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 for some nonunit elements then either or 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 M 2 is universal.


Introduction
Throughout this article, all rings are commutative with nonzero identity and all modules are unital. Let N denote the set of positive integers. For m 2 N, let ½1, m ¼ fn 2 Nj1 n mg: Let R be a ring. An ideal I of R is said to be proper if I 6 ¼ R: The radical of I is denoted by ffiffi I p ¼ fx 2 Rjx n 2 I for some n 2 Ng: We denote by MinðIÞ 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 2 R and abc 2 I, then ab 2 I or ac 2 I or bc 2 I: In this case, ffiffi I p ¼ P is a prime ideal with P 2 I or ffiffi I p ¼ P 1 \ P 2 where P 1 , P 2 are incomparable prime ideals with P 1 P 2 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 , :::, a nþ1 2 R and Q nþ1 i¼1 a i 2 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], ð ffiffi I p Þ n 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 Q r i¼1 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 Q r i¼1 I i of proper n-absorbing ideals of R. Ahmed et al. called AF-dimðRÞ (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-dimðRÞ ¼ 1: An FAF-ring (finite absorbing factorization ring) is a ring such that AF-dimðRÞ < 1: Recall that a general ZPI-ring is a ring whose proper ideals can be written as a product of prime ideals. Therefore, AF-dimðRÞ measures, in some sense, how far R is from being a general ZPI-ring, cf. [1,Proposition 3]. By dimðRÞ, 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 OAideal) if whenever we take nonunit elements a, b, c 2 R with abc 2 I, then ab 2 I or c 2 I: In this case, ffiffi I p ¼ P is a prime ideal, cf. [25,Theorem 2.3]. And if R is a ring in which exists an OAideal 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 Q n i¼1 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 OAfactorization.
We call R a 1-absorbing prime factorization ring (OAF-ring) if every proper ideal has an OAfactorization. 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 OAideal, 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 OAFrings 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.

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 I 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.
(2) Let R be local. Then, M is the maximal ideal of R.
()) Let I be an OA-ideal such that I is not a prime ideal. Since I is proper, we infer that I M: Since I is not prime, there are a, b 2 M n I such that ab 2 I: To prove that M 2 I, it suffices to show that xy 2 I for all x, y 2 M: Let x, y 2 M: Then, xyab 2 I: Since xy, a, b 2 M, b 6 2 I and I is an OA-ideal, it follows that xya 2 I: Again, since x, y, a 2 M, a 6 2 I and I is an OA-ideal, we have that xy 2 I: (() Clearly, if I is a prime ideal, then I is an OA-ideal. Now let M 2 I M: Then, I is proper. Let a, b, c 2 M be such that abc 2 I: Then, ab 2 M 2 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 2 R be such that abc 2 I: We have to show that ab 2 I or ac 2 I or bc 2 I: First let a or b or c be a unit of R. Without restriction let a be unit of R. Since abc 2 I, we infer that bc 2 I: Now let a, b, and c be nonunits. Then, ab 2 I or c 2 I: If c 2 I, then ac 2 I: The in particular statement is clear. w Proposition 2.2. Let R be an OAF-ring and I be a proper ideal of R. Then, R/I is an OAF-ring.
It suffices to show that J i =I is an OA-ideal for each i 2 ½1, m: Let i 2 ½1, m and let a, b, c 2 R be such that a, b, c are three nonunit elements of R/I and a b c 2 J i =I: Clearly, a, b, c are nonunit elements of R and abc 2 J i : Since J i is an OA-ideal of R, we get that ab 2 J i or c 2 J i which implies that a b 2 J i =I or c 2 J i =I: Therefore, R/I is an OAF-ring. w Proposition 2.3. Let S be a multiplicatively closed subset of R n 0. If R is an OAF-ring, then S À1 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 À1 R: Then, J ¼ S À1 I for some proper ideal I of R with I \ S ¼ ;: [25,Theorem 2.18]. Thus, S À1 R is an OAF-ring. The in particular statement is clear. w Let R be a ring. Then, R is said to be a p-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. Proof. This is an immediate consequence of Lemma 2.1(1).
w In the light of the above remark, we give the next result.
Corollary 2.5. Let R 1 and R 2 be two rings and R ¼ R 1 Â R 2 be their direct product. The following statements are equivalent.
w Let R be a ring. Then, R is called a von Neumann regular ring if for each x 2 R there is some y 2 R with x ¼ 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.
(2) R is a Noetherian von Neumann regular ring.
(3) R is a finite direct product of fields.
Proof. Observe that the polynomial ring R½X is never local, since X and 1 À X are nonunit elements of R½X, but their sum is a unit. Consequently, R½X is an OAF-ring if and only if R½X 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.
w 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 M 2 is divided. The following statements are equivalent.
(1) Each two principal OA-ideals which contain M 2 are comparable.
(2) For each OA-ideal I of R, we have that I is a prime ideal or I ¼ M 2 : Proof. (1) ) (2): Let I be an OA-ideal of R such that I is not a prime ideal of R. Then, M 2 I & M by Lemma 2.1 (2). Assume that M 2 & I: Let x 2 I n M 2 and let y 2 M n I: Then, x, y 6 2 M 2 , and thus, M 2 xR, yR (since M 2 is divided). It follows that xR and yR are (principal) OA-ideals of R by Lemma 2.1(2). Since y 6 2 xR and xR and yR are comparable, we infer that xR & yR: Consequently, there is some z 2 M such that x ¼ yz, and hence x 2 M 2 , a contradiction. Therefore, I ¼ M 2 : (2) ) (1): This is obvious.
w Let R be a ring. An ideal I of R is called 2-generated if I ¼ xR þ yR for some (not necessarily distinct) x, y 2 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.
(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 p-ring, then R is a general ZPI-ring. Let R be an arithmetical p-ring and let M be a maximal ideal of R. It is straightforward to show that R M is a p-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 dimðR M Þ 1 by [21, page 205]. We infer that dimðRÞ 1, and thus R is a general ZPI-ring by [ (1) ) (2) ) (3): This is obvious.
(3) ) (1): It is sufficient to show that R is a p-ring. If R is not local, then R is a p-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, M 2 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 p-ring.

Preparational results
From Lemma 2.1(3), we have that jMinðIÞj ¼ 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 MinðIÞ is finite.
MinðI i Þ, and thus jMinðIÞj n: w 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 I, there is some ideal L of R such that J ¼ 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 2 P and let xR ¼ Q n i¼1 I i be an OA-factorization. There is some j 2 ½1, n such that I j P: By Lemma 2.1(2), we have that P ¼ I j , and hence xR ¼ PJ for some ideal J of R. We infer that xR ¼ PðxR : PÞ: Now let I be an ideal of R such that I P: Then, I ¼ P y2I yR ¼ P y2I PðyR : PÞ ¼ P P y2I ðyR : PÞ, and thus, P is a multiplication ideal.
w 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 dimðRÞ ! 1 and every proper principal ideal of R has an OA-factorization. Then, R is an integral domain and if dimðRÞ ! 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 Minð0Þ is finite and each P 2 Minð0Þ is principal.
Claim: Every proper principal ideal of R/N has an OA-factorization. Let I be a proper principal ideal of R/N.
It suffices to show that I i =N is an OAideal of R/N for each i 2 ½1, n: Let i 2 ½1, n: If I i is a prime ideal of R, then N 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 I i M: Note that R/N is local with maximal ideal M/N. Since ðM=NÞ 2 ¼ M 2 =N I i =N 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 p-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 ¼ M 2 [ [ Q2Minð0Þ Q: Next we prove that M 2 xR for each x 2 R n L: Let x 2 R n L: Without restriction let x be a nonunit. Note that xR cannot be a product of more than one OAideal, and hence, xR is an OA-ideal. By Lemma 2.1(2), we have that M 2 xR: Now we show that P M 2 for each P 2 Minð0Þ: Let P 2 Minð0Þ: Assume that P 6 M 2 : Let w 2 R n P: Then, P þ wR 6 L by the prime avoidance lemma, and thus there is some v 2 ðP þ wRÞ n L: It follows that M 2 vR P þ wR: Since P is a nonmaximal prime ideal, we have that R/P has no simple R/P-submodules, and hence \ y2RnP ðP þ yRÞ ¼ P: (Note that if \ y2RnP ðP þ yRÞ 6 ¼ P, then \ y2RnP ðP þ yRÞ=P is a simple R/P-submodule of R/P.) This implies that M 2 \ y2RnP ðP þ yRÞ ¼ P, and thus P ¼ M, a contradiction.
Let Q 2 Minð0Þ: By the prime avoidance lemma, there is some z 2 M n L: by Nakayama's lemma), and hence R is an integral domain.
Case 2: dimðRÞ ! 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 2 Q n [ P2Minð0Þ P: Since R is reduced, we have that \ L2Minð0Þ L ¼ 0: If y 2 R is nonzero with xy ¼ 0, then y 6 2 L and xy 2 L for some L 2 Minð0Þ, and hence x 2 L, a contradiction. We infer that x is a regular element of R. Let xR ¼ Q n i¼1 I i be an OA-factorization. Then, I j Q for some j 2 ½1, 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 Q and Q 6 ¼ M, we have that I j is a prime ideal by Lemma 2.1 (2). Consequently, P I j for some P 2 Minð0Þ: Since I j is regular, we infer that P & I j , and hence P ¼ PI j (since I j is principal). It follows (e.g. from Nakayama's lemma) that P ¼ 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 dimðRÞ ! 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 2 L be nonzero. Let zR ¼ Q m k¼1 J k be an OA-factorization. Then, J ' L for some ' 2 ½1, m: Since R is an integral domain, zR is invertible, and hence, J ' is invertible. Therefore, J ' is nonzero and principal (since R is local). Since L 6 ¼ M, it follows from Lemma 2.1(2) that J ' is a prime ideal.
Case 3: dimðRÞ ! 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 dimðR=NÞ ! 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 dimðR=NÞ ! 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 2 Q such that Q ¼ qR þ N: Let qR ¼ Q n i¼1 I i be an OA-factorization. Then, I j Q for some j 2 ½1, n: Since Q 6 ¼ M, we infer by Lemma 2.1(2) that I j is a prime ideal of R. Therefore, Q ¼ qR þ N I j Q, and hence I j ¼ Q.
Assume that Q 6 ¼ qR: Then, qR ¼ QJ for some proper ideal J of R. It follows that q 2 qR ¼ ðqR þ NÞJ qJ þ N, and thus qð1 À aÞ 2 N for some a 2 J: Since a is a nonunit of R, we obtain that q 2 N: This implies that Q ¼ qR þ N ¼ N, a contradiction. We infer that Q ¼ qR. Since N & Q and N is a prime ideal of R, we have that N ¼ NQ. Consequently, N ¼ 0 (e.g. by Nakayama's lemma, since N is principal), and thus R ffi R=N is a unique factorization domain. w 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, dimðRÞ 2 and each nonmaximal prime ideal of R is principal.
Proof. First we show that dimðR P Þ 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 ¼ J P for some 2-generated ideal J of R with J P: Let J ¼ Q n i¼1 J i be an OA-factorization. Then, , n is such that J i P, then J i is a prime ideal of R by Lemma 2.1 (2), and thus ðJ i Þ 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 dimðR P Þ 1: This implies that dimðRÞ 2: It remains to show that every nonmaximal prime ideal of R is principal. Without restriction let dimðRÞ ! 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. w 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, dimðRÞ 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, dimðRÞ 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 ¼ 0: Let Q be a nonmaximal prime ideal of R.
Assume that Q 6 M 2 : Since dimðRÞ 2 by Proposition 3.4, there is some prime ideal P of R such that Q P and dimðR=PÞ ¼ 1: Next we show that M 2 P þ yR for each y 2 R n P: Let y 2 R n P and set J ¼ P þ yR: Without restriction let J & M: Note that J is 2-generated by Proposition 3.4. Since J 6 M 2 , J cannot be a product of more than one OA-ideal, and thus, J is an OA-ideal of R. Since P & J & M, we have that J is not a prime ideal of R, and thus M 2 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 ¼ \ x2RnP ðP þ xRÞ: (Observe that if \ x2RnP ðP þ xRÞ 6 ¼ P, then \ x2RnP ðP þ xRÞ=P is a simple R/P-submodule of R/P.) Therefore, M 2 \ x2RnP ðP þ xRÞ ¼ P, and hence P ¼ M, a contradiction. We infer that Q M 2 : There is some z 2 M n 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 M 2 for each nonmaximal prime ideal L of R, and thus zR is not a nonmaximal prime ideal. Consequently, Q & M 2 & zR by Lemma 2.1(2), and hence Q ¼ zQ. Since Q is principal by Proposition 3.4, it follows (e.g. by Nakayama's lemma) that Q ¼ 0: Assume that M 2 is principal. Then, M is invertible, and hence M is principal (since D is local). Note that D is a DVR (since dimðDÞ ¼ 1), and hence, D is a unique factorization domain, a contradiction.
We infer that M 2 is not principal. We show that D is atomic. Let y 2 D be a nonzero nonunit. Then, yD ¼ Q n i¼1 I i for some principal OA-ideals I i . There are nonzero nonunits x i 2 D such that y ¼ Q n i¼1 x i and I j ¼ x jD for each j 2 ½1, n: Let i 2 ½1, 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 I i : Since M 2 is not principal, we have that x i 6 2 M 2 : Therefore, x i is irreducible. Finally, let z 2 D be irreducible. Then, zD ¼ Q m j¼1 J j for some principal OA-ideals J j . Since zD is maximal among the proper principal ideals of D, we obtain that zD ¼ J j for some j 2 ½1, n: (() 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 ¼ xD for some nonzero nonunit x 2 D: Observe that x ¼ Q n i¼1 x i for some irreducible elements x i 2 D: It follows that Q n i¼1 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 6 ¼ M: Then, D is a unique factorization domain by Proposition 3.3, and hence, P is principal. Therefore, \ n2N P n is a prime ideal of D by [5, Theorem 2.2(1)]. Since \ n2N P n & P, we infer that \ n2N P n ¼ 0: Now let P ¼ M. Assume that \ n2N M n 6 ¼ 0 and let x 2 \ n2N 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 xD, and hence, M 2m xD M 4m M 2m : This implies that xD ¼ M 2m ¼ M 4m ¼ x 2 D, and thus, x is a unit of D, a contradiction. Therefore, \ n2N M n ¼ 0: w Lemma 3.7. Let R be a local ring with maximal ideal M such that M 2 is divided and such that either M is nilpotent or R is an integral domain with \ n2N M n ¼ 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 ¼ 0, and hence, R is a field and both statements are clearly satisfied. Now let M be not idempotent. There is some x 2 M n M 2 : In what follows, we freely use the fact that if N is an ideal of R and z 2 R such that N zR, then N ¼ zðN : zRÞ, and hence, N ¼ zJ for some ideal J of R.
Next we prove that M 2 ¼ xM and xR is an OA-ideal of R. Since x 6 2 M 2 and M 2 is divided, we have that M 2 xR M: Therefore, xR is an OA-ideal by Lemma 2.1(2). Since M 2 & xR, there is some proper ideal J of R with M 2 ¼ xJ, and thus M 2 xM: Obviously, xM M 2 , and hence M 2 ¼ xM: Now we show that R is an OAF-ring. Let I be a proper ideal of R. First let I ¼ 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 M n : Observe that I M n ¼ x nÀ1 M x nÀ1 R: Consequently, I ¼ x nÀ1 L ¼ ðxRÞ nÀ1 L for some proper ideal L of R. Assume that L M 2 : Note that L M 2 ¼ xM xR: This implies that L ¼ xA for some proper ideal A of R, and hence I ¼ x n A x n M ¼ M nþ1 , a contradiction. We infer that M 2 L (since M 2 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 2 M: First let y ¼ 0. If M is nilpotent, then x k ¼ 0 for some k 2 N, and thus yR ¼ ðxRÞ 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 ' 2 N such that y 2 M ' : Therefore, y ¼ x 'À1 z for some z 2 M: If z 2 M 2 , then z ¼ xv for some v 2 M, and hence y ¼ x ' v 2 M 'þ1 , a contradiction. We infer that z 6 2 M 2 , and thus M 2 zR M: It follows from Lemma 2.1(2) that zR is an OA-ideal of R. Consequently, yR ¼ ðxRÞ 'À1 ðzRÞ is a product of principal OA-ideals.

Characterization of OAF-rings and related concepts
First, we recall several definitions and discuss the factorization theoretical properties of local onedimensional OAF-domains. Let D be an integral domain with quotient field K. Then,D ¼ fx 2 Kj there is some nonzero c 2 D such that cx n 2 D for all n 2 Ng is called the complete integral closure of D. Let ðD :DÞ ¼ fx 2 D j xD Dg be the conductor of D inD: The domain D is called completely integrally closed if D ¼D and D is said to be seminormal if for all x 2 K such that x 2 , x 3 2 D, it follows that x 2 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 thatD is a DVR and ðD :DÞ 6 ¼ 0: For each subset, X K let X À1 ¼ fx 2 K j xX Dg and X v ¼ ðX À1 Þ À1 : An ideal I of D is called divisorial if I v ¼ 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.  (5) þ (6) ) (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 2 D, and hence, M 2 yD P: Therefore, P ¼ M, and thus, D is one-dimensional. It remains to show thatD is a DVR and ðD :   (1) R is an OAF-ring.
(3) ) (4): First let each OA-ideal of R be a prime ideal. Then, R is a p-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 2 M n 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 2 M n M 2 : Then, xR is an OA-ideal. We infer by Lemma 2.1(2) that M 2 xR: Consequently, M 2 is divided. It follows from Lemma 2.1 that M 2 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 \ n2N M n ¼ 0 by Lemma 3.6. It remains to show that M 2 is divided. Let x 2 R n 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 xR: (4) ) (1): Clearly, every general ZPI-ring is an OAF-ring. The rest follows from Lemma 3.7. w (1) Each proper principal ideal of R has an OA-factorization.
(2) R is a p-ring or an OAF-ring.
(3) R satisfies one of the following conditions.
(A) R is a p-ring.
(B) R is a local domain, M 2 is divided and \ n2N M n ¼ 0: (C) R is local, M 2 is divided and M is nilpotent.
Proof. (1) ) (2): If R is not local, then R is a p-ring by Remark 2.4 (2). Now let R be local. If dimðRÞ ! 2, then R is a unique factorization domain by Proposition 3.3, and hence, R is a p-ring. If dimðRÞ 1, then R is an OAF-ring by Theorem 4.2.
(2) () (3): This is an immediate consequence of Theorem 4.2 and the fact that every general ZPI-ring is a p-ring.  (2) ) (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.
w 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 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. (1) Every proper ideal of R is an OA-ideal.
(2) Every proper principal ideal of R is an OA-ideal.
(2) ) (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 2 R is a nonunit, then x 2 R is a prime ideal, and hence x 2 R ¼ xR and x ¼ 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 ¼ 0: Assume that M 2 6 ¼ 0: Then, R is an integral domain and there is some nonzero x 2 M 2 : It follow from Lemma 2.1(2) that x 2 R is a prime ideal or M 2 x 2 R: If x 2 R is a prime ideal, then x 2 R ¼ xR: If M 2 x 2 R, then M 2 x 2 R xR M 2 , and thus x 2 R ¼ xR: In any case, we have that x 2 R ¼ xR, and hence, x is a unit (since x is regular), a contradiction.

OA-factorization properties and trivial ring extensions
Let A be a ring and E be an A-module. Then, A / E, the trivial (ring) extension of A by E, is the ring whose additive structure is that of the external direct sum AE and whose multiplication is defined by ða, eÞðb, f Þ ¼ ðab, af þ beÞ for all a, b 2 A and all e, f 2 E: (This construction is also known by other terminology and other notation, such as the idealization AðþÞE:) 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 ¼ aE for each regular element a 2 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 ¼ A / E be the trivial ring extension of A by E.
(1) I / 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 / E where L is an OA-ideal of A. Proof.
(2) Let J be an OA-ideal of R. Our aim is to show that 0 / E J: Let e 2 E and let a 2 A be a nonunit regular element. Then, e ¼ af for some f 2 E and thus ða, 0Þð0, f Þð0, eÞ ¼ ð0, 0Þ 2 J: Since J is an OA-ideal, we conclude that ða, 0Þð0, f Þ ¼ ð0, eÞ 2 J or ð0, eÞ 2 J which implies that 0 / E J: Therefore, J ¼ L / E with L ¼ fb 2 Ajðb, gÞ 2 J for some g 2 Eg and L is an ideal of A by [7, Theorems 3.1 and 3.3 (1)]. Now the result follows from (1). w Corollary 5.2. Let A be an integral domain that is not a field, E be a divisible A-module and R ¼ A / E. Then, the OA-ideals of R have the form I / E where I is an OA-ideal of A.
Next, we study the transfer of the OAF-ring property to the trivial ring extension. (1) R is an OAF-ring if and only if one of the following conditions is satisfied.
(A) A is a general ZPI-ring, E is cyclic and the annihilator of E is a ðpossibly emptyÞ product of idempotent maximal ideals of A. (B) A is local, M 2 is divided, E ¼ 0 and either M is nilpotent or A is a domain with \ n2N M n ¼ 0: (C) A is local, M 2 ¼ 0, ME¼ aE for each nonzero a 2 M and ME ¼ Mx for each x 2 E n 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 ¼ 0 and ME ¼ 0: Proof. (1) ()) First let R be an OAF-ring. By Theorem 4.2, it follows that (a) R is a general ZPIring or (b) R is local with maximal ideal N, N 2 is divided and (N is nilpotent or R is a domain such that \ n2N N n ¼ 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 N 2 is divided. Observe that A is local with maximal ideal M and N ¼ M / E by [7, Theorem 3.2(1)]. If R is a domain such that \ n2N N n ¼ 0, then E ¼ 0 (for if z 2 E is nonzero, then ð0, zÞ is a nonzero zero-divisor of R), and hence A ffi R is a domain, M 2 is divided and \ n2N M n ¼ 0: Now let N be nilpotent. If E ¼ 0, then A ffi R, and thus M 2 is divided and M is nilpotent. From now on let E be nonzero. There is some k 2 N such that N k ¼ 0: Note that N 2 ¼ M 2 / ME and N k ¼ M k / M kÀ1 E, and thus M k ¼ 0: Since N 2 is divided, we have that 0 / E N 2 or N 2 0 / E: If 0 / E N 2 , then E ¼ ME, and hence E ¼ M k E ¼ 0, a contradiction. Therefore, N 2 0 / E, which implies that M 2 ¼ 0: Let a 2 M be nonzero. Then, ða, 0Þ 6 2 N 2 , and hence N 2 ða, 0ÞR ¼ aA / aE: Consequently, ME aE, and thus ME ¼ aE. Finally, let x 2 E n ME: Then, ð0, xÞ 6 2 N 2 : We infer that N 2 ð0, xÞR ¼ 0 / Ax: This implies that ME Ax: If ME 6 Mx, then bx 2 ME for some unit b 2 A, and hence x 2 ME, a contradiction. It follows that ME Mx, which clearly implies that ME ¼ Mx.
(() 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 OAFring by Theorem 4.2, and hence R ffi A is an OAF-ring. Now let condition (C) be satisfied. Set N ¼ M / 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 N 2 is divided. Since M 2 ¼ 0, we obtain that N 3 ¼ M 3 / M 2 E ¼ 0, and thus N is nilpotent. It remains to show that N 2 ða, xÞR for each ða, xÞ 2 R n N 2 : Let a 2 A and x 2 E be such that ða, xÞ 6 2 N 2 : Since N 2 ¼ 0 / ME, we have to show that 0 / ME ða, xÞR: 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 2 0 / ME: Then, z ¼ ð0, yÞ for some y 2 ME: Case 1: a is a nonzero nonunit. Since ME ¼ aE, there is some v 2 E such that y ¼ av. Observe that z ¼ ð0, avÞ ¼ ða, xÞð0, vÞ 2 ða, xÞR: Case 2: a ¼ 0. Then, x 2 E n ME (since ða, xÞ 6 2 N 2 ). Since ME ¼ Mx, there is some b 2 M such that y ¼ bx. It follows that z ¼ ð0, bxÞ ¼ ða, xÞðb, 0Þ 2 ða, xÞR: 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 ¼ 0: It follows that A is local with maximal ideal M and N ¼ M / E by [7, Theorem 3.2(1)]. Moreover, 0 ¼ N 2 ¼ M 2 / ME, and hence M 2 ¼ 0 and ME ¼ 0: Conversely, let A be local, M 2 ¼ 0 and ME ¼ 0: Set N ¼ M / E: Then, R is local with maximal ideal N by [7, Theorem 3.2(1)] and N 2 ¼ M 2 / ME ¼ 0: We infer by Proposition 4.5 that each proper ideal of R is an OA-ideal. (1) R is an OAF-ring.
(2) A is a field.
(3) Every proper ideal of R is an OA-ideal.
Proof. (1) ) (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 ¼ 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 ¼ 0: Since A is an integral domain, it follows that M ¼ 0, and hence A is a field.
(3) ) (1): This is obvious. w Remark 5.5. In general, if A is an OAF-ring and E is an A-module, then A / 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 / 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 ¼ 0. Set R ¼ A / E. The following statements are equivalent.
Proof. (1) ) (2): Assume that M 2 6 ¼ 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 ¼ M ¼ 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 ¼ M ¼ 0 (since M is idempotent). In any case, we obtain that M 2 ¼ 0, a contradiction. (1) A is a p-ring, E is cyclic and the annihilator of E is a ð possibly emptyÞ product of idempotent maximal ideals of A. (2) A is local, M 2 is divided, E ¼ 0 and either M is nilpotent or A is a domain with \ n2N M n ¼ 0: (3) A is local, M 2 ¼ 0, ME ¼ aE for each nonzero a 2 M and ME ¼ Mx for each x 2 E n ME: Proof. This can be proved along similar lines as Theorem 5.