a preassigned natural number. In the next corollary we prove that such construction is possible, and the integral domain can be chosen to be non-Noetherian.

Corollary 37.

Choose a field extension $F\subseteq L$, an indeterminate X and $n\in \mathbb{N}$. Let $$R=F+X^{2}L+X^{4}L+\cdots +X^{2n}L+X^{2n+2}L\left[X\right],$$ $$S=F+X^{3}L+X^{6}L+\cdots +X^{3n-3}L+X^{3n}L+X^{3n+2}L\left[X\right].$$

Then

1. R and S are Mori FAF-domains.

2. $\text{AF‐dim}(R)=2n+3$ and $\text{AF‐dim}(S)=2n+2$.

Proof.

We will prove the Corollary only for R, since the proof can be easily adapted for the case of S.

Let $I=X^{2n+2}L[X]$ and $P=X^{2}L+X^{4}L+\cdots +X^{2n}L+X^{2n+2}L[X]$. Then R is a pullback domain $R/I{\times }_{L[X]/I}L[X]$, and $\text{Spec}(R)=\{P,0\}\cup \{Q\cap R⏧Q\in \text{Spec}(L[X]),X\notin Q\}$ [34, Lemma 1.1.4.(3)]. It follows that $R\subseteq L[X]$ is a unibranched extension, so R is a one-dimensional domain of finite character. Moreover, for each $Q\in \text{Spec}(L[X])$ such that $X\notin Q$, $R_{Q\cap R}\cong L{[X]}_Q$ is a DVR [34, Lemma 1.1.4.(3)] and has AF-dimension 1. Thus, $\text{AF‐dim}(R)=\text{AF‐dim}(R_P)$ by [25, Corollary 31]. Let $V=L{[X]}_{\mathit{\text{XL}}[X]}$ and $M=P{R}_P=X^{2}L+X^{4}L+\cdots +X^{2n}L+X^{2n+2}V$. Then M is the maximal ideal of R_P, and $M^2=X^{2}M$. It also follows that R_P is a conducive domain by Theorem 4, since $X^{2n+2}\in R_P:V$. So R is locally conducive.

(1): Note that $X^{2n+2}\in R:L[X]=R:R^{*}$, so by Theorem 34 it remains to show that R is Mori. In fact, by (1) and (4) of Theorem 16, we only need to prove that for each nonzero ideal I of R_P, there exists a finitely generated ideal J of R_P such that $R_P:I=R_P:J$ and $J\subseteq I$. Let I be a nonzero ideal of R_P, and let $T=L+M$. Then T is a quasilocal overring of R_P with maximal ideal M. Moreover, T is a Noetherian ring by Eakin-Nagata theorem, so IT = JT for some finitely generated ideal $J\subseteq I$. Since $R_P:T=X^2{R}_P$, we have $R_P:\mathit{\text{HT}}=(R_P:T):H=X^2(R_P:H)$ for each ideal H of R_P. Therefore, $R_P:I=X^{-2}(R_P:\mathit{\text{IT}})=X^{-2}(R_P:\mathit{\text{JT}})=R_P:J$.

(2): Let $\mathcal{S}=\left\{I\text{is an ideal of}{R}_P⏧I\cancel{\subseteq }{M}^2\right\}$ and choose $I\in \mathcal{S}$. Then we have $\mathit{\text{IV}}=X^{r}V$ for some $r\le 2n+3$. Hence, $4n+4+i-r\ge 2n+2$ for each $i\in \mathbb{N}$, and $X^{4n+4+i}L\subseteq \mathit{\text{fR}}$ for every $f\in V$ with $\mathit{\text{fV}}=X^{r}V$. So $M^{2n+3}=X^{4n+4}M=X^{4n+6}L+X^{4n+8}L+\cdots +X^{6n+4}L+X^{6n+6}V\subseteq I$, and ${\omega }_{R_P}(I)\le 2n+3$. On the other hand, let $I=X^{2n+3}R$. Then $I\in \mathcal{S}$, and $M^{2n+2}\cancel{\subseteq }I$ since $X^{4n+4}\in M^{2n+2}\setminus I$. Thus, ${\omega }_{R_P}(I)=2n+3$, and $2n+3=\mathrm{\text{max}}\{{\omega }_{R_P}(I)⏧I\in \mathcal{S}\}$. Moreover, R_P is finite-absorbing by [25, Theorem 29.(2)], [1, Proposition 3.5] and (1). Therefore, $\text{AF‐dim}(R_P)=2n+3$ by Lemma 35. □

Remark 38.

1. The ring R in Corollary 37 is Noetherian exactly when $[L:F]<\infty$ [23, Theorem 4].

2. Unlike TAF-domains, FAF-domains are not necessarily Mori. Let V be a valuation domain of value group $\mathbb{Q}$ that contains a field of characteristic zero [37, Proposition 18.4, Corollary 18.5]. Then for a nonzero nonunit $a\in V$, set M = aV. It follows that V/M has characteristic zero, so it contains a field L (of characteristic zero). Now let $R=L{\times }_{V/M}V$ which is an APVD with maximal ideal M by Proposition 9. Since $\text{AF‐dim}(R)\le 3$ by Theorem 36.(1), R is an FAF-domain. But the value group of V is a nondiscrete subgroup of $\mathbb{R}$, so $V=M:M$ is not a DVR and R is not a Mori domain by Corollary 11, and $\text{AF‐dim}(R)=3$ by Proposition 19.

We now focus on the pullback properties of FAF-domains.

Proposition 39.

Let R be a Mori FAF-domain with $R:R^{*}\ne (0)$. Then there exists a Dedekind domain T, maximal ideals $N_1,\dots ,N_n$ of T, $a_1,\dots ,a_n\in \mathbb{N}$ and unibranched ring extensions $D_i\subseteq T/N_i^{a_i}$ such that $R={\pi }^{-1}({\prod }_{i=1}^n{D}_i)$ where $\pi :T\to {\prod }_{i=1}^n(T/N_i^{a_i})$ is the canonical projection.

Proof.

$R^{*}$ is a Dedekind domain by Theorem 17.(1). Therefore, letting $T=R^{*}$ and $I=R:R^{*}$, it follows that $R=R/I{\times }_{T/I}T$. Moreover, there exist only finitely many maximal ideals $N_1,\dots ,N_n$ of T that contains I, and $I={\prod }_{i=1}^n{N}_i^{a_i}$ for some $a_i\in \mathbb{N}$. Fix $i\in \{1,\dots ,n\}$ and let $M_i=N_i\cap R, D_i=R/(N_i^{a_i}\cap R)$. Then $D_i\subseteq T/N_i^{a_i}$ is a unibranched ring extension, and $R={\pi }^{-1}({\prod }_{i=1}^n{D}_i)$ [34, Lemma 1.1.6]. □

Proposition 40.

Let R be an integral domain that is not a field. Then the following are equivalent.

1. R is a Noetherian FAF-domain with $R:R\prime \ne (0)$.

2. There exists a Dedekind domain T, maximal ideals $N_1,\dots ,N_n$ of T, $a_1,\dots ,a_n\in \mathbb{N}$ and unibranched ring extensions $D_i\subseteq T/N_i^{a_i}$ such that $T/N_i^{a_i}$ is a finite D_i-module for each i, and $R={\pi }^{-1}({\prod }_{i=1}^n{D}_i)$ where $\pi :T\to {\prod }_{i=1}^n(T/N_i^{a_i})$ is the canonical projection.

3. $R\prime$ is a Dedekind domain, and $R/(R:R\prime )\to R\prime /(R:R\prime )$ is a unibranched ring extension that is also a finite module extension.

In particular, if (1) holds, then $R\prime$ is a finite R-module and $\text{AF‐dim}(R)=\underset{1\le i\le n}{\mathrm{\text{max}}}\{\text{AF‐dim}(R_{N_i\cap R})\}$.

Proof.

(1) $\Rightarrow$ (2): Suppose that R is a Noetherian FAF-domain with $R:R\prime \ne (0)$. By Proposition 39 there exists a Dedekind domain T, maximal ideals $N_1,\dots ,N_n$ of T, $a_1,\dots ,a_n\in \mathbb{N}$ and $D_i\subseteq T/N_i^{a_i}$ is a unibranched ring extension for each $i\in \{1,\dots ,n\}$ such that $R={\pi }^{-1}({\prod }_{i=1}^n{D}_i)$ where $\pi :T\to {\prod }_{i=1}^n(T/N_i^{a_i})$ is the canonical projection. Therefore, $R_{M_i}=D_i{\times }_{T/N_i^{a_i}}{T}_{N_i}$, where $M_i=N_i\cap R$ for each $i\in \{1,\dots ,n\}$ [34, Lemma 1.1.6]. Moreover, $R_{M_i}$ is a Noetherian conducive domain by Theorem 34. Hence, $T/N_i^{a_i}$ is a finite D_i-module [19, Theorem 6].

(2) $\Rightarrow$ (3): Suppose that (2) holds. For each $i\in \{1,\dots ,n\}$, since $D_i\to T/N_i^{a_i}$ is unibranched and finite, so is ${\prod }_{i=1}^n{D}_i\subseteq {\prod }_{i=1}^n(T/N_i^{a_i})\cong T/{\prod }_{i=1}^n{N}_i^{a_i}$. It follows that ${\prod }_{i=1}^n{N}_i^{a_i}$ is a common ideal of R and T, so $R\prime =R^{*}=T^{*}=T$ and $R:R\prime ={\prod }_{i=1}^n{N}_i^{a_i}$. Hence, (3) follows.

(3) $\Rightarrow$ (1): Assume (3). If $R:R\prime =(0)$, then $R\prime$ is a finite R-module, which is a contradiction. Therefore, $R:R\prime \ne (0)$, and R is a pullback domain $R/(R:R\prime ){\times }_{R\prime /(R:R\prime )}R\prime$. Therefore, R is Noetherian [34, Proposition 1.1.7] and $R\subseteq R\prime$ is unibranched [34, Lemma 1.1.4.(3)]. It follows that R is an FAF-domain by Theorem 34. □

Proposition 41.

Let R be an integral domain. Then the following are equivalent.

1. R is a Mori LAPVD with $R:R^{*}\ne (0)$.

2. There exists a Dedekind domain T, maximal ideals $N_1,\dots ,N_n$ of T, $a_1,\dots ,a_n\in \mathbb{N}$ and subfields F_i of $T/N_i^{a_i}$ for each $i\in \{1,\dots ,n\}$, such that $R={\pi }^{-1}({\prod }_{i=1}^n{F}_i)$ where $\pi :T\to {\prod }_{i=1}^n(T/N_i^{a_i})$ is the canonical projection.

3. $R^{*}$ is a Dedekind domain, $R/(R:R^{*})$ is an Artinian reduced ring and $R/(R:R^{*})\subseteq R^{*}/(R:R^{*})$ is a unibranched ring extension.

Proof.

(1) $\Rightarrow$ (2): Suppose that (1) holds. By Proposition 39, there exists a Dedekind domain T, maximal ideals $N_1,\dots ,N_n$ of T, $a_1,\dots ,a_n\in \mathbb{N}$ and unibranched ring extensions $D_i\subseteq T/N_i^{a_i}$ such that $R={\pi }^{-1}({\prod }_{i=1}^n{D}_i)$ where $\pi :T\to {\prod }_{i=1}^n(T/N_i^{a_i})$ is the canonical projection. Since $R_{M_i}=D_i{\times }_{T/N_i^{a_i}}{T}_{N_i}$, D_i must be a field by Proposition 9.

(2) $\Rightarrow$ (3): If (2) holds, then let $M_i=N_i\cap R$ for each $i\in \{1,\dots ,n\}$. Since $R:R^{*}=N_1^{a_1}\cdots N_n^{a_n}, {\left\{M_i\right\}}_{i=1}^n$ is the set of (distinct) prime ideal of R that contains $R:R^{*}$ [34, Lemma 1.1.4.(6)]. It then follows that $R:R^{*}=M_1\cap \cdots \cap M_n$, and $R/(R:R^{*})$ is an Artinian reduced ring and $R/(R:R^{*})\subseteq R^{*}/(R:R^{*})$ is a unibranched ring extension. By [34, Lemma 1.1.4.(10)], $R^{*}=T^{*}=T$ is Dedekind.

(3) $\Rightarrow$ (1): Assume that (3) is true. It is routine to see that $R:R^{*}\ne (0)$. Since R is the pullback domain $R/(R:R^{*}){\times }_{R^{*}/(R:R^{*})}{R}^{*}$, it follows that $R\subseteq R^{*}$ is unibranched [34, 1.1.4.(3)]. It also follows that R is an LAPVD by [34, Lemma 1.1.6] and Proposition 9. By Corollary 21, we obtain the promised result. □

In [1, Corollary 3.9], the authors gave a complete description of AF-dimension of $\mathbb{Z}[\sqrt{m}]$ when m is a square-free integer. On the other hand, [29, Theorem 2.5] establishes a characterization theorem of an order of a quadratic number fields being a GPVD. In the last topic of this section, we extend these results simultaneously.

Recall that given a square-free integer n, an order of $\mathbb{Q}\left(\sqrt{n}\right)$ is of the form $\mathbb{Z}[r{\omega }_n]$ for some $r\in \mathbb{N}$ where $${\omega }_n=\{\begin{array}{cc}\frac{1+\sqrt{n}}{2}\hfill & \text{if}n\equiv 1(\text{mod}4),\hfill \\ \hfill & \hfill \\ \sqrt{n}\hfill & \text{if}n\cancel{\equiv }1(\text{mod}4).\hfill \end{array}$$

Recall that for an integer a and a prime number p, the Kronecker symbol $({}_p^a)$ is defined as follows: $$\left(\begin{array}{c}a\\ 2\end{array}\right)=\{\begin{array}{cc}0\hfill & \text{if}a\equiv 0(\text{mod}2)\hfill \\ 1\hfill & \text{if}a\equiv \pm 1(\text{mod}8)\hfill \\ -1\hfill & \text{if}a\equiv \pm 3(\text{mod}8)\hfill \end{array}$$and when $p\ne 2, {(}_p^a)$ equals the Legendre symbol: $$\left(\begin{array}{c}a\\ p\end{array}\right)=\{\begin{array}{cc}0\hfill & \text{if}p⏧a\hfill \\ 1\hfill & \text{if}p\nmid a\text{and}{x}^2\equiv a(\text{mod}p)\text{has an integer solution}\hfill \\ -1\hfill & \text{otherwise}\hfill \end{array}$$

For a square-free integer $n\notin \{0,1\}$, $$d_n=\{\begin{array}{cc}n\hfill & \text{if}n\equiv 1(\text{mod}4)\hfill \\ 4n\hfill & \text{otherwise}\hfill \end{array}$$is called the discriminant of $\mathbb{Q}\left(\sqrt{n}\right)$.

Theorem 42.

[26, Proposition 5.16] Let n be a square-free integer and p a prime number. Then

1. If $(\begin{array}{c}{d}_n\\ p\end{array})=0$, then $p\mathbb{Z}[{\omega }_n]=P^2$ for some prime ideal P of $\mathbb{Z}[{\omega }_n]$.

2. If $(\begin{array}{c}{d}_n\\ p\end{array})=1$, then $p\mathbb{Z}[{\omega }_n]=P_1{P}_2$ for some distinct prime ideals P₁, P₂ of $\mathbb{Z}[{\omega }_n]$.

3. If $(\begin{array}{c}{d}_n\\ p\end{array})=-1$, then $p\mathbb{Z}[{\omega }_n]$ is a prime ideal of $\mathbb{Z}[{\omega }_n]$.

Theorem 43.

Let n be a square-free integer, $r\in \mathbb{N}\setminus \left\{1\right\}$ with prime factorization $r=p_1^{a_1}\cdots p_m^{a_m}$, and $R=\mathbb{Z}[r{\omega }_n]$. Then

1. $\text{AF‐dim}(R)<\infty$ if and only if $(\begin{array}{c}{d}_n\\ p_i\end{array})\ne 1$ for each $i\in \{1,\dots ,m\}$.

2. Suppose that $\text{AF‐dim}(R)<\infty$, and set

   ${\mathcal{F}}_1=\{i\in \{1,\dots ,m\}⏧(\begin{array}{c}{d}_n\\ p_i\end{array})=0\},$

   ${\mathcal{F}}_2=\{i\in \{1,\dots ,m\}⏧(\begin{array}{c}{d}_n\\ p_i\end{array})=-1\}.$

Then $\text{AF‐dim}(R)=\underset{i\in {\mathcal{F}}_1,j\in {\mathcal{F}}_2}{\mathrm{\text{max}}}\{2a_i+1,2a_j\}.$

Proof.

(1): Let $I=R:R\prime$. Note that $R\prime =\mathbb{Z}[{\omega }_n]$ is a Dedekind domain. Thus, we have the pullback

It follows that R is an FAF-domain if and only if the number of minimal prime ideals of I in R is the same as that of I in $R\prime$ (Proposition 40), which happens exactly when $\mathit{\text{pR}}\prime$ is a power of a prime ideal of $R\prime$ for each prime factor p of r in $\mathbb{Z}$. Theorem 42 then yields (1).

(2): Note that $\{1,\dots ,m\}={\mathcal{F}}_1\cup {\mathcal{F}}_2$ by (1). Fix $j\in \{1,\dots ,m\}$, and let N be the maximal ideal of $R\prime$ that contains $p_{j}R\prime$. Let $V=R{\prime }_N, S=R_{N\cap R}$ and M the maximal ideal of S. Then V is a DVR, and $M^2=p_{j}M$ since $M=p_{j}S+p_j^{a_j}V$. Choose $b\in N$ such that NV = bV. Suppose that $j\in {\mathcal{F}}_1$. Then $p_{j}R\prime =N^2$ by Theorem 42, so $N^{2a_j+2}V=p_j^{a_j+1}V\subseteq p_{j}M=M^2$. Hence, if I is an ideal of S such that $I\cancel{\subseteq }{M}^2$, then IV = fV for some $f\in I$ such that $\mathit{\text{fV}}=N^{t}V$ for some $t\le 2a_j+1$. In other words, $f=b^{t}u$ for some unit u of V. Then $p_j^{2a_j+1}\in f{p}_j^{a_j}V$ and $p_j^{3a_j+1}V\subseteq f{p}_j^{a_j}V$, so $M^{2a_j+1}=p_j^{2a_j}M\subseteq \mathit{\text{fM}}\subseteq I$. On the other hand, $J=p_j^{a_j}\mathit{\text{bS}}$ is an ideal of S such that $J\cancel{\subseteq }{M}^2$, and $p_j^{2a_j}\in M^{2a_j}\setminus J$. Thus, by Lemma 35, $\text{AF‐dim}(R_{N\cap R})=2a_j+1$.

Now consider the case when $j\in {\mathcal{F}}_2$. Then $N=p_{j}R\prime$ by Theorem 42. We have $N^{a_j+1}V=p_j^{a_j+1}V\subseteq p_{j}M=M^2$. Hence, if I is an ideal of S such that $I\cancel{\subseteq }{M}^2$, then $\mathit{\text{IV}}=p_j^{s}V$ for some $s\le a_j$. In other words, $f=b^{s}v$ for some unit v of V. Since $p_j^{2a_j}\in f{p}_j^{a_j}V$ and $p_j^{3a_j-1}V\subseteq f{p}_j^{a_j}V$, it follows that $M^{2a_j}=p_j^{2a_j-1}M\subseteq I$. We then claim that $V\ne S+N$. Indeed, if $V=S+N$, then $M=p_{j}S+p_j^{a_j}V=p_{j}S+p_j^{a_j}(S+N)=p^{j}S+p^{a_j}S+p^{a_j}N=p_{j}S$, so S is a DVR by Krull intersection theorem. Then S = V, which is a contradiction. Hence, $V\ne S+N$ and there exists a unit w of V such that $w\notin S+N$. Note also that $w^{-1}\notin S+N$ since S + N is an integral overring of S. Now $J\prime =p_j^{a_j}\mathit{\text{wS}}$ is an ideal of S not in M². Indeed, if $p_j^{a_j}w\in M^2$, then $p_j^{a_j}w=p_j^{2}s+p_j^{a_j+1}u$ for some $s\in S,u\in V$. We may assume that $s=p^{a}u\prime$ for some $a\in {\mathbb{N}}_0$ and $u\prime \in R$, where $u\prime$ is a unit of R. Then $p_j^{a_j}w-p_j^{2}s=p_j^{a_j+1}u$. Since p_j generate the maximal ideal of V, we must have $a=a_j-2$ and $w=u\prime +p_{j}u\in S+N$, a contradiction. Similarly, we deduce that $p_j^{a_j-1}{w}^{-1}\notin S$, so $p_j^{2a_j-1}\in M^{2a_j-1}\setminus J\prime$. Therefore, we have $\text{AF‐dim}(R_{N\cap R})=2a_j$ when $j\in {\mathcal{F}}_2$ by Lemma 35. The conclusion now follows from Proposition 40 and the pullback structure of R discussed in the proof of (1) of this theorem. □

The first part of the following corollary retrieves [29, Theorem 2.5].

Corollary 44.

Let $n\in \mathbb{Z}\setminus \left\{0\right\}$ be square-free, and $R=\mathbb{Z}[r{\omega }_n]$ for some $r\in \mathbb{N}$.

1. The following are equivalent.

   1. R is a GPVD.

   2. R is an LPVD.

   3. $\text{AF‐dim}(R)\le 2$.

   4. r is square-free and $(\begin{array}{c}{d}_n\\ p\end{array})=-1$ whenever p is a prime factor of r.

2. The following are equivalent.

   1. R is an LAPVD.

   2. $\text{AF‐dim}(R)\le 3$.

   3. r is square-free and $(\begin{array}{c}{d}_n\\ p\end{array})\ne 1$ whenever p is a prime factor of r.

3. The following are equivalent.

   1. R is locally conducive.

   2. $\text{AF‐dim}(R)<\infty$.

   3. $(\begin{array}{c}{d}_n\\ p\end{array})\ne 1$ whenever p is a prime factor of r.

Proof.

(1): Since R is Noetherian and $R:R\prime \ne (0), (a)\Rightarrow (b)$ follows from Lemma 26, and $(b)\Rightarrow (c)$ from Proposition 19. The equivalence of (c) and (d) follows from Theorem 43 and [1, Corollary 3.9]. $(d)\Rightarrow (a)$ can be deduced from Theorems 28 and 42.

(2): $(a)\Rightarrow (b)$ follows from Theorem 36.(2). The equivalence of (b) and (c) follows from Theorem 43 and [1, Corollary 3.9]. $(c)\Rightarrow (a)$ can be deduced from Proposition 41 and Theorem 42.

(3): Follows from Theorems 34 and 43.(1). □

5 Rings of the form $A+\mathit{\text{XB}}[X]$

How the polynomial ring $R[X]$ behaves when we manipulate the coefficient ring R has been a stimulating topic for ring theorists. Probably the most famous result in this direction is the celebrated Hilbert basis theorem which states that if R is a Noetherian ring, then so is $R[X]$. In this spirit, several researchers studied the structure of the rings of the form $A+\mathit{\text{XB}}[X]$ where $A\subseteq B$ is an extension of rings (see [63] and its reference list), and one of the main topics was the investigation of Krull dimension of $A+\mathit{\text{XB}}[X]$ under the assumption that A and B are integral domains. In this section, we take an opposite approach by restricting the Krull dimension of $A+\mathit{\text{XB}}[X]$ to one and studying the behavior of A and B. Specifically, we characterize when $A+\mathit{\text{XB}}[X]$ is a TAF-ring in terms of A and B. We first need the following well-known lemma.

Lemma 45.

[49, Exercise 1-6.1] Let $A\subseteq B$ an extension of rings and P a minimal prime ideal of A. Then $N\cap A=P$ for some prime ideal N of B.

Proof.

Let $S=A\setminus P$. Then S is a multiplicatively closed subset of B. Choose an ideal N maximal with respect to the property such that $N\cap S=\varnothing$. Such N is a prime ideal of B, so $N\cap A$ is a prime ideal of A contained in P. Since P is a minimal prime ideal of A, we have $N\cap A=P$. □

In the following lemma, a couple of well-known results concerning the prime ideals of rings of the form $A+\mathit{\text{XB}}[X]$ are collected. Recall that when we say $\text{Spec}(R)$ is Noetherian for a ring R, we mean that R satisfies the ascending chain condition on radical ideals.

Lemma 46.

Let $A\subseteq B$ an extension of rings and $R=A+\mathit{\text{XB}}[X]$. Then

1. $\text{Spec}(R)=S_1\cup S_2$, where

   $S_1=\{P+\mathit{\text{XB}}[X]⏧P\in \text{Spec}(A)\},$

   $S_2=\{Q\cap R⏧Q\in \text{Spec}(B[X]),X\notin Q\}.$

2. $\text{Spec}(R)$ is Noetherian if and only if $\text{Spec}(A)$ and $\text{Spec}(B)$ are Noetherian.

3. $1+\mathrm{\text{max}}\{\text{dim}(A),\text{dim}(B)\}\le \text{dim}(R)\le \text{dim}(A)+\text{dim}(B[X])$.

Proof.

(1): Since R is a pullback ring $A{\times }_{B[X]/\mathit{\text{XB}}[X]}B[X]$, (1) holds by [34, Lemma 1.1.4].

(2): [56, Proposition 6.1.(2)].

(3): Let $P\in \text{Spec}(A)$ and $M\in \text{Spec}(B)$. Choose a minimal prime ideal P₀ of A contained in P, and $N\in \text{Spec}(B)$ such that $P_0=N\cap A$ (such N exists by Lemma 45). Then $N[X]\cap R=P_0+\mathit{\text{XN}}[X]$ is a minimal prime ideal of R properly contained in $P+\mathit{\text{XB}}[X]$. Hence, $1+h{t}_A(P)\le h{t}_R(P+\mathit{\text{XB}}[X])\le \text{dim}(R)$. On the other hand, $M[X]\cap R=(M\cap A)+\mathit{\text{XM}}[X]⊊(M\cap A)+\mathit{\text{XB}}[X]$, so by (1) $1+h{t}_B(M)\le 1+h{t}_R(M[X]\cap R)\le \text{dim}(R)$. Thus, the first inequality of (3) follows. The second inequality follows from the fact that $\text{Spec}(R)$ is a quotient space of the disjoint union of $\text{Spec}(A)$ and $\text{Spec}(B[X])$ [34, Theorem 1.4]. □

Throughout this manuscript, whenever R is of the form $A+\mathit{\text{XB}}[X]$ for a ring extension $A\subseteq B$, the notations S₁ and S₂ will be used to denote the sets introduced in Lemma 46.(1). The following theorem, taken from [58, Corollaire 9], is crucial in the proof of Lemma 48.

Theorem 47.

Let $A\subseteq B$ be an extension of rings such that B is a strongly Laskerian ring. Suppose that there exists a nonzero common ideal I of A and B contained in only finitely many prime ideals of A, and these prime ideals are all maximal ideals of A. Then A is a strongly Laskerian ring.

Lemma 48.

Let $A\subseteq B$ be an extension of rings and $R=A+\mathit{\text{XB}}[X]$. Then R is a one-dimensional strongly Laskerian ring if and only if A is a zero-dimensional strongly Laskerian ring and B is an Artinian ring.

Proof.

Suppose that R is one-dimensional strongly Laskerian ring. Then both A and B are zero-dimensional rings by Lemma 46.(3). Since $A\cong R/\mathit{\text{XB}}[X]$ and R is strongly Laskerian, so is A. It then remains to show that B is Noetherian. We will use the argument similar to that of [25, Proposition 59]. Suppose that B is not Noetherian, and choose ${\left\{b_n\right\}}_{n\in \mathbb{N}}\subseteq B$ so ${\{(b_1,\dots ,b_n)B\}}_{n\in \mathbb{N}}$ is a strictly ascending chain of ideals of B. Set J be the ideal of R generated by ${\left\{b_n{X}^n\right\}}_{n\in \mathbb{N}}$. Then for each $n\in \mathbb{N}\setminus \left\{1\right\}$, $b_n{X}^n=(b_{n}X)X^{n-1}$ is a product of n elements of R, but no $(n-1)$-subproduct of it is in J. Thus, ${\omega }_R(J)\ge n-1$. Since n is chosen arbitrarily, we must have ${\omega }_R(J)=\infty$. However, since R is strongly Laskerian, it must be finite-absorbing [25, Lemma 19], so we have a contradiction. Therefore, B must be Noetherian.

Conversely, suppose that A is a zero-dimensional strongly Laskerian ring and B is an Artinian ring. Then $B[X]$ is a Noetherian ring, and $\mathit{\text{XB}}[X]$ is a nonzero common ideal of R and $B[X]$. From Lemma 46.(1) it also follows that every prime ideal of R containing $\mathit{\text{XB}}[X]$ is a maximal ideal, and there are only finitely many minimal prime ideals of $\mathit{\text{XB}}[X]$ in R. Hence, we conclude that $\mathit{\text{XB}}[X]$ is contained in only finitely many prime ideals of R. Thus, R is strongly Laskerian by Theorem 47, and it is one-dimensional by Lemma 46.(3). □

Recall that a commutative ring R is said to be von Neumann regular if it is reduced and its Krull dimension is zero, Bézout if every finitely generated ideal of R is principally generated, arithmetical if the set of ideals of R_M forms a chain under set inclusion for each maximal ideal M of R, and satisfies $(*)$ if each ideal of R whose radical is prime is a primary ideal of R [39–41]. It is known that $R[X]$ is a Bézout ring if and only if R is von Neumann regular [38, Theorem 18.7], [2, Theorem 6]. If $A⊊B$ is a ring extension, then even if A and B are both von Neumann regular rings, $A+\mathit{\text{XB}}[X]$ may not be Bézout; if $A⊊B$ are fields, then $A+\mathit{\text{XB}}[X]$ is not Bézout [34, Corollary 1.1.9.(1)]. On the other hand, $\mathbb{Z}+X\mathbb{Q}[X]$, being a pullback domain $\mathbb{Z}{\times }_{\mathbb{Q}}\mathbb{Q}[X]$, is a two-dimensional Bézout domain by Lemma 46.(3) and [34, Corollary 1.1.11], while $R[X]$ is a one-dimensional ring if $R[X]$ is Bézout. Hence, one may conjecture that $A+\mathit{\text{XB}}[X]$ is a one-dimensional Bézout ring if and only if A = B is a von Neumann regular ring, and this is indeed true as the next Proposition shows.

Proposition 49.

[54, Corollaries 3.4 and 4.8] Let $A\subseteq B$ rings, X an indeterminate, and $R=A+\mathit{\text{XB}}[X]$.

1. The following are equivalent.

   1. R is a one-dimensional reduced ring.

   2. Both A and B are von Neumann regular rings.

2. The following are equivalent.

   1. R_M is a Mori PVD for each maximal ideal M of R.

   2. R_M is a one-dimensional domain for each maximal ideal M of R.

   3. R satisfies $(*)$.

   4. R is a one-dimensional LPVR.

   5. R is a one-dimensional locally divided ring.

   6. $A\subseteq B$ is a unibranched extension of von Neumann regular rings.

3. The following are equivalent.

   1. R is a one-dimensional strongly Laskerian reduced ring.

   2. Both A and B are semi-quasilocal von Neumann regular rings, i.e., finite product of fields.

4. The following are equivalent.

   1. R is a one-dimensional Bézout ring.

   2. R_M is a DVR for each maximal ideal M of R.

   3. R is a one-dimensional arithmetical ring.

   4. A = B is a von Neumann regular ring.

5. The following are equivalent.

   1. R is a TAF-ring.

   2. R is an FAF-ring.

   3. R is strongly Laskerian and locally divided.

   4. R is finite-absorbing and locally divided.

   5. $A\subseteq B$ is a unibranched extension of semi-quasilocal von Neumann regular rings, i.e., finite products of fields.

   6. There exist fields ${\left\{F_i\right\}}_{i=1}^n$ and ${\left\{L_i\right\}}_{i=1}^n$ such that $A\cong {\prod }_{i=1}^n{F}_i, B\cong {\prod }_{i=1}^n{L}_i$ and $F_i\subseteq L_i$ for each $i\in \{1,\dots ,n\}$.

6. (cf. [1, Corollary 4.4]) The following are equivalent.

   1. R is a general ZPI-ring.

   2. R is a strongly Laskerian Bézout ring.

   3. R is a TAF-ring and A = B.

   4. A = B is a semi-quasilocal von Neumann regular ring, i.e., finite product of fields.

Proof.

(1): $(a)\Rightarrow (b)$: Suppose that R is one-dimensional reduced ring. Then both A and B are zero-dimensional by Lemma 46.(3). Now, A, being a subring of a reduced ring R, is a reduced ring. Similarly, if $b\in \mathit{\text{Nil}}(B)$, then $\mathit{\text{bX}}\in \mathit{\text{Nil}}(R)=(0)$, so b = 0 and we conclude that $\mathit{\text{Nil}}(B)=(0)$. Since both A and B are reduced rings with Krull dimension zero, they must be von Neumann regular.

$(b)\Rightarrow (a)$: Suppose that (b) holds. Then $\text{dim}(A)=0$ and $\text{dim}(B[X])=1$ [37, Corollary 30.3], so we have $\text{dim}(R)=1$ by Lemma 46.(3). On the other hand, since B is reduced, so is $B[X]$. Hence, R is reduced since it is a subring of $B[X]$.

(2): $(a)\Rightarrow (b)$: By Proposition 19, a Mori PVD has Krull dimension at most 1. Since R does not have any maximal ideal of height 0, (a) implies (b).

$(b)\Rightarrow (c)\Rightarrow (e)$: Follows from the fact that R satisfies $(*)$ if and only if, R_M is either a zero-dimensional ring or a one-dimensional domain for each maximal ideal M of R [41, Theorem 1].

$(e)\Rightarrow (f)$: Suppose that (e) holds. Now choose a maximal ideal M of R and a minimal prime ideal I of R contained in M. Since $\text{dim}(A)=0$ by Lemma 46.(3), $I\notin S_1$. Hence, $I\in S_2$, and $I=N[X]\cap R$ for some $N\in \text{Spec}(B)$. Let $P=N\cap A$, and we have $I=P+\mathit{\text{XN}}[X]$. Since $X\notin I$, we have $\mathit{\text{Nil}}(R)R_M\subseteq I{R}_M\subseteq X{R}_M$ since R_M is locally divided. Therefore, $\mathit{\text{Nil}}(R)\subseteq \mathit{\text{XR}}$. If $a\in \mathit{\text{Nil}}(A)$, then $a\in \mathit{\text{Nil}}(R)\subseteq \mathit{\text{XR}}$, so we must have a = 0. Hence, $\mathit{\text{Nil}}(A)=(0)$ and A is reduced. Similarly, if $b\in \mathit{\text{Nil}}(B)$, then $\mathit{\text{bX}}\in \mathit{\text{Nil}}(R)\subseteq \mathit{\text{XR}}=\mathit{\text{XA}}+X^{2}B[X]$, so $\mathit{\text{bX}}\in \mathit{\text{XA}}$ and $b\in A$. Thus, $b\in \mathit{\text{Nil}}(A)=(0)$ and we conclude that $\mathit{\text{Nil}}(B)=(0)$. Then R is reduced, being a subring of a reduced ring $B[X]$. Hence, R is a one-dimensional reduced ring, and both A and B are von Neumann regular by (1).

It remains to show that $A\subseteq B$ is unibranched. Let P be a prime ideal of A. Then by Lemma 45 there exists a prime ideal of B that contracts to P. On the other hand, suppose that there exist two prime ideals M, N of B that contract to P. Then $M\prime =M[X]\cap R=P+\mathit{\text{XM}}[X]$ and $N\prime =N[X]\cap R=P+\mathit{\text{XN}}[X]$ are incomparable prime ideals of R contained in the maximal ideal $P+\mathit{\text{XB}}[X]$ of R. Since R is locally divided, this is a contradiction [13, Proposition 2.1.(d)]. Hence, there exists exactly one prime ideal of B that contracts to P. Consequently, $A\subseteq B$ is unibranched.

$(f)\Rightarrow (a)$: Suppose that (f) holds, and choose a maximal ideal M of R. If $M\in S_1$, then $M=P+\mathit{\text{XB}}[X]$ for some $P\in \text{Spec}(A)$. Now, let N the unique maximal ideal of B such that $N\cap A=P$, and $S=A\setminus P$. Then both A_P and $B_N=B_S$ are fields. Since R_M is a pullback domain $A_P{\times }_{B_N}B{[X]}_S$ [33, Proposition 1.9], R_M is a Mori PVD by [7, Proposition 2.6] and Corollary 13. If $M\in S_2$, then $M=Q\cap R$ for some maximal ideal Q of $B[X]$ such that $X\notin Q$, and $R_M\cong B{[X]}_Q$. Since B is von Neumann regular, $B{[X]}_Q$ is a DVR as mentioned in the proof of $(1)\Rightarrow (2)$ of [2, Theorem 6]. In particular, R_M is a Mori PVD. Therefore, (a) follows.

$(a)\Rightarrow (d)$: Follows from definition.

$(d)\Rightarrow (e)$: Follows from [8, Lemma 1.(a)].

(3): Follows from (1) and Lemma 48.

(4): $(a)\Rightarrow (c)$: Follows from [47, Theorem 2].

$(c)\Rightarrow (d)$: Suppose that R is a one-dimensional arithmetical ring. Then by (1), $A\subseteq B$ is a unibranched extension of von Neumann regular rings. Now, let P be a maximal ideal of A, N the unique maximal ideal of B such that $N\cap A=P$, and $S=A\setminus P$. Then both A_P and $B_N=B_S$ are fields, so $R_S=A_P+X{B}_S[X]$ is a Prüfer domain, and we must have A_P = B_S [34, Proposition 1.1.8.(1)]. Since B/A is an A-module, and ${(B/A)}_P=B_N/A_P=0$ for each prime ideal P of A, we have $B/A=0$ and A = B.

$(d)\Rightarrow (a)$: Follows from [38, Theorem 18.7].

$(b)\Rightarrow (c)$: Trivial.

$(d)\Rightarrow (b)$: Follows from [2, Theorem 6].

(5): $(a)\Rightarrow (b)$: Trivial.

$(b)\Rightarrow (c)$: Follows from [25, Proposition 33].

$(c)\iff (d)$: Follows from [25, Lemma 20.(2)].

$(d)\Rightarrow (e)$: Suppose that R is a finite-absorbing locally divided ring. Then R satisfies $(*)$ [25, Proposition 25]. Hence, by (2), $A\subseteq B$ is a unibranched extension of von Neumann regular rings. On the other hand, since R is finite-absorbing, so is A [25, Lemma 18.(4)], and A has only finitely many minimal prime ideals [6, Theorem 2.5]. Since A is zero-dimensional, A is semilocal, and so is B since $A\subseteq B$ is unibranched.

$(e)\Rightarrow (f)$: Assume (e), and let $M_1,\dots ,M_n$ (respectively, $N_1,\dots ,N_n$) be maximal ideals of A (respectively, B) such that $N_i\cap A=M_i$ for each $i\in \{1,\dots ,n\}$. Then by the Chinese remainder theorem, $A\subseteq B$ implies ${\prod }_{i=1}^{n}A/M_i\subseteq {\prod }_{i=1}^{n}B/N_i$. Hence, letting $F_i=A/M_i$ and $L_i=B/N_i$, we may assume that $F_i\subseteq L_i$ for each $i\in \{1,\dots ,n\}$.

$(f)\Rightarrow (e)$: It is well-known that a finite product of fields is semi-quasilocal von Neumann regular rings. Hence, if (f) is true, then $A\subseteq B$ is an extension of semi-quasilocal von Neumann regular rings. It is also a unibranched extension since $F_i\subseteq L_i$ is unibranched for each $i\in \{1,\dots ,n\}$.

$(e)\Rightarrow (a)$: Suppose that (e) is true. Then R is a strongly Laskerian LPVR by (2) and (3). Hence, R is a TAF-ring by Theorem 32.

(6): $(a)\Rightarrow (b)$: Suppose that R is a general ZPI-ring. Then R has Krull dimension one [37, p.469], and R is a strongly Laskerian arithmetical ring [25, Corollary 36], so A = B is a von Neumann regular ring by (4). Therefore, R is strongly Laskerian by (5), and must be a Bézout ring by (4).

$(b)\iff (c)$: Assume (b). Since every Bézout ring is locally divided, R must be a one-dimensional TAF-ring by (2) and (5). Hence, by (4) we have A = B.

$(c)\iff (d)$: Follows from (5).

$(c)\Rightarrow (a)$: If R is a TAF-ring and A = B, then R is a general ZPI-ring [1, Corollary 4.4]. □

In a univariate polynomial ring, several statements of Proposition 49 are equivalent.

Theorem 50.

The following are equivalent for a ring R.

1. R is a von Neumann regular ring.

2. $R[X]$ is a Bézout ring.

3. $R[X]$ is an arithmetical ring.

4. $R[X]$ satisfies $(*)$.

5. $R{[X]}_M$ is a DVR for each maximal ideal M of $R[X]$.

6. $R{[X]}_M$ is Mori PVD for each maximal ideal M of $R[X]$.

7. $R{[X]}_M$ is a PVD for each maximal ideal M of $R[X]$.

8. $R[X]$ is a LPVR.

9. $R{[X]}_M$ is a TAF-domain for each maximal ideal M of $R[X]$.

10. $R{[X]}_M$ is a TAF-ring for each maximal ideal M of $R[X]$.

11. $R{[X]}_M$ is a FAF-ring for each maximal ideal M of $R[X]$.

12. $R{[X]}_M$ is a one-dimensional domain for each maximal ideal M of $R[X]$.

13. $R[X]$ is a locally divided ring.

14. $R[X]$ is a one-dimensional reduced ring.

Proof.

$(1)\Rightarrow (2)$: [38, Theorem 18.7].

$(2)\Rightarrow (3)$: [47, Theorem 2].

$(5)\Rightarrow (6)\Rightarrow (7)\Rightarrow (8)$: Well-known.

$(8)\Rightarrow (13)$: [8, Lemma 1.(a)].

$(6)\iff (9)$: Proposition 19.

$(9)\Rightarrow (10)\Rightarrow (11)\Rightarrow (12)\Rightarrow (13)$: Follows from definition.

$(1)\iff (4)\iff (13)$: [25, Theorem 60].

$(1)\iff (3)\iff (5)\iff (14)$: Proposition 49.(1). □

We end this paper with an example showing how we can construct an integral domain whose overrings are FAF-domains using the $A+\mathit{\text{XB}}[X]$ construction.

Corollary 51.

Let $A\subseteq B$ be an extension of integral domains, X an indeterminate, and $R=A+\mathit{\text{XB}}[X]$. Then the following are equivalent.

1. $A\subseteq B$ is an algebraic field extension.

2. Every overring of R is a TAF-domain.

3. Every overring of R is a FAF-domain.

Proof.

$(1)\Rightarrow (2)$: Suppose that $A\subseteq B$ is an algebraic field extension. Then R is a TAF-domain by Proposition 49.(5). Moreover, $R\prime =R^{*}=B[X]$ [34, Lemma 1.1.4.(9)]. Hence, every overring of R is a TAF-domain by Proposition 19 and Corollary 24.

$(2)\Rightarrow (3)$: Trivial.

$(3)\Rightarrow (1)$: Suppose that every overring of R is a FAF-domain. Then A and B are fields by Proposition 49.(5). To prove that $A\subseteq B$ is algebraic, notice that if there is an element t of B that is transcendental over A, then $A[t]+\mathit{\text{XB}}[X]$ is a two-dimensional overring of R by Lemma 46.(3), which contradicts our assumption since every FAF-domain has Krull dimension at most one [1, Theorem 4.1]. □

Acknowledgments

The author wishes to thank the anonymous referee whose careful reading and helpful suggestions, including the current form of Proposition 15, greatly improved the presentation of this manuscript.

Additional information

Funding

This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT)(No. 2022R1C1C2009021).