On Strongly Regular Rings and Generalizations of Semicommutative Rings

International Mathematical Forum, Vol. 7, 2012, no. 16, 777-790 On Strongly Regular Rings and Generalizations of Semicommutative Rings Tikaram Subedi Department of Mathematics North Eastern Hill University, Permanent Campus Shillong-793022, Meghalaya, India tsubedi2010@gmail.com Ardeline M. Buhphang Department of Mathematics North Eastern Hill University, Permanent Campus Shillong-793022, Meghalaya, India ardeline17@gmail.com Abstract. Some properties of a ring R in which l(a) is a GW-ideal of R for every a R are given. Further, it is shown that a ring R is strongly regular if and only if R is a left quasi duo ring whose maximal essential right ideals are YJ-injective if and only if R is a left N duo ring whose maximal essential left ideals are YJ-injective if and only if R is a left N duo left (resp., right) SF-ring. Mathematics Subject Classification: 16E50; 16D25; 16D40; 16D50 Keywords: Von Neumann regular rings, strongly regular rings, GP-V -rings; SF-rings, generalised weak ideal 1 Introduction Throughout this paper, R denotes an associative ring with identity and all our modules are unital. The symbols J(R), I(R), N(R), respectively stand for the Jacobson radical, the set of all idempotent elements and the set of all nilpotent elements of R. By an ideal of R, we mean a two sided ideal of R. As usual, R is a reduced ring if N(R) = 0. R is a left (resp., right) duo ring if every left (resp., right) ideal of R is an ideal. R is a left quasi duo (resp., MELT ) ring if every maximal (resp., maximal essential) left ideal of R is an

778 T. Subedi and A. M. Buhphang ideal. Right quasi duo rings and MERT rings are defined similarly. A ring R is left (resp., right) N duo ([6]), if for every a N(R), Ra (resp., ar) is an ideal of R. A ring R is a ZI-ring ([1]) or a semicommutative ring ([6]) if for every a, b R such that ab = 0, we have arb = 0. It is well known that R is a ZI-ring if and only if l(a) (resp., r(a)) is an ideal of R for every a R. A ring R is abelian if every idempotent of R is central. R is (von Neumann) regular if for every a R, there exists some b R such that a = aba. R is strongly regular if for every a R, there exists some b R such that a = a 2 b. It is known that a ring R is strongly regular if and only if R is a reduced regular ring. Following [2], a ring R is left weakly regular (resp., right weakly regular) if for every left (resp., right) ideal I of R, I = I 2 and R is weakly regular if it is both left and right weakly regular. A regular ring is clearly weakly regular, but a weakly regular ring need not be regular (for example, see [2], Remark 6). A left R-module M is Y J injective ([8])(resp., Wnil-injective ([6])) if for every 0 a R (resp., N(R)), there exists a positive integer n such that a n 0 and every left R-homomorphism from Ra n to M extends to one from R to M. Right YJ-injective modules and right Wnil-injective modules are similarly defined. YJ- injective modules defined here are same as GP-injective modules in [1]. R is a left (resp., right) GP-V-ring if every simple left R-module is YJinjective ([8]). R is a left (resp., right) GP-V -ring if every simple singular left R-module is YJ-injective ([8]). A regular ring is a left (resp., right) GP-V-ring but a left (resp., right) GP-V-ring need not necessarily be regular (see, [8]). R is a left (resp., right) SF-ring ([3]), if simple left (resp., right) R-modules are flat. It is well known that regular rings are left (resp., right) SF-rings. Ramamurthy in [3] initiated the study of left (resp., right) SF-rings and of the question whether a left (resp., right) SF-ring is necessarily regular. Since then, left (resp., right) SF-rings have been extensively studied by many authors and the regularity of left (resp., right) SF-rings which satisfy certain additional conditions is proved (cf. for example, [3], [4], [7], [9]-[12]) but the question remains open. Definition 1.1 Following [12], a left ideal L of a ring R is a generalized weak ideal (GW -ideal), if for every a L, there exists some n > 0 such that a n R L. A right ideal K of R is defined similarly to be a generalized weak ideal. Example 1.2 There exists a ring R in which (1) Every left ideal of R is a GW-ideal. (2) Every right ideal of R is a GW-ideal. (3) R is not left N duo.

On strongly regular rings 779 (4) R is not right N duo. (5) R is left quasi duo. (6) R is right quasi duo. (7) R is not a ZI-ring. Proof. Take R = a a 1 a 2 a 3 0 a a 4 a 5 0 0 a a 6 0 0 0 a : a, a i Z 2, i = 1, 2, 3, 4, 5, 6 Then every non-unit of R is nilpotent and hence it follows that every left (resp., right) ideal of R is a GW-ideal. Let M be a maximal left ideal of R and s M, r R such that sr / M. Then M + Rsr = R and hence u + vsr = 1 for some u M, v R. Now, s is not a unit so that s = 0 a 1 a 2 a 3 0 0 a 4 a 5 0 0 0 a 6. for some a i Z 2, i = 1, 2, 3, 4, 5, 6. It is easy to see that (vsr) 4 = 0, that is (1 u) 4 = 0. Then from u M, we get 1 M, a contradiction to M R. Hence M is an ideal of R. Therefore R is left quasi duo. Similarly R is right quasi duo. Let x = Rx = duo. Let y = 0 1 0 0 0 b 0 0 0 0 0 1. Then x2 = 0 so x N(R), but : b Z 2 is not an ideal of R so that R is not left N. Then y2 = 0 and

780 T. Subedi and A. M. Buhphang yr = N duo. Let d = 0 0 0 c a a 1 a 2 a 3 0 a a 4 a 5 0 0 a a 6 0 0 0 a l(d) = 0 0 0 1 0 0 0 1 0 0 0 1 : c Z 2 is not an ideal of R, hence R is not right. Then it is easy to see that d = 0 if and only if a = 0, a 4 = 0, a 1 = a 2. Thus 0 b 1 b 1 b 2 0 0 0 b 3 0 0 0 b 4 Now L is not an ideal as 0 1 1 0 L, 0 1 1 0 : b i Z 2, i = 1, 2, 3, 4 0 0 1 0 0 0 1 0 It follows that R is not a ZI-ring. R, but = 0 0 1 0 = L (say). / L. Example 1.3 There exists a ring R such that R is left N duo but not for every a R, l(a) is a GW-ideal of R. {( ) a b Proof. Take R = : a, b, c Z 0 c 2 }. Then {( ) ( )} 0 0 0 1 N(R) =,. ( ) 0 1 It is easy to see that R is an ideal of R. Therefore R is a left N duo 0 0

On strongly regular rings 781 (( 0 0 ring. Now, l 0 1 )) = {( x 0 0 0 ) } : x Z 2 is not a GW-ideal of R. The purpose of this paper is to study those rings R satisfying l(a) is a GW-ideal of R for all a R. Also we prove the regularity of some classes of left (resp., right) SF-rings. 2 YJ-injective modules and GW-ideals Proposition 2.1 The following conditions are equivalent for a ring R: (1) R is abelian. (2) l(e) is a GW-ideal of R for every e I(R). (3) r(e) is a GW-ideal of R for every e I(R). Proof. Clearly (1) = (2) and (1) = (3). (2) = (1). Let e I(R) and x R. Since 1 e l(e) and l(e) is a GW-ideal, there exists a positive integer n such that (1 e) n x l(e) which implies that xe = exe. Again, 1 e I(R) and e l(1 e). Since l(1 e) is a GW-ideal, there exists a positive integer m such that e m x l(1 e). This gives ex = exe. Therefore ex = xe. Hence R is abelian. Similarly (3) = (1). Since a ring R is strongly regular if and only if R is an abelian regular ring, the following corollary follows. Corollary 2.2 The following conditions are equivalent: (1) R is strongly regular. (2) R is a regular ring such that l(e) is a GW-ideal of R for every e I(R). (3) R is a regular ring such that r(e) is a GW-ideal of R for every e I(R). Lemma 2.3 Let R be an abelian ring and a R. Then (1) If M is a maximal left ideal of R such that l(a) + Ra M, then M is essential. (2) If K is a maximal right ideal of R such that r(a) + ar K, then K is essential.

782 T. Subedi and A. M. Buhphang Proof. (1). If M is not essential, then there exists 0 e I(R) such that M = l(e). Since a M = l(e), we have ae = 0. It follows from R is abelian that ea = 0. This shows that e l(a) M = l(e), whence e = e 2 = 0, a contradiction to e 0. Therefore M is an essential left ideal of R. Similarly we can prove (2). Lemma 2.4 Let a R. Then (1) If M is a maximal left ideal of R such that Rr(a) + Ra M, then M is essential. (2) If K is a maximal right ideal of R such that l(a)r + ar K, then K is essential. Proof. (1). If M is not essential, then there exists 0 e I(R) such that M = l(e). Now, a M = l(e), so e r(a) Rr(a) M = l(e). This yields e = e 2 = 0, a contradiction. This proves that M is an essential left ideal of R. Similarly we can prove (2). Theorem 2.5 Let R be a reduced ring. The following conditions are equivalent: (1) R is strongly regular. (2) Every maximal essential left ideal of R is YJ-injective. (3) Every maximal essential right ideal of R is YJ-injective. Proof. Clearly (1) = (2) and (1) = (3). (2) = (1). Since R is reduced, l(w) = r(w) = r(w m ) for every w R and for every positive integer m. Let 0 u R such that l(u) + Ru R, then there exists a maximal left ideal M of R containing l(u) + Ru. Since a reduced ring is abelian, by Lemma 2.3, M is essential, so that by hypothesis, M is YJ-injective. Therefore there exists a positive integer n such that u n 0 and every left R-homomorphism from Ru n to M extends to one from R to M. It follows that there exists some v M such that u n = u n v. This gives 1 v r(u n ) = r(u) M which yields 1 M, a contradiction. Therefore for all a R, l(a) + Ra = R. This proves that R is strongly regular. Similarly (3) = (1). Proposition 2.6 Let R be a ring such that l(a) is a GW-ideal of R for every a R. If R satisfies one of the following conditions, then R is reduced. (1) Every simple singular left R-module is Wnil-injective.

On strongly regular rings 783 (2) Every simple singular right R-module is Wnil-injective. (3) R is a left GP-V -ring. (4) R is a right GP-V -ring. (5) Every maximal essential right ideal of R is Wnil-injective. (6) Every maximal essential right ideal of R is YJ-injective. Proof. (1). If R not reduced, then there exists some 0 b R such that b 2 = 0. There exists a maximal left ideal M of R such that l(b) M. By Proposition 2.1 and Lemma 2.3, M is essential and so R/M is a simple singular left R-module. By hypothesis, R/M is Wnil-injective, so every left R-homomorphism from Rb to R/M extends to one from R to R/M. Define f : Rb R/M by f(rb) = r + M for every r R. Since l(b) M, f is well-defined. It follows that there exists c R such that 1 bc M. If bc / M, then M + Rbc = R which gives x + ybc = 1 for some x M, y R. Since cyb l(b) and l(b) is a GW-ideal of R, there exists a positive integer n such that (cyb) n c l(b). Therefore (1 x) n+1 = (ybc) n+1 = yb(cyb) n c M. It follows that 1 M, a contradiction to M R. Hence R is reduced. (2). Suppose 0 b R such that b 2 = 0. There exists a maximal right ideal K of R such that r(b) K. By Proposition 2.1 and Lemma 2.3, K is essential. It follows that R/K is Wnil-injective and there exists some c R such that 1 cb K. Now l(b) is a GW-ideal of R and b l(b), so there exists a positive integer n such that (cb) n c l(b). This implies that (bc) n+1 = b(cb) n c l(b). Therefore (cb) n+1 r(b) K. Again 1 cb K, so (cb) n (cb) n+1 = (1 cb)(cb) n K. It follows from (cb) n+1 K that (cb) n K. Also, (cb) n 1 (cb) n = (1 cb)(cb) n 1 K, whence (cb) n 1 K. Proceeding in this manner, we get cb K. Therefore 1 K, a contradiction to K R. This proves that R is reduced. (5). Suppose 0 b R such that b 2 = 0. There exists a maximal left ideal M of R such that l(b) M. We claim that l(b)r M. If it is not true, then sr / M for some s l(b), r R. This implies that M + Rsr = R which

784 T. Subedi and A. M. Buhphang yields x + tsr = 1 for some x M, t R. Since l(b) is a GW-ideal of R and rts l(b), there exists a positive integer n such that (rts) n r l(b). Therefore (1 x) n+1 = (tsr) n+1 = ts(rts) n r l(b). It follows that 1 M, a contradiction. Hence l(b)r M R. So, there exists a maximal right ideal K of R such that l(b)r K. By Proposition 2.1 and Lemma 2.4, K is essential, so by hypothesis, K is Wnil-injective. Hence it follows that there exists some c K such that b = cb which implies that 1 c l(b) K, whence 1 K, a contradiction. This proves that R is reduced. Also (3) = (1), (4) = (2) and (6) = (5). This completes the proof. Corollary 2.7 ([1], Lemma 3) Let R be a ZI, left or a right GP-V -ring. Then R is a reduced ring. By Proposition 2.6 and Theorem 2.5, the following theorem follows. Theorem 2.8 Let R be a ring such that l(a) is a GW-ideal of R for every a R. The following conditions are equivalent. (1) R is strongly regular. (2) Every maximal essential right ideal of R is YJ-injective. Theorem 2.9 Let R be a left GP-V -ring. If l(a) is a GW-ideal of R for every a R, then R is a reduced left weakly regular ring. Proof. By Proposition 2.6, R is reduced. If R is not left weakly regular, then there exists b R and l(b)+rbr R, so there exists a maximal left ideal M of R such that l(b) + RbR M. By Proposition 2.1 and Lemma 2.3, M is essential. Therefore by hypothesis, R/M is YJ-injective. Thus there exists a positive integer n such that b n 0 and every left R-homomorphism from Rb n to R/M extends to one from R to R/M. Define f : Rb n R/M by f(rb n ) = r+m for every r R. Since R is reduced, l(b) = l(b n ) which yields f is well-defined. It follows that there exists some c R such that 1 b n c M. Now b n c RbR M, whence 1 M, a contradiction. This proves that R is left weakly regular. Theorem 2.10 Let R be a right GP-V -ring. If l(a) is a GW-ideal of R for every a R, then R is a reduced left weakly regular ring.

On strongly regular rings 785 Proof. R is reduced by Proposition 2.6. If R is not left weakly regular, then there exists b R and l(b) + RbR R. Since R is reduced, l(b) = r(b). Let K be a maximal right ideal of R such that r(b) + RbR K. Following the dual of the proof of Theorem 2.9, we get a contradiction. So R is left weakly regular. Since a reduced ring is left weakly regular if and only if it is right weakly regular, the following corollary follows. Corollary 2.11 Let R be a left or a right GP-V -ring. If l(a) is a GW-ideal of R for every a R, then R is a reduced weakly regular ring. Corollary 2.12 ([1], Theorem 4) Let R be a ZI, left or a right GP-V -ring, then R is a reduced weakly regular ring. Theorem 2.13 Let R be a ring such that l(a) is a GW-ideal of R for every a R. The following conditions are equivalent: (1) R is strongly regular. (2) R is an MELT left GP-V -ring. (3) R is an MELT right GP-V -ring. (4) R is an MERT right GP-V -ring. Proof. It is clear that (1) = (2), (3) and (4). (2) = (1). Let b R such that l(b) + Rb R. There exists a maximal left ideal M of R such that l(b) + Rb M. By Proposition 2.1 and Lemma 2.3, M is essential, so R is a left GP-V -ring implies R/M is YJ-injective. Therefore there exists a positive integer n such that b n 0 and every left R-homomorphism from Rb n to R/M extends to that from R to R/M. Define f : Rb n R/M by f(rb n ) = r +M for every r R. By Proposition 2.6, R is reduced, so l(b n ) = l(b) and f is well-defined. It follows that there exists some c R such that 1 b n c M. Since R is an MELT ring, b n c M. Therefore 1 M, a contradiction. Thus l(d) + Rd = R for all d R and R is strongly regular. Similarly, we can prove (4) = (1). (3) = (1). R is reduced by Proposition 2.6, so l(w) = r(w) = l(w m ) for every w R and for every positive integer m. If R is not regular, then there exists b R such that r(b) + Rb R. There exists a maximal left ideal M of R such that r(b) + Rb M. By Lemma 2.4, M is essential. Since R is an MELT ring and b M, we have RbR M. Hence r(b) + RbR M R. Therefore there exists a maximal right ideal K of R such that r(b)+rbr K. By Lemma 2.3, K is essential. Thus R/K is YJ-injectve. It follows that there exists a positive

786 T. Subedi and A. M. Buhphang integer n and c R such that b n 0 and 1 cb n K. But cb n RbR K, hence 1 K which is a contradiction. Therefore R is regular. Since a reduced regular ring is strongly regular, it follows that R is strongly regular. Corollary 2.14 ([8], Theorem 2.3) If R is a ZI-ring, then the following statements are equivalent: (1) R is strongly regular. (2) R is an MELT left GP-V-ring. (3) R is an MERT right GP-V-ring. (4) R is an MELT left GP-V -ring. (5) R is an MERT right GP-V -ring. Theorem 2.15 Let R be a left quasi duo ring. If every maximal essential right ideal of R is YJ-injective, then R is strongly regular. Proof. Let 0 a R such that a 2 = 0. There exists a maximal left ideal M of R containing l(a). If l(a)r M, then sr / M for some s l(a), r R. This yields M + Rsr = R and so x + ysr = 1 for some x M, y R. Since R is left quasi duo, and s M, we have ysr M. Therefore 1 x M, whence 1 M, a contradiction. Thus l(a)r M R and therefore there exists a maximal right ideal K of R such that l(a)r K. By Lemma 2.4, K is essential so that by hypothesis, K is YJ-injective. Following the proof of (5) of Proposition 2.6, we get a contradiction. So R is reduced. Therefore by Theorem 2.5, R is strongly regular. Definition 2.16 Following [12], a left ideal L of a ring R is a weak ideal (W -ideal), if for every 0 a L, there exists some n > 0 such that a n 0 and a n R L. A right ideal K of R is defined similarly to be a weak ideal. Lemma 2.17 ([4], Lemma 3.14) Let L be a left ideal of R. Then R/L is a flat left R-module if and only if for each a L, there exists some b L such that a = ab. Lemma 2.18 ([4], Remark 3.13) A reduced left (resp., right) SF-ring is strongly regular. Theorem 2.19 Let R be a ring such that l(a) is a W-ideal of R for every a R. The following conditions are equivalent: (1) R is strongly regular.

On strongly regular rings 787 (2) R is a left SF-ring. (3) Every maximal essential left ideal of R is YJ-injective. Proof. It is known that (1) = (2) and (1) = (3). (2) = (1). Suppose 0 b R such that b 2 = 0. Let x r(b) and r R. Then b l(x). Since l(x) is a W-ideal of R and b 2 = 0, we have br l(x). This proves that r(b) is a left ideal of R. Therefore there exists a maximal left ideal M of R such that r(b) M. Since R is a left SF-ring and b r(b) M, by Lemma 2.17, there exists c M such that b = bc, that is 1 c r(b) M and so 1 M, a contradiction. This proves that R is reduced and therefore by Lemma 2.18, R is strongly regular. (3) = (1). Let b R, b 2 = 0. Following the proof of (2) = (1), r(b) is a left ideal of R. There exists a maximal left ideal M of R containing r(b). By Lemma 2.4, M is essential, hence by hypothesis, M is YJ-injective. It follows that there exists some c M such that b = bc. This implies that 1 c r(b) M, whence 1 M, a contradiction. Therefore R is reduced and thus by Theorem 2.5, R is strongly regular. 3 Some results on left N duo rings Proposition 3.1 Let R be a semiprime left N duo ring, then R is reduced. Proof. Let a R, a 2 = 0. Since R is left N duo, Ra is an ideal of R and hence Ra = RaR. This gives RaRa = Ra 2 = 0. It follows from R is semiprime that Ra = 0, so a = 0. This proves that R is reduced. Corollary 3.2 A left N duo ring is left weakly regular if and only if it is right weakly regular. Proposition 3.3 Let R be a left N duo ring. following conditions, then R is reduced. If R satisfies one of the (1) Every simple left R-module is Wnil-injective. (2) Every simple right R-module is Wnil-injective. Proof. (1). Let 0 a R, a 2 = 0 and M be a maximal left ideal of R containing l(a). By hypothesis, R/M is Wnil-injective, hence every left R-homomorphism from Ra to R/M extends to one from R to R/M. Define f : Ra R/M by f(ra) = r + M for every r R. Then f is well-defined and there exists b R such that 1 ab M. Since R is left N duo, Ra is an ideal of R, so ab Ra l(a) M, whence 1 M, a contradiction. This proves that R is reduced.

788 T. Subedi and A. M. Buhphang (2). Let 0 a R, a 2 = 0. There exists a maximal right ideal K of R containing r(a). Now, R/K is Wnil-injective. It is clear that there exists some b R such that 1 ba K. Suppose ba / K, then K + bar = R. This gives x + bay = 1 for some x K, y R. As R is left N duo, Ra is an ideal of R. Therefore there exists some t R such that ayb = ta. Thus (1 x) 2 = (bay) 2 = b(ayb)ay = bta 2 y = 0. It follows that 1 K, a contradiction to K R. Hence R is reduced. Corollary 3.4 A left N duo, left or a right GP-V-ring is reduced. Lemma 3.5 ([4], Proposition 3.2) Let R be a left (resp., right) SF-ring and I be an ideal of R. Then R/I is also a left (resp., right) SF-ring. Lemma 3.6 ([4], Theorem 4.10) A left quasi duo left (resp., right) SF-ring is strongly regular. Theorem 3.7 Let R be a left N duo ring. equivalent: The following conditions are (1) R is strongly regular. (2) Every maximal essential left ideal of R is YJ-injective. (3) R is a left SF-ring. (4) R is a right SF-ring. Proof. It is known that (1) = (2), (3) and (4). (2) = (1). Suppose R is not reduced, then there exists 0 a R such that a 2 = 0. Then r(a) K for some maximal right ideal K of R. We claim that Rr(a) K. If it is not true, then rs / K for some r R, s r(a). This implies that K + rsr = R which yields x + rst = 1 for some x K, t R and so ax + arst = a. As R is left N duo, Ra is an ideal of R so that ar = za for some z R. Hence arst = zast = 0 and therefore ax = a. This implies 1 x r(a) K, whence 1 K, a contradiction to K R. Therefore Rr(a) K R. Let M be a maximal left ideal of R containing Rr(a). By Lemma 2.4, M is essential. Hence by hypothesis, M is YJ-injective, so there exists some b M such that a = ab, that is 1 b r(a) M implying 1 M, a contradiction. Therefore R is reduced. Hence by Theorem 2.5, we have R is strongly regular. (3) = (1). Let 0 a R such that a 2 = 0. Let x r(a), r R. Since R is left N duo, Ra is an ideal of R. Hence ar = ta for some t R. Therefore arx = tax = 0. This proves that r(a) is a left ideal of R. Following the proof

On strongly regular rings 789 of (2) = (1) of Theorem 2.19, we get a contradiction. This proves that R is reduced so that by Lemma 2.18, R is strongly regular. (4) = (1). Suppose a 2 J(R) such that a / J(R). We claim that l(a)r R. If it is not true, then a = at i r i where t i l(a), r i R. Since a / J(R), at k / J(R) for some k. Therefore at k / M for some maximal right ideal M of R. This implies that M + at k R = R and hence x + at k y = 1 for some x M, y R. Now, (at k ) 2 = a(t k a)t k = 0 which implies that at k N(R). Since R is left N duo, Rat k is an ideal of R. So at k y = sat k for some s R. Therefore x + sat k = 1 which yields xat k = at k, whence at k M, a contradiction to at k / M. Hence a J(R) which is again a contradiction. This proves that l(a)r R. Hence there exists a maximal right ideal K of R such that l(a)r K. Since a 2 J(R) K and R is right SF, by dual of Lemma 2.17, there exists some c K such that a 2 = ca 2. So a ca l(a) K. Since ca K, we have a K. Hence, again there exists some d K such that a = da so that 1 d l(a) K, whence 1 K. This contradiction shows that R/J(R) is a reduced ring. Hence by Lemma 3.5 and Lemma 2.18, R/J(R) is strongly regular. Therefore R/J(R) is left duo and hence R is left quasi duo. Thus by Lemma 3.6, R is strongly regular. References [1] N.K Kim, S.B. Nam and J.Y. Kim, On simple singular GP-injective modules, Communications in Algebra, 27(5)(1999), 2087-2096. [2] V.S. Ramamurthy, Weakly regular rings, Canad. Math. Bull., 16(3)(1973), 317-321. [3] V.S. Ramamurthy, On the injectivity and flatness of certain cyclic modules, Proc. Amer. Math. Soc., 48(1975), 21-25. [4] M. B. Rege, On von Neumann regular rings and SF-rings, Math. Japonica, 31(6)(1986), 927-936. [5] X.Song and X.Yin, Generalizations of V-rings, Kyungpook Math. J., 45(2005), 357-362. [6] J.Wei and Libin Li, Nilpotent elements and reduced rings, Turk. J. Math., 34(2010), 1-13. [7] Y. Xiao, One sided SF-rings with certain chain conditions, Canad. Math. Bull., 37(2)(1994), 272-277. [8] G. Xiao, On GP-V-rings and characterizations of strongly regular rings, Northeast. Math. J., 18 (4)(2002), 291-297.

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