# GF-Regular Modules

Authors: Areej M. Abduldaim; Sheng Chen
Journal: Journal of Applied Mathematics
 (2013)

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## Abstract

We introduced and studiedGF-regular modules as a generalization of π-regular rings to modules as well as regular modules (in the sense of Fieldhouse). An R-module M is called GF-regular if for each x∈M and r∈R, there exist t∈R and a positive integer n such that rntrnx=rnx. The notion of G-pure submodules was introduced to generalize pure submodules and proved that an R-module M is GF-regular if and only if every submodule of M is G-pure iff M𝔐  is a GF-regular R𝔐-module for each maximal ideal 𝔐 of  R. Many characterizations and properties of GF-regular modules were given. An R-module M is GF-regular  iff R/annx  is a π-regular ring for each 0≠x∈M iff R/annM is a π-regular ring for finitely generated module M. If M is a GF-regular module, then JM=0.


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## Body

## 1. Introduction

Throughout this paper, unless otherwise stated,R is a commutative ring with nonzero identity and all modules are left unitary. For an R-module M, the annihilator of x∈M in R is annR(x)={r∈R:rx=0}. The symbol □ stands for the end of the proof if the proof is given or the end of the statement when the proof is not given.Recall that a ringR is said to be regular (in the sense of von Neumann) if for each r∈R, there exists t∈R such that rtr=r [1]. The concept of regular rings was extended firstly to π-regular rings by McCoy [2], recall that a ring R is π-regular if for each r∈R, there exist t∈R and a positive integer n such that rntrn=rn [2] and secondly to modules in several nonequivalent ways considered by Fieldhouse [3], Ware [4], Zelmanowitz [5], and Ramamurthi and Rangaswamy [6]. In [7], Jayaraman and Vanaja have studied generalizations of regular modules (in the sense of Zelmanowitz) by Ramamurthi [8] and Mabuchi [9]. Following [10], we denoted Fieldhouse’ regular modules by F-regular. An R-module M is called F-regular if each submodule of M is pure [3].Dissimilar to the generalizations that have been studied in [7, 9] and [8], in this paper a new generalization of π-regular rings to modules and F-regular modules was introduced, called GF-reular (generalized F-regular) modules. An R-module M is called GF-regular if for each x∈M and r∈R, there exist t∈R and a positive integer n such that rntrnx=rnx. A ring R is called GF-regular if R is GF-regular as an R-module. On the other hand, GF-regular modules are also a generalization of π-regular rings. Thus, R is a π-regular ring if and only if R is a GF-regular R-module. Furthermore, we introduced a new class of submodules, named, G-pure submodules as a generalization of pure submodules. A submodule P of an R-module M is said to be G-pure if for each r∈R, there exists a positive integer n such that P∩rnM=rnP. Recall that a submodule P of an R-module M is pure if P∩IM=IP for each ideal I of R [11]. We find that the relationship between GF-regular modules and G-pure submodules is an analogous relationship between F-regular modules and pure submodules.In Section3.1 of this paper, after the concept of GF-regular modules was introduced, we obtained several characteristic properties of GF-regular modules. For instance, it was proved that the following are equivalent for an R-module M: (1) M is GF-regular; (2) every submodule of M is G-pure; (3) R/ann(x) is a π-regular ring for each 0≠x∈M; (4)  and for each x∈Mand r∈R, there exist t∈R and a positive integer n such that rn+1tx=rnx. It is also shown that if M is a finitely generated R-module, then M is GF-regular if and only if R/ann(M) is a π-regular ring.Section3.2 was devoted to investigate the relationship between GF-regular modules with the localization property and semisimple modules. For example, we proved that M is a GF-regular R-module if and only if M𝔐 is a GF-regular R𝔐-module for every maximal ideal 𝔐 of R if and only if M𝔐 is a semisimple R𝔐-module for every maximal ideal 𝔐 of R.Finally, in Section3.3 we studied some properties of the Jacobson radical, J(M), of GF-regular modules. Thus we proved that if M is a GF-regular R-module, then J(M)=0, and also we get that if J(R) is a reduced ideal of a ring R and M is a GF-regular R-module, then J(R)·M=0.


## 2. The Notion ofGF-Regular Modules and General Results

We start by recalling that anR-module M is F-regular if each submodule of M is pure [3], and a ring R is π-regular if for each r∈R, there exist t∈R and a positive integer n such that rntrn=rn [2].Definition 1.
AnR-module M is called GF-regular if for each x∈M and r∈R, there exist t∈R and a positive integer n such that rntrnx=rnx. A ring R is GF-regular if and only if R is GF-regular as an R-module.The following gives another characterization forGF-regular modules.Proposition 2.
AnR-module M is GF-regular if and only if R/ann(x) is a π-regular ring for each 0≠x∈M.Proof.
Suppose thatM is a GF-regular R-module, so for each x∈M and r∈R, there exist t∈R and a positive integer n such that rntrnx=rnx; hence, (rntrn-rn)∈ann(x) which means that r-ntr-n=r-n; therefore, R/ann(x) is a π-regular ring. Conversely, suppose that R/ann(x) is a π-regular ring for each 0≠x∈M, thus for each r-∈R/ann(x), there exist t-∈R/ann(x) and a positive integer n such that r-ntr-n=r-n; hence, rntrn-rn∈ann(x) which implies that (rntrn-rn)x=0; therefore, M is a GF-regular R-module.It is clear that everyF-regular module is GF-regular, but the converse may not be true in general; for example, by applying Proposition 2 to the Z-module Z4, we can easily see that it is GF-regular; however, Z4 is not an F-regular Z-module. In fact, the Z-module Zn is GF-regular for each positive integer n [12], while it is not F-regular for some positive integer n. On the other hand, the Z-module Q is not GF-regular because for each 0≠x∈Q we have that annZ(x)=0, but Z/annZ≃Z which is not a π-regular ring [12].Remark 3.
(1)
IfR is a π-regular ring, then every R-module is GF-regular.
(2)
Every module over Artinian ringR is GF-regular (because every Artinian ring is   π-regular [12]).
(3)
A ringR is π-regular if and only if R is GF-regular as an R-module.
(4)
Every submodule of aGF-regular module is GF-regular module. In particular, every ideal of a π-regular ring R is GF-regular R-module. Furthermore, it follows from (1) that if I is an ideal of a π-regular ring R, then the R-module R/I is GF-regular.
(5)
The converse of (1) is true if the module is free, that is, any freeR-module M is      GF-regular if and only if R is a π-regular ring. For if, M is a free R-module, then ann(x)=0 for each 0≠x∈M, so R≃R/ann(x) is a π-regular ring.
(6)
If anR-module M is GF-regular and it contains a nontorsion element, then R is a   π-regular ring. In particular, if M is a GF-regular R-module and R is not a π-regular ring, then M is a torsion R-module.Now from Proposition2 and Remark 3(3), we conclude the following.Corollary 4.
The following statements are equivalent for a ring:(1)
R is a π-regular ring;(2)
R/ann(r) is a π-regular ring for each 0≠r∈R.We have seen previously that everyF-regular R-module is GF-regular. In the following we consider some conditions such that the converse is true.Remark 5.
(1)
LetR be a reduced ring. An R-module M is F-regular if and only if M is a GF-regular R-module.
(2)
AnR-module M is F-regular if and only if M is a GF-regular R-module and L(R/ann(x))=0 for each0≠x∈M, where L(R/ann(x)) is the prime radical of the ring R/ann(x).Now, we describeGF-regular modules over the ring of integers Z.Proposition 6.
AZ-module M is GF-regular if and only if M is a torsion Z-module.Proof.
IfM is a GF-regular Z-module, then by Remark 3(6) M is a torsion      Z-module. Conversely, if M is a torsion Z-module, then annZ(x)=nZ for some positive integer n; hence, Z/annZ(x)≃Zn is a π-regular ring for each positive integern [12], which implies that M is a GF-regular Z-module.Proposition 7.
Every homomorphic image of aGF-regular R-module is GF-regular.Proof.
LetM, M′ be two R-modules such that M is GF-regular and let  f∶M→M′ be an R-epimorphism. For every y∈M′, there exists x∈M such that f(x)=y. It is clear that ann(x)⊆ann(y). Define α∶R/ann(x)→R/ann(y) by α(r+ann(x))=r+ann(y) for each r∈R. It is an easy matter to check that α is well defined R-epimorphism. Since R/ann(x) is a π-regular ring, then R/ann(y) is also a π-regular ring [12]. Therefore, M′ is a GF-regular R-module.Corollary 8.
The following statements are equivalent for anR-module M:(1)
M/N is a GF-regular R-module for every nonzero submodule N of M.(2)
M/Rx is a GF-regular R-module for every 0≠x∈M.Another characterization of aGF-regular R-module is given in the next result.Proposition 9.
AnR-module M is GF-regular if and only if for each x∈M and r∈R, there exist t∈R and a positive integer n such that rn+1tx=rnx.Proof.
Suppose thatM is a GF-regular R-module, so for each x∈M and r∈R, there exist s∈R and a positive integer n such that rnsrnx=rnx, then we can take t=srn-1∈R and hence rn+1tx=rnx. Conversely, for each x∈M and r∈R, there exist s∈R and a positive integer n such that rn+1sx=rnx. Now, rnsnrnx=rn+1ssn-1rn-1x=rnsn-1rn-1x=rn+1ssn-2rn-2x=rnsn-2rn-2x=⋯=rn+1sx=rnx (after n times), thus rntrnx=rnx where t=sn which implies that M is a GF-regular R-module.


## 3. Main Results

### 3.1.GF-Regular Modules and Purity

Recall that a submoduleP of an R-module M is pure in M if each finite system of equations
(1)Pi=∑jrijxj,rij∈R,Pj∈P,1≤j≤m,
which is solvable in M, is solvable in P [13]. It is not difficult to prove that P is pure in M if and only if for each ideal I of R, P∩IM=IP [11]. This motivates us to introduce the following definition as a generalization of pure submodules.Definition 10.
A submoduleP of an R-module M is called G-pure if for each r∈R, there exists a positive integer n such that P∩rnM=rnP.It is clear that every pure module isG-pure.The following theorem gives another characterization ofGF-regular modules in terms of G-pure submodules.Theorem 11.
AnR-module M is GF-regular if and only if every submodule of M is G-pure.Proof.
Suppose thatM is a GF-regular R-module and let P be any submodule of M. For each r∈R and for some positive integer n, let x∈P∩rnM, then there exists y∈M such that x=rny. Since M is GF-regular, then there exists t∈R such that rny=rntrny. Put e=trn, then rny=erny which implies that x=ex, but x∈P, so x=ex∈rnP and hence P∩rnM⊆rnP. On the other hand, it is clear that rnP⊆P∩rnM, thus P⋂rnM=rnP which means that P is a G-pure submodule.
Conversely, assume that every submodule isG-pure and let x∈M and p∈R such that Rpnx=P which is a G-pure submodule of M for some positive integer n, then P∩rnM=rnP for each r∈R. In particular, if r=p we get rnx∈P∩rnM⊆rnP=rnRrnx which implies that there exists t∈R such that rntrnx=rnx, so M is a GF-regular R-module.Corollary 12.
AnR-module M is GF-regular if and only if for each x∈M, there exist p∈R and a positive integer n such that Rpnx is a G-pure submodule.Remark 13.
Fieldhouse in [11] proved that for a submodule P of an R-module M, if M/P is a flat R-module, then P is pure. On the other hand, if M is flat and P is pure, then M/P is flat. So, immediately we have that for a flat R-module, if M/P is a flat R-module for each submodule P of M, then M is GF-regular R-module. It is not difficult to prove that in case of F-regular modules the converse of the latest statement is true; however, we do not know whether it is true for GF-regular modules or not.Remark 14.
In [14],  Mao proved that a right R-module N is GP-flat if and only if there exists an exact sequence 0→K→M→N→0 with M free such that for any r∈R, there exists a positive integer n satisfying K∩Mrn=Krn, where (1) a right R-module N is said to be generalized P-flat (GP-flat for short) if for any r∈R, there exists a positive integer n (depending on r) such that the sequence 0→N⊗Rrn→N⊗R is exact [15], (2) a right R-module N is P-flat [16] or torsion-free [15] if for any r∈R, the sequence 0→N⊗Rr→N⊗R is exact. Obviously, every flat module is P-flat [16] and every P-flat module is GP-flat [14].According to the above remark we get the following.Corollary 15.
AnR-module N is GP-flat if and only if there exists an exact sequence 0→P→M→N→0 with P is a submodule of a free R-module M such that P is a G-pure submodule.Corollary 16.
For every submoduleP of a free R-module M, if there exists an exact sequence 0→P→M→N→0 such that P is a G-pure submodule in M, then N is a GP-flat R-module if and only if M is GF-regular.Now, we recall that (1) anR-module M is p-injective if for every principal ideal I of R, every R-homomorphism of I into M extends to one of R into M [17]. A ring R is called p-injective if R is p-injective as an R-module. (2) An R-module M is called       YJ-injective if for any 0≠r∈R, there exists a positive integer n such that rn≠0 and any R-homomorphism of Rrn into M extends to one of R into M. A ring R is called YJ-injective if R is YJ-injective as an R-module [18]. YJ-injective modules are called GP-injective modules by some other authors [19–22]. (3) An R-module M is called WGP-injective (weak GP-injective) if for any r∈R, there exists a positive integer n such that every R-homomorphism of Rrn into M extends to one of R into M (rn may be zero). A ring R is called WGP-injective if R is WGP-injective as an R-module [23–25]. (4) A ring R is called p.p. if every principal ideal of R is projective. And R is called GPP-ring if for any r∈R, there exists a positive integer n (depending on r) such that Rrn is projective [26, 27].Note thatp-injectivity implies YJ-injectivity (or GP-injectivity) and     WGP-injectivity, as well as the concept of p.p. rings implies the concept of GPP-rings. However, the notion of YJ-injective (or GP-injective) modules is not the same notion of WGP-injective modules.It is known that a ringR is π-regular if and only if every R-module is WGP-injective [12, 22], so from all the above we conclude the following theorem.Theorem 17.
The following statements are equivalent for a ringR.(1)
R is a π-regular ring.(2)
R/ann(r) is a π-regular ring for each 0≠r∈R.(3)
Any freeR-module is GF-regular.(4)
EveryR-module is WGP-injective.We end this section by the following two related results.Proposition 18.
LetM be an R-module. If R/ann(M) is a π-regular ring, then M is a GF-regular R-module.Proof.
We have thatann(M)⊆ann(x) for each x∈M, so there exists an obvious R-epimorphism φ∶R/ann(M)→R/ann(x) defined by φ(r+ann(M))=r+ann(x). Since R/ann(M) is a π-regular ring, then R/ann(x) is a π-regular ring [12]; therefore, M is a GF-regular R-module.In case of finitely generated modules, the converse of Proposition18 is true.Proposition 19.
LetM be an R-module. If M is a finitely generated GF-regular R-module, then R/ann(M) is a π-regular ring.Proof.
Let{x1,x2,…,xn} be a finite set of generators of M. Put N=ann(M), and Ni=ann(xi), 1≤i≤k, then N=∩iNi. Now define φ∶R/N→⊕∑i=1nR/Ni by φ(r+N) = (r+N1,r+N2,…,r+Nn) for each r+N∈R/N. It is easily checked that φ is a ring monomorphism. Thus, R/N can be identified with a subring T of ⊕∑i=1nR/Ni. In fact
(2)T={(r+N1,r+N2,…,r+Nn)∶r∈R}.
We will show now thatT, and hence R/N is a π-regular ring. Since M is a GF-regular R-module, then R/Ni is a π-regular ring, thus for each r∈R and 1≤i≤k, there exist ti∈R and a positive integer n such that rntirn+Ni=rn+Ni; this means that rntirnxi=rnxi. Define t by the relation 1-trn=∏i=1k(1-tirn), then rn(1-trn)xi=rn∏i=1k(1-tirn)xi=∏i=1k(rn-rntirn)xi=0 which implies that for each i, rn+Ni=rntrn+Ni, so T is a π-regular ring and hence R/N is a π-regular ring.


### 3.2.GF-Regular Modules and Localization

In this section we study the localization property and semisimple modules withGF-regular modules and we give some characterizations of  GF-regular modules in the sense of them.Theorem 20.
LetM be an R-module. M is a GF-regular R-module if and only if M𝔐 is a GF-regular R𝔐-module for each maximal ideal 𝔐 in R.Proof.
LetM be a GF-regular R-module, and let 𝔐 be any maximal ideal in R. Let x/t∈M𝔐 and r/t1∈R𝔐, where x∈M, r∈R and t, t1∈R-𝔐. So there exist k∈R and a positive integer n such that rnkrnx=rnx. Hence, (r/t1)n(x/t)=rnx/t1nt=(rnkrnx/t1nt)(t1n/t1n)=(rn/t1n)(kt1n/1)(rn/t1n)(x/t) = (r/t1)n(kt1n/1)(r/t1)n, where  kt1n/1∈R𝔐, then M𝔐 is GF-regular R𝔐-module.
Conversely, suppose thatM𝔐 is a GF-regular R𝔐-module. Let P be a submodule of M and let 𝔐 be a maximal ideal of R. By Theorem 11, P𝔐 is a G-pure submodule of M𝔐; therefore, P𝔐∩(Rrn)𝔐M𝔐=(Rrn)𝔐P𝔐 for each r∈R and for some positive integer n. But by [28], we have that P𝔐∩(Rrn)𝔐M𝔐=P𝔐∩(RrnM)𝔐=(P∩RrnM)𝔐 and (RrnP)𝔐=(Rrn)𝔐P𝔐, then (RrnM∩P)𝔐=(RrnP)𝔐, again by [28], we get that RrnM∩P=RrnP, which implies that P is a G-pure submodule of M and by Theorem 11      M, is a GF-regular R-module.Recall that anR-module M is simple if 0 and M are the only submodules of M, and an R-module M is said to be semisimple if M is a sum of simple modules (may be infinite). A ring R is semisimple if it is semisimple as an R-module [29]. It is known that over any ring R, a semisimple module is F-regular [4, 30], consequently it is GF-regular. Furthermore, it is known that over a local ring, every F-regular module is semisimple [31]. We can generalize the latest statement as the following.Proposition 21.
EveryGF-regular module over local ring is semisimple.Proof.
Let𝔐 be the only maximal ideal of R. Since M is GF-regular, then for each 0≠x∈M we have that R/ann(x) is GF-regular local ring which implies that R/ann(x) is a field [12]; hence, ann(x) is a maximal ideal, so 𝔐=ann(x) for each 0≠x∈M. Therefore, 𝔐=ann(x)=ann(M). On the other hand, R/𝔐≃R/ann(M) is a field, which implies that M is a vector space over the field R/ann(M) which is a simple ring. Then M is a semisimple module over the ring R/ann(M). Thus, M is a semismple R-module [29].As an immediate result from Theorem20 and Proposition 21, we get the following.Corollary 22.
LetM be an R-module. M is GF-regular if and only if M𝔐 is a semisimple R𝔐-module for each maximal ideal 𝔐 of R.We mentioned before that everyF-regular R-module is GF-regular; the following gives us another condition such that the converse is true.Corollary 23.
LetR be a local ring. An R-module M is F-regular if and only if M is a GF-regular R-module.Corollary 24.
AnR-module M=N⨁K is GF-regular if and only if N and K are GF-regular R-modules.Proof.
Assume thatN and K are GF-regular R-modules, then for each maximal ideal 𝔐 in R, each of N𝔐 and K𝔐 is a semisimple module (Proposition 21); hence, it is an easy matter to check that N𝔐+K𝔐 is a semisimple module, so M𝔐=N𝔐⨁K𝔐 is a GF-regular module. Thus, M is a GF-regular module (Theorem 20). The other direction is obtained directly from Proposition 7.Finally we can summarize that the conditions under whichF-regular modules coincide with GF-regular modules and the characterizations of GF-regular modules, of Section 2 with those of this section, in the following Proposition 25 and Theorem 26, respectively:Proposition 25.
AnR-module M is GF-regular if and only if M is an F-regular module, if any of the following conditions are satisfied.(1)
Ris a local ring.(2)
Ris a reduced ring.(3)
The prime radical of the ringR/ann(x)is zero for each0≠x∈M.Theorem 26.
The following statements are equivalent for a ringR.(1)
M is a GF-regular R-module.(2)
R/ann(x) is a π-regular ring for each 0≠x∈M(3)
For eachx∈M and r∈R, there exist t∈R and positive integer n such that rn+1x=rnx.(4)
Every submodule ofM is G-pure.(5)
For eachx∈M, there exist p∈R and a positive integer n such that Rpnx is a G-pure submodule.(6)
N is a GP-flat R-module, if for every submodule P of a free R-module M there exists an exact sequence 0→P→M→N→0 such that P is a G-pure submodule in M.(7)
IfM is a finitely generated R-module, then R/ann(M) is a π-regular ring.(8)
M𝔐 is a GF-regular R𝔐-module for each maximal ideal  𝔐 in R.(9)
M𝔐 is a semisimple R𝔐-module for each maximal ideal 𝔐 of R.


### 3.3. The Jacobson Radical ofGF-Regular Modules

LetM be an R-module. A submodule N of M is said to be small in M if for each submodule K of M such that N+K=M, we have K=M [32]. The Jacobson radical of a ring R will be denoted by J(R). The following submodules of M are equal: (1) the intersection of all maximal submodules of M, (2) the sum of all the small submodules of M, and (3) the sum of all cyclic small submodules of M. This submodule is called the Jacobson radical of M and will be denoted by J(M) [29, 32].It is appropriate now to note that for each elementr∈R it may happen that rn=0. But some cases demand that rn must be nonzero element. For this purpose we introduce the following concept.Definition 27.
AnR-module M is called SGF-regular if for each 0≠x∈M and r∈R, there exist t∈R and a positive integer n with rn≠0 such that rntrnx=rnx. A ring R is called SGF-regular if it is SGF-regular as an R-module.It is clear thatSGF-regularity implies GF-regularity and they are coincide if R is a reduced ring.Proposition 28.
LetM be an SGF-regular R-module, then J(R).M=0.Proof.
For each0≠x∈M and for each 0≠r∈R, there exist t∈R and a positive integer n with rn≠0 such that rntrnx=rnx, then rnx(rnx-1)=0. If r∈J(R), then rn∈J(R) and (rnt-1) is invertible, so rnx=0, but we have that rn≠0 and x≠0; hence, rx=0 which implies that J(R)·M=0.Recall that anR-module M is faithful if for every r∈R such that rM=0 implies r=0 [29], or equivalently, an R-module M is called faithful if ann(M)=0 [33].Corollary 29.
IfM is a faithful SGF-regular R-module, then J(R)=0.Corollary 30.
LetR be a reduced ring and M be a GF-regular R-module, then J(R)·M=0.Corollary 31.
LetR be any ring such that J(R) is a reduced ideal of R and let M be a GF-regular R-module, then J(R)·M=0.Corollary 32.
LetR be a reduced ring. If M is a faithful GF-regular R-module, then J(R)=0.It is suitable to mention that, in general, not every module contains a maximal submodule; for example,Q as Z-module has no maximal submodule. So we have the next two results, but first we need Lemma 33 which is proved in [29].Lemma 33.
AnR-module M is semisimple if and only if each submodule of M is direct summand.Proposition 34.
LetM be a GF-regular R-module, then J(M)=0.Proof.
SinceM is a GF-regular R-module, then M𝔐 is a semisimple R𝔐-module for each maximal ideal 𝔐 of R (Corollary 22). Since each cyclic submodule of M𝔐 is direct summand (Lemma 33), then it cannot be small; therefore, the Jacobson radical of a semisimple module is zero, so J(M𝔐)=0 for each maximal ideal 𝔐 of R. On the other hand, J(M)𝔐⊆J(M𝔐) [28], thus J(M)𝔐=0 for each maximal ideal 𝔐 of R, and hence J(M)=0 [28].Corollary 35.
Every nonzeroGF-regular R-module M contains a maximal submodule.Proof.
Suppose not, thenJ(M)=M, but J(M)=0 (Proposition 34), so M=0 which is a contradiction.Corollary 36.
LetM be a GF-regular R-module, then for each 0≠x∈M, there exist a maximal submodule 𝔐 such that x∉𝔐.Proof.
Ifx∈P, for each maximal submodule 𝔐 of M, then x∈J(M)=0 which implies that x=0.Corollary 37.
LetM be a GF-regular R-module, then every proper submodule of M contained in a maximal submodule.Proof.
LetN be a proper submodule of M. Since M is a GF-regular R-module, then M/N≠0 is GF-regular (Proposition 7), so M/N contains a maximal submodule (Corollary 35), which means that there exists a submodule K of M such that N⊆K, K/N is a maximal submodule of M/N; therefore, K is a maximal submodule of M and contains N.Corollary 38.
Every simple submodule of aGF-regular R-module is direct summand.Proof.
LetN be a simple submodule of a GF-regular R-module M, then N is cyclic; say N=Rx, then there exists a maximal submodule 𝔐 of M such that x∉𝔐 (Corollary 37). It is clear that M=𝔐+Rx. Now, if Rx∩𝔐≠(0), then Rx∩𝔐=Rx because Rx is a simple submodule. Thus, x∈𝔐 which is a contradiction, so  M=Rx⨁𝔐.


## 3.1.GF-Regular Modules and Purity

Recall that a submoduleP of an R-module M is pure in M if each finite system of equations
(1)Pi=∑jrijxj,rij∈R,Pj∈P,1≤j≤m,
which is solvable in M, is solvable in P [13]. It is not difficult to prove that P is pure in M if and only if for each ideal I of R, P∩IM=IP [11]. This motivates us to introduce the following definition as a generalization of pure submodules.Definition 10.
A submoduleP of an R-module M is called G-pure if for each r∈R, there exists a positive integer n such that P∩rnM=rnP.It is clear that every pure module isG-pure.The following theorem gives another characterization ofGF-regular modules in terms of G-pure submodules.Theorem 11.
AnR-module M is GF-regular if and only if every submodule of M is G-pure.Proof.
Suppose thatM is a GF-regular R-module and let P be any submodule of M. For each r∈R and for some positive integer n, let x∈P∩rnM, then there exists y∈M such that x=rny. Since M is GF-regular, then there exists t∈R such that rny=rntrny. Put e=trn, then rny=erny which implies that x=ex, but x∈P, so x=ex∈rnP and hence P∩rnM⊆rnP. On the other hand, it is clear that rnP⊆P∩rnM, thus P⋂rnM=rnP which means that P is a G-pure submodule.
Conversely, assume that every submodule isG-pure and let x∈M and p∈R such that Rpnx=P which is a G-pure submodule of M for some positive integer n, then P∩rnM=rnP for each r∈R. In particular, if r=p we get rnx∈P∩rnM⊆rnP=rnRrnx which implies that there exists t∈R such that rntrnx=rnx, so M is a GF-regular R-module.Corollary 12.
AnR-module M is GF-regular if and only if for each x∈M, there exist p∈R and a positive integer n such that Rpnx is a G-pure submodule.Remark 13.
Fieldhouse in [11] proved that for a submodule P of an R-module M, if M/P is a flat R-module, then P is pure. On the other hand, if M is flat and P is pure, then M/P is flat. So, immediately we have that for a flat R-module, if M/P is a flat R-module for each submodule P of M, then M is GF-regular R-module. It is not difficult to prove that in case of F-regular modules the converse of the latest statement is true; however, we do not know whether it is true for GF-regular modules or not.Remark 14.
In [14],  Mao proved that a right R-module N is GP-flat if and only if there exists an exact sequence 0→K→M→N→0 with M free such that for any r∈R, there exists a positive integer n satisfying K∩Mrn=Krn, where (1) a right R-module N is said to be generalized P-flat (GP-flat for short) if for any r∈R, there exists a positive integer n (depending on r) such that the sequence 0→N⊗Rrn→N⊗R is exact [15], (2) a right R-module N is P-flat [16] or torsion-free [15] if for any r∈R, the sequence 0→N⊗Rr→N⊗R is exact. Obviously, every flat module is P-flat [16] and every P-flat module is GP-flat [14].According to the above remark we get the following.Corollary 15.
AnR-module N is GP-flat if and only if there exists an exact sequence 0→P→M→N→0 with P is a submodule of a free R-module M such that P is a G-pure submodule.Corollary 16.
For every submoduleP of a free R-module M, if there exists an exact sequence 0→P→M→N→0 such that P is a G-pure submodule in M, then N is a GP-flat R-module if and only if M is GF-regular.Now, we recall that (1) anR-module M is p-injective if for every principal ideal I of R, every R-homomorphism of I into M extends to one of R into M [17]. A ring R is called p-injective if R is p-injective as an R-module. (2) An R-module M is called       YJ-injective if for any 0≠r∈R, there exists a positive integer n such that rn≠0 and any R-homomorphism of Rrn into M extends to one of R into M. A ring R is called YJ-injective if R is YJ-injective as an R-module [18]. YJ-injective modules are called GP-injective modules by some other authors [19–22]. (3) An R-module M is called WGP-injective (weak GP-injective) if for any r∈R, there exists a positive integer n such that every R-homomorphism of Rrn into M extends to one of R into M (rn may be zero). A ring R is called WGP-injective if R is WGP-injective as an R-module [23–25]. (4) A ring R is called p.p. if every principal ideal of R is projective. And R is called GPP-ring if for any r∈R, there exists a positive integer n (depending on r) such that Rrn is projective [26, 27].Note thatp-injectivity implies YJ-injectivity (or GP-injectivity) and     WGP-injectivity, as well as the concept of p.p. rings implies the concept of GPP-rings. However, the notion of YJ-injective (or GP-injective) modules is not the same notion of WGP-injective modules.It is known that a ringR is π-regular if and only if every R-module is WGP-injective [12, 22], so from all the above we conclude the following theorem.Theorem 17.
The following statements are equivalent for a ringR.(1)
R is a π-regular ring.(2)
R/ann(r) is a π-regular ring for each 0≠r∈R.(3)
Any freeR-module is GF-regular.(4)
EveryR-module is WGP-injective.We end this section by the following two related results.Proposition 18.
LetM be an R-module. If R/ann(M) is a π-regular ring, then M is a GF-regular R-module.Proof.
We have thatann(M)⊆ann(x) for each x∈M, so there exists an obvious R-epimorphism φ∶R/ann(M)→R/ann(x) defined by φ(r+ann(M))=r+ann(x). Since R/ann(M) is a π-regular ring, then R/ann(x) is a π-regular ring [12]; therefore, M is a GF-regular R-module.In case of finitely generated modules, the converse of Proposition18 is true.Proposition 19.
LetM be an R-module. If M is a finitely generated GF-regular R-module, then R/ann(M) is a π-regular ring.Proof.
Let{x1,x2,…,xn} be a finite set of generators of M. Put N=ann(M), and Ni=ann(xi), 1≤i≤k, then N=∩iNi. Now define φ∶R/N→⊕∑i=1nR/Ni by φ(r+N) = (r+N1,r+N2,…,r+Nn) for each r+N∈R/N. It is easily checked that φ is a ring monomorphism. Thus, R/N can be identified with a subring T of ⊕∑i=1nR/Ni. In fact
(2)T={(r+N1,r+N2,…,r+Nn)∶r∈R}.
We will show now thatT, and hence R/N is a π-regular ring. Since M is a GF-regular R-module, then R/Ni is a π-regular ring, thus for each r∈R and 1≤i≤k, there exist ti∈R and a positive integer n such that rntirn+Ni=rn+Ni; this means that rntirnxi=rnxi. Define t by the relation 1-trn=∏i=1k(1-tirn), then rn(1-trn)xi=rn∏i=1k(1-tirn)xi=∏i=1k(rn-rntirn)xi=0 which implies that for each i, rn+Ni=rntrn+Ni, so T is a π-regular ring and hence R/N is a π-regular ring.


## 3.2.GF-Regular Modules and Localization

In this section we study the localization property and semisimple modules withGF-regular modules and we give some characterizations of  GF-regular modules in the sense of them.Theorem 20.
LetM be an R-module. M is a GF-regular R-module if and only if M𝔐 is a GF-regular R𝔐-module for each maximal ideal 𝔐 in R.Proof.
LetM be a GF-regular R-module, and let 𝔐 be any maximal ideal in R. Let x/t∈M𝔐 and r/t1∈R𝔐, where x∈M, r∈R and t, t1∈R-𝔐. So there exist k∈R and a positive integer n such that rnkrnx=rnx. Hence, (r/t1)n(x/t)=rnx/t1nt=(rnkrnx/t1nt)(t1n/t1n)=(rn/t1n)(kt1n/1)(rn/t1n)(x/t) = (r/t1)n(kt1n/1)(r/t1)n, where  kt1n/1∈R𝔐, then M𝔐 is GF-regular R𝔐-module.
Conversely, suppose thatM𝔐 is a GF-regular R𝔐-module. Let P be a submodule of M and let 𝔐 be a maximal ideal of R. By Theorem 11, P𝔐 is a G-pure submodule of M𝔐; therefore, P𝔐∩(Rrn)𝔐M𝔐=(Rrn)𝔐P𝔐 for each r∈R and for some positive integer n. But by [28], we have that P𝔐∩(Rrn)𝔐M𝔐=P𝔐∩(RrnM)𝔐=(P∩RrnM)𝔐 and (RrnP)𝔐=(Rrn)𝔐P𝔐, then (RrnM∩P)𝔐=(RrnP)𝔐, again by [28], we get that RrnM∩P=RrnP, which implies that P is a G-pure submodule of M and by Theorem 11      M, is a GF-regular R-module.Recall that anR-module M is simple if 0 and M are the only submodules of M, and an R-module M is said to be semisimple if M is a sum of simple modules (may be infinite). A ring R is semisimple if it is semisimple as an R-module [29]. It is known that over any ring R, a semisimple module is F-regular [4, 30], consequently it is GF-regular. Furthermore, it is known that over a local ring, every F-regular module is semisimple [31]. We can generalize the latest statement as the following.Proposition 21.
EveryGF-regular module over local ring is semisimple.Proof.
Let𝔐 be the only maximal ideal of R. Since M is GF-regular, then for each 0≠x∈M we have that R/ann(x) is GF-regular local ring which implies that R/ann(x) is a field [12]; hence, ann(x) is a maximal ideal, so 𝔐=ann(x) for each 0≠x∈M. Therefore, 𝔐=ann(x)=ann(M). On the other hand, R/𝔐≃R/ann(M) is a field, which implies that M is a vector space over the field R/ann(M) which is a simple ring. Then M is a semisimple module over the ring R/ann(M). Thus, M is a semismple R-module [29].As an immediate result from Theorem20 and Proposition 21, we get the following.Corollary 22.
LetM be an R-module. M is GF-regular if and only if M𝔐 is a semisimple R𝔐-module for each maximal ideal 𝔐 of R.We mentioned before that everyF-regular R-module is GF-regular; the following gives us another condition such that the converse is true.Corollary 23.
LetR be a local ring. An R-module M is F-regular if and only if M is a GF-regular R-module.Corollary 24.
AnR-module M=N⨁K is GF-regular if and only if N and K are GF-regular R-modules.Proof.
Assume thatN and K are GF-regular R-modules, then for each maximal ideal 𝔐 in R, each of N𝔐 and K𝔐 is a semisimple module (Proposition 21); hence, it is an easy matter to check that N𝔐+K𝔐 is a semisimple module, so M𝔐=N𝔐⨁K𝔐 is a GF-regular module. Thus, M is a GF-regular module (Theorem 20). The other direction is obtained directly from Proposition 7.Finally we can summarize that the conditions under whichF-regular modules coincide with GF-regular modules and the characterizations of GF-regular modules, of Section 2 with those of this section, in the following Proposition 25 and Theorem 26, respectively:Proposition 25.
AnR-module M is GF-regular if and only if M is an F-regular module, if any of the following conditions are satisfied.(1)
Ris a local ring.(2)
Ris a reduced ring.(3)
The prime radical of the ringR/ann(x)is zero for each0≠x∈M.Theorem 26.
The following statements are equivalent for a ringR.(1)
M is a GF-regular R-module.(2)
R/ann(x) is a π-regular ring for each 0≠x∈M(3)
For eachx∈M and r∈R, there exist t∈R and positive integer n such that rn+1x=rnx.(4)
Every submodule ofM is G-pure.(5)
For eachx∈M, there exist p∈R and a positive integer n such that Rpnx is a G-pure submodule.(6)
N is a GP-flat R-module, if for every submodule P of a free R-module M there exists an exact sequence 0→P→M→N→0 such that P is a G-pure submodule in M.(7)
IfM is a finitely generated R-module, then R/ann(M) is a π-regular ring.(8)
M𝔐 is a GF-regular R𝔐-module for each maximal ideal  𝔐 in R.(9)
M𝔐 is a semisimple R𝔐-module for each maximal ideal 𝔐 of R.


## 3.3. The Jacobson Radical ofGF-Regular Modules

LetM be an R-module. A submodule N of M is said to be small in M if for each submodule K of M such that N+K=M, we have K=M [32]. The Jacobson radical of a ring R will be denoted by J(R). The following submodules of M are equal: (1) the intersection of all maximal submodules of M, (2) the sum of all the small submodules of M, and (3) the sum of all cyclic small submodules of M. This submodule is called the Jacobson radical of M and will be denoted by J(M) [29, 32].It is appropriate now to note that for each elementr∈R it may happen that rn=0. But some cases demand that rn must be nonzero element. For this purpose we introduce the following concept.Definition 27.
AnR-module M is called SGF-regular if for each 0≠x∈M and r∈R, there exist t∈R and a positive integer n with rn≠0 such that rntrnx=rnx. A ring R is called SGF-regular if it is SGF-regular as an R-module.It is clear thatSGF-regularity implies GF-regularity and they are coincide if R is a reduced ring.Proposition 28.
LetM be an SGF-regular R-module, then J(R).M=0.Proof.
For each0≠x∈M and for each 0≠r∈R, there exist t∈R and a positive integer n with rn≠0 such that rntrnx=rnx, then rnx(rnx-1)=0. If r∈J(R), then rn∈J(R) and (rnt-1) is invertible, so rnx=0, but we have that rn≠0 and x≠0; hence, rx=0 which implies that J(R)·M=0.Recall that anR-module M is faithful if for every r∈R such that rM=0 implies r=0 [29], or equivalently, an R-module M is called faithful if ann(M)=0 [33].Corollary 29.
IfM is a faithful SGF-regular R-module, then J(R)=0.Corollary 30.
LetR be a reduced ring and M be a GF-regular R-module, then J(R)·M=0.Corollary 31.
LetR be any ring such that J(R) is a reduced ideal of R and let M be a GF-regular R-module, then J(R)·M=0.Corollary 32.
LetR be a reduced ring. If M is a faithful GF-regular R-module, then J(R)=0.It is suitable to mention that, in general, not every module contains a maximal submodule; for example,Q as Z-module has no maximal submodule. So we have the next two results, but first we need Lemma 33 which is proved in [29].Lemma 33.
AnR-module M is semisimple if and only if each submodule of M is direct summand.Proposition 34.
LetM be a GF-regular R-module, then J(M)=0.Proof.
SinceM is a GF-regular R-module, then M𝔐 is a semisimple R𝔐-module for each maximal ideal 𝔐 of R (Corollary 22). Since each cyclic submodule of M𝔐 is direct summand (Lemma 33), then it cannot be small; therefore, the Jacobson radical of a semisimple module is zero, so J(M𝔐)=0 for each maximal ideal 𝔐 of R. On the other hand, J(M)𝔐⊆J(M𝔐) [28], thus J(M)𝔐=0 for each maximal ideal 𝔐 of R, and hence J(M)=0 [28].Corollary 35.
Every nonzeroGF-regular R-module M contains a maximal submodule.Proof.
Suppose not, thenJ(M)=M, but J(M)=0 (Proposition 34), so M=0 which is a contradiction.Corollary 36.
LetM be a GF-regular R-module, then for each 0≠x∈M, there exist a maximal submodule 𝔐 such that x∉𝔐.Proof.
Ifx∈P, for each maximal submodule 𝔐 of M, then x∈J(M)=0 which implies that x=0.Corollary 37.
LetM be a GF-regular R-module, then every proper submodule of M contained in a maximal submodule.Proof.
LetN be a proper submodule of M. Since M is a GF-regular R-module, then M/N≠0 is GF-regular (Proposition 7), so M/N contains a maximal submodule (Corollary 35), which means that there exists a submodule K of M such that N⊆K, K/N is a maximal submodule of M/N; therefore, K is a maximal submodule of M and contains N.Corollary 38.
Every simple submodule of aGF-regular R-module is direct summand.Proof.
LetN be a simple submodule of a GF-regular R-module M, then N is cyclic; say N=Rx, then there exists a maximal submodule 𝔐 of M such that x∉𝔐 (Corollary 37). It is clear that M=𝔐+Rx. Now, if Rx∩𝔐≠(0), then Rx∩𝔐=Rx because Rx is a simple submodule. Thus, x∈𝔐 which is a contradiction, so  M=Rx⨁𝔐.



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