EXTENSIONS OF STRONGLY Π-REGULAR RINGS

EXTENSIONS OF STRONGLY Π-REGULAR RINGS H. Chen, K. Kose and Y. Kurtulmaz ABSTRACT An deal I of a rng R s strongly π-regular f for any x I there exst n N and y I such that x n = x n+1 y. We prove that every strongly π-regular deal of a rng s a B-deal. An deal I s perodc provded that for any x I there exst two dstnct m, n N such that x m = x n. Furthermore, we prove that an deal I of a rng R s perodc f and only f I s strongly π-regular and for any u UI), u 1 Z[u]. Key Words: strongly π-regular deal; B-deal; perodc deal. 2010 Mathematcs Subject Classfcaton: 16E50, 16U50, 16E20. 1. INTRODUCTION A rng R s strongly π-regular f for any x R there exst n N, y R such that x n = x n+1 y. For nstance, all artnan rngs and all algebrac algebra over a fled. Such rngs are extensvely studed by many authors from very dfferent vew ponts cf. [1], [3-4], [7], [9-12] and [14]). We say that an deal I of a rng R s strongly π-regular provded that for any x I there exst n N, y I such that x n = x n+1 y. Many propertes of strongly π-regular rngs were extended to strongly π-regular deals n [6]. Recall that a rng R has stable range one provded that ar + br = R wth a, b R mples that there exsts a y R such that a + by R s nvertble. The stable range one condton s especally nterestng because of Evans Theorem, whch states that a module cancels from drect sums whenever has stable range one. For general theory of stable range condtons, we refer the reader to [6]. An deal I of a rng R s a B-deal provded that ar + br = R wth a 1 + I, b R mples that there exsts a y R such that a + by R s nvertble. An deal I s a rng R s stable provded that ar + br = R wth a I, b R mples that there exsts a y R such that a + by R s nvertble. As s well known, every B-deal of a rng s stable, but the converse s not true. In [1, Theorem 4], Ara proved that every strongly π-regular rng has stable range one. Ths was extended to deals,.e., every strongly π-regular deal of a rng s stable cf. [5]). The man purpose of ths note s to extend these results, and show that every strongly π- regular deal of a rng s a B-deal. An deal I of a rng R s perodc provded that for any

2 H. Chen, K. Kose and Y. Kurtulmaz x I there exsts two dstnct m, n N such that x m = x n. Furthermore, we show that an deal I of a rng R s perodc f and only f I s strongly π-regular and for any u UI), u 1 Z[u]. Several new propertes of such deals are also obtaned. Throughout, all rngs are assocatve wth an dentty and all modules are untary modules. UR) denotes the set of all nvertble elements n the rng R and UI) = 1+I ) UR). 2. STRONGLY Π-REGULAR IDEALS The am of ths secton s to nvestgate more elementary propertes of strongly π-regular deals and construct more related examples. For any x R, we defne σ x : R R gven by σ x r) = xr for all r R. Theorem 2.1. Let I be an deal of a rng R. Then the followng are equvalent: 1) I s strongly π-regular. 2) For any x I, there exsts n 1 such that R = kerσ n x ) mσ n x ). Proof. 1) 2) Let x I. In vew of [6, Proposton 13.1.15], there exst n N, y I such that x n = x n+1 y and xy = yx. It s easy to check that σx n = σx n+1 σ y. If a kerσx n ) mσx n ), then a = σx n r) and σx n a) = 0. Ths mples that x 2n r = σx 2n r) = 0, and so a = x n r = x n+1 yr = yx n+1 r = y n x 2n r = 0. Hence, kerσx n ) mσx n ) = 0. For any r R, we see that r = r σx n y n r) ) + σx n y n r), and then R = kerσx n ) + mσx n ), as requred. 2) 1) Wrte 1 = a + b wth a kerσx n ) and b mσx n ). For any x I. σx n 1) = σx n b), and so x n x 2n R. Thus, I s strongly π-regular. Corollary 2.2. Let I be a strongly π-regular deal of a rng R, and let x I. Then the followng are equvalent: 1) σ x s a monomorphsm. 2) σ x s an epmorphsm. 3) σ x s an somorphsm. Proof. 1) 2) In vew of Theorem 2.1, there exsts n 1 such that R = kerσ n x ) mσ n x ). Snce σ x s a monomorphsm, so s σ n x. Hence, kerσ n x ) = 0, and then R = mσ n x ). Ths mples that σ x s an epmorphsm. 2) 3) Snce R = kerσ n x ) mσ n x ), t follows from R = mσ n x ) that kerσ n x ) = 0. Hence, σ x s a monomorphsm. Therefore σ x s an somorphsm. 3) 1) s trval. Proposton 2.3. Let I be an deal of a rng R. Then the followng are equvalent: 1) I s strongly π-regular. 2) For any x I, RxR s strongly π-regular.

EXTENSIONS OF STRONGLY Π-REGULAR RINGS 3 Proof. 1) 2) Let x I. For any a RxR, there exsts an element b I such that a n = a n+1 b for some n N. Hence, a n = a n+1 ab 2 ). As ab 2 RxR, we see that RxR s strongly π-regular. 2) 1) For any x I, RxR s strongly π-regular, and so there exsts a y RxR such that x n = x n+1 y. Clearly, y I, and therefore I s strongly π-regular. The ndex of a nlpotent element n a rng s the least postve nteger n such that x n = 0. The ndex I) of an deal I of a rng R s the supremum of the ndces of all nlpotent elements of I. An deal I of a rng R s of bounded ndex f I) <. It s well known that I) n f and only f I contans no drect sums of n + 1 nonzero parwse somorphc rght deals cf. [9, Theorem 7.2]). Theorem 2.4. Let R be a rng, and let Then I s a strongly π-regular deal of R. I = {a R RaR) < }. Proof. Let x, y I and z R. Then RxzR, RzxR RxR. Ths mples that RxzR and RzxR are strongly π-regular of bounded ndex. Hence, xz, zx I. Obvously, Rx y)r RxR+RyR. For any a Rx y)r, a = c+d where c RxR and d RyR. Snce RxR s strongly π-regular, there exsts some n N such that c n = c n+1 r for a r R. Let RyR s of bounded ndex m. Then c n = c nm+1 s for a s R. Hence, a nm+1 s a n RyR. As RyR s strongly π-regular, we can fnd k N and d RyR such that a nm+1 s a n) k = a nm+1 s a n) k+1 d, Hence, d = d a nm+1 s a n) d, d a nm+1 s a n) = a nm+1 s a n) d. a nm+1 s a n ) a nm+1 s a n ) 2 d ) k = a nm+1 s a n) k 1 a nm+1 s a n )d ) k = a nm+1 s a n) k 1 a nm+1 s a n )d ) = 0. Therefore a nm+1 s a n) m = a nm+1 s a n) m+1 t. As a result, a nm a nm+1 R. Hence, we can fnd a r R such that a nm = a nm+1 ar). Therefore I s a strongly π-regular deal of R. Corollary 2.5. Let R be a rng of bounded ndex. Then s the maxmal strongly π-regular deal of R. I = {a R RaR s strongly π-regular} Proof. Snce R s of bounded ndex, so s RaR for any a R. In vew of Theorem 2.4, I = {a R RaR s strongly π-regular} s a strongly π-regular deal of R. Thus we complete the proof by Proposton 2.3. Example 2.6. Let V be an nfnte-dmensonal vector space over a feld F, let R = End F V ), and let I = {σ R dm F σv ) < }. Then I s strongly π-regular, whle R s not strongly π-regular.

4 H. Chen, K. Kose and Y. Kurtulmaz Proof. Clearly, I s an deal of the rng R. We have the descendng chan σv ) σ 2 V ). As dm F σv ) <, we can fnd some n N such that σ n V ) = σ n+1 V ). Snce V s a projectve rght F -module, we can fnd some τ R such that the followng dagram V V τ σ n σ n+1 σ n+1 V ) commutates,.e., σ n+1 τ = σ n. Hence, σ n = σ n+1 στ 2 ). Therefore I s a strongly π-regular deal of R. Let ɛ be an element of R such that εx ) = x +1 where {x 1, x 2, } s the bass of V. If R s strongly π-regular, there exsts some m N such that ε m R = ε m+1 R, and so ε m V ) = ε m+1 V ). As ε m x ) = x +m for all, we see that ε m V ) = x F >m x F = ε m+1 V ). Ths gves a contradcton. Therefore R s not a strongly π-regular >m+1 rng. Example 2.7. Let V be an nfnte-dmensonal ) vector ) space over a feld F, let R = R R 0 R End F V ), and let S =. Then I = s a strongly π-regular deal of R, 0 R 0 0 whle S s not a strongly π-regular rng. Proof. By the dscusson n Example 2.6, R s not strongly π-regular. Hence, S s not strongly π-regular. As I 2 = 0, one easly checks that I s a strongly π-regular deal of the rng S. An deal I of a rng R s called a gsr-deal f for any a I there exsts some nteger n 2 such that ara = a n Ra n. For nstance, every deal of strongly regular rngs s a gsr-deal. Example 2.8. Every gsr-deal of a rng s strongly π-regular. Proof. Let I be a gsr-deal of a rng R. Gven x 2 = 0 n I/ I JR) ), then x 2 I JR). As I s a gsr-deal, we see that xrx = x 2 Rx 2 JR),.e., RxR) 2 JR). As JR) s semprme, t follows that RxR JR), and so x JR). That s, x = 0. Ths mples that I/ I JR) ) s reduced. For any dempotent e I/ I JR) ) and any a R/JR), t follows from ea1 e) ) 2 = 0 that ea1 e) = 0, thus ea = eae. Lkewse, ae = eae. Ths mples that ea = ae. As a result, every dempotent n I/ I JR) ) s central. For any x I JR), there exsts some y R such that x 2 = x 2 yx 2, and then x 2 1 yx 2 ) = 0. Ths mples that x 2 = 0. Assume that x 2 = 0. As I s a gsr-deal, we see that xrx = x 2 Rx 2 = 0. That s, RxR) 2 JR), and so x JR). Therefore I JR) = {x I x 2 = 0}. Let x I. Then there exsts some n 2 such that xrx = x n Rx n. Hence, x 2 = x 2 yx 2. As x 2 y I s an dempotent, we see that x 2 x 6 y 2 I JR). By the precedng dscusson, we get x 2 x 6 y 2) 2 = 0. Ths mples that x 4 = x 5 r for some r I. Thus I s strongly π-regular. 3. STABLE RANGE CONDITION For any x, y R, wrte x y = x + y + xy. We use x [n] to stand for } x {{ x } n 1) n

EXTENSIONS OF STRONGLY Π-REGULAR RINGS 5 and x [0] = 0. The followng result was frstly observed n [1, Lemma 1], we nclude a smple proof to make the paper s self-contaned. Lemma 3.1. Let x, y j R, and let p, q j Z1 m, 1 j n). If p = q j = 1, j then ) ) p x q j y j = q j )x y j ); If j,jp p = q j = 0, then ) ) p x q j y j = j j p q j )x y j ).,j Proof. For any p, q j Z, one easly checks that p q j )x y j ) = ) ) p x q j y j +,j j ) ) ) ) q j p x + p q j y j. Therefore the result follows. j j Lemma 3.2. Let I be a strongly π-regular deal of a rng R. Then for any x I, there exsts some n N such that x [n] = x [n+1] y = z x [n+1] for y, z I. Proof. Let x I. Then x x 2 I. Snce I s a strongly π-regular deal, there exsts some n N such that x x 2 ) n = x x 2 ) n+1 s = s x x 2 ) n+1. Clearly, x x [2] = x x 2. Thus, x x [2] ) n = x x [2] ) n+1 s = x x [2] ) 2n t, ) where t = s n. Snce =0 1) n n = 0, t follows from Lemma 3.1 that n ) 1) n x [n ] x [2] ) []) = x x [2]) n. Thus, =0 x x [2] ) n = n =0 1) n ) x [n+] = x [n] + n =1 1) n ) x [n+]. ) Let u = =1 1) n n +1 x []. Then u x [n] = x [n] u. Snce =1 1) n n +1 usng Lemma 2.1 agan, x x [2]) n = x [n] x [n] u. Thus, we get x [n] x [n] u = x [n] x [n] u ) 2 t = x [n] x [n] u ) x [n] x [n] u ) t 0) = x [2n] x [2n] u x [2n] u + x [2n] u [2]) t 0) = x [2n] t u t u t + u [2] y + u + u u [2]) x [2n] = x [2n] u 2 t) x [2n]. Let v = x [2n] u 2 t) x [2n]. Then x [n] = x [n] u + v = x [n] u + v 0 ) u + v = x [n] u [2] + v u u ) + v. = x [n] u [n+1] + n v u [] u []) =0 ) = 1, by

6 H. Chen, K. Kose and Y. Kurtulmaz Further, Hence Further, we see that v u [] u [] = x [2n] u 2 t) x [2n]) u [] u [] = x [2n] u 2 t) x [2n] + 0 ) u [] u [] = x [2n] u 2 t) u [] x [2n] u [] x [n] = x [n] u [n+1] + n x [2n] u 2 t) u [] x [2n] u []) =0 n x [2n] u 2 t) u [] x [2n] u []) = x [2n] n u 2 t) u [] u [] ) + 0 ) x [2n]. =0 As n =1 1) +1 n It s easy to check that x [n] v, where v = ) = 1, we see that =0 u [n+1] = ) n n =1 1) +1 x []) [n+1] = C 1 n x [1+22+ +nn] 1+ + n=n+1 = C 1 n x [n] x [1+2+ +n 1)n]. 1+ + n=n+1 = ) n n )n+1 =1 1) +1 = 1, and so u [n+1] = C 1 n x [1+2+ +n 1)n]. Therefore C 1 n 1+ + n=n+1 1+ + n=n+1 x [n] = x [2n] v + x [2n] n u 2 t) u [] u [] ) ) x [2n] =0 = x [2n] v + n u 2 t) u [] u [] )) 0 ) =0 = x [2n] v + n u 2 t) u [] u [] ) ) =0 Let y = x [n 1] v + n u 2 t) u [] u [] ) ). Then x [n] = x [n+1] y wth y I. Lkewse, =0 x [n] = z x [n+1] for a z I, as requred. Theorem 3.3. Every strongly π-regular deal of a rng s a B-deal. Proof. Let I be a strongly π-regular deal of a rng R. Let a 1+I. Then a 1 I. In vew of Lemma 2.2, we can fnd some n N, b, c 1+I such that a 1) [n] = a 1) [n+1] b 1) = c 1) a 1) [n+1]. One easly checks that a 1) [n] = a n 1 and a 1) [n+1] = a n+1 1. Therefore a n = a n+1 b = ca n+1, and so a n a n+1 R Ra n+1. Accordng to [6, Proposton 13.1.2], a 1 + I s strongly π-regular. Accordng to [6, Theorem 13.1.7], I s a B-deal. Corollary 3.4. Let I be a strongly π-regular deal of a rng R, and let A be a fntely generated projectve rght R-module. If A = AI, then for any rght R-modules B and C, A B = A C mples that B = C. Proof. For any x I, we have n N and y R such that x n = x n+1 y and xy = yx. Hence x n = x n zx n, where z = y n. Let g = zx n and e = g + 1 g)x n g. Then e Rx s

EXTENSIONS OF STRONGLY Π-REGULAR RINGS 7 an dempotent. In addton, we have 1 e = 1 g) 1 x n g ) = 1 g) 1 x n) Rx. Set f = 1 e. Then there exsts an dempotent f I such that f Rx and 1 f Rx. Therefore I s an exchange deal of R. In vew of Theorem 3.3, I s a B-deal. Therefore we complete the proof by [6, Lemma 13.1.9]. Corollary 3.5. Let I be a strongly π-regular deal of a rng R, and let a, b 1 + I. If ar = br, then a = bu for for some u UR). Proof. Wrte ax = b and a = by. As a, b 1 + I, we see that x, y 1 + I. In vew of Theorem 3.3, I s a B-deal. Snce yx + 1 yx) = 1, there exsts an element z R such that u := y + 1 yx)z UR). Therefore bu = b y + 1 yx)z ) = by = a, as requred. Corollary 3.6. Let I be a strongly π-regular deal of a rng R, and let A M n I) be regular. Then A s the product of an dempotent matrx and an nvertble matrx. Proof. By vrtue of Theorem 3.3, I s a B-deal. As A M n I) s regular, we have a B M n I) such that A = ABA. Snce AB + I n AB) = I n, we get A + I n AB) ) B + I n AB)I n B) = I n where A + I n AB) I n + M n I). Thus, we can fnd a Y M n R) such that U := A + I n AB) + I n AB)I n B)Y GL n R). Therefore A = ABA = AB A + I n AB) + I n AB)I n B)Y ) = ABU, as requred. Let A s an algebra over a feld F. An element a of an algebra A over a feld F s sad to be algebrac over F f a s the root of some non-constant polynomal n F [x]. An deal I of A s sad to be an algebrac deal of A f every element n I s algebrac over F. Proposton 3.7. Let A s an algebra over a feld F, and let I be an algebrac deal of A. Then I s a B-deal. Proof. For any a I, a s the root of some non-constant polynomal n F [x]. So we can fnd a m,, a n F such that a n a n + a n 1 a n 1 + + a m a m = 0, where a m 0. Hence, a m = a n a n + + a m+1 a m+1 )a 1 m = a m+1 a n a n m 1 + + a m+1 )a 1 m. Set b = a n a n m 1 + + a m+1 )a 1 m. Then a m = a m+1 b. Therefore I s strongly π-regular, and so we complete the proof by Theorem 3.3. In the proof of Theorem 3.3, we show that for any a 1+I, there exsts some n N such that a n a n+1 b for a b 1+I f I s a strongly π-regular deal. A natural problem asks that f the converse of the precedng asserton s true. The answer s negatve from the followng counterexample. Let p Z be a prme and set Z p) = {a/b b Zp a/b n lowest terms)}. Then Z p) s a local rng wth maxmal pz p). Thus, the Jacobson radcal pz p) satsfes the condton above. Choose p/p + 1) pz p). Then p/p + 1) J Z p) ) s not nlpotent. Ths shows that pz p) s not strongly π-regular. 4. PERIODIC IDEALS An deal I of a rng R s perodc provded that for any x I there exst dstnct m, n N such that x m = x n. We note that an deal I of a rng R s perodc f and only f for any a I, there exsts a potent element p I such that a p s nlpotent and ap = pa.

8 H. Chen, K. Kose and Y. Kurtulmaz Lemma 4.1. Let I be an deal of a rng R. If I s perodc, then for any x 1 + I there exst m N, ft) Z[t] such that x m = x m+1 fx). Proof. For any a I, there exsts some n N such that a n = a n+1 a m n 1) where m n + 1. For any x 1 + I, we see that x 1 I. As n the proof n Lemma 3.2, we can fnd a ft) R[t] such that x 1) [n] = x 1) [n+1] fx) 1). One easly checks that x 1) [n] = x n 1 and x 1) [n+1] = x n+1 1. Therefore x n = x n+1 fx), as requred. Lemma 4.2. Let R be a rng, and let c R. If there exst a monc ft) Z[t] and some m N such that mc = 0 and fc) = 0, then there exst s, t Ns t) such that c s = c t. Proof. Clearly, Zc {0, c,, m 1)c}. Wrte ft) = t k + b 1 t k 1 + + b k 1 t + b k Z[t]. Then c k+1 = b 1 c k b k 1 c 2 b k c. Ths mples that {c, c 2, c 3,, c l, } {c, c 2, c 3,, c k, 0, c,, m 1)c, c 2,, m 1)c 2,, c k,, m 1)c k }. That s, {c, c 2, c 3,, c k, } s a fnte set. Hence, we can fnd some s, t N, s t such that c s = c t, as desred. As s well known, a rng R s perodc f and only f for any x R, there exsts n N and ft) Z[t] such that x n = x n+1 fx). We extend ths result to perodc deals. Lemma 4.3. Let I be an deal of a rng R. If for any x I, there exst n N and ft) Z[t] such that x n = x n+1 fx), then I s perodc. Proof. Let x I. If x s nlpotent, then we can fnd some n N such that x n = x n+1 = 0. Thus, we may assume that x I s not nlpotent. By hypothess, there exsts n N and gt) Z[t] such that x n = x n+1 gx). Thus, x n = x n+1 fx), where fx) = x gx) ) 2 Z[t]. In addton, f0) = 0. Let e = x n fx) ) n. Then 0 e = e 2 R and x n = x n e. Set S = ere and α = ex = xe. Then fα) = efx). Further, α n fα) ) n = e, α n = x n, α n = α n+1 fα). Thus, e = α n fα) ) n = α n+1 fα) ) n+1 = α n fα) ) n αfα) = eαfα) = αfα) n S. Wrte ft) = a 1 t + + a n t n. Then α a 1 α + + a n α n) = e. Ths mples that α 1 ) n+1 a 1 α 1 ) n 1 a n e = 0. Let gt) = t n+1 a 1 t n 1 a n Z[t]. Then gt) s a monc polynomal such that gα 1 ) = 0. Let T = {me S m Z}. Then T s a subrng of S. For any me I, by hypothess, there exsts a gt) Z[t] such that me) p = me) p+1 gme) me) p+1 T. Ths mples that T s strongly π-regular. Construct a map ϕ : Z T, m me. Then Z/Kerϕ = T. As Z s not strongly π-regular, we see that Kerϕ 0. Hence, T = Z q for some q N. Thus, qe = 0. As a result, qα 1 = 0. In vew of Lemma 4.2, we can fnd some s, t Ns t) such that α 1 ) s = α 1 ) t. Ths mples that α s = α t. Hence, x ns = x st, as asserted. Theorem 4.4. Let I be an deal of a rng R. Then I s perodc f and only f 1) I s strongly π-regular. 2) For any u UI), u 1 Z[u]. Proof. Suppose that I s perodc. Then I s strongly π-regular. For any u UI), t follows by Lemma 4.1 that there exst m N, ft) Z[t] such that u m = u m+1 fu). Hence, ufu) = 1, and so u 1 Z[u]. Suppose that 1) and 2) hold. For any x I, there exst m N and y I such that x m = x m yx m, y = yx m y and xy = yx from [6, Proposton 13.1.15]. Set u = 1 x m y + x m.

EXTENSIONS OF STRONGLY Π-REGULAR RINGS 9 Then u 1 = 1 x m y + y. Hence, u UI). By hypothess, there exsts an gt) Z[t] such that ugu) = 1. Further, x m = x m y 1 x m y + x m) = x m yu. Hence, x m u 1 = x m y, and so x m = x m yx m = x 2m gu) = x 2m x m gu) ) 2. Wrte gu) ) 2 = b0 + b 1 u + + b n u n Z[u]. For any 0, t s easy to check that x m u = x m 1 x m y + x m) Z[x]. Ths mples that x m gu) ) 2 Z[x]. Accordng to Lemma 4.3, I s perodc. It follows by Theorem 4.4 and Theorem 3.3 that every perodc deal of a rng s a B-deal. Corollary 4.5. Let I be a strongly π-regular deal of a rng R. If UI) s torson, then I s perodc. Proof. For any u UI), there exsts some m N such that u m = 1. Hence, u 1 = u m 1 Z[u]. Accordng to Theorem 4.4, we complete the proof. ) ) Z Z 0 Z Example 4.6. Let R = and I =. Then I s a nlpotent deal of 0 Z 0 0 ) ) m 1 1 1 1 R; hence, I s strongly π-regular. Clearly, UI), but 0 for any 0 1 0 1 m N. Thus, UI) s torson. The example above shows that the converse of Corollary 4.6 s not true. But we can derve the followng. Proposton 4.7. Let I be an deal of a rng R. If charr) 0, then I s perodc f and only f 1) I s strongly π-regular. 2) UI) s torson. Proof. Suppose that I s perodc. Then I s strongly π-regular. Let x UI). Then x s not nlpotent. By vrtue of Lemma 4.1, there exst m N, ft) Z[t] such that x m = x m+1 fx). As n the proof of Lemma 4.3, we have a monc polynomal gt) Z[t] such that gα 1 ) = 0. As charr) 0, we assume that charr) = q 0. Then qα 1 = 0. Accordng to Lemma 4.2, we can fnd two dstnct s, t N such that α 1 ) s = α 1 ) t. Smlarly to Lemma 4.3, x ns = x st, and so x s torson. Therefore UI) s torson. The converse s true by Corollary 4.5. REFERENCES [1] P. Ara, Strongly π-regular rngs have stable range one, Proc. Amer. Math. Soc., 1241996), 3293 3298. [2] A. Badaw; A.Y.M. Chn and H.V. Chen, On rngs wth near dempotent elements, Internat. J. Pure Appl. Math., 12002), 253 259. [3] G. Borooah; A.J. Desl and T.J. Dorsey, Strongly clean matrx rngs over commutatve local rngs, J. Pure Appl. Algebra, 2122008), 281 296. [4] M. Chacron, On a theorem of Hersten, Canadan J. Math., 211969), 1348-1353. [5] H. Chen and M. Chen, On strongly π-regular deals, J. Pure Appl. Algebra, 1952005), 21-32.

10 H. Chen, K. Kose and Y. Kurtulmaz [6] H. Chen, Rngs Related Stable Range Condtons, Seres n Algebra 11, World Scentfc, Hackensack, NJ, 2011. [7] A.J. Desl and T.J. Dorsey, A note on completeness n the theory of strongly clean rngs, Preprnt. [8] X. Du and Y. Yang, The adjont semgroup of a π-regular rngs, Acta Sc. Nat. Unv. Jlnenss 32001), 35 37. [9] K.R. Goodearl, Von Neumann Regular Rngs, Ptman, London, San Francsco, Melbourne, 1979. [10] N.K. Km and Yang Lee, On strong π-regularty and π-regularty, Comm. Algebra, 392011), 4470 4485. [11] M Ohor, On strongly π-regular rngs and perodc rngs, Math. J. Okayama Unv., 271985), 49 52. [12] F. T. Shrley, Regular and Strongly Π-Regular Rngs, Unversty of Texas at Austn, 1984. [13] L.W. Whte, Strongly Π-Regular Rngs and Matrx Rngs over Regular Rngs, Unversty of Texas at Austn, 1987. [14] H.P. Yu, On strongly π-regular rngs of stable range one, Bull. Austral. Math. Soc., 511995), 433 437. H. Chen Department of Mathematcs Hangzhou Normal Unversty Hangzhou 310034, Chna E-mal: huanynchen@yahoo.cn H. Kose Department of Mathematcs Ah Evran Unversty, Krsehr, Turkey E-mal: handankose@gmal.com Y. Kurtulmaz Department of Mathematcs Blkent Unversty, Ankara, Turkey E-mal: yosum@fen.blkent.edu.tr