STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS *

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1 ANALELE ŞTIINŢIFICE ALE UNIVERSITĂŢII AL.I. CUZA DIN IAŞI (S.N.) MATEMATICĂ, Tomul...,..., f... DOI: /aicu STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS * BY YOSUM KURTULMAZ Abstract. Let R be an arbitrary ring with identity. An element a R is strongly J-clean if there exist an idempotent e R and element w J(R) such that a = e + w and ew = ew. A ring R is strongly J-clean in case every element in R is strongly J-clean. In this note, we investigate the strong J-cleanness of the skew triangular matrix ring T n(r,σ) over a local ring R, where σ is an endomorphism of R and n = 2,3,4. Mathematics Subject Classification 2010: 15B33, 15B99, 16L99. Key words: strongly J-clean ring, skew triangular matrix ring, local ring. 1. Introduction Throughout this paper all rings are associative with identity unless otherwise stated. Let R be aring. J(R) and U(R) will denote, respectively, the Jacobson radical and the group of units in R. An element a R is strongly clean ifthereexistanidempotente R andaunitu R suchthat a = e+u and eu = ue. A ring R is strongly clean if every element in R is strongly clean. Many authors have studied such rings from very different points of view (cf. [1-9]). An element a R is strongly J-clean provided that there exist an idempotent e R and element w J(R) such that a = e+w and ew = ew. A ringr is strongly J-clean in case every element in R is strongly J-clean. Strong J-cleanness over commutative rings is studied in [1] and deduced the strong J-cleanness of T n (R) for a large class of local rings R, where T n (R) denotes the ring of all upper triangular matrices over R. * This paper is dedicated to my mother Gönül Ünalan.

2 2 YOSUM KURTULMAZ 2 Let σ be an endomorphism of R preserving 1 and T n (R,σ) be the set of all upper triangular matrices over the rings R. For any (a ij ),(b ij ) T n (R,σ), we define (a ij ) + (b ij ) = (a ij + b ij ), and (a ij )(b ij ) = (c ij ) where c ij = n k=i a ikσ k i( ) b kj. Then Tn (R,σ) is a ring under the preceding addition and multiplication. It is clear that T n (R,σ) will be T n (R) only when σ is the identity morphism. Let a R and the maps l a : R R and r a : R R denote, respectively, the abelian group endomorphisms given by l a (r) = ar and r a (r) = ra for all r R. Thus, l a r b is an abelian group endomorphism such that (l a r b )(r) = ar rb for any r R. Strong cleanness of T n (R,σ) for several n was studied in [3]. In this article, we investigate the strong J-cleanness of T n (R,σ) over a local ring R for n = 2,3,4 and then extend strong cleanness to such properties. In this direction we show that T 2 (R,σ) is strongly J-clean if and only if for any a 1+J(R),b J(R), l a r σ(b) : R R is surjective and R/J(R) = Z 2. Furtherifl a r σ(b) andl b r σ(a) aresurjectiveforanya 1+J(R),b J(R), then T 3 (R,σ) is strongly J-clean if and only if R/J(R) = Z 2. The necessary condition for T 3 (R,σ) to be strongly J-clean is also discussed. In addition to these, if l a r σ(b) and l b r σ(a) are surjective for any a 1 + J(R), b J(R), then T 4 (R,σ) is strongly J-clean if and only if R/J(R) = Z The case n = 2 By [Theorem 4.4, 2], the triangular matrix ring T 2 (R) over a local ring R is strongly J-clean if and only if R is bleached and R/J(R) = Z 2. We extend this result to the skew triangular matrix ring T 2 (R,σ) over a local ring R. Remark 2.1 will be used in the sequel without reference to. Remark 2.1. Note that iffor any ringr, R/J(R) = Z 2, then 2 J(R), 1 + J(R) = U(R) and 1 + U(R) = J(R). For if, f is the isomorphism R/J(R) = Z 2 then f(1+j(r)) = 1+2Z. Hence f(2+j(r)) = 2+2Z = 0+2Z. So 2+J(R) = 0+J(R), that is 2 J(R). 1+J(R) U(R). Let u U(R). Then f(u+j(r)) = 1+2Z = f(1+j(r)). Hence u 1 J(R) and so u 1+J(R). Thus, U(R) 1+J(R) and U(R) = 1+J(R). Lemma 2.2. Let R be a ring and let σ be an endomorphism of R. If T n (R,σ) is strongly J-clean for some n N, then so is R. Proof. Let e = diag(1,0,...,0) T n (R,σ). Then R = et n (R,σ)e. From Corollary 3.5 in [2], R is strongly J-clean.

3 3 STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS 3 Theorem 2.3. Let R be a local ring, and let σ be an endomorphism of R. Then the following are equivalent: (1) T 2 (R,σ) is strongly J-clean. (2) If a 1+J(R),b J(R), then l a r σ(b) : R R is surjective and R/J(R) = Z 2 Proof. (1) (2) From Lemma 2.2, R is strongly J-clean and by Lemma 4.2 in [2], R/J(R) = Z 2. By Remark 2.1, ( 1 + ) J(R) = U(R). Let a v a 1 + J(R),b J(R),v R. Then A = T 0 b 2 (R,σ). By ( ) e x hypothesis, there exists an idempotent E = T 0 f 2 (R,σ) such that A E J ( T 2 (R,σ) ) and AE = EA. Since R is local, all idempotents in R are 0 and 1. Thus, we see that e = 1,f = 0; otherwise, A E J ( T 2 (R,σ) ) ( ) 1 x. So E =. As AE = EA, we get v + xσ(b) = ax. 0 0 Hence, ax xσ(b) = v for some x R. As a result, l a r σ(b) : R R is surjective. ( ) a v (2) (1) Let A = T 0 b 2 (R,σ). Case 1. If a,b J(R), then A J ( T 2 (R,σ) ) is strongly J-clean. Case 2. If a,b 1 + J(R), then A I 2 J ( T 2 (R,σ) ) ; hence, A = I 2 +(A I 2 ) T 2 (R,σ) is strongly J-clean. Case 3. If a 1+J(R),b J(R), by hypothesis, l a r σ(b) : R R is ( ) 1 x surjective. Thus, ax xσ(b) = v for some x R. Choose E = 0 0 T 2 (R,σ). Then E 2 = E T 2 (R,σ), AE = EA and A E J ( T 2 (R,σ) ). That is, A T 2 (R,σ) is strongly J-clean. Case 4. Ifa J(R),b 1+J(R), thena+1 1+J(R),b+1 J(R)and byhypothesis, l a+1 r σ(b+1) : R Rissurjective. Thusax xσ(b) = v for ( ) 0 x some x R. Choose E = T (R,σ). Then E 2 = E T 2 (R,σ), AE = EA and A E J ( T 2 (R,σ) ). Hence, A T 2 (R,σ) is strongly J-clean. Therefore A T 2 (R,σ) is strongly J-clean. Corollary 2.4. Let R be a local ring, and let σ be an endomorphism of R. Then the following are equivalent:

4 4 YOSUM KURTULMAZ 4 (1) T 2 (R,σ) is strongly J-clean. (2) R/J(R) = Z 2 and T 2 (R,σ) is strongly clean. Proof. (1) (2) It is clear. ( ) a v (2) (1) Let a 1 + J(R),b J(R),v R. Then A = 0 b ( ) e x T 2 (R,σ). By hypothesis, there exists an idempotent E = 0 f T 2 (R,σ) such that A E J ( T 2 (R,σ) ) and AE = EA. Since R is local, we see that e = 0,f = 1; otherwise, A E J ( T 2 (R,σ) ) ( ) 0 x. So E =. 0 1 It follows from AE = EA that v + xσ(b) = ax, and so ax v = xσ(b). Therefore l a r σ(b) : R R is surjective. By Theorem 2.3, T 2 (R,σ) is strongly J-clean as R/J(R) = Z 2. Corollary 2.5. Let R be a ring, and R/J(R) = Z 2. If J(R) is nil, then T 2 (R,σ) is strongly J-clean. Proof. Clearly R is local. Let a 1 + J(R),b J(R). Then we can ( find some n N such that ) b n = 0. For any v R, we choose ) x = la 1+l a 2r b + +l a nr b n 1 )( (v). It can be easily checked that (la r b (x) ( =(l a r b la 1+l a 2r b + +l a nr b n 1) (v) = (v+a 1 vb+ +a n+1 vb n 1) a 1 vb+ +a n vb n) = v. Hence, l a r b : R R is surjective. Similarly, l a r σ(b) is surjective since σ(b) J(R). This completes the proof by Theorem 2.3. Example 2.6. Let Z 2 n = Z/2 n Z,n N, and let σ be an endomorphism of Z 2 n. Then, T 2 (Z 2 n,σ) is strongly J-clean. As Z 2 n is a local ring with the Jacobson radical 2Z 2 n. Obviously, J ( Z 2 n) is nil, and we are through by Corollary 2.5. Example 2.7. Let Z 4 = Z/4Z, let ( ) a b R = { a,b Z 0 a 4 }, ( ) ( ) a b a b and let σ : R R,. Then T 0 a 0 a 2 (R,σ) is strongly J- clean. Obviously, σ is an endomorphism of R. It is easy to check that

5 5 STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS 5 ( ) a b J(R) = { a Z 0 a 2,b Z 4 }, and then R/J(R) = Z 2 is a field. Thus, R is a local ring. In addition, ( J(R) ) 4 = 0, thus J(R) is nil. Therefore we obtain the result by Corollary The case n = 3 We now extend Theorem 2.3. to the case of 3 3 skew triangular matrix rings over a local ring. Theorem 3.1. Let R be a local ring. If l a r σ(b) and l b r σ(a) are surjective for any a 1+J(R),b J(R), then T 3 (R,σ) is strongly J-clean if and only if R/J(R) = Z 2. Proof. ( ) We noted in Remark 2.1, in this case we have σ(j(r)) J(R), σ(u(r)) U(R), 1 + J(R) = U(R) and 1+U(R) = J(R) and we use them in the sequel intrinsically.let A = (a ij ) T 3 (R,σ). We divide the proof into six cases. Case 1. If a 11,a 22,a J(R), then A = I 3 + (A I 3 ), and so A I 3 J(T 3 (R,σ)). Then A T 3 (R,σ) is strongly J-clean. Case 2. If a 11 J(R),a 22,a J(R), then we have an e 12 R such that a 11 e 12 e 12 σ(a 22 ) = a 12. Further, we have some e 13 R such that a 11 e 13 e 13 σ 2 (a 33 ) = e 12 σ(a 23 ) a 13. Choose 0 e 12 e 13 E = T 3 (R,σ) Then E 2 = E, and A = E+(A E), where A E J(T 3 (R,σ)). Furthermore, 0 e 12 σ(a 22 ) e 12 σ(a 23 )+e 13 σ 2 (a 33 ) EA = 0 a 22 a 23, 0 0 a 33 0 a 11 e 12 +a 12 a 11 e 13 +a 13 AE = 0 a 22 a 23, 0 0 a 33 and so EA = AE. That is, A T 3 (R,σ) is strongly J-clean. Case 3. If a 11 1+J(R),a 22 J(R),a 33 1+J(R), then we have an e 12 R such that a 11 e 12 e 12 σ(a 22 ) = a 12. Further, we have some e 23 R

6 6 YOSUM KURTULMAZ 6 such that a 22 e 23 e 23 σ(a 33 ) = a 23. Thus a 11 e 12 σ(e 23 ) + a 12 σ(e 23 ) = e 12 σ(a 22 )σ(e 23 ) = e 12 σ(a 23 ) e 12 σ(e 23 )σ 2 (a 33 ). Choose 1 e 12 e 12 σ(e 23 ) E = 0 0 e 23 T 3 (R,σ) Then E 2 = E, and A = E+(A E), where A E J(T 3 (R,σ)). Furthermore, a 11 a 12 +e 12 σ(a 22 ) a 13 +e 12 σ(a 23 ) e 12 σ(e 23 )σ 2 (a 33 ) EA = 0 0 e 23 σ(a 33 ), 0 0 a 33 a 11 a 11 e 12 a 11 e 12 σ(e 23 )+a 12 σ(e 23 )+a 13 AE = 0 0 a 22 e 23 +a 23, 0 0 a 33 and so EA = AE. Thus, A T 3 (R,σ) is strongly J-clean. Case 4. If a 11,a 22 1+J(R),a 33 J(R), then we find some e 23 R such that a 22 e 23 e 23 σ(a 33 ) = a 23. Thus, there exists e 13 R such that a 11 e 13 e 13 σ 2 (a 33 ) = a 13 a 12 σ(e 23 ). Choose 1 0 e 13 E = 0 1 e 23 T 3 (R,σ). Then E 2 = E, and A = E+(A E), where A E J(T 3 (R,σ)). Furthermore, a 11 a 12 a 13 +e 13 σ 2 (a 33 ) EA = 0 a 22 a 23 +e 23 σ(a 33 ), a 11 a 12 a 11 e 13 +a 12 σ(e 23 ) AE = 0 a 22 a 22 e 23, and so EA = AE. Therefore A T 3 (R,σ) is strongly J-clean. Case 5. If a 11 1+J(R),a 22,a 33 J(R), then we have some e 12 R such that a 11 e 12 e 12 σ(a 22 ) = a 12. Further, there exists e 13 R such that a 11 e 13 e 13 σ 2 (a 33 ) = a 13 +e 12 σ(e 23 ). Choose 1 e 12 e 13 E = T 3 (R,σ).

7 7 STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS 7 Then E 2 = E, and A = E +(A E), where A E J(T 3 (R,σ)). Hence a 11 a 12 +e 12 σ(a 22 ) a 13 +e 12 σ(a 23 )+e 13 σ 2 (a 33 ) EA =, a 11 a 11 e 12 a 11 e 13 AE =, and so EA = AE. Thus A T 3 (R,σ) is strongly J-clean. Case 6. If a 11 J(R),a 22 1+J(R),a 33 J(R), then we find some e 23 R such that a 22 e 23 e 23 σ(a 33 ) = a 23. Hence there is e 12 R such that a 11 e 12 e 12 σ(a 22 ) = a 12. It is easy to verify that e 12 σ(a 23 )+e 12 σ(e 23 )σ 2 (a 33 ) = e 12 σ(a 22 e 23 ) = a 11 e 12 σ(e 23 )+a 12 σ(e 23 ). Choose 0 e 12 e 12 σ(e 23 ) E = 0 1 e 23 T 3 (R,σ). Then E 2 = E, and A = E + (A E), where A E J(T 3 (R,σ)). In addition, 0 e 12 σ(a 22 ) e 12 σ(a 23 )+e 12 σ(e 23 )σ 2 (a 33 ) EA = 0 a 22 a 23 +e 23 σ(a 33 ), 0 a 11 e 12 +a 12 a 11 e 12 σ(e 23 )+a 12 σ(e 23 ) AE = 0 a 22 a 22 e 23 and so EA = AE. Consequently, A T 3 (R,σ) is strongly J-clean. Case 7. If a 11,a 22 J(R),a 33 1+J(R), then we find e 23 R such that a 22 e 23 e 23 σ(a 33 ) = a 23. Further, we have an e 13 R such that a 11 e 13 e 13 σ 2 (a 33 ) = a 13 a 12 σ(e 23 ). Choose 0 0 e 13 E = 0 0 e 23 T 3 (R,σ)

8 8 YOSUM KURTULMAZ 8 Then E 2 = E, and A = E+(A E), where A E J(T 3 (R,σ)). Furthermore, 0 0 e 13 σ 2 (a 33 ) EA = 0 0 e 23 σ(a 33 ), 0 0 a a 11 e 13 +a 12 σ(e 23 )+a 13 AE = 0 0 a 22 e 23 +a 23, 0 0 a 33 and so EA = AE. As a result, A T 3 (R,σ) is strongly J-clean. Case 8. If a 11,a 22,a 33 J(R), then A = 0+A, where A J(T 3 (R,σ)). Hence, A T 3 (R,σ) is strongly J-clean. Thus, T 3 (R,σ) is strongly J-clean. ( ) Similar to Theorem 2.3, we easily complete the proof. Corollary 3.2. Let R be a ring, and R/J(R) = Z 2. If J(R) is nil, then T 3 (R,σ) is strongly J-clean. Proof. Obviously R is local. Let a U(R),b J(R). Then we can find some n N such that b n = 0; hence, ( σ(b) ) n = 0. For any v R, we choose x = ( l a 1+l a 2r σ(b) + +l a nr σ(b) n 1) (v). It can be easily checked that (l a r σ(b) ) (x) = (la r σ(b) )( la 1 +l a 2r σ(b) + +l a nr σ(b) n 1) (v) = ( v+a 1 vσ(b)+ +a n+1 vσ(b) n 1) (a 1 vσ(b)+ +a n vσ(b) n) = v. Thus, l a r σ(b) : R R is surjective. Likewise, l b r σ(a) : R R is surjective. Consequently, T 3 (R,σ) is strongly J-clean by Theorem A characterization We will consider the necessary and sufficient conditions under which the skew triangular matrix ring T 3 (R,σ) is strongly J-clean. Lemma 4.1. Let R be a local ring. If T 3 (R,σ) is strongly J-clean, then l a r σ(b), l a r σ 2 (b), l b r σ(a) and l b r σ 2 (a) are surjective for any a 1+J(R),b J(R). Proof. Let a 1 + J(R),b J(R). Clearly, T 2 (R,σ) is strongly J- clean. By Theorem 2.3, l a r σ(b) is surjective. As 1 b 1 + J(R) and 1 a J(R), we get l 1 b r σ(1 a) : R R is surjective. For any v R, we have an x R such that (1 b)x xσ(1 a) = v. Thus, bx xσ(a) = v and so l b r σ(a) : R R is surjective.

9 9 STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS 9 Let v R and let b 0 v A = 0 b 0 T 3 (R,σ). 0 0 a We have an idempotent E = (e ij ) T 3 (R,σ) such that A E J ( T 3 (R,σ) ) and EA = AE. This implies that e 11,e 22,e 33 R are all idempotents. As a 1+J(R),b J(R), we have e 11 = 0,e 22 = 0 and e 33 = 1; otherwise, A E J(T 3 (R,σ)). As E = E, we have 0 0 e 13 E = 0 0 e 23, for some e 13,e 23 R. Observing that 0 0 be 13 +v 0 0 e 13 σ 2 (a) 0 0 be 23 = AE = EA = 0 0 e 23 σ(a), 0 0 a 0 0 a we have be 13 e 13 σ 2 (a) = v. Thus, l b r σ 2(a) : R R is surjective. Since 1 a J(R) and 1 b 1 + J(R), we have, l 1 a r σ 2 (1 b) : R R is surjective. Thus, we can find some x R such that (1 a)x xσ 2 (1 b) = v. This implies that ax xσ 2 (b) = v, hence l a r σ 2 (b) is surjective. Theorem 4.2. Let R be a local ring and let σ be an endomorphism of R. Then the following are equivalent: (1) T 3 (R,σ) is strongly J-clean. (2) R/J(R) = Z 2, and l a r σ(b) and l b r σ(a) are surjective for any a 1+J(R),b J(R). Proof. (1) (2) is obvious from Lemma 4.1. (2) (1) Clear from Theorem 3.1. Corollary 4.3. Let R be a local ring and let σ be an endomorphism of R.Then the following are equivalent: (1) T 2 (R,σ) is strongly J-clean.

10 10 YOSUM KURTULMAZ 10 (2) T 3 (R,σ) is strongly J-clean. (3) R/J(R) = Z 2 and l a r σ(b) is surjective for any a 1+J(R),b J(R). Proof. (1) (3) is proved by Theorem 2.3. (2) (3) is obvious from Theorem The case n = 4 We now extend the preceding discussion to the case of 4 4 skew triangular matrix rings over a local ring. Theorem 5.1. Let R be a local ring. If l a r σ(b) and l b r σ(a) are surjective for any a 1+J(R), b J(R), then T 4 (R,σ) is strongly J-clean if and only if R/J(R) = Z 2. Proof. ( ) As R/J(R) = Z 2, σ(j(r)) J(R). Let a 11 a 12 a 13 a 14 A = 0 a 22 a 23 a a 33 a 34 T 4(R,σ). a 44 We show the existence of e 11 e 12 e 13 e 14 E = 0 e 22 e 23 e e 33 e 34 T 4(R,σ), e 44 such that E 2 = E,AE = EA and A E J(T 4 (R,σ)). One can easily derive from E 2 = E that (a) e 12 = e 11 e 12 +e 12 σ(e 22 ) (b) e 13 = e 11 e 13 +e 12 σ(e 23 )+e 13 σ 2 (e 33 ) (c) e 23 = e 22 e 23 +e 23 σ(e 33 ) and from AE = EA that (d) a 11 e 12 e 12 σ(a 22 ) = e 11 a 12 a 12 σ(e 22 )

11 11 STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS 11 (e) a 11 e 13 e 13 σ 2 (a 33 ) = e 11 a 13 +e 12 σ(a 23 ) a 12 σ(e 23 ) a 13 σ 2 (e 33 ) (f) a 22 e 23 e 23 σ(a 33 ) = e 22 a 23 a 23 σ(e 33 ) Case 1. If a 22 J(R), a 11 1+J(R) then e 22 = 0, e 11 = 1. Hence, (d) implies that a 11 e 12 e 12 σ(a 22 ) = a 12 and by assumption there exists e 12 R such that (l a11 r σ(a22 ))(e 12 ) = a 12. (A) If a 33 1+J(R), then e 33 = 1. From (f), a 22 e 23 e 23 σ(a 33 ) = a 23 and (b) implies that e 13 = e 12 σ(e 23 ). (B) If a 33 J(R), then e 33 = 0. By (c), e 23 = 0. From (e), we have a 11 e 13 e 13 σ 2 (a 33 ) = a 13 +e 12 σ(a 23 ) a 12 σ(e 23 ) and by assumption there exists e 13 R such that (l a11 r σ(a33 ))(e 13 ) = a 13 +e 12 σ(a 23 ) a 12 σ(e 23 ). Case 2. If a 22 1+J(R), a 11 1+J(R), then e 22 = 1, e 11 = 1. By (a) implies that e 12 = 0. (C) If a 33 1+J(R), then e 33 = 1. From (b), we have e 13 = 0 and (c) implies that e 23 = 0. (D) If a 33 J(R), then e 33 = 0. By (f), we have a 22 e 23 e 23 σ(a 33 ) = a 23, and (e) gives rise to a 11 e 13 e 13 σ 2 (a 33 ) = a 13 +e 12 σ(a 23 ) a 12 σ(e 23 ) and by assumption there exists e 13 R such that (l a11 r σ(a33 ))(e 13 ) = a 13 +e 12 σ(a 23 ) a 12 σ(e 23 ). Case 3. If a J(R), a 11 J(R), then e 22 = 1, e 11 = 0. By (d), a 11 e 12 e 12 σ(a 22 ) = a 12 and there exists e 12 R such that (l a11 r σ(a22 ))(e 12 ) = a 12. (E) If a 33 1+J(R), then e 33 = 1. From (c), we have e 23 = 0. Then from (e), we have a 11 e 13 e 13 σ 2 (a 33 ) = e 12 σ(a 23 ) a 13 (F) If a 33 J(R), then e 33 = 0. From (f), we have a 22 e 23 e 23 σ(a 33 ) = a 23 and there exists e 23 R such that (l a22 r σ(a33 ))(e 23 ) = a 23. Then (b) implies that e 13 = e 12 σ(e 23 ). Case 4. If a 22 J(R), a 11 J(R), then e 22 = 0, e 11 = 0. Hence, (a) implies that e 12 = 0. (G) If a 33 1+J(R), then e 33 = 1. From (f), a 22 e 23 e 23 σ(a 33 ) = a 23 and there exists e 23 R such that (l a22 r σ(a33 ))(e 23 ) = a 23. So (e) gives us a 11 e 13 e 13 σ 2 (a 33 ) = a 12 σ(e 23 ) a 13. Hence there exists e 13 R such that (l a11 r σ 2 (a 33 ))(e 13 ) = a 12 σ(e 23 ) a 13. (H) If a 33 J(R), then e 33 = 0. From (c), we have e 23 = 0 and by (b) we obtain e 13 = 0. Similar to preceding calculations from E 2 = E we have (1) e 14 = e 11 e 14 +e 12 σ(e 24 )+e 13 σ 2 (e 34 )+e 14 σ 3 (e 44 )

12 12 YOSUM KURTULMAZ 12 (2) e 24 = e 22 e 24 +e 23 σ(e 34 )+e 24 σ 2 (e 44 ) (3) e 34 = e 33 e 34 +e 34 σ(e 44 ) and from AE = EA we have (4) a 11 e 14 e 14 σ 3 (a 44 ) = a 12 σ(e 24 ) a 13 σ 2 (e 34 ) a 14 σ 3 (e 44 )+e 11 a 14 + e 12 σ(a 24 )+e 13 σ 2 (a 34 ) (5) a 22 e 24 e 24 σ 2 (a 44 ) = a 23 σ(e 34 ) a 24 σ 2 (e 44 )+e 22 a 24 +e 23 σ(a 34 ) (6) a 33 e 34 e 34 σ(a 44 ) = a 34 σ(e 44 )+e 33 a 34 +e 34 σ(a 44 ) To complete the proof we only need to show the existence of e 14,e 24 and e 34 in R satisfying preceding conditions (1)-(6). Case 1. If a 44 J(R), a J(R), then e 44 = 0 and e 33 = 1, otherwise A E J(T 4 (R,σ)). By (6), a 33 e 34 e 34 σ(a 44 ) = a 34 and by hypothesis there exists e 34 such that (l a33 r σ(a44 ))(e 34 ) = a 34. Then by (5), a 22 e 24 e 24 σ 2 (a 44 ) = a 23 σ(e 34 ) + e 22 a 24 + e 23 σ(a 34 ). There are two possibilities: (A) If a J(R), then e 22 = 1 otherwise A E J(T 4 (R,σ)). Then there exists e 24 R such that (l a22 r σ 2 (a 44 ))(e 24 ) = a 24 a 23 σ(e 34 )+ e 23 σ(a 34 ). From(4), a 11 e 14 e 14 σ 3 (a 44 ) = a 12 σ(e 24 ) a 13 σ 2 (e 34 )+e 11 a 14 + e 12 σ(a 24 ) + e 13 σ 2 (a 34 ). If a 11 U(R), then e 11 = 1, otherwise A E J(T 4 (R,σ)). Hence, there exists e 14 R such that (l a11 r σ 3 (a 44 ))(e 14 ) = a 12 σ(e 24 ) a 13 σ 2 (e 34 )+a 14 +e 12 σ(a 24 )+e 13 σ 2 (a 34 ). If a 11 J(R), then e 11 = 0 and by (1), e 14 = e 12 σ(e 24 )+e 13 σ 2 (e 34 ). (B) If a 22 J(R), then e 22 = 0 otherwise A E J(T 4 (R,σ)). By (2), e 24 = e 23 σ(e 34 ). From equation (4), a 11 e 14 e 14 σ 3 (a 44 ) = a 12 σ(e 24 ) a 13 σ 2 (e 34 ) + e 11 a 14 + e 12 σ(a 24 ) + e 13 σ 2 (a 34 ). If a 11 U(R), then e 11 = 1. By hypothesis, there exists e 14 R such that (l a11 r σ 3 (a 44 ))(e 14 ) = a 12 σ(e 24 ) a 13 σ 2 (e 34 )+a 14 +e 12 σ(a 24 )+e 13 σ 2 (a 34 ). If a 11 J(R), then e 11 = 0 and by (1), e 14 = e 12 σ(e 24 )+e 13 σ 2 (e 34 ). Case 2. If a 44 1+J(R),a 33 1+J(R), then e 44 = e 33 = 1. Then by (3), e 34 = 0. Again there are two possibilities: (C) If a 22 U(R), then e 22 = 1 and by (2), e 24 = 0. If a 11 U(R), then e 11 = 1 and by (1), e 14 = 0. If a 11 J(R), then e 11 = 0. Then by equation (4), a 11 e 14 e 14 σ 3 (a 44 ) = e 12 σ(a 24 ) + e 13 σ 2 (a 34 ). Hence, there exists e 14 J(R) such that (l a11 r σ 3 (a 44 ))(e 14 ) = e 12 σ(a 24 )+e 13 σ 2 (a 34 ) (D) If a 22 J(R), then e 22 = 0 and by (5), a 22 e 24 e 24 σ 2 (a 44 ) = a 24 + e 23 σ(a 34 ). So, there exists e 24 R such that (l a22 r σ(a34 ) (e 24) =

13 13 STRONGLY J-CLEAN SKEW TRIANGULAR MATRIX RINGS 13 a 24 +e 23 σ(a 34 ). If a 11 J(R), then e 11 = 0. From equation (4), a 11 e 14 e 14 σ 3 (a 44 ) = a 12 σ(e 24 ) a 14 +e 12 σ(a 24 )+e 13 σ(a 34 ). By assumption, there exists e 14 R such that (l a11 r σ 3 ) = a 12 σ(e 24 ) a 14 + e 12 σ(a 24 ) + (a 44 ) e 13 σ(a 34 ). If a 11 U(R), then e 11 = 1. By equation (1), e 14 = e 12 σ(e 24 ). Case 3. If a J(R),a 33 J(R). In this case e 33 = 0 and e 44 = 1. By (6), a 33 e 34 e 34 σ(a 44 ) = a 34. Hence, there exists e 34 R such that (l a33 r σ(a44 ))(e 34 ) = a 34. Using (5), a 22 e 24 e 24 σ 2 (a 44 ) = e 22 a 24 +e 23 σ(a 34 ) a 23 σ(e 34 ) a 24. Then there are two possibilities: (E) Ifa 22 1+J(R), thene 22 = 1and from(2), e 24 = e 23 σ(e 34 ). Then by (4), a 11 e 14 e 14 σ 3 (a 44 ) = e 11 a 14 +e 12 σ(a 24 )+e 13 σ 2 (a 34 ) a 12 σ(e 24 ) a 13 σ 2 (e 34 ) a 14. If a 11 J(R), then e 11 = 0. So there exists e 14 R such that (l a11 r σ 3 (a 44 ))(e 14 ) = e 12 σ(a 24 )+e 13 σ 2 (a 34 ) a 12 σ(e 24 ) a 13 σ 2 (e 34 ) a 14. If a 11 U(R), then e 11 = 1 and by (1), e 14 = e 12 σ(e 24 ) e 13 σ 2 (e 34 ). (F) If a 22 J(R), then e 22 = 0 and by hypothesis there exists e 24 R suchthat (l a22 r σ 2 (a 44 ))(e 24 ) = a 24 +e 23 σ(a 34 ) a 23 σ(e 34 ). From equation (4), a 11 e 14 e 14 σ 3 (a 44 ) = e 11 a 14 + e 12 σ(a 24 ) + e 13 σ 2 (a 34 ) a 12 σ(e 24 ) a 13 σ 2 (e 34 ) a 14. If a 11 J(R), then e 11 = 0. From (4) and by hypothesis, thereexists e 14 R suchthat(l a11 r σ 3 (a 44 ))(e 14 ) = e 12 σ(a 24 )+e 13 σ 2 (a 34 ) a 12 σ(e 24 ) a 13 σ 2 (e 34 ) a 14. If a 11 U(R), then e 11 = 1 and by (1), e 14 = e 12 σ(e 24 ) e 13 σ 2 (e 34 ). Case 4. If a 44 J(R),a 33 J(R). In this case e 33 = e 44 = 0. By (3), e 34 = 0. (G) If a 22 J(R), then e 22 = 0. By (2), e 24 = 0. If a 11 J(R), then e 11 = 0 and from (1), e 14 = 0. If a 11 U(R), then e 11 = 1. Hence, equation (4) becomes a 11 e 14 e 14 σ 3 (a 44 ) = a 14 + e 12 σ(a 24 ) + e 13 σ 2 (a 34 ). By hypothesis there exists e 14 R such that (l a11 r σ 3 (a 44 ))(e 14 ) = a 14 + e 12 σ(a 24 )+e 13 σ 2 (a 34 ). (H) If a 22 1+J(R), then e 22 = 1 and from (5), a 22 e 24 e 24 σ 2 (a 44 ) = a 24 + e 23 σ(a 34 ). By assumption, there exists e 24 R such that (l a22 r σ 2 (a 44 ))(e 24 ) = a 24 + e 23 σ(a 34 ). If a 11 U(R), then e 11 = 1 and by (4), a 11 e 14 e 14 σ 3 (a 44 ) = a 12 σ(e 24 )+a 14 +e 12 σ(a 24 )+e 13 σ 2 (a 34 ). Hence, thereexistse 14 Rsuchthat(l a11 r σ 3 (a 44 ))(e 14 ) = a 12 σ(e 24 )+ a 14 + e 12 σ(a 24 ) + e 13 σ 2 (a 34 ). If a 11 J(R), then e 11 = 0 and from (1), e 14 = e 12 σ(e 24 ). Thus, we always find e 14,e 24 and e 34 in R. ( ) Analogous to Theorem 2.3 we easily obtain the result.

14 14 YOSUM KURTULMAZ 14 Acknowledgements. I would like to thank the referee for his/her careful reading and valuable comments. REFERENCES 1. Chen, H. On strongly J-clean rings, Comm. Algebra, 38 (2010), Chen, H. On uniquely clean rings, Comm. Algebra, 39 (2011), Chen, H.; Kose, H.; Kurtulmaz, Y. Strongly clean triangular matrix rings with endomorphisms, arxiv: Diesl, A.J. Classes of strongly clean rings, Thesis (Ph.D.)-University of California, Berkeley Li, Y. Strongly clean matrix rings over local rings, J. Algebra, 312 (2007), Nicholson, W.K.; Zhou, Y. Rings in which elements are uniquely the sum of an idempotent and a unit, Glasg. Math. J., 46 (2004), Nicholson, W.K.; Zhou, Y. Clean rings: a survey, Advances in ring theory, , World Sci. Publ., Hackensack, NJ, Yang, X.; Zhou, Y. Some families of strongly clean rings, Linear Algebra Appl., 425 (2007), Yang, X.; Zhou, Y. Strong cleanness of the 2 2 matrix ring over a general local ring, J. Algebra, 320 (2008), Received: 5.VI.2013 Revised: 18.VI.2013 Accepted: 21.VI.2013 Bilkent University, Department of Mathematics, Bilkent, Ankara, TURKEY yosum@fen.bilkent.edu.tr