Commutative rings with one-absorbing factorization

Abstract

Let R be a commutative ring with nonzero identity. Yassine et al. defined the concept of 1-absorbing prime ideals as follows: a proper ideal I of R is said to be a 1-absorbing prime ideal if whenever $\mathit{xyz}\in I$ for some nonunit elements $x,y,z\in R,$ then either $xy\in I$ or $z\in I.$ We use the concept of 1-absorbing prime ideals to study those commutative rings in which every proper ideal is a product of 1-absorbing prime ideals (we call them OAF-rings). Any OAF-ring has dimension at most one and local OAF-domains (D, M) are atomic such that M² is universal.

Keywords:

• OA-ideal
• OAF-ring
• trivial ring extension

2020 MATHEMATICS SUBJECT CLASSIFICATION:

• Primary: 13B99
• Secondary: 13A15
• 13G05
• 13B21

1. Introduction

Throughout this article, all rings are commutative with nonzero identity and all modules are unital. Let $\mathbb{N}$ denote the set of positive integers. For $m\in \mathbb{N},$ let $[1,m]=\{n\in \mathbb{N}|1\le n\le m\}.$ Let R be a ring. An ideal I of R is said to be proper if $I\ne R.$ The radical of I is denoted by $\sqrt{I}=\{x\in R|x^n\in I$ for some $n\in \mathbb{N}\}.$ We denote by $\text{Min}(I)$ the set of minimal prime ideals over the ideal I. The concept of prime ideals plays an important role in ideal theory and there are many ways to generalize it.

In [9], Badawi introduced and studied the concept of 2-absorbing ideals which is a generalization of prime ideals. An ideal I of R is a 2-absorbing ideal if whenever $a,b,c\in R$ and $\mathit{abc}\in I,$ then $ab\in I$ or $ac\in I$ or $bc\in I.$ In this case, $\sqrt{I}=P$ is a prime ideal with $P^2\subseteq I$ or $\sqrt{I}=P_1\cap P_2$ where P₁, P₂ are incomparable prime ideals with $P_1{P}_2\subseteq I,$ cf. [9, Theorem 2.4]. In [8], Anderson and Badawi introduced the concept of n-absorbing ideals as a generalization of prime ideals where n is a positive integer. An ideal I of R is called an n-absorbing ideal of R, if whenever $a_1,a_2,\dots ,a_{n+1}\in R$ and ${\prod }_{i=1}^{n+1}{a}_i\in I,$ then there are n of the a_i’s whose product is in I. In this case, due to Choi and Walker [13, Theorem 1], ${(\sqrt{I})}^n\subseteq I.$

In [23], Mukhtar et al. studied the commutative rings whose ideals have a TA-factorization. A proper ideal is called a TA-ideal if it is a 2-absorbing ideal. By a TA-factorization of a proper ideal I, we mean an expression of I as a product ${\prod }_{i=1}^r{J}_i$ of TA-ideals. Mukhtar et al. prove that any TAF-ring has dimension at most one and the local TAF-domains are atomic pseudo-valuations domains. Recently in [1], Ahmed et al. studied commutative rings whose proper ideals have an n-absorbing factorization. Let I be a proper ideal of R. By an n-absorbing factorization of I we mean an expression of I as a product ${\prod }_{i=1}^r{I}_i$ of proper n-absorbing ideals of R. Ahmed et al. called AF-$\text{dim}(R)$ (absorbing factorization dimension) the minimum positive integer n such that every ideal of R has an n-absorbing factorization. If no such n exists, set AF-$\text{dim}(R)=\infty .$ An FAF-ring (finite absorbing factorization ring) is a ring such that AF-$\text{dim}(R)<\infty .$ Recall that a general ZPI-ring is a ring whose proper ideals can be written as a product of prime ideals. Therefore, AF-$\text{dim}(R)$ measures, in some sense, how far R is from being a general ZPI-ring, cf. [1, Proposition 3]. By $\dim (R),$ we denote the Krull dimension of R.

In [25], Yassine et al. introduced the concept of a 1-absorbing prime ideal which is a generalization of a prime ideal. A proper ideal I of R is a 1-absorbing prime ideal (our abbreviation OA-ideal) if whenever we take nonunit elements $a,b,c\in R$ with $\mathit{abc}\in I,$ then $ab\in I$ or $c\in I.$ In this case, $\sqrt{I}=P$ is a prime ideal, cf. [25, Theorem 2.3]. And if R is a ring in which exists an OA-ideal that is not prime, then R is a local ring, that is a ring with one maximal ideal.

Let I be a proper ideal of R. By an OA-factorization of I, we mean an expression of I as a product ${\prod }_{i=1}^n{J}_i$ of OA-ideals. The aim of this note is to study the commutative rings whose proper ideals (resp., proper principal ideals, resp., proper 2-generated ideals) have an OA-factorization.

We call R a 1-absorbing prime factorization ring (OAF-ring) if every proper ideal has an OA-factorization. An OAF-domain is a domain which is an OAF-ring. Our article consists of five sections (including the introduction).

In the next section, we characterize OA-ideals (Lemma 2.1) and we prove that if I is an OA-ideal, then I is a primary ideal. We also show that the OAF-ring property is stable under factor ring (resp., fraction ring) formation (Propositions 2.2 and 2.3). Furthermore, we investigate OAF-rings with respect to direct products (Corollary 2.5) and polynomial ring extensions (Corollary 2.6). We prove that the general ZPI-rings are exactly the arithmetical OAF-rings (Theorem 2.8).

The third section consists of a collection of preparational results which will be of major importance in the fourth section. For instance, we show that the Krull dimension of an OAF-ring is at most one (Theorem 3.5).

The fourth section contains the main results of our article. Among other results, we provide characterizations of OAF-rings (Theorem 4.2), rings whose proper principal ideals have an OA-factorization (Corollary 4.3) and rings whose proper (principal) ideals are OA-ideals (Proposition 4.5).

In the last section, we study the transfer of the various OA-factorization properties to the trivial ring extension.

2. Characterization of OA-ideals and simple facts

We start with a characterization of OA-ideals. Recall that a ring R is a Q-ring (cf. [3]) if every proper ideal of R is a product of primary ideals.

Lemma 2.1.

Let R be a ring with Jacobson radical M and I be an ideal of R.

1. If R is not local, then I is an OA-ideal if and only if I is a prime ideal.

2. If R is local, then I is an OA-ideal if and only if I is a prime ideal or $M^2\subseteq I\subseteq M.$

3. Every OA-ideal is a primary TA-ideal. In particular, every OAF-ring is both a Q-ring and a TAF-ring.

Proof.

(1) This follows from [25, Theorem 2.4].

(2) Let R be local. Then, M is the maximal ideal of R.

($\Rightarrow$) Let I be an OA-ideal such that I is not a prime ideal. Since I is proper, we infer that $I\subseteq M.$ Since I is not prime, there are $a,b\in M\setminus I$ such that $ab\in I.$ To prove that $M^2\subseteq I,$ it suffices to show that $xy\in I$ for all $x,y\in M.$ Let $x,y\in M.$ Then, $\mathit{xyab}\in I.$ Since $xy,a,b\in M,b\notin I$ and I is an OA-ideal, it follows that $\mathit{xya}\in I.$ Again, since $x,y,a\in M,a\notin I$ and I is an OA-ideal, we have that $xy\in I.$

($\Leftarrow$) Clearly, if I is a prime ideal, then I is an OA-ideal. Now let $M^2\subseteq I\subseteq M.$ Then, I is proper. Let $a,b,c\in M$ be such that $\mathit{abc}\in I.$ Then, $ab\in M^2\subseteq I.$ Therefore, I is an OA-ideal.

(3) Let I be an OA-ideal. It is an immediate consequence of (1) and (2) that I is a primary ideal. Now let $a,b,c\in R$ be such that $\mathit{abc}\in I.$ We have to show that $ab\in I$ or $ac\in I$ or $bc\in I.$

First let a or b or c be a unit of R. Without restriction let a be unit of R. Since $\mathit{abc}\in I,$ we infer that $bc\in I.$

Now let a, b, and c be nonunits. Then, $ab\in I$ or $c\in I.$ If $c\in I,$ then $ac\in I.$ The in particular statement is clear. □

Proposition 2.2.

Let R be an OAF-ring and I be a proper ideal of R. Then, R/I is an OAF-ring.

Proof.

Let J be a proper ideal of R which contains I. Let $J={\prod }_{i=1}^m{J}_i$ be an OA-factorization. Then, $J/I={\prod }_{i=1}^m(J_i/I).$ It suffices to show that $J_i/I$ is an OA-ideal for each $i\in [1,m].$ Let $i\in [1,m]$ and let $a,b,c\in R$ be such that $\overline{a},\overline{b},\overline{c}$ are three nonunit elements of R/I and $\overline{a}\overline{b}\overline{c}\in J_i/I.$ Clearly, a, b, c are nonunit elements of R and $\mathit{abc}\in J_i.$ Since J_i is an OA-ideal of R, we get that $ab\in J_i$ or $c\in J_i$ which implies that $\overline{a}\overline{b}\in J_i/I$ or $\overline{c}\in J_i/I.$ Therefore, R/I is an OAF-ring. □

Proposition 2.3.

Let S be a multiplicatively closed subset of $R\setminus \mathbf{0}$. If R is an OAF-ring, then $S^{-1}R$ is an OAF-ring. In particular, R_M is an OAF-ring for every maximal ideal M of R.

Proof.

Let J be a proper ideal of $S^{-1}R.$ Then, $J=S^{-1}I$ for some proper ideal I of R with $I\cap S=\varnothing .$ Let $I={\prod }_{i=1}^m{I}_i$ be an OA-factorization. Then, $J={\prod }_{i=1}^m(S^{-1}{I}_i)$ where each $S^{-1}{I}_i$ which is proper is an OA-ideal by [25, Theorem 2.18]. Thus, $S^{-1}R$ is an OAF-ring. The in particular statement is clear. □

Let R be a ring. Then, R is said to be a π-ring if every proper principal ideal of R is a product of prime ideals. We say that R is a unique factorization ring (in the sense of Fletcher, cf. [4]) if every proper principal ideal of R is a product of principal prime ideals. A unique factorization domain is an integral domain which is a unique factorization ring.

Remark 2.4.

Let R be a non local ring.

1. R is a general ZPI-ring if and only if R is an OAF-ring.

2. R is a π-ring if and only if each proper principal ideal of R has an OA-factorization.

3. R is a unique factorization ring if and only if each proper principal ideal of R is a product of principal OA-ideals.

Proof.

This is an immediate consequence of Lemma 2.1(1). □

In the light of the above remark, we give the next result.

Corollary 2.5.

Let R₁ and R₂ be two rings and $R=R_1\times R_2$ be their direct product. The following statements are equivalent.

1. R is an OAF-ring.

2. R is a general ZPI-ring.

3. R₁ and R₂ are general ZPI-rings.

Proof.

This follows from Remark 2.4(1) and [21, Exercise 6(g), page 223]. □

Let R be a ring. Then, R is called a von Neumann regular ring if for each $x\in R$ there is some $y\in R$ with $x=x^{2}y.$ The ring R is von Neumann regular if and only if R is a zero-dimensional reduced ring (see [19, Theorem 3.1, page 10]).

Corollary 2.6.

Let R be a ring. The following statements are equivalent.

1. $R[X]$ is an OAF-ring.

2. R is a Noetherian von Neumann regular ring.

3. R is a finite direct product of fields.

Proof.

Observe that the polynomial ring $R[X]$ is never local, since X and $1-X$ are nonunit elements of $R[X],$ but their sum is a unit. Consequently, $R[X]$ is an OAF-ring if and only if $R[X]$ is a general ZPI-ring by Remark 2.4(1). The rest is now an easy consequence of [2, Theorem 6 and Corollary 6.1], [21, Exercise 10, page 225] and Hilbert’s basis theorem. □

Let R be a ring and I be an ideal of R. Then, I is called divided if I is comparable to every ideal of R (or equivalently, I is comparable to every principal ideal of R).

Lemma 2.7.

Let R be a local ring with maximal ideal M such that M² is divided. The following statements are equivalent.

1. Each two principal OA-ideals which contain M² are comparable.

2. For each OA-ideal I of R, we have that I is a prime ideal or $I=M^2.$

Proof.

(1) $\Rightarrow$ (2): Let I be an OA-ideal of R such that I is not a prime ideal of R. Then, $M^2\subseteq I\subset M$ by Lemma 2.1(2). Assume that $M^2\subset I.$ Let $x\in I\setminus M^2$ and let $y\in M\setminus I.$ Then, $x,y\notin M^2,$ and thus, $M^2\subseteq xR,yR$ (since M² is divided). It follows that xR and yR are (principal) OA-ideals of R by Lemma 2.1(2). Since $y\notin xR$ and xR and yR are comparable, we infer that $xR\subset yR.$ Consequently, there is some $z\in M$ such that x = yz, and hence $x\in M^2,$ a contradiction. Therefore, $I=M^2.$

(2) $\Rightarrow$ (1): This is obvious. □

Let R be a ring. An ideal I of R is called 2-generated if $I=xR+yR$ for some (not necessarily distinct) $x,y\in R.$ Note that every principal ideal of R is 2-generated. We say that R is a chained ring if each two ideals of R are comparable under inclusion. Moreover, R is said to be an arithmetical ring if R_M is a chained ring for each maximal ideal M of R.

Theorem 2.8.

Let R be a ring. The following statements are equivalent.

1. R is a general ZPI-ring

2. R is an arithmetical OAF-ring.

3. R is an arithmetical ring and each proper principal ideal of R has an OA-factorization.

Proof.

First, we show that if R is an arithmetical π-ring, then R is a general ZPI-ring. Let R be an arithmetical π-ring and let M be a maximal ideal of R. It is straightforward to show that R_M is a π-ring. Moreover, R_M is a chained ring, and hence every 2-generated ideal of R_M is principal. Therefore, every proper 2-generated ideal of R_M is a product of prime ideals of R_M. Consequently, R_M is a general ZPI-ring by [22, Theorem 3.2]. This implies that $\dim (R_M)\le 1$ by [21, page 205]. We infer that $\dim (R)\le 1,$ and thus R is a general ZPI-ring by [16, Theorems 39.2, 46.7, and 46.11].

(1) $\Rightarrow$ (2) $\Rightarrow$ (3): This is obvious.

(3) $\Rightarrow$ (1): It is sufficient to show that R is a π-ring. If R is not local, then R is a π-ring by Remark 2.4(2). Therefore, we can assume that R is local with maximal ideal M. Since R is local, we have that R is a chained ring. Therefore, M² is divided and each two OA-ideals of R are comparable. We infer by Lemma 2.7 that each OA-ideal of R is a product of prime ideals. Now it clearly follows that R is a π-ring. □

3. Preparational results

From Lemma 2.1(3), we have that $|\text{Min}(I)|=1$ for every OA-ideal I of R. In view of this remark, we obtain the following result.

Proposition 3.1.

Let R be a ring and I be a proper ideal of R. If I has an OA-factorization, then $\text{Min}(I)$ is finite.

Proof.

Let $I={\prod }_{i=1}^n{I}_i$ be an OA-factorization. It follows that $\text{Min}(I)\subseteq {\cup }_{i=1}^n\text{Min}(I_i),$ and thus $|\text{Min}(I)|\le n.$ □

Let R be a ring and I be an ideal of R. Then, I is called a multiplication ideal of R if for each ideal J of R with $J\subseteq I,$ there is some ideal L of R such that J = IL.

Lemma 3.2.

Let R be a local ring such that each proper principal ideal of R has an OA-factorization. Then, each nonmaximal minimal prime ideal of R is principal.

Proof.

Let P be a nonmaximal minimal prime ideal of R. By [2, Theorem 1], it is sufficient to show that P is a multiplication ideal.

Let $x\in P$ and let $xR={\prod }_{i=1}^n{I}_i$ be an OA-factorization. There is some $j\in [1,n]$ such that $I_j\subseteq P.$ By Lemma 2.1(2), we have that P = I_j, and hence xR = PJ for some ideal J of R. We infer that $xR=P(xR:P).$

Now let I be an ideal of R such that $I\subseteq P.$ Then, $I=\sum _{y\in I}yR=\sum _{y\in I}P(yR:P)=P\sum _{y\in I}(yR:P),$ and thus, P is a multiplication ideal. □

The next result is a generalization of [16, Theorem 46.8] and its proof is based on the proof of the same result.

Proposition 3.3.

Let R be a local ring with maximal ideal M such that $\dim (R)\ge 1$ and every proper principal ideal of R has an OA-factorization. Then, R is an integral domain and if $\dim (R)\ge 2$, then R is a unique factorization domain.

Proof.

Let N be the nilradical of R. It follows from Proposition 3.1 and Lemma 3.2 that $\text{Min}(\mathbf{0})$ is finite and each $P\in \text{Min}(\mathbf{0})$ is principal.

Claim: Every proper principal ideal of R/N has an OA-factorization. Let I be a proper principal ideal of R/N. Then, $I=(xR+N)/N$ for some $x\in M.$ Let $xR={\prod }_{i=1}^n{I}_i$ be an OA-factorization. We infer that $I=(xR)/N=({\prod }_{i=1}^n{I}_i)/N={\prod }_{i=1}^n(I_i/N).$ It suffices to show that $I_i/N$ is an OA-ideal of R/N for each $i\in [1,n].$ Let $i\in [1,n].$ If I_i is a prime ideal of R, then $N\subseteq I_i,$ and hence, $I_i/N$ is a prime ideal of R/N. Now let I_i be not a prime ideal of R. By Lemma 2.1(2), we have that $M^2\subseteq I_i\subseteq M.$ Note that R/N is local with maximal ideal M/N. Since ${(M/N)}^2=M^2/N\subseteq I_i/N\subseteq M/N,$ it follows by Lemma 2.1(2) that $I_i/N$ is an OA-ideal of R/N. This proves the claim.

Case 1: R is one-dimensional. We prove that R is an integral domain. If every OA-ideal of R is a prime ideal, then R is π-ring, and hence, R is an integral domain by [16, Theorem 46.8]. Now let not every OA-ideal of R be a prime ideal. It follows from Lemma 2.1(2) that M is not idempotent. Set $L=M^2\cup \underset{Q\in \text{Min}(\mathbf{0})}{\cup }Q.$ Next we prove that $M^2\subseteq xR$ for each $x\in R\setminus L.$ Let $x\in R\setminus L.$ Without restriction let x be a nonunit. Note that xR cannot be a product of more than one OA-ideal, and hence, xR is an OA-ideal. By Lemma 2.1(2), we have that $M^2\subseteq xR.$

Now we show that $P\subseteq M^2$ for each $P\in \text{Min}(\mathbf{0}).$ Let $P\in \text{Min}(\mathbf{0}).$ Assume that $P⊈M^2.$ Let $w\in R\setminus P.$ Then, $P+wR⊈L$ by the prime avoidance lemma, and thus there is some $v\in (P+wR)\setminus L.$ It follows that $M^2\subseteq vR\subseteq P+wR.$ Since P is a nonmaximal prime ideal, we have that R/P has no simple R/P-submodules, and hence $\underset{y\in R\setminus P}{\cap }(P+yR)=P.$ (Note that if $\underset{y\in R\setminus P}{\cap }(P+yR)\cancel{=}P,$ then $\underset{y\in R\setminus P}{\cap }(P+yR)/P$ is a simple R/P-submodule of R/P.) This implies that $M^2\subseteq \underset{y\in R\setminus P}{\cap }(P+yR)=P,$ and thus P = M, a contradiction.

Let $Q\in \text{Min}(\mathbf{0}).$ By the prime avoidance lemma, there is some $z\in M\setminus L.$ We infer that $Q\subset M^2\subset zR.$ Consequently, Q = zQ. Since Q is principal, it follows that $Q=\mathbf{0}$ (e.g. by Nakayama’s lemma), and hence R is an integral domain.

Case 2: $\dim (R)\ge 2$ and R is reduced. We show that R is a unique factorization domain. There is some nonmaximal nonminimal prime ideal Q of R. By the prime avoidance lemma, there is some $x\in Q\setminus \underset{P\in \text{Min}(\mathbf{0})}{\cup }P.$ Since R is reduced, we have that $\underset{L\in \text{Min}(\mathbf{0})}{\cap }L=\mathbf{0}.$ If $y\in R$ is nonzero with xy = 0, then $y\notin L$ and $xy\in L$ for some $L\in \text{Min}(\mathbf{0}),$ and hence $x\in L,$ a contradiction. We infer that x is a regular element of R. Let $xR={\prod }_{i=1}^n{I}_i$ be an OA-factorization. Then, $I_j\subseteq Q$ for some $j\in [1,n].$ Since x is regular, I_j is invertible, and hence, I_j is a regular principal ideal (because invertible ideals of a local ring are regular principal ideals). Since $I_j\subseteq Q$ and $Q\cancel{=}M,$ we have that I_j is a prime ideal by Lemma 2.1(2). Consequently, $P\subseteq I_j$ for some $P\in \text{Min}(\mathbf{0}).$ Since I_j is regular, we infer that $P\subset I_j,$ and hence P = PI_j (since I_j is principal). It follows (e.g. from Nakayama’s lemma) that $P=\mathbf{0}$ (since P is principal). We obtain that R is an integral domain.

To show that R is a unique factorization domain, it suffices to show by [4, Theorem 2.6] that every nonzero prime ideal of R contains a nonzero principal prime ideal. Since $\dim (R)\ge 2$ and R is local, we only need to show that every nonzero nonmaximal prime ideal of R contains a nonzero principal prime ideal. Let L be a nonzero nonmaximal prime ideal of R and let $z\in L$ be nonzero. Let $zR={\prod }_{k=1}^m{J}_k$ be an OA-factorization. Then, $J_{\ell }\subseteq L$ for some $\ell \in [1,m].$ Since R is an integral domain, zR is invertible, and hence, $J_{\ell }$ is invertible. Therefore, $J_{\ell }$ is nonzero and principal (since R is local). Since $L\cancel{=}M,$ it follows from Lemma 2.1(2) that $J_{\ell }$ is a prime ideal.

Case 3: $\dim (R)\ge 2.$ We have to show that R is a unique factorization domain. Note that R/N is a reduced local ring with maximal ideal M/N and $\dim (R/N)\ge 2.$ Moreover, each proper principal ideal of R/N has an OA-factorization by the claim. It follows by Case 2 that R/N is a unique factorization domain, and thus N is the unique minimal prime ideal of R. Since R/N is a unique factorization domain and $\dim (R/N)\ge 2,$ R/N possesses a nonzero nonmaximal principal prime ideal. We infer that there is some nonminimal nonmaximal prime ideal Q of R such that Q/N is a principal ideal of R/N. Consequently, there is some $q\in Q$ such that $Q=qR+N.$ Let $qR={\prod }_{i=1}^n{I}_i$ be an OA-factorization. Then, $I_j\subseteq Q$ for some $j\in [1,n].$ Since $Q\cancel{=}M,$ we infer by Lemma 2.1(2) that I_j is a prime ideal of R. Therefore, $Q=qR+N\subseteq I_j\subseteq Q,$ and hence I_j = Q.

Assume that $Q\cancel{=}qR.$ Then, qR = QJ for some proper ideal J of R. It follows that $q\in qR=(qR+N)J\subseteq qJ+N,$ and thus $q(1-a)\in N$ for some $a\in J.$ Since a is a nonunit of R, we obtain that $q\in N.$ This implies that $Q=qR+N=N,$ a contradiction. We infer that Q = qR. Since $N\subset Q$ and N is a prime ideal of R, we have that N = NQ. Consequently, $N=\mathbf{0}$ (e.g. by Nakayama’s lemma, since N is principal), and thus $R\cong R/N$ is a unique factorization domain. □

Proposition 3.4.

Let R be a local ring with maximal ideal M such that each proper 2-generated ideal of R has an OA-factorization. Then, $\dim (R)\le 2$ and each nonmaximal prime ideal of R is principal.

Proof.

First we show that $\dim (R_P)\le 1$ for each nonmaximal prime ideal P of R. Let P be a nonmaximal prime ideal and let I be a proper 2-generated ideal of R_P. Observe that I = J_P for some 2-generated ideal J of R with $J\subseteq P.$ Let $J={\prod }_{i=1}^n{J}_i$ be an OA-factorization. Then, $I=J_P={\prod }_{i=1}^n{(J_i)}_P={\prod }_{i=1,J_i\subseteq P}^n{(J_i)}_P.$ If $i\in [1,n]$ is such that $J_i\subseteq P,$ then J_i is a prime ideal of R by Lemma 2.1(2), and thus ${(J_i)}_P$ is a prime ideal of R_P. We infer that I is a product of prime ideals of R_P. It follows from [22, Theorem 3.2], that R_P is a general ZPI-ring. It is an easy consequence of [21, page 205] that $\dim (R_P)\le 1.$

This implies that $\dim (R)\le 2.$ It remains to show that every nonmaximal prime ideal of R is principal. Without restriction let $\dim (R)\ge 1.$ It follows from Proposition 3.3 that R is either a one-dimensional domain or a two-dimensional unique factorization domain. In any case, we have that each nonmaximal prime ideal of R is principal. □

In the next result, we will prove a generalization of the fact that every OAF-ring has Krull dimension at most one.

Theorem 3.5.

Let R be a ring such that every proper 2-generated ideal of R has an OA-factorization. Then, $\dim (R)\le 1.$

Proof.

If every OA-ideal of R is a prime ideal, then R is a general ZPI-ring by [22, Theorem 3.2], and hence, $\dim (R)\le 1$ by [21, page 205]. Now let not every OA-ideal of R be a prime ideal. We infer by Lemma 2.1 that R is local and the maximal ideal of R is not idempotent. Let M be the maximal ideal of R. It suffices to show that if Q is a nonmaximal prime ideal of R, then $Q=\mathbf{0}.$ Let Q be a nonmaximal prime ideal of R.

Assume that $Q⊈M^2.$ Since $\dim (R)\le 2$ by Proposition 3.4, there is some prime ideal P of R such that $Q\subseteq P$ and $\dim (R/P)=1.$ Next we show that $M^2\subseteq P+yR$ for each $y\in R\setminus P.$ Let $y\in R\setminus P$ and set $J=P+yR.$ Without restriction let $J\subset M.$ Note that J is 2-generated by Proposition 3.4. Since $J⊈M^2,$ J cannot be a product of more than one OA-ideal, and thus, J is an OA-ideal of R. Since $P\subset J\subset M,$ we have that J is not a prime ideal of R, and thus $M^2\subseteq J$ by Lemma 2.1(2). Moreover, R/P is an integral domain that is not a field. Consequently, R/P does not have any simple R/P-submodules, which implies that $P=\underset{x\in R\setminus P}{\cap }(P+xR).$ (Observe that if $\underset{x\in R\setminus P}{\cap }(P+xR)\cancel{=}P,$ then $\underset{x\in R\setminus P}{\cap }(P+xR)/P$ is a simple R/P-submodule of R/P.) Therefore, $M^2\subseteq \underset{x\in R\setminus P}{\cap }(P+xR)=P,$ and hence P = M, a contradiction. We infer that $Q\subseteq M^2.$

There is some $z\in M\setminus M^2$ (since M is not idempotent). Since zR is a product of OA-ideals, we have that zR is an OA-ideal of R. As shown before, $L\subseteq M^2$ for each nonmaximal prime ideal L of R, and thus zR is not a nonmaximal prime ideal. Consequently, $Q\subset M^2\subset zR$ by Lemma 2.1(2), and hence Q = zQ. Since Q is principal by Proposition 3.4, it follows (e.g. by Nakayama’s lemma) that $Q=\mathbf{0}.$ □

Lemma 3.6.

Let D be a local domain with maximal ideal M. Then, each proper principal ideal of D has an OA-factorization if and only if D is atomic and each irreducible element generates an OA-ideal. If these equivalent conditions are satisfied, then $\underset{n\in \mathbb{N}}{\cap }{P}^n=\mathbf{0}$ for each height-one prime ideal P of D.

Proof.

($\Rightarrow$) Let each proper principal ideal of D have an OA-factorization. If D is a unique factorization domain, then D is atomic and each irreducible element generates a prime ideal. Now let D be not a unique factorization domain. Then, $\dim (D)=1$ by Proposition 3.3.

Assume that M² is principal. Then, M is invertible, and hence M is principal (since D is local). Note that D is a DVR (since $\dim (D)=1$), and hence, D is a unique factorization domain, a contradiction.

We infer that M² is not principal. We show that D is atomic. Let $y\in D$ be a nonzero nonunit. Then, $yD={\prod }_{i=1}^n{I}_i$ for some principal OA-ideals I_i. There are nonzero nonunits $x_i\in D$ such that $y={\prod }_{i=1}^n{x}_i$ and I_j = x_jD for each $j\in [1,n].$ Let $i\in [1,n].$ If I_i is a prime ideal, then x_i is a prime element, and thus x_i is irreducible. Now let I_i not be a prime ideal. It follows from Lemma 2.1(2) that $M^2\subseteq I_i.$ Since M² is not principal, we have that $x_i\notin M^2.$ Therefore, x_i is irreducible.

Finally, let $z\in D$ be irreducible. Then, $zD={\prod }_{j=1}^m{J}_j$ for some principal OA-ideals J_j. Since zD is maximal among the proper principal ideals of D, we obtain that zD = J_j for some $j\in [1,n].$

($\Leftarrow$) Let D be atomic such that each irreducible element generates an OA-ideal. Let I be a proper principal ideal of D. Without restriction let I be nonzero. Then, I = xD for some nonzero nonunit $x\in D.$ Observe that $x={\prod }_{i=1}^n{x}_i$ for some irreducible elements $x_i\in D.$ It follows that ${\prod }_{i=1}^n{x}_{i}D$ is an OA-factorization of I.

Now let the equivalent conditions be satisfied and let P be a height-one prime ideal of D. First let $P\cancel{=}M.$ Then, D is a unique factorization domain by Proposition 3.3, and hence, P is principal. Therefore, $\underset{n\in \mathbb{N}}{\cap }{P}^n$ is a prime ideal of D by [5, Theorem 2.2(1)]. Since $\underset{n\in \mathbb{N}}{\cap }{P}^n\subset P,$ we infer that $\underset{n\in \mathbb{N}}{\cap }{P}^n=\mathbf{0}.$

Now let P = M. Assume that $\underset{n\in \mathbb{N}}{\cap }{M}^n\cancel{=}\mathbf{0}$ and let $x\in \underset{n\in \mathbb{N}}{\cap }{M}^n$ be nonzero. Then, xD is a product of m OA-ideals of D for some positive integer m. We infer by Lemma 2.1(2) that $M^{2m}\subseteq xD,$ and hence, $M^{2m}\subseteq xD\subseteq M^{4m}\subseteq M^{2m}.$ This implies that $xD=M^{2m}=M^{4m}=x^{2}D,$ and thus, x is a unit of D, a contradiction. Therefore, $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$ □

Lemma 3.7.

Let R be a local ring with maximal ideal M such that M² is divided and such that either M is nilpotent or R is an integral domain with $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}$. Then, R is an OAF-ring and every proper principal ideal of R is a product of principal OA-ideals.

Proof.

If M is idempotent, then $M=\mathbf{0},$ and hence, R is a field and both statements are clearly satisfied. Now let M be not idempotent. There is some $x\in M\setminus M^2.$ In what follows, we freely use the fact that if N is an ideal of R and $z\in R$ such that $N\subseteq zR,$ then $N=z(N:zR),$ and hence, N = zJ for some ideal J of R.

Next we prove that $M^2=xM$ and xR is an OA-ideal of R. Since $x\notin M^2$ and M² is divided, we have that $M^2\subseteq xR\subseteq M.$ Therefore, xR is an OA-ideal by Lemma 2.1(2). Since $M^2\subset xR,$ there is some proper ideal J of R with $M^2=xJ,$ and thus $M^2\subseteq xM.$ Obviously, $xM\subseteq M^2,$ and hence $M^2=xM.$

Now we show that R is an OAF-ring. Let I be a proper ideal of R. First let $I=\mathbf{0}.$ If M is nilpotent, then I is obviously a product of OA-ideals. If R is an integral domain, then I is an OA-ideal. Now let I be nonzero. In any case, there is a largest positive integer n such that $I\subseteq M^n.$ Observe that $I\subseteq M^n=x^{n-1}M\subseteq x^{n-1}R.$ Consequently, $I=x^{n-1}L={(xR)}^{n-1}L$ for some proper ideal L of R. Assume that $L\subseteq M^2.$ Note that $L\subseteq M^2=xM\subseteq xR.$ This implies that L = xA for some proper ideal A of R, and hence $I=x^{n}A\subseteq x^{n}M=M^{n+1},$ a contradiction. We infer that $M^2\subseteq L$ (since M² is divided). It follows from Lemma 2.1(2) that L is an OA-ideal. In any case, I is a product of OA-ideals.

Finally, we prove that every proper principal ideal of R is a product of principal OA-ideals. Let $y\in M.$ First let y = 0. If M is nilpotent, then $x^k=0$ for some $k\in \mathbb{N},$ and thus $yR={(xR)}^k$ is a product of principal OA-ideals. If R is an integral domain, then yR is a principal OA-ideal. Now let y be nonzero. There is some greatest $\ell \in \mathbb{N}$ such that $y\in M^{\ell }.$ Therefore, $y=x^{\ell -1}z$ for some $z\in M.$ If $z\in M^2,$ then z = xv for some $v\in M,$ and hence $y=x^{\ell }v\in M^{\ell +1},$ a contradiction. We infer that $z\notin M^2,$ and thus $M^2\subseteq zR\subseteq M.$ It follows from Lemma 2.1(2) that zR is an OA-ideal of R. Consequently, $yR={(xR)}^{\ell -1}(zR)$ is a product of principal OA-ideals. □

4. Characterization of OAF-rings and related concepts

First, we recall several definitions and discuss the factorization theoretical properties of local one-dimensional OAF-domains. Let D be an integral domain with quotient field K. Then, $\widehat{D}=\{x\in K|$ there is some nonzero $c\in D$ such that $c{x}^n\in D$ for all $n\in \mathbb{N}\}$ is called the complete integral closure of D. Let $(D:\widehat{D})=\{x\in D|x\widehat{D}\subseteq D\}$ be the conductor of D in $\widehat{D}.$ The domain D is called completely integrally closed if $D=\widehat{D}$ and D is said to be seminormal if for all $x\in K$ such that $x^2,x^3\in D,$ it follows that $x\in D.$ Note that every completely integrally closed domain is seminormal. We say that D is a finitely primary domain of rank one if D is a local one-dimensional domain such that $\widehat{D}$ is a DVR and $(D:\widehat{D})\cancel{=}\mathbf{0}.$ For each subset, $X\subseteq K$ let $X^{-1}=\{x\in K|xX\subseteq D\}$ and $X_v={(X^{-1})}^{-1}.$ An ideal I of D is called divisorial if I_v = I. Moreover, D is called a Mori domain if D satisfies the ascending chain condition on divisorial ideals. It is well known that every unique factorization domain and every Noetherian domain is a Mori domain (see [14, Corollary 2.3.13] and [11, page 57]). We say that D is half-factorial if D is atomic and each two factorizations of each nonzero element of D into irreducible elements are of the same length. Finally, D is called a C-domain if the monoid of nonzero elements of D (i.e. $D\setminus \mathbf{0}$) is a C-monoid. For the precise definition of C-monoids, we refer to [14, Definition 2.9.5].

Let D be a local domain with quotient field K and maximal ideal M. Set $(M:M)=\{x\in K|xM\subseteq M\}.$ Then, $(M:M)$ is called the ring of multipliers of M. Moreover, M² is said to be universal if $M^2\subseteq uD$ for each irreducible element $u\in D.$

Theorem 4.1.

Let D be a local domain with maximal ideal M such that D is not a field. The following statements are equivalent.

1. D is an OAF-domain.

2. D is a TAF-domain.

3. D is one-dimensional and every proper principal ideal has an OA-factorization.

4. D is one-dimensional and atomic and every irreducible element generates an OA-ideal.

5. D is atomic such that M² is universal.

6. $(M:M)$ is a DVR with maximal ideal M.

7. D is a seminormal finitely primary domain of rank one.

If these equivalent conditions are satisfied, then D is a half-factorial C-domain and a Mori domain.

Proof.

(1) $\Rightarrow$ (2): This follows from Lemma 2.1(3).

(1) $\Rightarrow$ (3): By Theorem 3.5, D is one-dimensional. The rest of assertion (3) is clear.

(2) $\iff$ (5) $\iff$ (6): This follows from [23, Theorem 4.3].

(3) $\iff$ (4): This is an immediate consequence of Lemma 3.6.

(4) $\Rightarrow$ (5): Let $y\in D$ be an irreducible element. Since yD is an OA-ideal and $\sqrt{yD}=M,$ we deduce from Lemma 2.1(2) that $M^2\subseteq yD.$ Hence, M² is universal.

(5) + (6) $\Rightarrow$ (1): It follows from [6, Theorem 5.1] that M² is comparable to every principal ideal of D, and thus M² is divided. Since $(M:M)$ is a DVR with maximal ideal M, we have that $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$ Consequently, D is an OAF-domain by Lemma 3.7.

(5) + (6) $\Rightarrow$ (7): First we show that D is finitely primary of rank one. Let P be a nonzero prime ideal of D. Then, P contains an irreducible element $y\in D,$ and hence, $M^2\subseteq yD\subseteq P.$ Therefore, P = M, and thus, D is one-dimensional. It remains to show that $\widehat{D}$ is a DVR and $(D:\widehat{D})\cancel{=}\mathbf{0}.$ Since $(M:M)$ is a DVR, we have that $(M:M)$ is completely integrally closed. Observe that $D\subseteq (M:M)\subseteq \widehat{D},$ and hence $\widehat{D}\subseteq \widehat{(M:M)}=(M:M).$ Therefore, $\widehat{D}=(M:M)$ is a DVR. Since $M\widehat{D}=M(M:M)\subseteq M\subseteq D$ and $M\cancel{=}\mathbf{0},$ we infer that $(D:\widehat{D})\cancel{=}\mathbf{0}.$

Next we show that D is seminormal. Let V be the group of units of $\widehat{D}.$ Let K be the field of quotients of D and let $x\in K$ be such that $x^2,x^3\in D.$ Then, $x^2,x^3\in \widehat{D}.$ Since $\widehat{D}$ is a DVR, $\widehat{D}$ is seminormal, and thus $x\in \widehat{D}.$ In particular, $x\in M$ or $x\in V.$ If $x\in M,$ then $x\in D.$ Now let $x\in V.$ Note that $V\cap D$ is the group of units of D (by [24, Corollary 1.4] and [12, Proposition 2.1]), and thus x² and x³ are units of D. Therefore, $x=x^{-2}{x}^3$ is a unit of D, and hence $x\in D.$

(7) $\Rightarrow$ (6): By [15, Lemma 3.3.3], we have that M is the maximal ideal of $\widehat{D}.$ If $x\in \widehat{D},$ then $xM\subseteq M$ (since M is an ideal of $\widehat{D}$). It is straightforward to show that $(M:M)\subseteq \widehat{D}.$ We infer that $(M:M)=\widehat{D}$ is a DVR.

Now let the equivalent statements of Theorem 4.1 be satisfied. It remains to show that D is a half-factorial C-domain and a Mori domain. It follows from [6, Theorem 6.2] that D is a half-factorial domain. Obviously, V is a subgroup of finite index of V and $VM\subseteq \widehat{D}M=(M:M)M\subseteq M.$ It follows from [18, Corollary 2.8] and [14, Corollary 2.9.8] that D is a C-domain. Moreover, D is a Mori domain by [18, Proposition 2.5.1]. □

We want to point out that a local one-dimensional OAF-domain need not be Noetherian. Let $K\subseteq L$ be a field extension such that $[L:K]=\infty$ and let $D=K+XL[\mathrm{}[X]\mathrm{}].$ Then, D is a local one-dimensional domain with maximal ideal $M=XL[\mathrm{}[X]\mathrm{}]$ and $(M:M)=L[\mathrm{}[X]\mathrm{}]$ is a DVR with maximal ideal M. Consequently, D is an OAF-domain by Theorem 4.1. Since $[L:K]=\infty ,$ it follows that D is not Noetherian.

An integral domain D is called a Cohen-Kaplansky domain if D is atomic and D has only finitely many irreducible elements up to associates. It follows from [6, Example 6.7] that there exists a local half-factorial Cohen-Kaplansky domain with maximal ideal M for which M² is not universal. We infer by Theorem 4.1 that the aforementioned domain is not an OAF-domain.

Theorem 4.2.

Let R be a ring with Jacobson radical M. The following statements are equivalent.

1. R is an OAF-ring.

2. Each proper 2-generated ideal of R has an OA-factorization.

3. $\dim (R)\le 1$ and each proper principal ideal has an OA-factorization.

4. R satisfies one of the following conditions.

   1. R is a general ZPI-ring.

   2. R is a local domain, M² is divided and $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$

   3. R is local, M² is divided and M is nilpotent.

Proof.

(1) $\Rightarrow$ (2): This is obvious.

(2) $\Rightarrow$ (3): This is an immediate consequence of Theorem 3.5.

(3) $\Rightarrow$ (4): First let each OA-ideal of R be a prime ideal. Then, R is a π-ring. By [16, Theorems 39.2, 46.7, and 46.11], R is a general ZPI-ring. Now let there be an OA-ideal of R which is not a prime ideal. It follows from Lemma 2.1 that R is local with maximal ideal M and M is not idempotent. Note that if $x\in M\setminus M^2,$ then xR cannot be a product of more than one OA-ideal, and hence xR is an OA-ideal.

Case 1: R is zero-dimensional. Let $x\in M\setminus M^2.$ Then, xR is an OA-ideal. We infer by Lemma 2.1(2) that $M^2\subseteq xR.$ Consequently, M² is divided. It follows from Lemma 2.1 that $M^2\subseteq I$ for each OA-ideal I of R. Since 0 is a product of OA-ideals, we have that 0 contains a power of M. This implies that M is nilpotent.

Case 2: R is one-dimensional. It follows from Proposition 3.3 that R is an integral domain, and hence $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}$ by Lemma 3.6. It remains to show that M² is divided. Let $x\in R\setminus M^2.$ Without restriction let x be a nonunit. Then, xR is an OA-ideal. By Lemma 2.1(2), we have that $M^2\subseteq xR.$

(4) $\Rightarrow$ (1): Clearly, every general ZPI-ring is an OAF-ring. The rest follows from Lemma 3.7. □

Corollary 4.3.

Let R be a ring with Jacobson radical M. The following statements are equivalent.

1. Each proper principal ideal of R has an OA-factorization.

2. R is a π-ring or an OAF-ring.

3. R satisfies one of the following conditions.

   1. R is a π-ring.

   2. R is a local domain, M² is divided and$\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$

   3. R is local, M² is divided and M is nilpotent.

Proof.

(1) $\Rightarrow$ (2): If R is not local, then R is a π-ring by Remark 2.4(2). Now let R be local. If $\dim (R)\ge 2,$ then R is a unique factorization domain by Proposition 3.3, and hence, R is a π-ring. If $\dim (R)\le 1,$ then R is an OAF-ring by Theorem 4.2.

(2) $\Rightarrow$ (1): This is obvious.

(2) $\iff$ (3): This is an immediate consequence of Theorem 4.2 and the fact that every general ZPI-ring is a π-ring. □

Corollary 4.4.

Let R be a ring with Jacobson radical M. The following statements are equivalent.

1. Each proper principal ideal of R is a product of principal OA-ideals.

2. R satisfies one of the following conditions.

   1. R is a unique factorization ring.

   2. R is a local domain, M² is divided and $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$

   3. R is local, M² is divided and M is nilpotent.

Proof.

(1) $\Rightarrow$ (2): If R is not local, then R is a unique factorization ring by Remark 2.4(3). If R is local, then the statement follows from Corollary 4.3 and the fact that every local π-ring is a unique factorization ring ([4, Corollary 2.2]).

(2) $\Rightarrow$ (1): Obviously, if R is a unique factorization ring, then each proper principal ideal of R is a product of principal OA-ideals. The rest is an immediate consequence of Lemma 3.7. □

In Lemma 2.1, we saw that if R is a local ring with maximal ideal M and I is an ideal of R such that $M^2\subseteq I,$ then I is an OA-ideal of R. Now we will give a characterization of the rings for which every proper (principal) ideal is an OA-ideal.

Proposition 4.5.

Let R be a ring with Jacobson radical M. The following statements are equivalent.

1. Every proper ideal of R is an OA-ideal.

2. Every proper principal ideal of R is an OA-ideal.

3. R is local and $M^2=\mathbf{0}.$

Proof.

(1) $\Rightarrow$ (2): This is obvious.

(2) $\Rightarrow$ (3): Assume that R is not local. Then, every proper principal ideal of R is a prime ideal by Lemma 2.1(1). Consequently, R is an integral domain. If $x\in R$ is a nonunit, then $x^{2}R$ is a prime ideal, and hence $x^{2}R=xR$ and x = 0. Therefore, R is a field, a contradiction. This implies that R is local with maximal ideal M. We infer by Lemma 2.1(2) that 0 is a prime ideal or $M^2=\mathbf{0}.$

Assume that $M^2\cancel{=}\mathbf{0}.$ Then, R is an integral domain and there is some nonzero $x\in M^2.$ It follow from Lemma 2.1(2) that $x^{2}R$ is a prime ideal or $M^2\subseteq x^{2}R.$ If $x^{2}R$ is a prime ideal, then $x^{2}R=xR.$ If $M^2\subseteq x^{2}R,$ then $M^2\subseteq x^{2}R\subseteq xR\subseteq M^2,$ and thus $x^{2}R=xR.$ In any case, we have that $x^{2}R=xR,$ and hence, x is a unit (since x is regular), a contradiction.

(3) $\Rightarrow$ (1): This is an immediate consequence of Lemma 2.1(2). □

5. OA-factorization properties and trivial ring extensions

Let A be a ring and E be an A-module. Then, $A\propto E,$ the trivial (ring) extension of A by E, is the ring whose additive structure is that of the external direct sum $A\oplus E$ and whose multiplication is defined by $(a,e)(b,f)=(ab,af+be)$ for all $a,b\in A$ and all $e,f\in E.$ (This construction is also known by other terminology and other notation, such as the idealization $A(+)E.$) The basic properties of trivial ring extensions are summarized in the textbooks [17, 19]. Trivial ring extensions have been studied or generalized extensively, often because of their usefulness in constructing new classes of examples of rings satisfying various properties (cf. [7, 10, 20]). We say that E is divisible if E = aE for each regular element $a\in A.$

We start with the following lemma.

Lemma 5.1.

Let A be a ring, I be an ideal of A and E be an A-module. Let $R=A\propto E$ be the trivial ring extension of A by E.

1. $I\propto E$ is an OA-ideal of R if and only if I is an OA-ideal of A.

2. Assume that A contains a nonunit regular element and E is a divisible A-module. Then, the OA-ideals of R have the form $L\propto E$ where L is an OA-ideal of A.

Proof.

(1) This follows immediately from [25, Theorem 2.20].

(2) Let J be an OA-ideal of R. Our aim is to show that $\mathbf{0}\propto E\subseteq J.$ Let $e\in E$ and let $a\in A$ be a nonunit regular element. Then, e = af for some $f\in E$ and thus $(a,0)(0,f)(0,e)=(0,0)\in J.$ Since J is an OA-ideal, we conclude that $(a,0)(0,f)=(0,e)\in J$ or $(0,e)\in J$ which implies that $\mathbf{0}\propto E\subseteq J.$ Therefore, $J=L\propto E$ with $L=\{b\in A|(b,g)\in J$ for some $g\in E\}$ and L is an ideal of A by [7, Theorems 3.1 and 3.3(1)]. Now the result follows from (1). □

Corollary 5.2.

Let A be an integral domain that is not a field, E be a divisible A-module and $R=A\propto E$. Then, the OA-ideals of R have the form $I\propto E$ where I is an OA-ideal of A.

Next, we study the transfer of the OAF-ring property to the trivial ring extension.

Theorem 5.3.

Let A be a ring with Jacobson radical M, E be an A-module and $R=A\propto E.$

1. R is an OAF-ring if and only if one of the following conditions is satisfied.

   1. A is a general ZPI-ring, E is cyclic and the annihilator of E is a $($possibly empty$)$ product of idempotent maximal ideals of A.

   2. A is local, M² is divided, $E=\mathbf{0}$ and either M is nilpotent or A is a domain with $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$

   3. A is local, $M^2=\mathbf{0}$, ME = aE for each nonzero $a\in M$ and ME = Mx for each $x\in E\setminus ME.$

   In particular, if R is an OAF-ring, then A is an OAF-ring.

2. Every proper ideal of R is an OA-ideal if and only if A is local, $M^2=\mathbf{0}$ and $ME=\mathbf{0}.$

Proof.

(1) ($\Rightarrow$) First let R be an OAF-ring. By Theorem 4.2, it follows that (a) R is a general ZPI-ring or (b) R is local with maximal ideal N, N² is divided and (N is nilpotent or R is a domain such that $\underset{n\in \mathbb{N}}{\cap }{N}^n=\mathbf{0}$). If R is a general ZPI-ring, then condition (A) is satisfied by [7, Theorem 4.10].

From now on let R be local with maximal ideal N such that N² is divided. Observe that A is local with maximal ideal M and $N=M\propto E$ by [7, Theorem 3.2(1)]. If R is a domain such that $\underset{n\in \mathbb{N}}{\cap }{N}^n=\mathbf{0},$ then $E=\mathbf{0}$ (for if $z\in E$ is nonzero, then $(0,z)$ is a nonzero zero-divisor of R), and hence $A\cong R$ is a domain, M² is divided and $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$

Now let N be nilpotent. If $E=\mathbf{0},$ then $A\cong R,$ and thus M² is divided and M is nilpotent. From now on let E be nonzero. There is some $k\in \mathbb{N}$ such that $N^k=\mathbf{0}.$ Note that $N^2=M^2\propto ME$ and $N^k=M^k\propto M^{k-1}E,$ and thus $M^k=\mathbf{0}.$ Since N² is divided, we have that $\mathbf{0}\propto E\subseteq N^2$ or $N^2\subseteq \mathbf{0}\propto E.$ If $\mathbf{0}\propto E\subseteq N^2,$ then E = ME, and hence $E=M^{k}E=\mathbf{0},$ a contradiction. Therefore, $N^2\subseteq \mathbf{0}\propto E,$ which implies that $M^2=\mathbf{0}.$

Let $a\in M$ be nonzero. Then, $(a,0)\notin N^2,$ and hence $N^2\subseteq (a,0)R=aA\propto aE.$ Consequently, $ME\subseteq aE,$ and thus ME = aE. Finally, let $x\in E\setminus ME.$ Then, $(0,x)\notin N^2.$ We infer that $N^2\subseteq (0,x)R=\mathbf{0}\propto Ax.$ This implies that $ME\subseteq Ax.$ If $ME⊈Mx,$ then $bx\in ME$ for some unit $b\in A,$ and hence $x\in ME,$ a contradiction. It follows that $ME\subseteq Mx,$ which clearly implies that ME = Mx.

($\Leftarrow$) Next we prove the converse. If condition (A) is satisfied, then R is a general ZPI-ring by [7, Theorem 4.10], and thus R is an OAF-ring. If condition (B) is satisfied, then A is an OAF-ring by Theorem 4.2, and hence $R\cong A$ is an OAF-ring. Now let condition (C) be satisfied. Set $N=M\propto E.$ Then, R is local with maximal ideal N by [7, Theorem 3.2(1)]. By Theorem 4.2, it suffices to show that N is nilpotent and N² is divided. Since $M^2=\mathbf{0},$ we obtain that $N^3=M^3\propto M^{2}E=\mathbf{0},$ and thus N is nilpotent. It remains to show that $N^2\subseteq (a,x)R$ for each $(a,x)\in R\setminus N^2.$ Let $a\in A$ and $x\in E$ be such that $(a,x)\notin N^2.$ Since $N^2=\mathbf{0}\propto ME,$ we have to show that $\mathbf{0}\propto ME\subseteq (a,x)R.$ If a is a unit of A, then (a, x) is a unit of R by [7, Theorem 3.7] and the statement is clearly true. Let $z\in \mathbf{0}\propto ME.$ Then, $z=(0,y)$ for some $y\in ME.$

Case 1: a is a nonzero nonunit. Since ME = aE, there is some $v\in E$ such that y = av. Observe that $z=(0,av)=(a,x)(0,v)\in (a,x)R.$

Case 2: a = 0. Then, $x\in E\setminus ME$ (since $(a,x)\notin N^2$). Since ME = Mx, there is some $b\in M$ such that y = bx. It follows that $z=(0,bx)=(a,x)(b,0)\in (a,x)R.$

The in particular statement now follows from Theorem 4.2.

(2) First let every proper ideal of R be an OA-ideal. By Proposition 4.5, we have that R is local with maximal ideal N and $N^2=\mathbf{0}.$ It follows that A is local with maximal ideal M and $N=M\propto E$ by [7, Theorem 3.2(1)]. Moreover, $\mathbf{0}=N^2=M^2\propto ME,$ and hence $M^2=\mathbf{0}$ and $ME=\mathbf{0}.$

Conversely, let A be local, $M^2=\mathbf{0}$ and $ME=\mathbf{0}.$ Set $N=M\propto E.$ Then, R is local with maximal ideal N by [7, Theorem 3.2(1)] and $N^2=M^2\propto ME=\mathbf{0}.$ We infer by Proposition 4.5 that each proper ideal of R is an OA-ideal. □

Corollary 5.4.

Let A be an integral domain, E be a nonzero A-module and $R=A\propto E$. The following statements are equivalent.

1. R is an OAF-ring.

2. A is a field.

3. Every proper ideal of R is an OA-ideal.

Proof.

(1) $\Rightarrow$ (2): It follows from Theorem 5.3(1) that A is a general ZPI-ring and the annihilator of E is a product of idempotent maximal ideals of A or that A is local with maximal ideal M such that $M^2=\mathbf{0}.$

First, let A be a general ZPI-ring such that the annihilator of E is a product of idempotent maximal ideals of A. Note that A is a Dedekind domain, and thus, the only proper idempotent ideal of A is the zero ideal. Since E is nonzero, the annihilator of E is a proper ideal of A, and hence A must possess an idempotent maximal ideal. We infer that the zero ideal is a maximal ideal of A, and thus A is a field.

Now let A be local with maximal ideal M such that $M^2=\mathbf{0}.$ Since A is an integral domain, it follows that $M=\mathbf{0},$ and hence A is a field.

(2) $\Rightarrow$ (3): Set $M=\mathbf{0}.$ Then, A is local with maximal ideal M, $M^2=\mathbf{0},$ and $ME=\mathbf{0}.$ Now the statement follows from Theorem 5.3(2).

(3) $\Rightarrow$ (1): This is obvious. □

Remark 5.5.

In general, if A is an OAF-ring and E is an A-module, then $A\propto E$ need not be an OAF-ring. Indeed, let A be an OAF-domain that is not a field and let E be a nonzero A-module. By Corollary 5.4, $A\propto E$ is not an OAF-ring.

Corollary 5.6.

Let A be a local ring with maximal ideal M and E be a nonzero A-module such that $ME=\mathbf{0}$. Set $R=A\propto E$. The following statements are equivalent.

1. R is an OAF-ring.

2. $M^2=\mathbf{0}.$

3. Every proper ideal of R is an OA-ideal.

Proof.

(1) $\Rightarrow$ (2): Assume that $M^2\cancel{=}\mathbf{0}.$ By Theorem 5.3(1), A is a local general ZPI-ring and M is idempotent (since the annihilator of E is a nonempty product of idempotent maximal ideals of A and M is the only maximal ideal of A). We infer by [21, Corollary 9.11] that A is a Dedekind domain or each proper ideal of A is a power of M (because local rings are indecomposable). If A is a Dedekind domain, then clearly $M^2=M=\mathbf{0}$ (since M is idempotent and a Dedekind domain has no nonzero proper idempotent ideals). Moreover, if every proper ideal of A is a power of M, then again $M^2=M=\mathbf{0}$ (since M is idempotent). In any case, we obtain that $M^2=\mathbf{0},$ a contradiction.

(1) $\Leftarrow$ (2) $\iff$ (3): This follows from Theorem 5.3. □

Example 5.7.

Let A be a local principal ideal ring with maximal ideal M such that A is not a field and $M^2=\mathbf{0}$ (e.g. $A=\mathbb{Z}/4\mathbb{Z}$). Set $R=A\propto A.$ Then, R is an OAF-ring, and yet not every proper ideal of R is an OA-ideal.

Proof.

Since $M\cancel{=}\mathbf{0},$ it follows from Theorem 5.3(2) that not every proper ideal of R is an OA-ideal. By Theorem 5.3(1), it remains to show that M = aA for each nonzero $a\in M$ and M = Mx for each $x\in A\setminus M.$ Note that M = zA for some $z\in M.$ If $a\in M$ is nonzero, then a = zb for some $b\in A.$ Clearly, $b\notin M,$ and thus b is a unit of A, which clearly implies that M = zA = aA. Finally, if $x\in A\setminus M,$ then x is a unit of A, and thus M = Mx. □

Remark 5.8.

Let A be a ring with Jacobson radical M, E be an A-module and $R=A\propto E.$ Then, each proper principal ideal of R has an OA-factorization if and only if one of the following conditions is satisfied.

1. A is a π-ring, E is cyclic and the annihilator of E is a $($ possibly empty$)$ product of idempotent maximal ideals of A.

2. A is local, M² is divided, $E=\mathbf{0}$ and either M is nilpotent or A is a domain with $\underset{n\in \mathbb{N}}{\cap }{M}^n=\mathbf{0}.$

3. A is local, $M^2=\mathbf{0},$ ME = aE for each nonzero $a\in M$ and ME = Mx for each $x\in E\setminus ME.$

Proof.

This can be proved along similar lines as Theorem 5.3(1) by using Corollary 4.3 and [7, Theorems 3.2(1) and 4.10]. □

Acknowledgments

We want to thank the referee for many helpful suggestions and comments which improved the quality of this paper. The first three authors dedicate this work to their Professor Faycal Lamrini for his retirement.

Additional information

Funding

The fourth-named author was supported by the Austrian Science Fund FWF, Project Number J4023-N35.