# The Unique MaximalGF-Regular Submodule of a Module

Authors: Areej M. Abduldaim; Sheng Chen
Journal: The Scientific World Journal
 (2013)

---

## Abstract

AnR-module A is called GF-regular if, for each a∈A and r∈R, there exist t∈R and a positive integer n such that rntrna=rna. We proved that each unitary R-module A contains a unique maximal GF-regular submodule, which we denoted by MGF(A). Furthermore, the radical properties of A are investigated; we proved that if A is an R-module and K is a submodule of A, then MGF(K)=K∩MGF(A). Moreover, if A is projective, then MGF(A) is a G-pure submodule of A and MGF(A)=M(R)·A.


---

## Body

## 1. Introduction

Throughout this paper,R is a commutative ring with identity and all modules are left unitary, unless otherwise stated. Recall that an element r∈R is said to be regular if there exists t∈R such that rtr=r; a ring R is called regular if and only if each element of R is regular. An ideal I of a ring R is regular if each of its elements is regular in R; indeed, a regular ideal I of R is itself a regular ring [1]. Brown and McCoy proved in [1] that each ring R contains a unique maximal regular ideal M(R) which satisfies the well-known radical properties. The ideal M(R) is called the regular radical of R.The concept of regularity was extended to modules in several ways and in [2] the notion of F-regular modules (in the sense of Fieldhouse [3]) was generalized to GF-regular modules. Let A be an R-module; an element a∈A is said to be GF-regular if for each r∈R there exist t∈R and a positive integer n such that rntrna=rna. An R-module A is called GF-regular if and only if all its elements are GF-regular; in particular, a ring R is GF-regular if and only if R is GF-regular as an R-module. On the other hand a ring R is π-regular if and only if R is a GF-regular R-module; 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. A submodule N of an R-module A is called GF-regular if each element of N is GF-regular and every submodule of a GF-regular module is a GF-regular module. Also, in [2] the concept of G-pure submodules was introduced; a submodule P of an R-module A is called G-pure if, for each r∈R, there exists a positive integer n such that P∩RrnA=RrnP.In this paper we show that each module contains a unique maximalGF-regular submodule, which we denote by MGF(A), and we show that MGF(A) satisfies some but not all of the usual radical properties.


## 2. Main Results

Theorem 1.
LetR be any ring. Every R-module contains a unique maximal GF-regular submodule.Proof.
LetR be any ring, let A be an R-module, and let
(1)G={N∣NisaGF-regularsubmoduleofA},
where G≠ϕ because (0) is a GF-regular submodule of A. Let {Ni} be an ascending chain in G and B=⋃i∈Λ‍Ni. Let b∈B; there exists j∈Λ such that b∈Nj, but Nj is a GF-regular submodule; then; for each r∈R, there exist t∈R and a positive integer n such that rntrnb=rnb; therefore b is a GF-regular element in B which implies that B is a GF-regular R-module. Now, by Zorn’s lemma, G contains a maximal element which we call MGF. To prove the uniqueness of MGF, assume that MGF1 and MGF2 be two maximal GF-regular submodules in A; then for any maximal ideal P of R each of MGF1p and MGF2p is semisimple over Rp [2, Proposition 21]. Now, let MGF1p∩MGF2p=Kp; then Kp⊆MGF1p and Kp⊆MGF2p; thus MGF1p=Kp+A1p and MGF2p=Kp+A2p, where A1p and A2p are two submodules of Ap [4]. Hence, MGF1p+MGF2p=A1p+Kp+A2p, but each of A1p, A2p, and Kp is a semisimple submodule; thus MGF1p+MGF2p is a semisimple submodule which implies that MGF1p+MGF2p is GF-regular [2]. So MGF1+MGF2 is a GF-regular submodule [2, Theorem 20]. Now, each of MGF1 and MGF2 is a maximal GF-regular submodule and hence MGF1+MGF2=MGF2=MGF1.Remark 2.
We denote the unique maximalGF-regular submodule of an R-module A by MGF(A). It is obvious that MGF(A) contains every GF-regular submodule of A; this means that MGF(A) is a GF-regular submodule which is not contained properly in any other GF-regular submodule. In fact, MGF(A) is the sum of all GF-regular submodules of A and MGF(A)=A if and only if A is a GF-regular module.Example 3.
(a) Since theZ-module Zn is GF-regular for each positive integer n [2], then MGF(Zn)=Zn.
(b) Each element in theZ-module Q is not GF-regular [2]; hence MGF(Q)=(0).
(c) Letp be a prime number and let A=Zp∞=⋃∀i‍Zpi be a Z-module. Let a∈⋃∀i‍Zpi; then there exists a positive integer m such that a∈Zpm, but Zn is a GF-regular Z-module for each positive integer n; hence a is a GF-regular element, so a∈MGF(Zp∞) which implies that MGF(Zp∞)=Zp∞.
(d) LetQ/Z be a Z-module; since Q/Z is a torsion Z-module, then Q/Z is a GF-regular Z-module [2, Proposition 6]. Since Q/Z=∑p‍Zp∞ for each prime number p, then MGF(Q/Z)=∑p‍Zp∞ for all primes p.Proposition 4.
LetA and B be R-modules, and let K be a submodule of A; then (a)
MGF(K)=K∩MGF(A),(b)
MGF(A⊕B)⊆MGF(A)⊕MGF(B).Proof.
(a) LetK be a submodule of A, and let k∈MGF(K); then k∈K and k is GF-regular in K which implies that k is GF-regular in A, thus k∈K∩MGF(A). Conversely, let k∈K and k∈MGF(A); therefore k is GF-regular in K which means that k∈MGF(K) and hence MGF(K)=K∩MGF(A).
(b) Letc∈MGF(A⊕B); then c=(a,b), where a∈A and b∈B. Since c is GF-regular, then each of a and b is GF-regular which means that a∈MGF(A) and b∈MGF(B); hence c∈MGF(A)⊕MGF(B).Proposition 5.
LetA and A′ be R-modules, and let f:A→A′ be an R-homomorphism; then f(MGF(A))⊆MGF(f(A)).Proof.
Ifa∈MGF(A), then R/ann(a) is a π-regular ring, but ann(a)⊆ann(f(a)); thus R/ann(f(a)) is an epimorphic image of R/ann(A); hence it is a π-regular ring. Therefore, f(a)∈MGF(f(a)) and f(MGF(A))⊆MGF(f(A)).Remark 6.
(a) Iff:A→A′ is an R-epimorphism, then f(MGF(A))≠MGF(A′) in general. In fact, let π:Z→Z4≃Z/4Z be the natural map, where Z and Z4 are Z-modules. It is easy to check that MGF(Z)=(0), f(MGF(Z))=(0), but MGF(Z4)≃Z2.
(b) It is shown in [1] that, for a ring R, M(R/M(R))=(0) which is not true in case of GF-regular modules; this means that MGF(A/MGF(A))≠(0) (as in (a)).Corollary 7.
For eachR-module A, M(R)·A⊆MGF(A).Proof.
For eacha∈A, let f:R→A be an R-homomorphism defined by f(r)=ra. Then f(M(R))⊆MGF(A) by Proposition 5, but M(R)·A=∑a‍f(M(R)); hence M(R)·A⊆MGF(A).LetJ(R) be the Jacobson radical of a ring R. Brown and McCoy proved in [1] that M(R)∩J(R)=(0). However, this is not true for GF-regular modules; for example, if A=Z4 is a Z-module, then MGF(A)≃Z2, J(A)≃Z2 and MGF(A)∩J(A)≠(0).Lemma 8.
LetA be an R-module and let P be a G-pure submodule of A. For any r∈R, there exists a positive integer n such that P=RrnP if and only if P⊆RrnA.Proof.
SinceP is G-pure in A, then, for each r∈R, there exists a positive integer n such that P∩RrnA=RrnP. If P=RrnP, then P∩RrnA=P, and hence P⊆RrnA. Conversely, if P⊆RrnA, then P∩RrnA=P, but P∩RrnA=RrnP; therefore P=RrnP.Lemma 9.
Letr∈J(R); if P is a finitely generated G-pure submodule of an R-module A such that P⊆RrnA for some positive integer n, then P=0.Proof.
By Lemma8 we get that P=RrnP and by Nakayama’s lemma [5], P=0.Theorem 10.
LetA be an R-module. If MGF(A) is a G-pure submodule of A, then MGF(A)∩J(R)·A=(0).Proof.
Letr∈MGF(A)∩J(R)·A, and let P=Rr. It is clear that P⊆MGF(A). Since MGF(A) is a GF-regular module, then P is a G-pure submodule in MGF(A) [2, Theorem 11]. But MGF(A) is G-pure in A; hence P is G-pure in A. Now, P⊆J(R)·A, so P=0 by Lemma 9. Therefore MGF(A)∩J(R)·A=(0).Recall thatM(R) is always a pure ideal in R. Hence M(R) is G-pure [2].Theorem 11.
LetA be a projective R-module; then (a)
MGF(A)=M(R)·A,(b)
MGF(A) is a G-pure submodule of A,(c)
MGF(A)∩J(A)=(0).Proof.
(a) By the dual basis lemma [4], for each a∈A we have that a=∑i‍fi(a)ai, where a,   ai∈A for all i and fi∈A*:=HomR(A,R). If a∈MGF(A), then the submodule Ra is GF-regular and fi(Ra) is a GF-regular ideal in R by Proposition 5, hence M(R). Thus MGF(A)⊆M(R)·A. We get the other direction of the inclusion by Corollary 7.
(b) First we claim that; for any two idealsK and L of R, (K∩L)A=KA∩LA; it is enough to show this locally; thus we may assume that A is free. It is clear that (K∩L)A⊆KA∩LA. On the other hand, let x∈KA∩LA; then
(2)x=∑‍rixi=∑‍sixiri∈K,si∈L.
By freeness, ri=si and x∈(K∩L)A. Now, let K be any ideal in R; then by (a) we get MGF(A)∩KA=M(R)∩KA=(M(R)∩K)A. But M(R) is G-pure ideal, so, for each r∈R, there exists a positive integer n such that M(R)∩Rrn=RrnM(R); hence MGF(A)∩RrnA=RrnM(R)·A=RrnMGF(A).
(c) SinceA is projective, then J(A)=J(R)·A [4] which implies that MGF(A)∩J(A)=(0) by Theorem 10.Corollary 12.
LetR be any ring, and let A be any R-module such that MGF(A) is a G-pure submodule and J(A)=J(R)·A; then MGF(A)∩J(A)=(0).Proof.
SinceJ(A)=J(R)·A, then by Theorem 10 we get that MGF(A)∩J(A)=(0).Remark 13.
IfA is a GF-regular R-module, then MGF(A)=A; hence J(R)·A=J(R)·A∩A=J(R)·A∩MGF(A)=(0). In fact, this shows that Theorem 10 is a generalization of [2, Proposition 28].
In [2] we noticed that every module over π-regular ring is GF-regular, but the converse need not be true in general. The next result shows how the converse may be true, but first we recall that if A is an R-module, then the trace of A is tr(A)=∑f∈A*‍f(A), where A*=Hom(A,R).Proposition 14.
LetA be a GF-regular R-module. If tr(A)=R, then R is a π-regular ring.Proof.
For eacha∈A and f∈A*=Hom(A,R), since Ra is a GF-regular submodule of A, then by [2, Proposition 7] we get that f(Ra) is a π-regular ideal. Thus f(Ra)⊆M(R), but f(Ra)⊆tr(A); hence tr(A)=R⊆M(R), which implies that M(R)=R and R is π-regular.



---

*Source: 750808-2013-09-15.xml*