On Weakly Regular Rings and SSF-rings

صخلملا ةفيعضلا ةمظتنملا تاقلحلا انسرد ثحبلا اذه يف , ىـنميلا ةذاشـلا اهتاساقم نوكت يتلا لا ةحطسم ةطيسب . طرشلا انسرد كلذكو (*) ةقلحلا ىلع R ققحي يذلاو L(a)⊆r(a) لكل a يف R . طرشلا ققحت يتلا ةقلحلا نأ انتبثأو (*) ةحطسم ةطيسبلا ىنميلا ةذاشلا اهتاساقم نوكت يتلاو , نأف ةمظتنم ةقلح نوكي ةقلحلا زكرم . ةقلحلا نإ اضيأ انتبثأو R ىـنميلا ةذاشلا اهتاساقم نوكت يتلا طرشلا ققحت و ةحطسم ةطيسبلا (*) نأف ةيوق ىنمي ةديقم وأ R ةفيعض ةمظتنمو ةلزتخم ةقلح .


1-Introduction
Through out this paper, R denotes an associative ring with identity, and all modules are unitary ring R-modules. For any non-empty subset X of ring R, r(X) and L(X) denote the right annihilator of X and the left annihilator of X, respectively. Z(R), J(R) will denote respectively the center of R and Jacobson redical of R. Recall that :-1-R is called reduced if R has no non-zero nilpotent elements . 2-R is said to be Von Numann regular (or just regular) if , aaRa for every aR, and R is called strongly regular if aa 2 R. Clearly every strongly regular ring is a regular reduced ring. 3-A ring R is said to be right (left) quesi duo ring [3], if every maximal right(left)ideal is a two-sided ideal of R. 4-Following [6], for any ideal of R, R/I is flat if and only if for each aI ,there exists bI such that a=ba. 5-A ring R is called weakly right duo [4] if for any aR, there exists a positive integer n such that a n R is a two-sided ideal of R.

Definition 2-1:-
A ring R is called a right (left) SSF-ring ,if every simple singular right (left) R-module is flat.
The following lemma is well-known, so we omit its proof.

Lemma 2-2:-
For any aZ(R) , if ara=a for some rR, then there exists bZ(R) such that a=aba. Proof:-see [7] We consider the condition(*) :R satisfies L(a) r(a) for any aR.

Proposition 2-3:-
If R satisfies (*), SSF-ring, then the center Z(R) of R is a Von Neumann regular ring. Proof:-First we will show that aR+r(a)=R for any aZ(R). If not, there exists a maximal right ideal M of R such that aR+r(a) M. Since aZ(R), aR+r(a) is an essential right ideal and so M must be an essential right ideal of R. Since R/M is flat and aM ,then there exists bM such that a=ba and hence (1-b) L(a) r(a) M implies lM, which is a contradiction. Therefore aR+r(a)=R for any a Z(R)and so we have a=ara for some rR. Applying Lemma (2-2) Z(R) is a Von Neumann regular ring.
Recall that a ring R is right (left) weakly regular if I 2 =I for each right (left) ideal I of R; equivalently. aaRaR (aRaRa) for every aR. R is weakly regular if it is both right and left weakly regular [5].

Lemma 2-4: -
If R satisfies (*), then RaR+r(a) is an essential right ideal. Proof :-see [7] Theorem 2-5: -If R satisfies (*), and SSF-ring, then R is a reduced weakly regular ring. Proof:-Let a 2 =0. Suppose that a0.By Lemma (2-4),r(a) is essential right ideal of R. Since a0,r(a) R. Thus there exists a maximal essential right ideal M of R containing r(a). Since R/M is flat and aM there exists bM such that a=ba and hence (1-b) L(a) r(a) M and so 1M, which is a contradiction. Hence a=0 and so R is reduced. Now, we will show that RaR+r(a)=R for any aR. Suppose that there exists bR such that RbR+r(b) R .Then there exists a maximal right ideal M of R containing RbR+r(b). By Lemma(2-4), M must be essential in R. Therefore R/M is flat . Then there exists cM such that b=cb and hence(1-c)L(b)r(b)M and so 1M, which is a contradiction. Therefore RaR+r(a)=R for any aR. Hence R is a right weakly regular ring. Since R is reduced, it also can be easily verified that R is a weakly regular ring.

Corollary 2-6: -
If R is a reduced and SSF-ring, then R is a weakly regular ring.

Lemma 2-7: -
Let R be a right quasi duo ring. Then R/J(R) is a reduced ring.

Proof:-see[6]
Proposition 2-8 :-Let R be a right quasi duo ring. The following statements are equivalent. a) R is a right weakly regular ring. b) R is a strongly regular ring. Proof :-see [6] Proposition2-9:-Every weakly right (left) duo ring is right (left)quasi-duo. Proof:-see [1] Proposition 2-10: -Let R be a right(left) quasi duo ring. If every simple singular right Rmodule is flat, then R/J(R) is strongly regular.
) for any R a  . Then by Lemma (2-4) ) (a r R a R + is an essential right ideal of R . Thus R/J(R) must be right essential in R . Therefore R/M is a simple singular right R-module and so R/M is flat, then there exists cM such that a=ca and hence (1-c) L(a) r(a) M and so IM, which is a contradiction. Therefore R/J(R) is right weakly regular since R/J(R) is reduced it also can be easily verified that R/J(R)is a weakly regular ring. By proposition(2-8), R is a strongly regular ring.
Corollary 2-11: -Let R be aweakly right duo, SSF-ring. Then R/J(R) is a right weakly regular ring. Proof:-By Proposition (2-9) R is a right quasi duo ring. Also by Proposition (2-10) R/J(R) is a right weakly regular ring.
A ring R is called strongly right bounded (briefly SRB) [2] if every non-zero right ideal contains a non-zero two-sided ideal of R.

Lemma 2-12:-
If R is a semi prime SRB ring, then R is reduced. Proof :-see [2] Theorem 2-13: -Let R be a SRB and SSF-ring. Then R is a reduced weakly regular. Proof: -By Corollary(2-6) and Lemma (2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12), it is enough to show that R is semi prime. Suppose that there exists a non-zero right ideal A of R such that A 2 =0. Then there exists 0aA such that a 2 =0. First observe that r(a) is an essential right ideal of R. If not there exists a non-zero right ideal K such that r(a)K is right essential in R. Since R is SRB, there is a non-zero ideal I of R such that I  K. Now aIaR  I r(a)  K = 0 . Hence Ir(a) K=0 ( aI  I). This is a contradiction. Thus r(a) must be a proper essential right ideal of R. Hence there exists a maximal right ideal M of R containing r(a). Clearly M is an essential right ideal of R, R/M is flat, then there exists cM such that a=ca. Now acaaRa A 2 =0 and so 1M, which is a contradiction. Therefore R must be semi-prime, hence R is a reduced weakly regular.