Ore Extensions of Extended Symmetric and Reversible Rings

Transcription

1 International Journal of Algebra, Vol. 3, 2009, no. 9, Ore Extensions of Extended Symmetric and Reversible Rings L moufadal Ben Yakoub and Mohamed Louzari Department of Mathematics, Abdelmalek Essaadi University B.P Tetouan, Morocco benyakoub@hotmail.com, mlouzari@yahoo.com Abstract Let σ be an endomorphism and δ an σ-derivation of a ring R. In this paper, we show that if R is (σ, δ)-skew Armendariz and aσ(b) = 0 implies ab = 0 for a, b R. Then R is symmetric (respectively, reversible) if and only if R is σ-symmetric (respectively, σ-reversible) if and only if R[x; σ, δ] is symmetric (respectively, reversible). As a consequence we obtain a generalization of [3, 9, 14]. Also, a number of related results are shown. Mathematics Subject Classification: 16U80; 16S36 Keywords: Armendariz rings; Ore extensions; (Extended) reversible rings; (Extended) symmetric rings Introduction Throughout this paper, R denotes an associative ring with unity. σ is a ring endomorphism, and δ an σ-derivation of R, that is, δ is an additive map such that δ(ab) =σ(a)δ(b)+δ(a)b for all a, b R. We denote R[x; σ, δ] the Ore extension whose elements are polynomials over R, the addition is defined as usual and the multiplication subject to the relation xa = σ(a)x + δ(a) for all a R. A ring R is called symmetric if abc = 0 implies acb = 0 for all a, b, c R. A ring R is called reversible if ab = 0 implies ba = 0 for all a, b R. Reduced rings (i.e., rings with no nonzero nilpotent elements) are symmetric by Anderson and Camillo [1, Theorem 1.3]. Commutative rings are clearly symmetric, symmetric rings are clearly reversible. Polynomial rings over reversible rings need not to be reversible, and polynomial rings over symmetric rings need not to be symmetric (see [11] and [18]). Rege and Chhawchharia [17] called a ring R an Armendariz if whenever polynomials f = n i=0 a ix i,g= m j=0 b jx j R[x] satisfy fg = 0, then a i b j = 0 for each i, j. The term Armendariz was introduced

2 424 L. Ben Yakoub and M. Louzari by Rege and Chhawchharia [17]. This nomenclature was used by them since it was Armendariz [2, Lemma 1], who initially showed that a reduced ring always satisfies this condition. According to Krempa [13], an endomorphism σ of a ring R is called to be rigid if aσ(a) = 0 implies a = 0 for all a R. A ring R is called σ-rigid if there exists a rigid endomorphism σ of R. Note that any rigid endomorphism is injective and σ-rigid rings are reduced rings by Hong et al. [7]. Properties of σ-rigid rings have been studied in [5, 7, 8, 13]. In [8], Hong et al. defined a ring R with an endomorphism σ to be σ-skew Armendariz if whenever polynomials f = n i=0 a ix i,g= m j=0 b jx j R[x; σ] satisfy fg = 0 then a i σ i (b j ) = 0 for each i, j. According to Hong et al. [9]. A ring R is said to be σ- Armendariz, if whenever polynomials f = n i=0 a ix i,g= m j=0 b jx j R[x; σ] satisfy fg = 0 then a i b j = 0 for each i, j. From Hashemi and Moussavi [6], a ring R is called a (σ, δ)-skew Armendariz ring if for p = n i=0 a ix i and q = m j=0 b jx j in R[x; σ, δ], pq = 0 implies a i x i b j x j = 0 for each i, j. By Hashemi and Moussavi [5], a ring R is σ-compatible if for each a, b R, aσ(b) = 0 if and only if ab = 0. Moreover, R is said to be δ-compatible if for each a, b R, ab = 0 implies aδ(b) = 0. If R is both σ-compatible and δ- compatible, we say that R is (σ, δ)-compatible. A ring R is σ-rigid if and only if R is (σ, δ)-compatible and reduced [5, Lemma 2.2]. From [3], a ring R is called right (respectively, left) σ-reversible if whenever ab = for a, b R, bσ(a) =0 (respectively, σ(b)a = 0). Also, by [14], a ring R is called right (respectively, left) σ-symmetric if whenever abc = 0 for a, b, c R, acσ(b) = 0 (respectively, σ(b)ac = 0). In this paper, we study the transfert of symmetry (σ-symmetry) and reversibility (σ-reversibility) from R to R[x; σ, δ] and vice versa. We show, that if R is (σ, δ)-skew Armendariz and aσ(b) = 0 implies ab = 0 for a, b R. Then R is symmetric (respectively reversible) if and only if R is σ-symmetric (respectively σ-reversible) if and only if if R[x; σ, δ] is symmetric (respectively, reversible). As a consequence we obtain a generalization of [9, Theorem 3.6], [10, Proposition 3.4], [11, Proposition 2.4], [3, Corollary 2.11] and [14, Theorem 2.10]. A connection between reversibility (respectively, symmetry) and σ-reversibility (respectively, σ-symmetry) of a ring is given. We also investigate connections to other related conditions. Examples to illustrate results are included. 1 Preliminaries and Examples We begin with the following definition. Definition 1.1. Let σ be an endomorphism of a ring R. We say that R satisfies the condition (C σ ) if whenever aσ(b) =0with a, b R, then ab =0. In the Ore extension R[x; σδ], we have for n 0, x n a = n i=0 f i n(a)xi where End(R, +) will denote the map which is the sum of all possible words in f n i

3 Ore extensions of extended symmetric and reversible rings 425 σ, δ built with i letters σ and n i letters δ. (In particular, f n n = σn,f n 0 = δn ), [15, Lemma 4.1]. Lemma 1.2. Let R be an (σ, δ)-skew Armendariz ring satisfying the condition (C σ ) and n 1 a natural number. If ab =0then σ n (a)b = δ n (a)b =0 for all a, b R. Proof. It suffices to show the result for n = 1. Let f = σ(a)x + δ(a) and g = b such that ab = 0. Then fg = σ(a)xb + δ(a)b = σ(ab)x + σ(a)δ(b) + δ(a)b = σ(ab)x + δ(ab) = 0. By Proposition 2.1, we have σ(a)b = δ(a)b =0. Lemma 1.3. Let R be an (σ, δ)-skew Armendariz reversible ring satisfying the condition (C σ ).Ifab =0, then af j i (b) =0for all i, j (i j). Proof. Let a, b R such that ab = 0, since R is reversible we have ba = 0 and so σ n (b)a = δ n (b)a = 0, by Lemma 1.2. Also, we have aσ n (b) =aδ n (b) = 0 for all n N. Thusaf j i (b) = 0 for all i, j. In the next, we show some connections between (σ, δ)-skew Armendariz rings, σ-rigid rings and rings with the condition (C σ ). Example 1.4. Let Z be the ring of integers and Z 4 be the ring of integers modulo 4. Consider the ring R = {( ) } a b a Z, b Z 0 a 4. (( )) ( ) a b a b Let σ : R R be an endomorphism defined by σ =. {( ) } 0 a 0 a a 0 Take the ideal I = a 4Z of R. Consider the factor ring 0 a R/I = {( ) } a b a, b 4Z. 0 a (i) R/I is not σ-skew Armendariz. In fact, ( ) ( ) (R/I)[x; σ], but σ (( ) ( ) x) =0 0 2 ( ) ( ) a b a b (ii) R/I satisfies the condition (C σ ).LetA =,B = 0 a 0 a R/I. If Aσ(B) =0then aa =0and ab = ba =0, so that AB =0.

4 426 L. Ben Yakoub and M. Louzari Example 1.5. Consider a ring of polynomials over Z 2, R = Z 2 [x]. Let σ : R R be an endomorphism defined by σ(f(x)) = f(0). Then: (i) R does not satisfy the condition (C σ ). Let f = 1+x, g = x R, we have fg =(1+x)x 0, however fσ(g) =(1+x)σ(x) =0. (ii) R is σ-skew Armendariz [8, Example 5]. {( ) } a t Example 1.6. Consider the ring R = a Z,t Q, where 0 a Z and Q are the set of all integers and all rational numbers, respectively. The (( ring )) R is commutative, ( ) let σ : R R be an automorphism defined by a t a t/2 σ =. 0 a 0 ( a ) (( )) ( ) 0 t 0 t 0 t (i) R is not σ-rigid. σ =0, but 0,ift ( ) 0 ( 0 ) a t b x (ii) R satisfies the condition (C σ ).Let and R such that 0 a 0 b ( a t 0 a ) (( b x σ 0 b )) =0, ( hence ab )( =0=ax/2+tb, ) soa =0or b =0. In each case, ax + tb =0, hence a t b x =0. We have also the same for the converse. 0 a 0 b (iii) R is σ-skew Armendariz [7, Example 1]. Examples 1.4 and 1.5 shows that the (σ, δ)-skew Armendariz property of a ring and the condition (C σ ) are independent of each other. From [5, Lemma 2.2], [4, Lemma 2.5] and Example 1.6. The class of σ-rigid rings is strictly included in the class of (σ, δ)-skew Armendariz rings satisfying the condition (C σ ). 2 Reversibility and Symmetry of Ore Extensions From Isfahani and Moussavi [16], a ring R is called skew Armendariz, if for f = n i=0 a ix i, g = m j=0 b jx j R[x; σ, δ], fg = 0 implies a 0 b j = 0 for all j. Obviously, (σ, δ)-skew Armendariz rings satisfying the condition (C σ ) are skew Armendariz. But the converse is not true by Example 2.2. Proposition 2.1. Let R be a ring, σ an endomorphism, and δ an σ-derivation of R. If one of the following conditions is satisfied. (i) R is (σ, δ)-skew Armendariz and satisfies the condition (C σ ); (ii) R is skew Armendariz and R[x; σ, δ] is symmetric;

5 Ore extensions of extended symmetric and reversible rings 427 Then, for f = n i=0 a ix i and g = m j=0 b jx j R[x; σ, δ], fg = 0 implies a i b j =0for all i, j. Proof. Let f = n i=0 a ix i, g = m j=0 b jx j R[x; σ, δ] such that fg =0. (i) Since R is (σ, δ)-skew Armendariz, then a i x i b j x j = 0 for all i, j. Or a i x i b j x j = a i i l=0 f i l (b j)x j+l = a i σ i (b j )x i+j +p(x) = 0 where p(x) is a polynomial of degree strictly less than i + j. Thusa i σ i (b j ) = 0, and by the condition (C σ ) we have a i b j = 0 for all i, j. (ii) We have (a 0 + a 1 x + + a n x n )(b 0 + b 1 x + + b m x m )=a 0 (b 0 + b 1 x + + b m x m )+(a 1 x + + a n x n )(b 0 + b 1 x + + b m x m ) = 0, or a 0 b j = 0 for all j, because R is skew Armendariz. So that (a 1 x+ +a n x n )(b 0 +b 1 x+ +b m x m )= (a a n x n 1 )x(b 0 + b 1 x + + b m x m ) = 0. Since R[x; σ, δ] is symmetric, then we have (a 1 x + + a n x n )(b 0 + b 1 x + + b m x m )x = 0, hence a 1 b j =0 for all j. Continuing this process, we have a i b j = 0 for all i, j. Proposition 2.1 shows that, σ-skew Armendariz rings satisfying the condition (C σ ) are σ-armendariz. Example 2.2. Consider the ring R = Z 2 [x]. Let σ : R R be an endomorphism defined by σ(f(x)) = f(0). Then (i) R[y; σ] is not reversible (so is not symmetric): Letf = ay, g = b R[y; σ] with a = 1+x and b = x, then fg = ayb = aσ(b)y =0. But, gf = bay = x(1+x)y 0. (ii) R does not satisfy the condition (C σ )(see Example 1.5). (iii) R is skew Armendariz: Since R is σ-skew Armendariz, by [8, Example 5]. Then [16, Theorem 2.2] implies that R is skew Armendariz. (iv) R is not σ-armendariz by [9, Example 1.9]. By Example 2.2, the conditions R[x; σ, δ] is symmetric and the condition (C σ ) in Proposition 2.1 are not superfluous. Lemma 2.3. Let R be an (σ, δ)-skew Armendariz ring satisfying the condition (C σ ). Then, for f = n i=0 a ix i,g = m j=0 b jx j and h = p k=0 c kx k R[x; σ, δ], iffgh =0then a i b j c k =0for all i, j, k. Proof. Let f = n i=0 a ix i, g = m j=0 b jx j and h = p k=0 c kx k R[x; σ, δ]. Note that if fg = then a i g = 0 for all i. Suppose that fgh = 0 then a i (gh) =0 for all i, so(a i g)h = 0 for all i. Then by Proposition 2.1, we have a i b j c k =0 for all i, j, k. Theorem 2.4. Let R be an (σ, δ)-skew Armendariz ring satisfying the condition (C σ ). Then (1) R is reversible if and only if R[x; σ, δ] is reversible; (2) R is symmetric if and only if R[x; σ, δ] is symmetric.

6 428 L. Ben Yakoub and M. Louzari Proof. Note that any subring of symmetric (respectively, reversible) ring is again symmetric (respectively, reversible). Conversely, let f = n i=0 a ix i,g = m j=0 b jx j and h = p k=0 c kx k R[x; σ, δ]. (1) If fg = 0 then a i b j = 0 for all i, j (by Proposition 2.1). Since R is reversible, we have b j a i = 0 for all i, j, Lemma 1.3 implies that b j fk l(a i) = 0 for all i, j. Thus gf = n m j i=0 j=0 k=0 b jf j k (a i)x i+k =0. (2) If fgh = 0 then a i b j c k = 0 for all i, j, k, by Lemma 2.3. Since R is symmetric, a i c k b j = 0 for all i, j, k. Also, R is reversible then by Lemma 1.3, we have a i fs i(c kft k(b j)) = 0 for all i, j, k, s, t with (s i, t k). Thus fhg =0. In the next we give an example of a symmetric (so reversible) ring which satisfies all the hypothesis of Theorem 2.4. Consider the following subring of the triangular ring T n (R). Let R be a ring and let T (R, n) := a 1 a 2... a n 1 a n 0 a 1 a 2... a n a 1... a n a 1 a i R, with n 2, be a subset of the triangular matrix ring T n (R). Then T (R, n) is a ring with addition point-wise and usual matrix multiplication. We can denote elements of T (R, n) by(a 1,a 2,,a n ). For an endomorphism σ and an σ-derivation δ of R, the natural extension σ : T (R, n) T (R, n) defined by σ((a i )) = σ((a i )) is an endomorphism of T (R, n) and δ : T (R, n) T (R, n) defined by δ((a i )) = δ((a i )), is an σ-derivation of T (R, n). Since R[x]/(x n ) T (R, n), (n 1), then by [10, Theorem 2.3], if R is a reduced ring then T (R, n) is symmetric. Example 2.5. Let R be a ring, σ an endomorphism of R and δ be a σ- derivation of R. Suppose that R is σ-rigid. Consider the ring T (R, 3) = a b c 0 a b 0 0 a a, b, c R. T (R, 3) is symmetric (because R is reduced). We show that T (R, 3) is (σ, δ)- skew Armendariz. Let p T (R, 3)[x; σ, δ], p can be expressed by the form of (p 1,p 2,p 3 ) for some p i s in R[x; σ, δ]. Let p =(p 1,p 2,p 3 ) and q =(q 1,q 2,q 3 ) be elements of T (R, 3)[x; σ, δ]. Assume that pq =0then pq =(p 1 q 1,p 1 q 2 +

7 Ore extensions of extended symmetric and reversible rings 429 p 2 q 1,p 1 q 3 + p 2 q 2 + p 3 q 1 )=0. So we have the following system of equations p 1 q 1 = 0 ; (1) p 1 q 2 + p 2 q 1 = 0 ; (2) p 1 q 3 + p 2 q 2 + p 3 q 1 = 0. (3) Since R[x; σ, δ] is reduced by [13, Theorem 3.3], we see that q 1 p 1 =0from Eq.(1). Multiplying p 1 to Eq.(2) from the right hand side, we have p 1 q 2 p 1 + p 2 q 1 p 1 =0. Thus p 1 q 2 p 1 =0and so p 1 q 2 =0. Hence p 2 q 1 =0. Multiplying p 1 to Eq.(3) from the right hand side, then p 1 q 3 p 1 + p 2 q 2 p 1 + p 3 q 1 p 1 =0. Then p 1 q 3 p 1 =0and so p 1 q 3 =0and hence Eq.(3) becomes p 2 q 2 + p 3 q 1 = 0. (4) Multiplying p 2 to Eq.(4) from the right hand side we have p 2 q 2 p 2 + p 3 q 1 p 2 =0, then we have p 2 q 2 =0, and so p 3 q 1 =0. Let p = n i=0 A ix i and q = m j=0 B jx j T (R, 3)[x; σ, δ], where A i = a i b i c i 0 a i b i 0 0 a i and B j = a j b j c j 0 a j b j 0 0 a j for 0 i n, 0 j m. Assume that pq =0. We claim that A i x i B j x j =0for 0 i n, 0 j m. By the preceding expressions of p and q, we can write p 1 = n i=0 a ix i, p 2 = n i=0 b ix i,p 3 = n i=0 c ix i,q 1 = m j=0 a j xj,q 2 = m j=0 b j xj and q 3 = m j=0 c j xj. Then a i a j =0, a i b j =0, b i a j =0, a i c j =0, b i b j =0and c i a j =0for all i, j by the preceding results and [7, Proposition 6]. Then A i B j =0for all i, j. Since T (R, 3) is (σ, δ)-compatible, we have A i f j l (B j)=0for all i, j, l with j l by [4, Lemma 2.4]. Thus A i x i B j x j = i l=0 A if j l (B j)x l+j =0for all i, j. Therefore T (R, 3) is (σ, δ)-skew Armendariz. By Theorem 2.4, T (R, 3)[x; σ,δ] is symmetric (so reversible). According to Hong et al. [9, Theorem 1.8], if R is σ-armendariz then R is σ-skew Armendariz. But the converse is note true by [9, Example 1.9]. Theorem 2.6. R is σ-armendariz if and only if R is σ-skew Armendariz and satisfies the condition (C σ ). Proof. ( ). By [9, Proposition 1.3(ii) and Theorem 1.8]. ( ). Let f = a 0 +a 1 x+ +a n x n and g = b 0 +b 1 x+ +b m x m R[x; σ], such that fg =0, since R is σ-skew Armendariz then a i σ i (b j ) = 0 for all i, j. Since R satisfies the condition (C σ ) then a i b j = 0 for all i, j. Therefore R is σ-armendariz. We clearly obtain the following corollaries of Theorem 2.4. Corollary 2.7 ([9, Theorem 3.6]). Let R be an σ-armendariz ring. Then (1) R is reversible if and only if R[x; σ] is reversible. (2) R is symmetric if and only if R[x; σ] is symmetric.

8 430 L. Ben Yakoub and M. Louzari Corollary 2.8 ([10, Proposition 3.4] and [11, Proposition 2.4]). Let R be an Armendariz ring. Then (1) R is reversible if and only if R[x] is reversible. (2) R is symmetric if and only if R[x] is symmetric. 3 σ-reversibility and σ-symmetry of Ore Extensions In the next Lemma we give a relationship between σ-reversibility (respectively, σ-symmetry) and reversibility (respectively, symmetry). Lemma 3.1. Let R be a ring and σ an endomorphism of R. IfR satisfies the condition (C σ ). Then (1) R is reversible if and only if R is σ-reversible; (2) R is symmetric if and only if R is σ-symmetric. Proof. (1) Let a, b R, ab = 0 implies bσ(a) = 0, with the condition (C σ ), we have ba =0. SoR is reversible. Conversely, let a, b R, suppose that ab =0. Ifbσ(a) 0 (i.e., R is not right σ-reversible), the reversibility of R gives σ(a)b 0. Also, R satisfies (C σ ), then σ(ab) 0. Contradiction. Now, if σ(b)a 0 (i.e., R is not left σ-reversible), the condition (C σ ) gives σ(ba) 0. Contradiction, because R is reversible. (2) Let a, b, c R such that abc = 0. We have bca = 0 (by reversibility), so caσ(b) = 0 (by right σ-reversibility), then σ(b)ca = 0 (by reversibility), and so σ(b)ac = 0. So we have the left σ-symmetry. With the same method, we obtain the right σ-symmetry. Conversely, if abc = 0, by right σ-symmetry we have acσ(b) = 0, then acb = 0 by the condition (C σ ). Example 3.2. Let Z 2 is the ring of integers modulo 2, take R = Z 2 Z 2 with the usual addition and multiplication. R is commutative, and so R is symmetric (so reversible). Now, consider σ : R R defined by σ((a, b)) = (b, a). Then, we have: (i) R is not σ-reversible (so not σ-symmetric): (1, 0)(0, 1) = 0, but (0, 1)σ((1, 0)) = (0, 1)(0, 1) = (0, 1) 0. (ii) R does not satisfy the condition (C σ ): (1, 0)σ((1, 0)) = 0, but (1, 0) 2 = (1, 0) 0. By Example 3.2, we see that the condition (C σ ) in Lemma 3.1, is not superfluous. Theorem 3.3. Let R be an (σ, δ)-skew Armendariz ring satisfying the condition (C σ ). The following statements are equivalent: (1) R is reversible;

9 Ore extensions of extended symmetric and reversible rings 431 (2) R is σ-reversible; (3) R is right σ-reversible; (4) R[x; σ, δ] is reversible. Proof. (1) (4). By Theorem 2.4. (1) (2) and (2) (3). Immediately from Lemma 3.1. (3) (1). Let a, b R, ifab = 0 then bσ(a) = 0 (right σ-reversibility), so ba = 0 (condition (C σ )). Theorem 3.4. Let R be an (σ, δ)-skew Armendariz ring satisfying the condition (C σ ). The following statements are equivalent: (1) R is symmetric; (2) R is σ-symmetric; (3) R is right σ-symmetric; (4) R[x; σ, δ] is symmetric. Proof. As of Theorem 3.3. The next corollaries are direct consequences of Theorems 2.6, 3.3 and 3.4. Corollary 3.5 ([3, Corollary 2.11]). Let R be an σ-armendariz ring. The following statements are equivalent: (1) R is reversible; (2) R is σ-reversible; (3) R is right σ-reversible; (4) R[x; σ] is reversible. Corollary 3.6 ([14, Theorem 2.10]). Let R be an σ-armendariz ring. The following statements are equivalent: (1) R is symmetric; (2) R is σ-symmetric; (3) R is right σ-symmetric; (4) R[x; σ] is symmetric. ACKNOWLEDGEMENTS. The authors express their gratitude to Professor Laiachi EL Kaoutit for valuable remarks and helpful comments. The second author wishes to thank Professor Amin Kaidi of University of Almería for his generous hospitality. This work was supported by the project PCI Moroccan-Spanish A/011421/07. References [1] D.D. Anderson and V. Camillo, Semigroups and rings whose zero products commute, Comm. Algebra, 27(6) (1999),

10 432 L. Ben Yakoub and M. Louzari [2] E.P. Armendariz, A note on extensions of Baer and p.p.-rings, J. Australian Math. Soc., 18 (1974), [3] M. Baser, C.Y. Hong and T.K. Kwak, On extended reversible rings, Algebra Colloq., 16(1) (2009), [4] L. Ben Yakoub and M. Louzari, On quasi-baer rings of Ore extensions, East-West J. of Math., 8(2) (2006), [5] E. Hashemi and A. Moussavi, Polynomial extensions of quasi-baer rings, Acta. Math. Hungar., 107(3) (2005), [6] E. Hashemi and A. Moussavi, On (σ, δ)-skew Armendariz rings,, J. Korean Math. Soc., 42(2) (2005), [7] C.Y. Hong, N.K. Kim and T.K. Kwak, Ore extensions of Baer and p.p.- rings, J. Pure Appl. Algebra, 151(3) (2000), [8] C.Y. Hong, N.K. Kim and T.K. Kwak, On Skew Armendariz Rings, Comm. Algebra, 31(1) (2003), [9] C.Y. Hong, T.K. Kwak and S.T. Rezvi, Extensions of generalized Armendariz rings, Algebra Colloq., 13(2) (2006), [10] C. Huh, H.K. Kim, N.K. Kim and Y. Lee, Basic examples and extensions of symmetric rings, J. Pure Appl. Algebra, 202 (2005), [11] N.K. Kim and Y. Lee, Extensions of reversible rings, J. Pure Appl. Algebra, 185 (2003), [12] N.K. Kim and Y. Lee, Armendariz rings and reduced rings, J. Algebra, 223 (2000), [13] J. Krempa, Some examples of reduced rings, Algebra Colloq., 3(4) (1996), [14] T.K. Kwak, Extensions of extended symmetric rings, Bull. Korean Math.Soc., (44) (2007), [15] T.Y. Lam, A. Leroy and J. Matczuk, Primeness, semiprimeness and the prime radical of Ore extensions, Comm. Algebra, 25(8) (1997), [16] A.R. Nasr-Isfahani and A. Moussavi, Ore extensions of skew Armendariz rings, Comm. Algebra, 36 (2008), [17] M.B. Rege and S. Chhawchharia, Armendariz rings, Proc. Japan. Acad. Ser. A Math. Sci., 73 (1997),

11 Ore extensions of extended symmetric and reversible rings 433 [18] Z. Wang and L. Wang, Polynomial rings over symmetric rings need not to be symmetric, Comm. Algebra, 34 (2006), Received: November, 2008