# n-absorbing ideal factorization in commutative rings

Hyun Seung choi  Department of Mathematics Education, Kyungpook National
University, 80 Daehakro, Bukgu, Daegu 41566, Korea<EMAIL_ADDRESS>
###### 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 $A+XB[X]$ 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 $n\geq 2$ if and only if $A$ and $B$ are both
Artinian reduced rings and the contraction map
$\textnormal{Spec}(B)\to\textnormal{Spec}(A)$ is a bijection. 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.

###### Key words and phrases:

almost pseudo-valuation domains, pseudo-valuation domains, Mori domains,
2-absorbing ideals, 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 [42], as
a generalization of valuation domains. Due to its several interesting
properties, pseudo-valuation domains have been extensively studied [30, 43],
and their generalizations were also considered. Among them are locally pseudo-
valuation domains (LPVDs) considered by Dobbs and Fontana [27], and almost
pseudo-valuation domains (APVDs) introduced by Badawi and Houston [13].

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
$\textnormal{Spec}(R^{*})\to\textnormal{Spec}(R)$ 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 [14, Corollary 3.7]. Motivated by the globalized pseudo-
valuation domains (GPVDs) introduced by Dobbs and Fontana [27], we provide
pullback descriptions of Mori GPVDs.

In section 4, we give a structure theorem of TAF-rings corresponding to [35,
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 [28, Theorem 2.5].

In section 5, we classify reduced rings of the form $A+XB[X]$ 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 $A+XB[X]$ is a TAF-ring exactly when
the contraction map $\textnormal{Spec}(B)\to\textnormal{Spec}(A)$ 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,
$\textnormal{Spec}(R)$).

######  Definition 1.

Let $R$ be an integral domain with quotient field $K$.

  1. (1)

An ideal $I$ of $R$ is strongly prime (respectively, strongly primary) if
given $x,y\in K$ with $xy\in I$, 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. (2)

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

  3. (3)

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

  4. (4)

$R$ is a Prüfer domain if $R_{M}$ is a valuation domain for each maximal ideal
$M$ of $R$.

  5. (5)

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

  6. (6)

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

  7. (7)

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

  8. (8)

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

  9. (9)

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

  10. (10)

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

  11. (11)

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

  12. (12)

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

  13. (13)

$R$ is locally conducive if $R_{M}$ is conducive for each maximal ideal $M$ of
$R$.

  14. (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 [9], [35] or [47].

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 [42, 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. (1)

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

  2. (2)

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

  3. (3)

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

  4. (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].

$(1)\Leftrightarrow(2)$: Follows from [5, Proposition 3.1 and Corollary 3.6]
and [13, Theorem 3.4].

$(1)\Leftrightarrow(3)$: Follows from [5, Proposition 3.11] and [13, Theorem
3.4].

$(1)\Leftrightarrow(4)$: Follows from [5, Theorem 2.1] and [13, Theorem 3.4].
∎

######  Theorem 4.

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

###### Proof.

This is just [26, Theorem 3.2]. ∎

An ideal $I$ of a ring $R$ is divided if $I\subsetneq aR$ 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 [13, 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 $R$ [35, 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 operations 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 [35, Chapter 32], which is based upon Krull’s work [49].

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

######  Theorem 6.

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

  1. (1)

$N\neq N^{2}$.

  2. (2)

$N$ is a principal ideal of $R$.

  3. (3)

$N$ is a divisorial ideal of $R$.

  4. (4)

$R$ is a divisorial domain.

Moreover, if $N=N^{2}$, then $\\{aN\mid a\in K\setminus\\{0\\}\\}$ is the set
of nondivisorial ideals of $R$.

Recall that when $R$ is a PVD, there exists a unique valuation overring $V$ of
$R$ such that $\textnormal{Spec}(R)=\textnormal{Spec}(V)$ as sets [42, Theorem
2.7]. Such $V$ is called the associated valuation overring of $R$.

######  Lemma 7.

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

  1. (1)

$R$ is an APVD.

  2. (2)

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

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

###### Proof.

(1)$\Rightarrow$(2): Let $R$ be an APVD with maximal ideal $M$. By [13,
Proposition 3.7], $R^{\prime}$ is a PVD, $M$ is an ideal of $R^{\prime}$, and
the radical of $M$ in $R^{\prime}$ is the maximal ideal of $R^{\prime}$. Let
$N$ (respectively, $N_{0}$) denote the maximal ideal of $R^{\prime}$
(respectively, $M:M$). Then $N_{0}$ is the radical of $M$ in $M:M$ by Theorem
3, so $M\subseteq N_{0}\cap R^{\prime}$. It follows that $N\subseteq N_{0}\cap
R^{\prime}$, and we have $N=N_{0}\cap R^{\prime}$ since $N_{0}\cap R$ is a
prime ideal of $R^{\prime}$. Now $N$ is a prime ideal of $M:M$ [42, Lemma
1.6]. It follows that $N$ is the maximal ideal of $M:M$. Since $M:M$ and $N:N$
are valuation overrings of $R^{\prime}$ with same maximal ideal $N$ by Theorem
3, we must have $N:N=M:M$ [35, Theorem 17.6]. Hence $M:M$ is the associated
valuation overring of $R^{\prime}$ [42, Theorem 2.7].

(2)$\Rightarrow$(1): Suppose that (2) holds. Since $R^{\prime}$ is a
quasilocal domain, so is $R$. Since $M:M$ is a valuation domain, it follows
that $R^{\prime}\subseteq M:M$ [51, Theorem 10.4] and $M$ 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 [42, 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
$\textnormal{dim}(R)=\textnormal{dim}(R^{\prime})$. ∎

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. (1)

$R$ is a PVD if and only if $M=N$.

  2. (2)

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

  3. (3)

Either $R=V$, or $M=R:V$ is a divisorial 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 $R=V$ [35, 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 $R\neq V$. 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 ideal of $R$. Therefore
$M=R:V$ is a divisorial ideal of $R$ [35, 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 $\\{(a,b)\in A\times B\mid u(a)=v(b)\\}$. The
pullback of our interest in this paper is 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 $B^{*}=D^{*}$ [32, Lemma 1.1.4.(10)]. Note that such $D$
can be identified to the ring $v^{-1}(A)$ [32, 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 $R=L\times_{V/M}V$ is a pullback. 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. [18, Theorem 1]).

Conversely, assume that $R$ is a pullback of the given form, and let
$\phi^{\prime}:R\to L$ the ring homomorphism induced by the canonical
projection map $\phi:V\to V/M$. Then $M=ker(\phi^{\prime})$ is an ideal of
$R$. In fact, $M$ is the maximal ideal of $R$ since $L\cong
R/ker(\phi^{\prime})$. It follows that $V$ is an overring of $R$, and $R$ is
an APVD by Theorem 3. Note that $M:M=V$ [32, 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 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 fractional ideal of $R$. Then
we have the following.

  1. (1)

Each nonzero ideal of $V$ is a divisorial fractional ideal of $R$.

  2. (2)

$J^{v}=JV$ for each nonprincipal ideal $J$ of $R$.

###### Proof.

If $I$ is an ideal of $V$, then $I^{v}\subseteq N:(N:I)$ [55, Proposition
1.20]. On the other hand, $N:(N:I)=I$ [20, Proposition 4.1]. Thus $(1)$
follows. For (2), we adapt the proof of [42, Proposition 2.14]. Choose a
nonprincipal ideal $J$ of $R$. Notice that if $x\in R:J$, then $xJ\subseteq
R$, and $xJ\neq R$ since $J$ cannot be a principal ideal of $R$. Hence,
$xJ\subseteq M$, and thereby $xJV\subseteq MV=M\subseteq R$ implies $x\in
R:JV$. Thus, $R:J\subseteq R:JV$. Since $JV$ is a divisorial fractional ideal
of $R$ by (1), we then have $JV=(JV)^{v}\subseteq J^{v}$. Since
$J^{v}\subseteq(JV)^{v}=JV$, (2) follows. ∎

######  Corollary 11.

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

  1. (1)

$R$ is a Mori APVD.

  2. (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 $M:M\neq R$. Then $M$ is a
divisorial ideal of $R$ by Corollary 8.(3). Hence $M:M$ is a Mori domain [15,
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 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 $R\neq V$. Then by Corollary 8.(3), $M$ is a divisorial ideal of
$R$, and so is $V$. Hence $N$ is a divisorial fractional ideal of $R$. Now,
let $\\{I_{\alpha}\\}_{\alpha\in\mathcal{A}}$ be a set of ascending chain of
divisorial 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 the ascending chain condition on principal ideals since every
nonunit of $R$ is a nonunit of $V$, while $R$ has the ascending chain property
on nonprincipal divisorial ideals since $V$ is a Noetherian ring. Hence, the
chain $\\{I_{\alpha}\\}_{\alpha\in\mathcal{A}}$ must be stationary, and $R$ is
Mori. ∎

######  Lemma 12.

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

  1. (1)

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

  2. (2)

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

  3. (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 [32, Lemma 1.1.4.(10)]. Since $R$ is conducive by Theorem 5, $R^{*}$
is completely integrally closed [40, 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 [35, 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 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 every nonzero nonunit of $R$ can be written
in at least one way as a finite product of irreducible elements of $R$. A
domain that satisfies the ascending chain condition on principal ideals is
atomic, but the converse fails in general [41]. 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. (1)

$R$ is a Mori PVD.

  2. (2)

$R$ is a PVD that satisfies the ascending chain condition on principal ideals.

  3. (3)

$R$ is an atomic PVD.

  4. (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. (1)

$R$ is an APVD that satisfies the ascending chain condition on principal
ideals.

  2. (2)

$R$ is an atomic APVD.

  3. (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 [10, 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, $M\neq N$. Thus there exists $a\in N\setminus M$, and we have
$M\subseteq aV$. Suppose that there exists a strictly ascending chain
$(f_{1})\subsetneq(f_{2})\subsetneq\cdots$ of nonzero principal ideals of $R$.
Then $f_{i}\in f_{i+1}M\subseteq af_{i+1}V$ for each $i\in\mathbb{N}$. Since
$V$ is one-dimensional, $f_{1}\in\bigcap\limits_{i\in\mathbb{N}}a^{i}V=(0)$
[35, Theorem 17.1.(3)], a contradiction. Hence $R$ must satisfy the ascending
chain condition on principal ideals. ∎

######  Proposition 15.

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

  1. (1)

Every overring of $R$ is Mori.

  2. (2)

$R^{\prime}$ is a valuation domain.

  3. (3)

Every overring of $R$ is a PVD.

###### Proof.

(1)$\Rightarrow$(2): Note that if every overring of $R$ is Mori, then
$R^{\prime}$ is a Dedekind domain [17, Theorem 3.4]. Since every integral
overring of $R$ is a PVD with maximal ideal $M$ [42, Theorem 1.7],
$R^{\prime}$ must be a DVR.

(2)$\Rightarrow$(1): If $R^{\prime}$ is a valuation domain, then by Lemmas 7
and 12, $R^{\prime}=M:M$ is a DVR with maximal ideal $M$. Let $T$ be an
overring of $R$. If $T=R^{\prime}$ or $T=K$, then it is clearly a Mori domain.
Suppose not. Since every overring of $R$ is comparable to $R^{\prime}$ [34,
Proposition 1.27 and Theorem 1.31], $T$ must be an integral overring of $R$.
On the other hand, $M$ is the maximal ideal of both $R$ and $R^{\prime}$ by
Theorem 3, so $M$ is also the maximal ideal of $T$. From Corollary 13 it then
follows that $T$ is Mori.

(2)$\Leftrightarrow$(3): [30, Corollaire 1.4]. ∎

##  3\. TAF-domains, Mori domains and locally pseudo-valuation domains

In 2007, Badawi [11] 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
$\prod\limits_{j\in\\{1,\dots,n+1\\}\setminus\\{i\\}}a_{j}\in I$. 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 [35, Chapter 39]. On the other hand, Mukhtar et.al.
[52] 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 [35, 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 $R$ [27,
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. (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. (2)

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

  3. (3)

If $R$ is Mori, then $R_{S}$ is Mori for each multiplicatively closed subset
$S$ of $R$.

  4. (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 [29, Proposition 2.6.11].

(2): This is exactly [33, Lemma 3.11].

(3): [29, Proposition 3.3.25].

(4): By (3), every Mori domain is locally Mori. The converse follows from [51,
Theorem 4.7] and [60, Corollary 5]. ∎

######  Theorem 17.

Let $R$ be an integral domain.

  1. (1)

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

  2. (2)

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

  3. (3)

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

###### Proof.

[15, Corollary 18], [19, Corollary 3.4], [16, Theorem 2.9] and [19, 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. (1)

$R$ is a Dedekind domain.

  2. (2)

$R$ is of finite character, and $R_{M}$ is a DVR for each maximal ideal $M$ of
$R$.

  3. (3)

$R$ is a Noetherian Prüfer domain.

  4. (4)

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

###### Proof.

(1)$\Leftrightarrow$(2): [35, Theorem 37.2].

(1)$\Leftrightarrow$(3): [35, Theorem 37.1].

(3)$\Rightarrow$(4): Let $R$ be a Noetherian Prüfer domain. Then $R$ is
integrally closed [35, 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 [51, Theorems 12.1, 12.4], which
is a DVR [26, 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. (1)

$R$ is a TAF-domain.

  2. (2)

$R$ is of finite character, and $R_{M}$ is a Mori PVD for each maximal ideal
$M$ of $R$.

  3. (3)

$R$ is a Mori LPVD.

  4. (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 $R$ [52,
Theorem 4.4]. Hence $R_{M}$ is an atomic PVD for each maximal ideal $M$ of $R$
by [52, Theorem 4.3], and Corollary 13 then yields (2) (cf. [48, Theorem
4.1]).

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

(3)$\Rightarrow$(4): Assume (3). Then $R$ is seminormal [27, Remarks 2.4], and
locally conducive [26, 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 [26, Corollary 2.5].
Hence $R_{M}$ has Krull dimension 1 by Theorem 17. It follows that $R_{M}$ is
a PVD [26, Corollary 2.6], and $R$ is of finite character by Theorem 16. Now
$R$ is a TAF-domain by [52, 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 [59, 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 [57, 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 [58, 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. (1)

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

  2. (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
$(R_{M})^{*}=MR_{M}:MR_{M}=R_{M}$ 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 $(M:M)_{R\setminus M}\subseteq MR_{M}:MR_{M}\subseteq
R_{M}:MR_{M}\subseteq R_{M}:JR_{M}=(R:J)_{R\setminus M}=(R:M)_{R\setminus
M}=(M:M)_{R\setminus M}$. Hence $(M:M)_{R\setminus M}=MR_{M}:MR_{M}$ is a DVR
by Corollary 11. On the other hand, $R_{M}$ is a one-dimensional quasilocal
domain by Lemma 12, so $(R_{M})^{*}\subsetneq K$ [26, Proposition 4.3.(ii)].
Since $(M:M)_{R\setminus M}\subseteq(R^{*})_{R\setminus
M}\subseteq(R_{M})^{*}\subsetneq K$, 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 [60, 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 [47, 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 $R$ [19, 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 $R=F+XL[X]$
is a TAF-domain [52, 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 $R_{M}$ [42, 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 extend Proposition 15. Recall from
[53] that an integral domain $R$ is an i-domain if the contraction map
$\textnormal{Spec}(T)\to\textnormal{Spec}(R)$ is injective for each overring
$T$ of $R$. On the other hand, as defined in [47, p.26], an integral domain
$R$ is an S-domain if $PR[X]$ 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. Then the following are equivalent.

  1. (1)

Each overring of $R$ is a Mori LPVD.

  2. (2)

Each overring of $R$ is a Mori domain.

  3. (3)

Each overring of $R$ is an LPVD.

  4. (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. (8)

$R$ is an $i$-domain.

  9. (9)

$R$ is a strong $S$-ring.

  10. (10)

$\textnormal{dim}(R[X])\leq 2$.

###### 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 [17, Propopsition 3.3, Theorem 3.4]. On the other hand, the equivalence of
(3), (6) and (8) follows from [27, Theorem 2.9].

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

(5)$\Leftrightarrow$(7): Follows from Proposition 19, [52, 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 [51, Theorem 12.5].

(5)$\Rightarrow$(8): Follows from [27, 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 $T_{N}\neq K$. 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 $R_{M}$ [27,
Corollary 2.10], which is a DVR by Corollary 13. Thus $T_{N}$ is an integral
overring of $R_{M}$ [34, 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
[50, Proposition 2.5], so $R$ is a strong $S$-ring [50, Corollary 4.7].

(9)$\Rightarrow$(10): Since $R$ has Krull dimension at most 1 by Proposition
19, the conclusion follows from [47, 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 $\textnormal{dim}(R[X])\leq
2$, then for each maximal ideal $M$ of $R$,
$\textnormal{dim}(R_{M}[X])=\textnormal{dim}(R[X]_{R\setminus M})\leq 2$ and
$R_{M}$ is a strong $S$-ring by [43, Theorem 2.5]. Now $(R_{M})^{\prime}$ is
the associated valuation overring of $R_{M}$ [43, 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. (1)

$R\subseteq T$ is a unibranched extension.

  2. (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 [27, p.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 [27, 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 $R_{M}$. Since
$(R_{M})^{*}=(T_{N})^{*}$, ∎

######  Lemma 26.

Let $R$ be a Mori domain. Then $R$ is a GPVD if and only if $R:R^{*}\neq(0)$
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 $N\cap
R=M$. Notice that $(R_{M})^{*}$ is the associated valuation overring of
$R_{M}$ by Lemma 12. Then by [16, 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. $R:R^{*}\neq(0)$ then follows from the definition of GPVDs.

Conversely, suppose that $R:R^{*}\neq(0)$ 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
$IR_{M_{i}}=R_{M_{i}}:(R^{*})_{R\setminus
M_{i}}=R_{M_{i}}:(R_{M_{i}})^{*}=M_{i}R_{M_{i}}$ 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 [14, Corollary 3.7].

######  Lemma 27.

The following are equivalent when $R$ is an LPVD.

  1. (1)

$R$ is strongly Laskerian.

  2. (2)

$R$ is finite-absorbing.

  3. (3)

$R$ is Mori.

###### Proof.

(1)$\Rightarrow$(2): This is [24, Lemma 19].

(2)$\Rightarrow$(3): Note that by [24, Lemma 30] and Proposition 19, we may
assume that $R$ is a PVD with maximal ideal $M$. But then the conclusion
follows from [24, 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 [52], the authors focus on
Noetherian TAF-domains. In particular, they characterized when a Noetherian
domain $R$ with $R:R^{\prime}\neq(0)$ is a TAF-domain [52, 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. [52, Corollary 4.10]) Let $R$ be an integral domain that is not a field.
Then the following are equivalent.

  1. (1)

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

  2. (2)

$R$ is a Mori GPVD.

  3. (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 $\pi^{-1}(\prod\limits_{i=1}^{n}K_{i})$ where
$\pi:T\to\prod\limits_{i=1}^{n}(T/N_{i})$ is the canonical map.

  4. (4)

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

###### Proof.

(1)$\Leftrightarrow$(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
$A=\bigcap\limits_{i=1}^{n}N_{i}=\bigcap\limits_{i=1}^{n}M_{i}$, and $R$ is
the pullback of the form stated in (2) (cf. [27, 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 [32, 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
$i\in\\{1,\dots,n\\}$ [32, 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. [32, Lemma 1.1.6]). Therefore $R_{M}$ is a PVD by [7,
Proposition 2.6]. On the other hand, if $R:T\not\subseteq M$, then there
exists unique prime ideal $N$ of $T$ such that $M=N\cap R$, and $R_{M}$ is
isomorphic to $T_{N}$[32, 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 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
$R:R^{*}\neq(0)$. 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 $N=\\{f\in R[X]\mid c(f)=R\\}$
and $N_{v}=\\{f\in R[X]\mid(c(f))^{v}=R\\}$ are multiplicatively closed
subsets of $R[X]$ [46, 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. (1)

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

  2. (2)

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

  3. (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 [23,
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 ideals
$\\{I_{i}\\}_{i\in\mathbb{N}}$ of $R$ induces a strictly ascending chain of
divisorial ideals $\\{I_{i}[X]_{N_{v}}\\}_{i\in\mathbb{N}}$ of $R[X]_{N_{v}}$
by [46, 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 [23, 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 $R$.

######  Lemma 30.

Let $R$ be a quasilocal ring with maximal ideal $M$.

  1. (1)

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

  2. (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. (3)

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

  4. (4)

Let $R$ be a PVR. Then every ideal of $R$ is comparable to $M^{2}$, and
$M^{2}=aM$ 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 $b\in M$, $a\not\in bM$ by
(2). Since $R$ is a PVR, we must have $bM\subseteq aR$, from which it follows
that $M^{2}\subseteq aR$. Then $M^{2}=aI$ for some ideal $I$ of $R$. Since
$M^{2}\neq aR$, $I\subseteq M$ and $aI\subseteq aM\subseteq M^{2}$. Therefore
$M^{2}=aM$. ∎

In the next lemma, we extend Corollary 13 and part of [52, 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. (1)

$R$ is a TAF-ring.

  2. (2)

$R$ is strongly Laskerian and every ideal of $R$ is comparable to $M^{2}$.

  3. (3)

$R$ is a strongly Laskerian PVR.

  4. (4)

$R$ is a PVR that satisfies the ascending chain condition on principal ideals.

  5. (5)

$R$ is an atomic PVR.

  6. (6)

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

###### Proof.

Note that strongly Laskerian rings satisfy the ascending chain condition on
principal ideals [44, Corollary 3.6.(b)], and every ring that satisfies the
ascending chain condition on principal ideals 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
$M^{m+1}=aM\subseteq I$, 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 $n\geq 2$, then $M^{n-2}M=M^{n-1}\not\subseteq
I$, so $I\subsetneq M^{n-2}M=M^{n-1}$ 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 $M^{n}\subseteq aR\subsetneq M^{n-1}$ and
$M^{m}\subseteq bR\subsetneq M^{m-1}$ for some $n,m\in\mathbb{N}$. If $n<m$,
then $bR\subsetneq M^{m-1}\subseteq M^{n}\subseteq aR$. It follows that
$bR=aI$ for some proper ideal $I$ of $R$, so $bR\subseteq aM$. On the other
hand, if $n\geq m$, then $aM\subseteq M^{n}\subseteq M^{m}\subseteq bR$.

$(5)\Rightarrow(1)$: 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 [24, Theorem 27]. Since every ideal of $R$ is comparable to
$M^{2}$ by Lemma 30.(3), $R$ is a TAF-ring [24, Proposition 33].

$(1)\Rightarrow(2)$: [24, Proposition 33].

$(2)\Rightarrow(3)$. Suppose that (2) holds. If $M=M^{2}$, then $M=\\{x\in
R\mid x\in xM\\}$ [21, Exercise 29.(d), Chapter IV, §2], so $M=(0)$ and $R$ is
a field. Hence we may assume that $M\neq M^{2}$. 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 $a,b\in M\setminus\\{0\\}$. 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 $bR=b_{1}\cdots
b_{m}R\subseteq M^{m}\subseteq M^{n+1}=a_{1}\cdots a_{n}M=aM$. On the other
hand, if $n\geq m$, then $aM=a_{1}\cdots a_{n}M=M^{n+1}\subseteq
M^{m+1}=b_{1}\cdots b_{m}M\subseteq b_{1}\cdots b_{m}R=bR$. Hence $M$ is
strongly prime, and $R$ is a PVR [8, Theorem 2]. ∎

Now we can derive the promised result.

######  Theorem 32.

(cf. [35, Theorem 39.2]) Let $R$ be a ring. Then the following are equivalent.

  1. (1)

$R$ is a TAF-ring.

  2. (2)

$R$ is strongly Laskerian, and for each maximal ideal $M$ of $R$, every
$M$-primary ideal of $R$ is comparable to $M^{2}$.

  3. (3)

$R$ is a strongly Laskerian LPVR.

  4. (4)

$R$ is a finite-absorbing LPVR.

  5. (5)

$R=R_{1}\times\cdots\times R_{r}$, where $R_{i}$ is either an atomic PVR or a
Mori LPVD for each $i\in\\{1,\dots,r\\}$.

###### Proof.

We may assume that $R$ is not a field.

(1)$\Rightarrow$(2): Follows from [24, 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)$\Leftrightarrow$(4): Note that every LPVR is locally divided [8, Lemma
1.(a)]. Hence (3)$\Leftrightarrow$(4) follows from [24, Lemma 20.(2)].

(4)$\Rightarrow$(1): By [6, Theorem 2.5] and [24, Corollary 32], we may assume
that $R$ is quasilocal, so Lemma 31 yields the conclusion.

(1)$\Leftrightarrow$(5): Follows from [52, 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 $\textnormal{AF-dim}(R)\leq
2$ (Proposition 19), we show that $\textnormal{AF-dim}(R)\leq 3$ 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. [24, Corollary 31], [1, Theorem 4.3]) Let $R$ be an integral domain and
$n\in\mathbb{N}$. Then the following are equivalent.

  1. (1)

$\textnormal{AF-dim}(R)\leq n$.

  2. (2)

$R$ is of finite character and $\textnormal{AF-dim}(R_{M})\leq n$ 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. (1)

$R$ is an FAF-domain.

  2. (2)

$R_{M}$ is an FAF-domain for each maximal ideal $M$ of $R$.

  3. (3)

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

  4. (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],
[17, 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
$f:\textnormal{Spec}(R^{*})\to\textnormal{Spec}(R)$ is bijective. Since $R$
has Krull dimension 1, the surjectivity of $f$ follows from [19, Proposition
1.1]. On the other hand, by Theorem 16.(3) $R_{M}$ is a Mori FAF-domain such
that $R_{M}:(R_{M})^{*}\supseteq(R:R^{*})_{R\setminus M}\neq(0)$. 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
$R_{M}:(R^{*})_{N}\supseteq(R:R^{*})_{M}\neq(0)$ 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 $N(R^{*})_{N}\cap R_{M}=MR_{M}$[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 $K$ [26, Proposition 4.3]. Hence
$(R_{M})^{*}$ is a one-dimensional valuation domain by [35, 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. ∎

[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. (1)

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

  2. (2)

Suppose that $M^{2}=aM$ for some $a\in M$, and $n=\max\\{\omega_{R}(I)\mid
I\textnormal{ is an ideal of $R$ such that }I\not\subseteq M^{2}\\}$ for some
$n\in\mathbb{N}$. Then $\textnormal{AF-dim}(R)=n$.

###### Proof.

(1): The statement follows from [24, Lemma 4].

(2): Assume first that $n=1$. If $I$ is an ideal of $R$ such that
$M^{2}\subseteq I\subseteq M$, then either $I=M^{2}$ or $I=M$. Therefore
$M^{2}=aM\subsetneq aR\subseteq M$ implies that $aR=M$. In other words, $R$ is
Noetherian by Cohen’s theorem, and must be a Dedekind domain [35, Theorem
38.1]. Thus $\textnormal{AF-dim}(R)=1$, and we are done.

Suppose that $n\geq 2$. Since there exists an ideal $I$ of $R$ such that
$I\not\subseteq M^{2}$ and $\omega_{R}(I)=n$ by our assumption, it follows
that $\textnormal{AF-dim}(R)\geq n$. Now choose an ideal $I_{0}$ of $R$
contained in $M^{2}$. Since $M^{2}=aM$, we have $I_{0}\subseteq aM$ and
$I_{0}=aI_{1}$ for some proper ideal $I_{1}$ 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
$M^{m}=aM^{m-1}$, so $M^{m-1}\subseteq I_{1}$ and $\omega_{R}(I_{1})\leq m-1$
by (1). Now, either $I_{1}\not\subseteq M^{2}$ and $I_{1}$ is an $n$-absorbing
ideal of $R$, or $I_{1}=aI_{2}$ for some proper ideal $I_{2}$ 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 $M^{2}=aM\subseteq aR$, we have $\omega_{R}(aR)\leq
2\leq n$ by (1). Therefore $I_{0}$ is a finite product of $n$-absorbing ideals
of $R$, and $\textnormal{AF-dim}(R)\leq n$. ∎

######  Theorem 36.

  1. (1)

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

  2. (2)

Every Mori LAPVD has AF-dimension at most 3.

  3. (3)

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

$\textnormal{AF-dim}(R)=\begin{cases}1&\text{if and only if }R\text{ is a
Pr{\"{u}}fer domain},\\\ 2&\text{if and only if }R\text{ is an LPVD that is
not a Pr{\"{u}}fer domain},\\\ 3&\text{if }R\text{ is an LAPVD that is not an
LPVD}.\end{cases}$

###### 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 $\bigcap\limits_{i=1}^{\infty}M^{i}=(0)$ [35,
Theorem 17.1]. Hence given a proper ideal $I$ of $R$, either $IV=(0)$ or
$M^{i}\subseteq IV$ for some $i$. The former yields that $I=(0)$, which is a
prime ideal. The latter gives $M^{i+1}\subseteq IVM=IM\subseteq I$. Note that
for each nonzero proper ideal $J$ of $R$, $\omega(J)\leq n$ if and only if
$M^{n}\subseteq J$ [24, Lemma 4]. Hence we must have $\omega_{R}(I)\leq i+1$,
and $R$ is finite-absorbing. If $I\not\subseteq M^{2}$, then $IV\not\subseteq
M^{2}$ and $M^{2}\subsetneq IV$, so $M^{3}\subseteq IM\subseteq I$ and
$\omega_{R}(I)\leq 3$. Therefore $\textnormal{AF-dim}(R)\leq 3$ 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 $\textnormal{AF-dim}(R)\leq 3$ 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 [29, 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 $n\geq 4$ 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

$\displaystyle R$ $\displaystyle=F+X^{2}L+X^{4}L+\cdots+X^{2n}L+X^{2n+2}L[X],$
$\displaystyle S$
$\displaystyle=F+X^{3}L+X^{6}L+\cdots+X^{3n-3}L+X^{3n}L+X^{3n+2}L[X].$

Then

  1. (1)

$R$ and $S$ are Mori FAF-domains.

  2. (2)

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

###### 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
$\textnormal{Spec}(R)=\\{P,0\\}\cup\\{Q\cap R\mid
Q\in\textnormal{Spec}(L[X]),X\not\in Q\\}$ [32, 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 $Q\in\textnormal{Spec}(L[X])$
such that $X\not\in Q$, $R_{Q\cap R}\cong L[X]_{Q}$ is a DVR [32, Lemma
1.1.4.(3)] and has AF-dimension 1. Thus $\textnormal{AF-
dim}(R)=\textnormal{AF-dim}(R_{P})$ by [24, Corollary 31]. Let
$V=L[X]_{XL[X]}$ and $M=PR_{P}=X^{2}L+X^{4}L+\cdots+X^{2n}L+X^{2n+2}V$. 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 $X^{2}\in R:L[X]=R:R^{*}$, 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
$R_{P}:HT=(R_{P}:T):H=X^{2}(R_{P}:H)$ for each ideal $H$ of $R_{P}$. Therefore
$R_{P}:I=X^{-2}(R_{P}:IT)=X^{-2}(R_{P}:JT)=R_{P}:J$.

(2): Let $\mathcal{S}=\\{I\textnormal{ is an ideal of }R_{P}\mid
I\not\subseteq M^{2}\\}$ and choose $I\in\mathcal{S}$. Then we have
$IV=X^{r}V$ for some $r\leq 2n+3$. Hence $4n+4+i-r\geq 2n+2$ for each
$i\in\mathbb{N}$, and $X^{4n+4+i}L\subseteq fR$ for every $f\in V$ with
$fV=X^{r}V$. 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 $\omega_{R_{P}}(I)\leq 2n+3$. On the other hand, let $I=X^{2n+3}R$.
Then $I\in\mathcal{S}$, and $M^{2n+2}\not\subseteq I$ since $X^{4n+4}\in
M^{2n+2}\setminus I$. Thus $\omega_{R_{P}}(I)=2n+3$, and
$2n+3=\max\\{\omega_{R_{P}}(I)\mid I\in\mathcal{S}\\}$. Moreover, $R_{P}$ is
finite-absorbing by [24, Theorem 29.(2)], [1, Proposition 3.5] and (1).
Therefore $\textnormal{AF-dim}(R_{P})=2n+3$ by Lemma 35. ∎

######  Remark 38.

  1. (1)

The ring $R$ in Corollary 37 is Noetherian exactly when $[L:F]<\infty$ [22,
Theorem 4].

  2. (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 [35, 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 $\textnormal{AF-dim}(R)\leq 3$ 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
$\textnormal{AF-dim}(R)=3$ by Proposition 19.

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

######  Proposition 39.

Let $R$ be a Mori FAF-domain with $R:R^{*}\neq(0)$. 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
$R=\pi^{-1}(\prod\limits_{i=1}^{n}D_{i})$ where
$\pi:T\to\prod\limits_{i=1}^{n}(T/N_{i}^{a_{i}})$ 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 $I=\prod\limits_{i=1}^{n}N_{i}^{a_{i}}$ 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 $R=\pi^{-1}(\prod\limits_{i=1}^{n}D_{i})$[32,
Lemma 1.1.6]. ∎

######  Proposition 40.

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

  1. (1)

$R$ is a Noetherian FAF-domain with $R:R^{\prime}\neq(0)$.

  2. (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 $R=\pi^{-1}(\prod\limits_{i=1}^{n}D_{i})$
where $\pi:T\to\prod\limits_{i=1}^{n}(T/N_{i}^{a_{i}})$ is the canonical
projection.

  3. (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
$\textnormal{AF-dim}(R)=\max\limits_{1\leq i\leq n}\\{\textnormal{AF-
dim}(R_{N_{i}\cap R})\\}$.

###### Proof.

(1)$\Rightarrow$(2): Suppose that $R$ is a Noetherian FAF-domain with
$R:R^{\prime}\neq(0)$. 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 $R=\pi^{-1}(\prod\limits_{i=1}^{n}D_{i})$
where $\pi:T\to\prod\limits_{i=1}^{n}(T/N_{i}^{a_{i}})$ 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 $i\in\\{1,\dots,n\\}$ [32, 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 [18, 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
$\prod\limits_{i=1}^{n}D_{i}\subseteq\prod\limits_{i=1}^{n}(T/N_{i}^{a_{i}})\cong
T/\prod\limits_{i=1}^{n}N_{i}^{a_{i}}$. It follows that
$\prod\limits_{i=1}^{n}N_{i}^{a_{i}}$ is a common ideal of $R$ and $T$, so
$R^{\prime}=R^{*}=T^{*}=T$ and
$R:R^{\prime}=\prod\limits_{i=1}^{n}N_{i}^{a_{i}}$. 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 $R:R^{\prime}\neq(0)$,
and $R$ is a pullback domain
$R/(R:R^{\prime})\times_{R^{\prime}/(R:R^{\prime})}R^{\prime}$. Therefore, $R$
is Noetherian [32, Proposition 1.1.7] and $R\subseteq R^{\prime}$ is
unibranched [32, 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. (1)

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

  2. (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
$R=\pi^{-1}(\prod\limits_{i=1}^{n}F_{i})$ where
$\pi:T\to\prod\limits_{i=1}^{n}(T/N_{i}^{a_{i}})$ is the canonical projection.

  3. (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
$R=\pi^{-1}(\prod\limits_{i=1}^{n}D_{i})$ where
$\pi:T\to\prod\limits_{i=1}^{n}(T/N_{i}^{a_{i}})$ 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 $R:R^{*}=N_{1}^{a_{1}}\cdots N_{n}^{a_{n}}$,
$\\{M_{i}\\}_{i=1}^{n}$ is the set of (distinct) prime ideal of $R$ that
contains $R:R^{*}$ [32, 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 [32, Lemma 1.1.4.(10)], $R^{*}=T^{*}=T$ is Dedekind.

(3)$\Rightarrow$(1): Assume that (3) is true. It is routine to see that
$R:R^{*}\neq(0)$. Since $R$ is the pullback domain
$R/(R:R^{*})\times_{R^{*}/(R:R^{*})}R^{*}$, it follows that $R\subseteq R^{*}$
is unibranched [32, 1.1.4.(3)]. It also follows that $R$ is an LAPVD by [32,
Lemma 1.1.6] and Proposition 9. By Corollary 21, we obtain the result
promised. ∎

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, [28, 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 $n$, a rank 2 free
$\mathbb{Z}$-submodule of a quadratic number field $\mathbb{Q}(\sqrt{n})$ is
called an order of $\mathbb{Q}(\sqrt{n})$. It is well-known that every
quadratic order of $\mathbb{Q}(\sqrt{n})$ is of the form
$\mathbb{Z}[r\omega_{n}]$ for some $r\in\mathbb{N}$ where

$\omega_{n}=\begin{cases}\cfrac{1+\sqrt{n}}{2}&\text{ if }n\equiv
1(\textnormal{mod }4),\\\ \\\ \sqrt{n}&\text{ if }n\not\equiv
1(\textnormal{mod }4).\end{cases}$

Recall that for an integer $a$ and a prime number $p$, the Kronecker symbol
$a\choose p$ is defined as follows:

${a\choose 2}=\begin{cases}0&\text{ if }a\equiv 0(\textnormal{mod }2)\\\
1&\text{ if }a\equiv\pm 1(\textnormal{mod }8)\\\ -1&\text{ if }a\equiv\pm
3(\textnormal{mod }8)\end{cases}$

when $p\neq 2$, $a\choose p$ equals the Legendre symbol:

${a\choose p}=\begin{cases}0&\text{ if }p\mid a\\\ 1&\text{ if }p\nmid a\text{
and }x^{2}\equiv a(\textnormal{mod }p)\text{ has an integer solution}\\\
-1&\text{ otherwise }\end{cases}$

For a square-free integer $n\not\in\\{0,1\\}$,

$d_{n}=\begin{cases}n&\text{ if }n\equiv 1(\textnormal{mod }4)\\\ 4n&\text{
otherwise }\end{cases}$

is called the discriminant of $\mathbb{Q}(\sqrt{n})$.

######  Theorem 42.

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

  1. (1)

If ${d_{n}\choose p}=0$, then $p\mathbb{Z}[\omega_{n}]=P^{2}$ for some prime
ideal $P$ of $\mathbb{Z}[\omega_{n}]$.

  2. (2)

If ${d_{n}\choose p}=1$, then $p\mathbb{Z}[\omega_{n}]=P_{1}P_{2}$ for some
distinct prime ideals $P_{1},P_{2}$ of $\mathbb{Z}[\omega_{n}]$.

  3. (3)

If ${d_{n}\choose p}=-1$, then $p\mathbb{Z}[\omega_{n}]$ is a prime ideal of
$\mathbb{Z}[\omega_{n}]$.

######  Theorem 43.

Let $n$ be a square-free integer, $r\in\mathbb{N}\setminus\\{1\\}$ with prime
factorization $r=p_{1}^{a_{1}}\cdots p_{m}^{a_{m}}$, and
$R=\mathbb{Z}[r\omega_{n}]$. Then

  1. (1)

$\textnormal{AF-dim}(R)<\infty$ if and only if ${d_{n}\choose p_{i}}\neq 1$
for each $i\in\\{1,\dots,m\\}$.

  2. (2)

Suppose that $\textnormal{AF-dim}(R)<\infty$, and set

$\displaystyle\mathcal{F}_{1}$
$\displaystyle=\\{i\in\\{1,\dots,m\\}\mid{d_{n}\choose p_{i}}=0\\},$
$\displaystyle\mathcal{F}_{2}$
$\displaystyle=\\{i\in\\{1,\dots,m\\}\mid{d_{n}\choose p_{i}}=-1\\}.$

Then $\textnormal{AF-
dim}(R)=\max\\{\max\limits_{i\in\mathcal{F}_{1}}\\{2a_{i}+1\\},\max\limits_{i\in\mathcal{F}_{2}}\\{2a_{i}\\}\\}.$

###### Proof.

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

${R}$${R/I\cong\mathbb{Z}/r\mathbb{Z}}$${R^{\prime}}$${R^{\prime}/I\cong
R^{\prime}/rR^{\prime}.}$$\scriptstyle{\iota}$$\scriptstyle{\pi}$

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 $pR^{\prime}$ 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 $I\not\subseteq M^{2}$, then $IV=fV$ for some $f\in I$
such that $fV=N^{t}V$ for some $t\leq 2a_{j}+1$. In other words, $f=b^{t}u$
for some unit $u$ of $V$. Then $p_{j}^{2a_{j}+1}\in fp_{j}^{a_{j}}V$ and
$p_{j}^{3a_{j}+1}V\subseteq fp_{j}^{a_{j}}V$, so
$M^{2a_{j}+1}=p_{j}^{2a_{j}}M\subseteq fM\subseteq I$. On the other hand,
$J=p_{j}^{a_{j}}bS$ is an ideal of $S$ such that $J\not\subseteq M^{2}$, and
$p_{j}^{2a_{j}}\in M^{2a_{j}}\setminus J$. Thus by Lemma 35, $\textnormal{AF-
dim}(R_{N\cap R})=2a_{j}+1$.

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 $I\not\subseteq M^{2}$, then
$IV=p_{j}^{s}V$ for some $s\leq a_{j}$. In other words, $f=b^{s}v$ for some
unit $v$ of $V$. Since $p_{j}^{2a_{j}}\subseteq fp_{j}^{a_{j}}V$ and
$p_{j}^{3a_{j}-1}\subseteq fp_{j}^{a_{j}}V$, it follows that
$M^{2a_{j}}=p_{j}^{2a_{j}-1}M\subseteq I$. We then claim that $V\neq S+N$.
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, $V\neq S+N$ and there exists a unit $w$ of $V$ such that
$w\not\in S+N$. Note also that $w^{-1}\not\in S+N$ since $S+N$ is an integral
overring of $S$. Now $J^{\prime}=p_{j}^{a_{j}}wS$ is an ideal of $S$ not in
$M^{2}$. 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 $p_{j}^{a_{j}-1}w^{-1}\not\in
S$, so $p_{j}^{2a_{j}-1}\in M^{2a_{j}-1}\setminus J^{\prime}$. Therefore, we
have $\textnormal{AF-dim}(R_{N\cap R})=2a_{j}$ 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 [28, Theorem 2.5].

######  Corollary 44.

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

  1. (1)

The following are equivalent.

    1. (a)

$R$ is a GPVD.

    2. (b)

$R$ is an LPVD.

    3. (c)

$\textnormal{AF-dim}(R)\leq 2$.

    4. (d)

$r$ is square-free and ${d_{n}\choose p}=-1$ whenever $p$ is a prime factor of
$r$.

  2. (2)

The following are equivalent.

    1. (a)

$R$ is an LAPVD.

    2. (b)

$\textnormal{AF-dim}(R)\leq 3$.

    3. (c)

$r$ is square-free and ${d_{n}\choose p}\neq 1$ whenever $p$ is a prime factor
of $r$.

  3. (3)

The following are equivalent.

    1. (a)

$R$ is locally conducive.

    2. (b)

$\textnormal{AF-dim}(R)<\infty$.

    3. (c)

${d_{n}\choose p}\neq 1$ whenever $p$ is a prime factor of $r$.

###### Proof.

(1): Since $R$ is Noetherian and $R:R^{\prime}\neq(0)$, $(a)\Rightarrow(b)$
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 Theorem 28 and Theorem 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 $A+XB[X]$

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 $A+XB[X]$ where
$A\subseteq B$ is an extension of rings (see [61] and its reference list), and
one of the main topics was the investigation of Krull dimension of $A+XB[X]$
under the assumption that $A$ and $B$ are integral domains. In this section,
we take an opposite approach by restricting the Krull dimension of $A+XB[X]$
to one and studying the behavior of $A$ and $B$. Specifically, we characterize
when $A+XB[X]$ is a TAF-ring in terms of $A$ and $B$. We first need the
following well-known lemma.

######  Lemma 45.

[47, 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 $N\cap
S=\emptyset$. 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 $A+XB[X]$ are collected. Recall that when we say
$\textnormal{Spec}(R)$ 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 $R=A+XB[X]$. Then

  1. (1)

$\textnormal{Spec}(R)=S_{1}\cup S_{2}$, where

$\displaystyle S_{1}$ $\displaystyle=\\{P+XB[X]\mid
P\in\textnormal{Spec}(A)\\},$ $\displaystyle S_{2}$ $\displaystyle=\\{Q\cap
R\mid Q\in\textnormal{Spec}(B[X]),X\not\in Q\\}.$

  2. (2)

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

  3. (3)

$1+\max\\{\textnormal{dim}(A),\textnormal{dim}(B)\\}\leq\textnormal{dim}(R)\leq\textnormal{dim}(A)+\textnormal{dim}(B[X])$.

###### Proof.

(1): Since $R$ is a pullback ring $A\times_{B[X]/XB[X]}B[X]$, (1) holds by
[32, Lemma 1.1.4].

(2): [54, Proposition 6.1.(2)].

(3): Let $P\in\textnormal{Spec}(A)$ and $M\in\textnormal{Spec}(B)$. Choose a
minimal prime ideal $P_{0}$ of $A$ contained in $P$, and
$N\in\textnormal{Spec}(B)$ such that $P_{0}=N\cap A$ (such $N$ exists by Lemma
45). Then $N[X]\cap R=P_{0}+XN[X]$ is a minimal prime ideal of $R$ properly
contained in $P+XB[X]$. Hence $1+ht_{A}(P)\leq
ht_{R}(P+XB[X])\leq\textnormal{dim}(R)$. On the other hand, $M[X]\cap R=(M\cap
A)+XM[X]\subsetneq(M\cap A)+XB[X]$, so by (1) $1+ht_{B}(M)\leq
1+ht_{R}(M[X]\cap R)\leq\textnormal{dim}(R)$. Thus the first inequality of (3)
follows. The second inequality follows from the fact that
$\textnormal{Spec}(R)$ is a quotient space of the disjoint union of
$\textnormal{Spec}(A)$ and $\textnormal{Spec}(B[X])$ [32, Theorem 1.4]. ∎

Throughout this manuscript, whenever $R$ is of the form $A+XB[X]$ for a ring
extension $A\subseteq B$, the notations $S_{1}$ and $S_{2}$ will be used to
denote the sets introduced in Lemma 46.(1). The following theorem, taken from
[56, 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 $R=A+XB[X]$. 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 $A\cong R/XB[X]$ 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 [24, Proposition 59].
Suppose that $B$ is not Noetherian, and choose
$\\{b_{n}\\}_{n\in\mathbb{N}}\subseteq B$ 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
$\\{b_{n}X^{n}\\}_{n\in\mathbb{N}}$. Then for each
$n\in\mathbb{N}\setminus\\{1\\}$, $b_{n}X^{n}=(b_{n}X)X^{n-1}$ is a product of
$n$ elements of $R$, but no $(n-1)$-subproduct of it is in $J$. Thus
$\omega_{R}(J)\geq n-1$. Since $n$ is chosen arbitrarily, we must have
$\omega_{R}(J)=\infty$. However, since $R$ is strongly Laskerian, it must be
finite-absorbing [24, 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 $XB[X]$ is a
nonzero common ideal of $R$ and $B[X]$. From Lemma 46.(1) it also follows that
every prime ideal of $R$ containing $XB[X]$ is a maximal ideal, and there are
only finitely many minimal prime ideals of $XB[X]$ in $R$. Hence we conclude
that $XB[X]$ 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 $R$ [37, 38, 39]. It is known that $R[X]$ is a Bézout ring if and
only if $R$ is von Neumann regular [36, Theorem 18.7], [2, Theorem 6]. If
$A\subsetneq B$ is a ring extension, then even if $A$ and $B$ are both von
Neumann regular rings, $A+XB[X]$ may not be Bézout; if $A\subsetneq B$ are
fields, then $A+XB[X]$ is not Bézout [32, 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 [32, Corollary 1.1.11], while $R[X]$ is a one-
dimensional ring if $R[X]$ is Bézout. Hence one may conjecture that $A+XB[X]$
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.

[52, Corollaries 3.4 and 4.8]) Let $A\subseteq B$ rings, $X$ an indeterminate,
and $R=A+XB[X]$.

  1. (1)

The following are equivalent.

    1. (a)

$R$ is a one-dimensional reduced ring.

    2. (b)

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

  2. (2)

The following are equivalent.

    1. (a)

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

    2. (b)

$R_{M}$ is a one-dimensional domain for each maximal ideal $M$ of $R$.

    3. (c)

$R$ satisfies $(*)$.

    4. (d)

$R$ is a one-dimensional LPVR.

    5. (e)

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

    6. (f)

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

  3. (3)

The following are equivalent.

    1. (a)

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

    2. (b)

Both $A$ and $B$ are semi-quasilocal von Neumann regular rings.

  4. (4)

The following are equivalent.

    1. (a)

$R$ is a one-dimensional Bézout ring.

    2. (b)

$R_{M}$ is a DVR for each maximal ideal $M$ of $R$.

    3. (c)

$R$ is a one-dimensional arithmetical ring.

    4. (d)

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

  5. (5)

The following are equivalent.

    1. (a)

$R$ is a TAF-ring.

    2. (b)

$R$ is an FAF-ring.

    3. (c)

$R$ is strongly Laskerian and locally divided.

    4. (d)

$R$ is finite-absorbing and locally divided.

    5. (e)

$A\subseteq B$ is a unibranched extension of semi-quasilocal von Neumann
regular rings.

    6. (f)

There exist fields $\\{F_{i}\\}_{i=1}^{n}$ and $\\{L_{i}\\}_{i=1}^{n}$ such
that $A\cong\prod\limits_{i=1}^{n}F_{i}$, $B\cong\prod\limits_{i=1}^{n}L_{i}$
and $F_{i}\subseteq L_{i}$ for each $i\in\\{1,\dots,n\\}$.

  6. (6)

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

    1. (a)

$R$ is a general ZPI-ring.

    2. (b)

$R$ is a strongly Laskerian Bézout ring.

    3. (c)

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

    4. (d)

$A=B$ is a semi-quasilocal von Neumann regular ring.

###### 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 $b\in Nil(B)$,
then $bX\in Nil(R)=(0)$, so $b=0$ and we conclude that $Nil(B)=(0)$. 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 $\textnormal{dim}(A)=0$
and $\textnormal{dim}(B[X])=1$ [35, Corollary 30.3], so we have
$\textnormal{dim}(R)=1$ 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 for each maximal ideal $M$ of $R$, $R_{M}$ is either a
zero-dimensional ring or a one-dimensional domain [39, 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
$\textnormal{dim}(A)=0$ by Lemma 46.(3), $I\not\in S_{1}$. Hence $I\in S_{2}$,
and $I=N[X]\cap R$ for some $N\in\textnormal{Spec}(B)$. Let $P=N\cap A$, and
we have $I=P+XN[X]$. Since $X\not\in I$, we have $Nil(R)R_{M}\subseteq
IR_{M}\subseteq XR_{M}$ since $R_{M}$ is locally divided. Therefore
$Nil(R)\subseteq XR$. If $a\in Nil(A)$, then $a\in Nil(R)\subseteq XR$, so we
must have $a=0$. Hence $Nil(A)=(0)$ and $A$ is reduced. Similarly, if $b\in
Nil(B)$, then $bX\in Nil(R)\subseteq XR=XA+X^{2}B[X]$, so $bX\in XA$ and $b\in
A$. Thus $b\in Nil(A)=(0)$ and we conclude that $Nil(B)=(0)$. 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 contracts to $P$. Then $M^{\prime}=M[X]\cap R=P+XM[X]$ and
$N^{\prime}=N[X]\cap R=P+XN[X]$ are incomparable prime ideals of $R$ contained
in the maximal ideal $P+XB[X]$ of $R$. Since $R$ is locally divided, this is a
contradiction [12, 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 $M=P+XB[X]$ for some $P\in\textnormal{Spec}(A)$.
Let $S=(A\setminus P)+XB[X]$ and choose $N\in\textnormal{Spec}(B)$ such that
$M=N\cap A$. Since $A\subseteq B$ is unibranched, $M_{0}=N[X]\cap R=P+XN[X]$
is the only prime ideal of $R$ properly contained in $M$, and $A_{P}$ is a
subfield of a field $B_{N}$ since the former is isomorphic to $A/P$ and the
latter to $B/N$. Therefore, $B[X]_{S}=B[X]_{N+XB[X]}=(B_{N}[X])_{N+XB[X]}$ is
a local PID, which is a DVR. Since $R_{M}$ is a pullback domain
$A_{P}\times_{B_{N}}B[X]_{S}$ [31, 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 $X\not\in Q$, 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 [45, 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 $A$ and $M=P+XB[X]$. Then there
exists unique maximal ideal $N$ of $B$ such that $N\cap A=P$, and $R_{M}$ is
the pullback domain $A_{P}\times_{B_{N}}B[X]_{N+XB[X]}$ as mentioned in the
proof of $(f)\Rightarrow(a)$ of (2). Since $R$ is arithmetical, $R_{M}$ is a
valuation domain, and we must have $A_{P}=B_{N}$ [32, 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 [36, 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 [24, Proposition 33].

$(c)\Leftrightarrow(d)$: Follows from [24, Lemma 20.(2)].

$(d)\Rightarrow(e)$: Suppose that $R$ is a finite-absorbing locally divided
ring. Then $R$ satisfies $(*)$ [24, 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 $A$ [24, 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 $(e)$. Since $A$ and $B$ are semilocal von Neumann
regular rings, they are Artinian reduced rings. Hence $A$ (respectively, $B$)
must be isomorphic to finite product of fields $F_{1}\times\cdots\times F_{n}$
(respectively, $L_{1},\dots,L_{m}$). Since $A\subseteq B$ is unibranched, we
may assume that $n=m$. In fact, for each $i\in\\{1,\dots,n\\}$, every maximal
ideal of $A$ (respectively, $B$) is of the form
$M_{i}=\prod\limits_{j=1}^{n}S_{j}$ where $S_{j}=F_{j}$ if $j\neq i$ and
$F_{i}=\\{0\\}$ (respectively, $N_{i}=\prod\limits_{j=1}^{n}T_{j}$ where
$T_{j}=L_{j}$ if $j\neq i$ and $T_{i}=\\{0\\}$). Hence by reindexing
$\\{F_{1}\dots,F_{n}\\}$ if necessary, we may assume that $M_{i}=N_{i}\cap A$
for each $i$. Then for a fixed $i\in\\{1,\dots,n\\}$, the ring $F_{i}\cong
A/M_{i}$ can be embedded to $L_{i}\cong B/N_{i}$. Hence we may assume that
$F_{i}\subseteq L_{i}$.

$(f)\Rightarrow(e)$: 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 [35, p.469], and $R$ is a strongly Laskerian arithmetical
ring [24, 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).

$(b)\Leftrightarrow(c)$: 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 $A=B$.

$(c)\Leftrightarrow(d)$: 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. (1)

$R$ is a von Neumann regular ring.

  2. (2)

$R[X]$ is a Bézout ring.

  3. (3)

$R[X]$ is an arithmetical ring.

  4. (4)

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

  5. (5)

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

  6. (6)

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

  7. (7)

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

  8. (8)

$R[X]$ is a LPVR.

  9. (9)

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

  10. (10)

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

  11. (11)

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

  12. (12)

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

  13. (13)

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

  14. (14)

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

###### Proof.

$(1)\Rightarrow(2)$: [36, Theorem 18.7].

$(2)\Rightarrow(3)$: [45, Theorem 2].

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

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

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

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

$(1)\Leftrightarrow(4)\Leftrightarrow(13)$: [24, Theorem 60].

$(1)\Leftrightarrow(3)\Leftrightarrow(5)\Leftrightarrow(14)$: 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 $A+XB[X]$ construction.

######  Corollary 51.

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

  1. (1)

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

  2. (2)

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

  3. (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,
$R^{\prime}=R^{*}=B[X]$ [32, 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 $A[t]+XB[X]$ 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]. ∎

## acknowledgement

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

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