On Rings whose Simple Singular R-Modules are GP-Injective

In this work we give a characterization of rings whose simple singular right R-modules are Gp-injective .We prove that if R is a quasi-duo ring whose simple singular right R-modules are Gp-injective, then any reduced right ideal of R is a direct summand. We also consider that a zero commutative ring with every simple singular left R-module is Gp-injective


Introduction:
Throughout this paper, R denotes an associative ring with identity, and all modules are unitary right R-modules. Recall that: (1) A right Rmodule M is called general right principally injective (briefly right Gpinjective) if for any 0aR there exists a positive integer n, such that a n 0 and any right R-homomorphism of a n R into M extends to one of R into M;(2) R is called reduced if R has no non-zero nilpotent elements; (3) R is right (left) quasi-duo ring if every maximal right (left) ideal of R is an ideal of R; (4) A ring R is called semi-prime if 0 is the only nilpotent ideal ; (5) for any element a in R we define a right annihilator of a by r(a)={xR:ax=0} and a left annihilator of a, l(a) is similarly defined.

Rings whose simple singular modules are GP-Injective:
In this section, we study rings whose simple singular right Rmodules are Gp-injective. We begin this section with the following result.

Proposition 2-1:
Let R be a qusi-duo ring, with every simple singular right R-modules is Gpinjective. Then any reduced right ideal of R is a direct summand. Proof: Let I=aR be a reduced principal right ideal of R. We shall show that aR+r(a)=R. if not, there exists a maximal right ideal M of R such that aR+r(a)M .Now, M is essential right ideal of R, if not, then there exists a non-zero right ideal L of R such that M  L=0 . Then aRLML M  L=0, implies that L  r(a) M, so M  L=L=0, and this is a contradiction.
So M must be essential right ideal of R. Therefore R/M is Gpinjective. Then there exists a positive integer n such that any Rhomomorphism of a n R into R/M extends to one of R into R/M . let f:a n R→R/M be defined by f(a n r)=r+M. f is a well-defined Rhomomorphism. Indeed , let r1,r2R such that a n r1=a n r2 . Then a n r1-a n r2=0, implies that a n (r1-r2)=0, so r1-r2r(a n ), since I is reduced. Therefore r(a n )=r(a) ,this implies that r1-r2r(a)M. Hence, r1+M=r2+M. Now R/M is Gp-injective,so there exists cR such that 1+M=f(a n ) =ca n +M. Hence , 1ca n M, since a n M and R is a quasi-duo ring , then ca n M and so 1M .
This contradicts M  R .
Therefore aR+r(a)=R. In particular ar+c=1, for some rR and cr(a), whence a 2 r=a. if we set d=ar 2 I , then a=a 2 d. clearly (a-ada) 2 =0, since I is reduced, thus a =ada , and hence I=eR, where e=ad is an idempotent element. Thus I is a direct summand.

Proposition 2-2:
Let R be a semi-prime ring with every simple singular right Rmodule is Gp-injective . Then every right ideal of R is an idempotent .
Proof:For any right ideal I of R, suppose there exists an element b in I , such that bI 2 . Then bR(bR) 2 . Since R is a semi-prime ring, then (bR) 2 is essential in bR. By zorn's lemma , the set of right ideals J such that (bR) 2 JbR has a maximal member L. Then bR/L is a simple singular, and therefore is Gp-injective. Now, let f:bR→bR/L is the canonical homomorphism defined by f(br)=br+L for all ring R , since bR/L is Gpinjective, so there exists cR, such that f(br)=(bc+L)br. Then f(b)=(bc+L)b=b+L , which implies that b+L=bcb+L. Hence; b-bcbL, whence it follows that bL. Thus bRL and this is a contradiction. Therefore I=I 2 .

3-Zero Commutative Rings
In this section we introduce the notion of a zero commutative ring in order to study the connection between rings whose simple singular right Rmodules are Gp-injective and other rings.

Definition3-1:
A ring R is called zero commutative (briefly ZC)if for a,bR , ab=0 if ba=0. We shall begin this section with the following result.

Lemma 3-2:
Let R be a ZC ring. Then RaR+l(a) is an essential left ideal of R. Proof:Given a R, assume that [RaR+l(a)]  I=0 , where I is a right ideal of R. Then aII  RaR=0, so Ir(a)l(a). Hence, I=0; where RaR+l(a)is an essential left ideal of R

Lemma 3-3:
Let R be a ZC ring with every simple singular left R-module is Gpinjective, then R is reduced.
Proof:Let a 2 =0 . suppose that a0. By lemma (3-2), l(a)is an essential left ideal of R. since a0, l(a)R. Thus, there exists a maximal essential left ideal M of R containing L(a), therefore R/M is Gp-injective. So any Rhomomorphism of Ra intoR/M extends to one of R into R/M. Let f:Ra→R/M be defined by f(ra)=r+M. Clearly, f is a well-defined R-homomorphism . Thus 1+M=f(a)=ac+M. Hence, 1-acM and so 1M, which is a contradiction. Hence a=0, and so R is reduced.

Definition3-4:
A ring R is said to be right weakly regular if for all a in R , there exists b in RaR such that a=ab . Now, we give the main result.

Proposition 3-5:
If R is ZC and every simple singular left R-module is Gp-injective, then R is a reduced weakly regular ring.
Proof:By Lemma (3-3) , R is a reduced ring. We shall show that RaR+l(a)=R for any aR. Suppose that there exists bR such that RbR+l(b)R . Then there exists a maximal left ideal M of R containing RbR+l(b). By Lemma (3-2), M must be essential in R. Therefore R/M is Gp-injective. So there exists a positive integer n such that any Rhomomrphism of Rb n into R/M extends to one of R into R/M. let f:Rb n →R/M be defined by f(rb n )=r+M . Since R is a reduced ring, f is a well-R-homomorphism. Now, R/M is Gp-injective , so there exists cR such that 1+M=f(b n )=b n c+M . Hence 1-b n cM and so 1M, which is a contradiction. Therefore RaR+l(a)=R for any aR. Hence R is a left weakly regular ring. Since R is reduced, then RaR+r(a)=R, implies that R is a right weakly regular ring. Therefore R is a weakly regular ring. Kimand Nam in [2] proved that. Rings whose simple right R-modules are Gp-injective are always semi-prime. But in general rings whose simple singular right R-modules are Gp-injective need not be semi-prime.

Proposition 3-6:
Let R be a ZC ring, and every simple singular left R-module is Gpinjective, then R is a semi-prime ring .
Proof: From Lemma (3-3), R is a reduced ring and then R is a semi-prime ring