On Simple GP – Injective Modules

In this paper, we study rings whose simple right R-module are GP-injective. We prove that ring whose simple right R-module is GP-injective it will be right -weakly regular ring. Also, proved that if R is N duo ring or R is NCI ring whose simple right Rmodule is GP-injective is S-weakly regular ring.


1-Introduction
Throughout in this paper, R is associative ring with identity and all modules are unitary. For a subset X of R, the left(right) annihilator of X in R is denoted by l(X)(r(X)). If X={a}, we usually abbreviate it to l(a)(r(a)). We write J(R),N(R), for the Jacobson radical, the set of nilpotent elements, the nil radical (that means the sum of all nil ideals), prime radical (that means the intersection of all prime ideals)respectively. .

A ring R is called NI if A ring R is 2-primal if
A ring R is said to be semiprimitive if J(R)=0 [1]. An element in the ring R said to be right (left) weakly regular if ) [12]. A right R-module M is called Generalized Principally injective (briefly, GPinjective) if for any there exists a positive integer n such that and any right R-homomorphism of into M extends to one of R into M [8]. Right GPinjective modules are called right YJ-injective modules by several authors [16].

Some Properties of Rings whose Simple Right R-module are GP-injective.
We give a different prove that proved by Kim, et. al. in [8].

Theorem 2.1
Let R be a ring whose every simple right R-module is GP-injective. Then R is semiprime.

Proof:
We shall show that is no nilpotent ideal in R, if not, suppose there exists with , , that means , there exists a maximal right ideal M of R containing r(a), R/M is GP-injective. Hence, there exists an appositive integer n=1 such that and any R-homomorphism of R into R/M extends to one of R into R/M, we define such that where . We have to show that f is well defined R-homomorphism, let where , , x+M=y+M, f(ax)=x+M=y+M=f(ay), f(ax)=f(ay),so f is well defined right R-homomorphism, since R/M is GP-injective, there exists such that 1+M=f(a)=(b+M)(a+M)=ba+M, 1+M=ba+M, , since , we get that which is a contradiction. Therefore, a=0. This shows that R issemiprime.
We give a different prove that is proved by Xue in [16].

Proposition 2.2
Let R be a ring whose simple right R-module is GP-injective. Then, R is semiprimitive.

Proof:
We shall show that if not, there exists then either or not, if not, there exists a maximal right ideal M of R containing R/M is GP-injective, there exists a positive integer n and such that any R-homomorphism of R into R/M extends to one of R into R/M,Let such that where , we have to show that f is well defined, let where then (x-y)+M=M, x+M=y+M, f( x)=x+M=y+M=f( y), f( x)=f( y),so f is well defined right R-homomorphism, since R/M is GP-injective, there exists such that 1+M=f( )=(b+M)( +M)=b +M, 1+M=b +M, since we get that which is a contradiction. That means in particular there exists and such that since so , is invertible, there exists such that must which is a contradiction with . Therefore, a=0, so, J(R)=0.This shows that R is semiprimitive.

Corollary 2.3
Let R be a ring whose simple right R-module is GP-injective. Then, .

Proof:
We shall show that since is the large nil ideal of R, It is clearly that J(R) containing every nil ideal, so but by Proposition 2.3. Thisshows that

Theorem 2.4
Let R be a ring whose simple right R-module is GP-injective. Then, the set is right weakly regular.

Proof:
We shall show that for all if not,suppose there exists such that then there exists a maximal right ideal M of R containing R/M is GP-injective, there exists appositive integer n=1 such that and any R-homomorphism of R into R/M extends to one of R into R/M. Now, let such that where . Note that f is well defined right R-homomorphism, since R/M is GP-injective there exists such that 1+M=f(a)=(b+M)(a+M)=ba+M, 1+M=ba+M, , since , we get that which is a contradiction.. Therefore, In particular, there exists and such that that is for all This shows that the set is right weakly regular.

Theorem 2.5
Let R be a ring without zero divisors whose simple right R-module is GPinjective. Then, R is a simple ring.

Proof:
We shall show that there is no two sided ideal of R, if not there exists a two sides ideal of R, RaR is a two sided ideal for some since ,RaR if there exists a maximal right ideal M of R containing R/M is GPinjective, there exists a positive integer n and such that any R-homomorphism of R into R/M extends to one of R into R/M, we define such that where , let r1,r2 such that r1= r2, )=0, since a is a non-zero divisor, so must , which is a contradiction. Therefore, for all that means R not containing any two sided ideal of R. This shows that R a simple ring.

Rings Whose Simple Right R-module are GP-injective and it relation with other Rings.
In this section, we give different conditions to the ring whose simple right Rmodule is GP-injective to get the reduced, S-weakly regular, regular, strongly regular ring.
A ring R is said to be N duo if aR=Ra, for all [15].

Theorem 3.1
Let R be N duo ring whose every simple right R-module is GP-injective. Then, R is a reduced ring.

Proof:
We shall show that N(R) =0, if not, there exists with if aR+r(a) , there exists a maximal right ideal M of R containing is GP-injective, there exists an appositive integer n=1 such that and any Rhomomorphism of R into R/M that extends to one of R into R/M. Let such that where, . Note that f is well defined right Rhomomorphism, since R/M is GP-injective there exists such that 1+M=f(a)=(b+M)(a+M)=ba+M, 1+M=b+M, , since R is N duo ring and we get aR=Ra, , so which is a contradiction. Therefore, In particular, there exists and such that for all This shows that is a reduced ring.
Call a ring R NCI if N(R) is containing a non-zero ideal of R whenever Clearly, NI ring is NCI [5].

Theorem 3.2
Let R be an NCI ring whose simple right R-module is GP-injective. then R is a reduced ring.

Proof:
We shall show that N(R) =0, if not, since R is an NCI ring, so N(R) is containing a non-zero ideal I, but I is nil ideal, It is clearly that J(R) containing every nil ideal, so I form proposition 2.4, I=0, that is mean N(R) mustbe an ideal, similarly N(R)=0.This is shows that is reduced ring.

Theorem 3.3
Let R be aring whose simple right R-module is GP-injective. Then, the following conditions are equivalent: 1-R is reduced ring.

Proof:
We shall prove that R is S-weakly regular when R is reduced, and the proof of the other condition that is clearly form Theorem 3.3.
We shall show that for all if not, there exists such that there exists a maximal right ideal M of R containing R/M is GP-injective, there exists an appositive integer n such that and any R-homomorphism of R into R/M extends to one of R into R/M. Let such that where, . let where since R is a reduced ring, then (x-y)+M=M, x+M=y+M, f( x)=x+M=y+M=f( y), f( x)=f( y),so f is well defined right R-homomorphism, since R/M is GP-injective there exists such that 1+M=f( )=(b+M)( +M)=b +M, 1+M=b+M. Now b M, it is true when , which is a contradiction. Therefore, for Now, when n=1, 1+M=f( )=(b+M)( +M)=b +M, 1+M=b +M, by multiply a+M in the left side and b+M in the right side, we have ba+M= +M, since M, then , we get that , since 1+M=b +M, 1-ba which is a contradiction. Therefore , for all a . In particular, there exists y, and such that y This shows that R is an S-weakly regular ring.
An element a of ring R is said to be a right regular element if the right annihilator ideal is zero (r(a)=0) [6]. A ring R is said to be MERT if and only if every maximal essential right ideal of R is an ideal [14]. A ring R is called Kasch if every simple right R-module embeds in R, equivalently, for every maximal right ideal M of R is a right annihilator of R [3]. Call a ring R a right SF-ring if each simple right Rmodule is flat [13]. A ring R is said to be regular if for all

Lemma 3.5 [6]
Let R be a semiprime ring with maximum condition on left and right annihilators. Then, every essential right ideal contains a regular element.

Lemma 3.7 [7]
Let R be a MERT ring, then the following conditions are equivalent: 1-R is regular. 2-R is right SF.

Theorem 3.8
Let R be a MERT ring whose every simple right module is GP-injective and satisfies maximum condition on left and right annihilators. Then R is Kasch ring and right SF-ring, hence R is regular ring.

Proof :
We shall prove that every maximal right ideal is direct summand, if not, suppose that M a maximal right ideal of R which is not a direct summand of R, then M is a maximal essential right ideal of R, by Theorem 2.1 and Lemma 3.5, we have M containing a non-zero divisor a, R/M is GP-injective, there exists a positive integer n and such that any R-homomorphism of R into R/M extends to one of R into R/M, Let such that where , since a is a non-zero divisor f is well defined right R-homomorphism, since R/M is GP-injective, there exists such that 1+M=f( )=b +M, , since M is essential right ideal and R is MERT ring, M is an ideal, so , we get that which is a contradiction. Therefore, M a direct summand. This shows that every maximal right ideal of R is a direct summand.
There exists J right ideal for any maximal right ideal M such that , in particular there exists such that m+j=1, so for all , jd=0, then but M is a maximal right ideal, we have M=r(j), for every maximal right ideal, This shows that R is a right kasch ring.
Also, md=d for all from Lemma 3.6, we get that R/M is flat right Rmodule and that is for all M maximal right ideal of R. This shows that R is a right SFring.
Since, R is MERT and right SF-ring, by using Lemma 3.7, we get that R is a regular ring.

Theorem 3.9
Let R be N duo ring and MERT whose simple right R-module is GP-injective. Then, R is a strongly regular ring.

Proof:
We shall show that , for all . If not then there exists , for some , there exists a maximal right ideal M of R containing M is either essential or direct summand, if M is not essential, then M=r(e) for some , by Theorem 3.1, R is a reduced ring, , e , but hence e=0, and M==r(e)=r(0)=R, M=R, which is a contradiction. Therefore, M is an essential right ideal of R. Thus, R/M is GP-injective, there exists a positive integer n such that and any R-homomorphism of R into R/M extends to one of R into R/M, Let be defined by where .Note that f is well defined right R-homomorphism, because R is a reduced ring. since R/M is GP-injective, there exists such that 1+M=f( )=b +M, , since M is an essential right ideal and R is MERT ring, M is an ideal, so , we get that which is also contradiction. Therefore, for all This shows that R is a strongly regular ring.

Finally, we give the following important result.
In [16] Xue proved , if every simple left R-module is GP-injective, then for any nonzero there exists a positive integer n=n(a) such that and [Proposition 2].
In the above proof, we have and , it is clear that proof leads us to because , we give a new proof that strengthens the above , hence A ring R is said to be right (left) s -weakly regular ring if, for every there exists a positive integer n, depending on such that [10].

Theorem 3.10
Let R be a ring whose simple right R-module is GP-injective. Then R is right -weakly regular ring.

Proof:
Let and is not a nilpotent element, if then there exists a maximal right ideal of containing by hypothesis is , we define such that where , we show that f is well defined, let where , , then (x-y)+M=M, x+M=y+M, f( x)=x+M=y+M=f( y), f( x)=f( y). So f is well defined right R-homomorphism, since R/M is GP-injective, there exists and such that 1+M=f( )=(b+M)( +M)=b +M, 1+M= +M, , since1+M=b +M, multiply by from left and b+M from right, we get +M= +M, , but so but we get that which is a contradiction. Therefore, In particular, there exists and such that and that is for all which is not nilpotent elements. When is nilpotent element, there exists a positive integer m such that so , for any This shows that is a right -