# PS-Modules over Ore Extensions and Skew Generalized Power Series Rings

Authors: Refaat M. Salem; Mohamed A. Farahat; Hanan Abd-Elmalk
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

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## Abstract

A rightR-module MR is called a PS-module if its socle, SocMR, is projective. We investigate PS-modules over Ore extension and skew generalized power series extension. Let R be an associative ring with identity, MR a unitary right R-module, O=Rx;α,δ Ore extension, MxO a right O-module, S,≤ a strictly ordered additive monoid, ω:S→EndR a monoid homomorphism, A=RS,≤,ω the skew generalized power series ring, and BA=MS,≤RS,≤,ω  the skew generalized power series module. Then, under some certain conditions, we prove the following: (1) If MR is a right PS-module, then MxO is a right PS-module. (2) If MR is a right PS-module, then BA is a right PS-module.


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## Body

## 1. Introduction

Throughout this paperR denotes an associative ring with identity and MR a unitary right R-module. According to Nicholson and Watters [1], MR is called aPS-module if every simple submodule is projective and equivalently if its socle, SocMR, 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 subsetX of R, denote(1)lMX=m∈M∣mX=0.Theorem 1 (see [3]).
The following statements are equivalent for a rightR-module MR:(1)
MR is a PS-module.(2)
IfL is a maximal right ideal of R, then either lML=0 or L=eR, where e2=e∈R.A left PS-moduleMR is defined analogously. A ring R is said to be aleft PS-ring if RR is a PS-module. Every semiprime ring is a PS-ring. Every PP-ring is a PS-ring (where a ring R is called PP-ring if every principal left ideal is projective). In particular every Baer ring is a PS-ring (where a ring R 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 lRJR=0, then R is a PS-ring. In fact JR⊂L for every maximal right ideal so lRL=0.The notion of PS-rings is not left-right symmetric (cf. [1]). A ring R isduo if each one-sided ideal of R is a two-sided ideal. As a generalization of left duo rings, a ring R is calledweakly left duo if for every r∈R there is a natural number nr such that Rrnr is a two-sided ideal of R. A ring R is weakly duo if it is weakly right and left duo. In [3], Weimin proved that a duo ring R 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 R is a PS-ring if and only if R is a right PS-ring.IfR is a PS-ring so also are Rx and Rx. The converse of this result is false by the following example.Example 2 (see [1], Example 3.2).
IfR=Z4, then Rx and Rx are PS-rings but R 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 rightR-module MR and the PS property of the right Ore extension module MxOMxRx;α,δ.Letα be an endomorphism of R and δ:R→R an α-derivation of R, that is, an additive map such that(2)δab=δab+αaδb,∀a,b∈R.In case α is the identity map, δ is called just a derivation of R.The Ore extensionO=Rx;α,δ is the set of all polynomials ∑i=0naixi with the usual sum and the following multiplication rule:(3)xa=αax+δa.We assume that 1 is the identity element of O=Rx;α,δ. This means that α1=1 and δ1=0. 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 integersi, j with j≥i≥0, fij∈EndR,+ will denote the map which is the sum of all possible words in α, δ built with i letters of α and j-i letters of δ. For instance,(4)f00=IdR,fjj=αj,f0j=δj,fj-1j=αj-1δ+αj-2δα+⋯+δαj-1.For any positive integer n and r∈R, we have(5)xnr=∑i=0nfinrxi(see [8], Lemma 4.1). This formula uniquely determines a general product of (left) polynomials in O and will be used freely in what follows.Given a rightR-module MR, MxO is a right O-module with the natural action of O on Mx applying the above twist whenever necessary. The verification that this defines a valid O-module structure on Mx is almost identical to the verification that O=Rx;α,δ is a ring and it is straightforward (see [9]).Definition 3 (see [9]).
Given a moduleMR, an endomorphism α:R→R and an α-derivation δ:R→R. One says that MR is α-compatible if for each m∈MR, r∈R, one has mr=0⇔mαr=0. Moreover, One says that MR is δ-compatible if for each m∈MR, r∈R, one has mr=0⇒mδr=0. If MR is both α-compatible and δ-compatible, one says that MR is α,δ-compatible.Note that ifMR is α-compatible (resp., δ-compatible), then MR is αi-compatible (resp., δi-compatible) for all i≥1. It is clear that MR is α-compatible (resp., δ-compatible), then so is any submodule of MR. A ring R is α,δ-compatible if and only if RR is an α,δ-compatible module.As an immediate consequence of Definition3, we obtain the following.Lemma 4.
LetMR be an α,δ-compatible module. For each m∈M and a∈R, one has the following:(1)
ma=0 if and only if mαna=0 for any positive integer n.(2)
Ifma=0, then mfija=0 for all j≥i≥0.Lemma 5 (see [10], Lemma 2.5).
LetMR be an α,δ-compatible module, mx=m0+⋯+mkxk∈Mx, and r∈R. If mxr=0, then mir=0 for each i.Definition 6 (see [10]).
Given a moduleMR, an endomorphism α:R→R and an α-derivation δ:R→R. One says that MR is α,δ-Armendariz if whenever mx=∑i=0kmixi∈Mx and fx=∑j=0najxj∈Rx;α,δ satisfy mxfx=0, one has mixiajxj=0 for all i, j.A ringR is called α,δ-Armendariz if RR is an α,δ-Armendariz module.Using Lemma5 it is easy to deduce that if MR is α,δ-compatible and α,δ-Armendariz, then for any mx=∑i=0kmixi∈Mx and fx=∑j=0najxj∈Rx;α,δ, mxfx=0 if and only if miaj=0 for all i, j.Theorem 7.
LetMR be an α,δ-compatible and α,δ-Armendariz module. If MR is a PS-module, then MxO is a PS-module.Proof.
LetL be a maximal right ideal of O. We will show that either lMxL=0 or L=hO, where h2=h∈O. Let I be the set of all coefficients of all polynomials in L and let J be the right ideal of R generated by I. If J=R, then there exist a1,…,an∈I and r1,…,rn∈R such that(6)1=a1r1+⋯+anrn.Suppose that φx=∑i=0kmixi∈lMxL and φ≠0, then for every gx=∑j=0najxj∈L, we have(7)φxgx=∑i=0kmixi∑j=0najxj=0.Since MR is α,δ-compatible and α,δ-Armendariz, it follows that(8)miaj=0,∀0≤i≤k,0≤j≤n.Consequently, for every a∈I, mia=0, 0≤i≤k. Hence we get(9)mi=mi1=mia1r1+⋯+anrn=mia1r1+⋯+mianrn=0,a contradiction. Then lMxL=0. Suppose that J≠R. We will show that J is a maximal right ideal of R. Let r∈R-J. If r∈L, then r∈I and so r∈J, a contradiction. Thus r∉L. Since L is a maximal right ideal of O,(10)O=L+rO.It follows that there exist gx=∑i=0naixi∈L and hx=∑j=0mbjxj∈O such that 1=a0+rb0. If a0=0, then 1=rb0∈rR and so R=J+rR. If a0≠0, then a0∈I⊆J which implies that R=J+rR. Hence J is a maximal right ideal of R. Since MR is a PS-module, it follows that either lMJ=0 or J=eR, where e2=e∈R. According to that we have the following two cases.
Case  1. Suppose that lMJ=0. We will show that lMxL=0. Let φx=∑i=0kmixi∈lMxL and φ≠0; then for every gx=∑j=0najxj∈L, we have(11)φxgx=∑i=0kmixi∑j=0najxj=0.Since MR is α,δ-compatible and α,δ-Armendariz, it follows that miaj=0 for all 0≤i≤k and 0≤j≤n. Consequently, for every a∈I, mia=0, 0≤i≤k. For any r∈J, there exist a1,…,an∈I and r1,…,rn∈R such that r=a1r1+⋯+anrn. Hence(12)mir=mia1r1+⋯+anrn=mia1r1+⋯+mianrn=0,which implies that mi∈lMJ=0. Thus φx=0, a contradiction. Hence lMxL=0.
Case  2. Suppose that J=eR, where e2=e∈R. We will show that L=eO, where e2=e∈O. To show that eO⊆L, we need to prove that e∈L. If e∉L, then O=L+eO. Thus there exist gx=∑i=0naixi∈L and hx=∑j=0mbjxj∈O such that 1=a0+eb0. If a0=0, then 1=eb0∈eR=J, a contradiction. If a0≠0, then a0∈I⊆J which implies that 1=a0+eb0∈J+eR=J, a contradiction. Therefore e∈L which implies that eO⊆L. Now we show that L⊆eO. Suppose that gx=∑i=0naixi∈L; then, for all 0≤i≤n, ai∈I⊆J=eR and so ai=eai. We have(13)egx=e∑i=0naixi=∑i=0neaixi=∑i=0naixi=gx;it follows that gx∈eO. Thus L⊆eO.IfMR=RR, we get the following.Corollary 8.
LetR be an α,δ-compatible and α,δ-Armendariz ring. If R is a right PS-ring, then Rx;α,δ is a right PS-ring.


## 3. PS-Modules over Skew Generalized Power Series Rings

LetS,≤ be an ordered commutative monoid. Unless stated otherwise, the operation of S will be denoted additively, and the identity by 0. Recall that S,≤ is artinian if every strictly decreasing sequence of elements of S is finite and that S,≤ is narrow if every subset of pairwise order-incomparable elements of S is finite. The following construction is due to Zhongkui [11].LetS,≤ be a strictly ordered monoid (i.e., if s,s′,t∈S and s<s′, then s+t<s′+t), R a ring, and ω:S→EndR a monoid homomorphism. Consider the set A=RS,≤,ω of all maps f:S→R whose support (suppf=s∈S∣fs≠0) is artinian and narrow.For everys∈S and f,g∈A, let(14)Xsf,g=u,v∈S×S∣u+v=s;fu≠0,gv≠0.It follows from ([12], 4.1) that Xsf,g is a finite set.This fact allows defining the operation of multiplication (convolution) as follows:(15)fgs=∑u,v∈Xsf,gfuωugv,and fgs=0 if Xsf,g=ϕ. With this operation and pointwise addition A=RS,≤,ω becomes a ring, which is called thering of skew generalized power series with coefficients in R and exponents in S.In [13], Zhao and Jiao generalized this construction to obtain the skew generalized power series modules over skew generalized power series rings as follows.LetMR be a right R-module; let B be the set of all maps φ:S→M such that suppφ=s∈S∣φs≠0} is artinian and narrow. With pointwise addition, B=MS,≤ is an abelian additive group. For each f∈A=RS,≤,ω and φ∈B, the set(16)Xsφ,f=u,v∈S×S∣u+v=s;φu≠0,fv≠0is finite (see [14], Lemma 1). This allows defining the scalar multiplication of the elements of B by scalars from A as follows:(17)φfs=∑u,v∈Xsφ,fφuωufv,and φfs=0 if Xsφ,f=ϕ. With this operation and pointwise addition, one can easily show that B is a right A-module, which is called themodule of skew generalized power series with coefficients in M and exponents in S.For everys∈S if we set ωs=IdR∈AutR⊂EndR, the identity map of R, then A=RS,≤,ω=RS,≤ is the ring of generalized power series in the sense of Ribenboim [12] and B=MS,≤ is the untwisted module of generalized power series in the sense of [15].For anyr∈R we associated the map cr∈A defined by(18)crx=r,ifx=0,0,ifx≠0.For anym∈M and s∈S, we define a map dms∈B by(19)dmsx=m,ifx=s,0,ifx≠s.Definition 9 (see [13]).
A rightR-module MR is called ω-compatible whenever ma=0 if and only if mωsa=0 for any s∈S, m∈M, and a∈R.Clearly,R is an ω-compatible ring if and only if RR is an ω-compatible R-module.Theorem 10.
LetS,≤ be a strictly totally ordered monoid which satisfies the condition 0≤s for every s∈S and let MR be an ω-compatible module. If MR is a PS-module, then BA is a PS-module.Proof.
LetL be a maximal right ideal of A. We will show that either lBL=0 or L=hA, where h2=h∈A. Since S,≤ is a strictly totally ordered monoid, suppf is a nonempty well-ordered subset of S, for every 0≠f∈A. We denote by πf the smallest element of support f.
For anys∈S, set(20)Is=fs∣f∈L,πf=s⊂R,I=⋃s∈SIs.
LetJ be the right ideal of R generated by I. If J=R, then there exist s1,…,sn∈S, f1,…,fn∈L, and r1,…,rn∈R such that(21)1=f1s1r1+⋯+fnsnrn,where fisi∈Isi and πfi=si, for every 1≤i≤n. We will show that lBL=0. Suppose that φ∈lBL and φ≠0. Then suppφ is a nonempty well-ordered subset of S. Let t=πφ; if(22)φtωtfisi≠0,forsome1≤i≤n,then (23)φfit+si=∑u,v∈Xt+siφ,fiφuωufiv=φtωtfisi≠0.This means that φfi≠0 for some 1≤i≤n, a contradiction. Thus(24)φtωtfisi=0,∀1≤i≤n.Since MR is an ω-compatible module, we get(25)φtfisi=0,∀1≤i≤n.Consequently(26)φt=φt1=φtf1s1r1+⋯+fnsnrn=φtf1s1r1+⋯+φtfnsnrn=0,a contradiction. Thus lBL=0. Suppose that J≠R. We will show that J is a maximal right ideal of R. Let r∈R-J. If cr∈L, then r=cr0∈I0⊂I and so r∈J, a contradiction. Therefore cr∉L. Since L is a maximal right ideal of A,(27)A=L+crA.It follows that there exist f∈L and g∈A such that c1=f+crg. Thus (28)1=c10=f0+crg0=f0+rω0g0=f0+rg0.If f0=0, then 1=rg0∈rR. So, R=J+rR.
Iff0≠0, then 0∈suppf. Since 0≤s for every s∈S, πf=0. Thus f0∈I0⊂I⊂J, which implies that R=J+rR.
HenceJ is a maximal right ideal of R. Since MR is a PS-module, it follows that either lMJ=0 or J=eR, where e2=e∈R. According to that we have the following two cases.
Case  1. Suppose that lMJ=0. We will show that lBL=0. Let φ∈lBL and φ≠0. Then suppφ is a nonempty well-ordered subset of S. Let s=πφ. For any r∈J, there exist s1,…,sn∈S, f1,…,fn∈L, and r1,…,rn∈R such that(29)r=f1s1r1+⋯+fnsnrn,where fisi∈Isi and πfi=si, for every 1≤i≤n. Since φ∈lBL, f1,…,fn∈L, we get φfi=0 for every 1≤i≤n. If(30)φsωsfisi≠0,forsome1≤i≤n,then (31)φfis+si=∑u,v∈Xs+siφ,fiφuωufiv=φsωsfisi≠0.This means that φfi≠0 for some 1≤i≤n, a contradiction. Thus(32)φsωsfisi=0,∀1≤i≤n.Since MR is an ω-compatible module, we get(33)φsfisi=0,∀1≤i≤n.Consequently(34)φsr=φsf1s1r1+⋯+fnsnrn=φsf1s1r1+⋯+φsfnsnrn=0.Therefore, φs∈lMJ=0 and πφ=s. Thus φ=0, a contradiction. Hence lBL=0.
Case  2. Suppose that J=eR, where e2=e∈R. We will show that L=ceA, where (ce)2=ce∈A. To show that ceA⊆L, we need to prove that ce∈L. If ce∉L, then A=L+ceA. Thus there exist f∈L and g∈A such that c1=f+ceg. Thus (35)1=c10=f0+ceg0=f0+eω0g0=f0+eg0.If f0=0, then 1=eg0∈eR=J, a contradiction.
Iff0≠0, then 0∈suppf. Since 0≤s for every s∈S, πf=0. Thus f0∈I0⊂I⊂J, which implies that f0∈J and J=eR. Hence 1=f0+eg0∈J+eR=J, a contradiction. Therefore ce∈L which implies that ceA⊆L.
Conversely, suppose thatf∈L and πf=s; then fs∈Is⊂I⊂J=eR and so fs=efs. We claim that fu=efu for any u∈suppf.
Suppose thatfv=efv for each v<u. Consider the following element fu∈A defined by(36)fux=fx,x<u,0,x≥u.Thus πf-fu=u. By hypothesis it is easy to see that fu=cefu∈ceA⊂L. Thus f-fu∈L. By analogy with the proof above, it follows that(37)f-fuu=ef-fuu,which implies that fu=efu. Thus our claim holds. Therefore(38)ceft=∑u,v∈Xtce,fceuωufv=ce0ω0ft=eft=ft.Hence f=cef∈ceA. Thus L=ceA and the result follows since ce is an idempotent of A.As a special case of the last result if we setMR=RR we get the following.Corollary 11.
LetS,≤ be a strictly totally ordered monoid which satisfies the condition that 0≤s for every s∈S and let R be an ω-compatible ring. If R is a right PS-ring, then A=RS,≤,ω is a right PS-ring.If we setωs=IdR, for every s∈S, we get the following as a corollary.Corollary 12 (see [6], Theorem 1).
LetS,≤ be a strictly totally ordered monoid which satisfies the condition that 0≤s for every s∈S. If MR is a PS-module, then MS,≤RS,≤ is a PS-module.IfMR=RR, we get the following as a corollary.Corollary 13 (see [5], Theorem 4).
LetS,≤ be a strictly totally ordered monoid which satisfies the condition that 0≤s for every s∈S. If R is a right PS-ring, then RS,≤ is a right PS-ring.



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