Abstract

A right -module is called a PS-module if its socle, , is projective. We investigate PS-modules over Ore extension and skew generalized power series extension. Let be an associative ring with identity, a unitary right -module, Ore extension, a right -module, a strictly ordered additive monoid, a monoid homomorphism, the skew generalized power series ring, and the skew generalized power series module. Then, under some certain conditions, we prove the following: (1) If is a right PS-module, then is a right PS-module. (2) If is a right PS-module, then is a right PS-module.

This paper is dedicated to Professor M. H. Fahmy on the occasion of his 68th birthday

1. Introduction

Throughout this paper denotes an associative ring with identity and a unitary right -module. According to Nicholson and Watters [1], is called a PS-module if every simple submodule is projective and equivalently if its socle, , is projective. Examples of PS-modules include nonsingular modules, regular modules in the sense of Zelmanowitz [2], and modules with zero socle. The class of PS-modules is closed under direct sums and submodules. In [3], Weimin proved that PS-modules are preserved by Morita equivalences and excellent extensions.

For any subset of , denote

Theorem 1 (see [3]). The following statements are equivalent for a right -module :(1) is a PS-module.(2)If is a maximal right ideal of , then either or , where .

A left PS-module is defined analogously. A ring is said to be a left PS-ring if is a PS-module. Every semiprime ring is a PS-ring. Every PP-ring is a PS-ring (where a ring is called PP-ring if every principal left ideal is projective). In particular every Baer ring is a PS-ring (where a ring is called Baer if every left (or right) annihilator is generated by an idempotent). A ring for which every simple singular module is injective is a PS-ring. If , then is a PS-ring. In fact for every maximal right ideal so .

The notion of PS-rings is not left-right symmetric (cf. [1]). A ring is duo if each one-sided ideal of is a two-sided ideal. As a generalization of left duo rings, a ring is called weakly left duo if for every there is a natural number such that is a two-sided ideal of . A ring is weakly duo if it is weakly right and left duo. In [3], Weimin proved that a duo ring is a PS-ring if and only if it is a right PS-ring. In [4], Dingguo generalized this result to weakly duo rings as follows: a weakly duo reduced ring is a PS-ring if and only if is a right PS-ring.

If is a PS-ring so also are and . The converse of this result is false by the following example.

Example 2 (see [1], Example 3.2). If , then and are PS-rings but is not PS-ring.

The motivation of this paper is to investigate the PS property of Ore extension modules and the skew generalized power series extension modules. These results generalize the corresponding results for polynomial rings, generalized power series rings, and modules [5, 6].

2. PS-Modules over Ore Extension Rings

This section is devoted to study the relationship between the PS property of a right -module and the PS property of the right Ore extension module .

Let be an endomorphism of and an -derivation of , that is, an additive map such thatIn case is the identity map, is called just a derivation of .

The Ore extension is the set of all polynomials with the usual sum and the following multiplication rule:We assume that is the identity element of . This means that and . This definition of noncommutative polynomial rings with identity was first introduced by Ore [7]. Ever since the appearance of Ore’s fundamental paper [7], Ore extensions have played an important role in noncommutative ring theory and many noncommutative ring theorists have investigated Ore extensions from different points of view such as ideal theory, order theory, Galois theory, and homological algebras.

For integers , with , will denote the map which is the sum of all possible words in , built with letters of and letters of . For instance,For any positive integer and , we have(see [8], Lemma 4.1). This formula uniquely determines a general product of (left) polynomials in and will be used freely in what follows.

Given a right -module , is a right -module with the natural action of on applying the above twist whenever necessary. The verification that this defines a valid -module structure on is almost identical to the verification that is a ring and it is straightforward (see [9]).

Definition 3 (see [9]). Given a module , an endomorphism and an -derivation . One says that is -compatible if for each , , one has . Moreover, One says that is -compatible if for each , , one has . If is both -compatible and -compatible, one says that is -compatible.

Note that if is -compatible (resp., -compatible), then is -compatible (resp., -compatible) for all . It is clear that is -compatible (resp., -compatible), then so is any submodule of . A ring is -compatible if and only if is an -compatible module.

As an immediate consequence of Definition 3, we obtain the following.

Lemma 4. Let be an -compatible module. For each and , one has the following:(1) if and only if for any positive integer .(2)If , then for all .

Lemma 5 (see [10], Lemma 2.5). Let be an -compatible module, , and . If , then for each .

Definition 6 (see [10]). Given a module , an endomorphism and an -derivation . One says that is -Armendariz if whenever and satisfy , one has for all , .

A ring is called -Armendariz if is an -Armendariz module.

Using Lemma 5 it is easy to deduce that if is -compatible and -Armendariz, then for any and , if and only if for all , .

Theorem 7. Let be an -compatible and -Armendariz module. If is a PS-module, then is a PS-module.

Proof. Let be a maximal right ideal of . We will show that either or , where . Let be the set of all coefficients of all polynomials in and let be the right ideal of generated by . If , then there exist and such thatSuppose that and , then for every , we haveSince is -compatible and -Armendariz, it follows thatConsequently, for every , , . Hence we geta contradiction. Then . Suppose that . We will show that is a maximal right ideal of . Let . If , then and so , a contradiction. Thus . Since is a maximal right ideal of ,It follows that there exist and such that . If , then and so . If , then which implies that . Hence is a maximal right ideal of . Since is a PS-module, it follows that either or , where . According to that we have the following two cases.
Case  1. Suppose that . We will show that . Let and ; then for every , we haveSince is -compatible and -Armendariz, it follows that for all and . Consequently, for every , , . For any , there exist and such that . Hencewhich implies that . Thus , a contradiction. Hence .
Case  2. Suppose that , where . We will show that , where . To show that , we need to prove that . If , then . Thus there exist and such that . If , then , a contradiction. If , then which implies that , a contradiction. Therefore which implies that . Now we show that . Suppose that ; then, for all , and so . We haveit follows that . Thus .

If , we get the following.

Corollary 8. Let be an -compatible and -Armendariz ring. If is a right PS-ring, then is a right PS-ring.

3. PS-Modules over Skew Generalized Power Series Rings

Let be an ordered commutative monoid. Unless stated otherwise, the operation of will be denoted additively, and the identity by 0. Recall that is artinian if every strictly decreasing sequence of elements of is finite and that is narrow if every subset of pairwise order-incomparable elements of is finite. The following construction is due to Zhongkui [11].

Let be a strictly ordered monoid (i.e., if and , then ), a ring, and a monoid homomorphism. Consider the set of all maps whose support (supp) is artinian and narrow.

For every and , letIt follows from ([12], 4.1) that is a finite set.

This fact allows defining the operation of multiplication (convolution) as follows:and if . With this operation and pointwise addition becomes a ring, which is called the ring of skew generalized power series with coefficients in and exponents in .

In [13], Zhao and Jiao generalized this construction to obtain the skew generalized power series modules over skew generalized power series rings as follows.

Let be a right -module; let be the set of all maps such that supp is artinian and narrow. With pointwise addition, is an abelian additive group. For each and , the setis finite (see [14], Lemma 1). This allows defining the scalar multiplication of the elements of by scalars from as follows:and if . With this operation and pointwise addition, one can easily show that is a right -module, which is called the module of skew generalized power series with coefficients in and exponents in .

For every if we set , the identity map of , then is the ring of generalized power series in the sense of Ribenboim [12] and is the untwisted module of generalized power series in the sense of [15].

For any we associated the map defined by

For any and , we define a map by

Definition 9 (see [13]). A right -module is called -compatible whenever if and only if for any , , and .

Clearly, is an -compatible ring if and only if is an -compatible -module.

Theorem 10. Let be a strictly totally ordered monoid which satisfies the condition for every and let be an -compatible module. If is a PS-module, then is a PS-module.

Proof. Let be a maximal right ideal of . We will show that either or , where . Since is a strictly totally ordered monoid, is a nonempty well-ordered subset of , for every . We denote by the smallest element of support .
For any , setLet be the right ideal of generated by . If , then there exist , , and such thatwhere and , for every . We will show that . Suppose that and . Then supp is a nonempty well-ordered subset of . Let ; ifthen This means that for some , a contradiction. ThusSince is an -compatible module, we getConsequentlya contradiction. Thus . Suppose that . We will show that is a maximal right ideal of . Let . If , then and so , a contradiction. Therefore . Since is a maximal right ideal of ,It follows that there exist and such that . Thus If , then . So, .
If , then . Since for every , . Thus , which implies that .
Hence is a maximal right ideal of . Since is a PS-module, it follows that either or , where . According to that we have the following two cases.
Case  1. Suppose that . We will show that . Let and . Then supp is a nonempty well-ordered subset of . Let . For any , there exist , , and such thatwhere and , for every . Since , , we get for every . Ifthen This means that for some , a contradiction. ThusSince is an -compatible module, we getConsequentlyTherefore, and . Thus , a contradiction. Hence .
Case  2. Suppose that , where . We will show that , where . To show that , we need to prove that . If , then . Thus there exist and such that . Thus If , then , a contradiction.
If , then . Since for every , . Thus , which implies that and . Hence , a contradiction. Therefore which implies that .
Conversely, suppose that and ; then and so . We claim that for any .
Suppose that for each . Consider the following element defined byThus . By hypothesis it is easy to see that . Thus . By analogy with the proof above, it follows thatwhich implies that . Thus our claim holds. ThereforeHence . Thus and the result follows since is an idempotent of .

As a special case of the last result if we set we get the following.

Corollary 11. Let be a strictly totally ordered monoid which satisfies the condition that for every and let be an -compatible ring. If is a right PS-ring, then is a right PS-ring.

If we set , for every , we get the following as a corollary.

Corollary 12 (see [6], Theorem 1). Let be a strictly totally ordered monoid which satisfies the condition that for every . If is a PS-module, then is a PS-module.

If , we get the following as a corollary.

Corollary 13 (see [5], Theorem 4). Let be a strictly totally ordered monoid which satisfies the condition that for every . If is a right PS-ring, then is a right PS-ring.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.