A Note on Skew Armendariz Rings Weixing Chen and Wenting Tong Author Queries AQ Au: Please provide Keywords. Order Now
0 0 Communications in Algebra, :, 00 Copyright Taylor & Francis, Inc. ISSN: 00- print/- online DOI: 0.0/AGB-000 A NOTE ON SKEW ARMENDARIZ RINGS # Weixing Chen Department of Mathematics, Nanjing University, Nanjing, China and Department of Mathematics, Shandong Institute of Business and Technology, Yantai, China Wenting Tong Department of Mathematics, Nanjing University, Nanjing, China In this note, we answer a question of Hong et al. (00) by proving that if is a monomorphism of a reduced ring R, and R is -skew Armendariz, then R is -rigid. Key Words:. AQ 000 Mathematics Subject Classification: Primary N0, P0. Throughout this note R denotes an associative ring with identity. For a ring endomorphism, Hong et al. (00) introduced -skew Armendariz rings, which are generalizations of -rigid rings and Armendariz rings, and investigated their properties. Recall that for a ring R with a ring endomorphism R R, a skew polynomial ring R x of R is the ring obtained by giving the polynomial ring over R with the new multiplication xr = r x for all r R. R is called an -skew Armendariz ring if for p x = m i=0 a ix i and q x = n j=0 b jx j in R x p x q x = 0 implies a i i b j = 0 for all 0 i m and 0 j n. According to Krempa (), an endomorphism of a ring R is called rigid if a a = 0 implies a = 0 for a R. In Hong et al. (000), a ring is called -rigid if there exists a rigid endomorphism of R. Note that any rigid endomorphism of a ring is a monomorphism and -rigid rings are reduced rings (i.e., rings without nonzero nilpotent elements) by Hong et al. (000, Proposition ). The main purpose of this note is to answer a question of Hong et al. (00). Question. Let be a monomorphism (or automorphism) of a (commultative) reduced ring R, and R be -skew Armendariz. Is R -rigid? The following theorem answers the above question in the affirmative. Received November, 00; Revised January, 00 # Communicated by A. Facchini. Address correspondence to Weixing Chen, Department of Mathematics, Shandong Institute of Business and Technology, Yantai 0, P.R. China; E-mail: cwxcwx00@.com
CHEN AND TONG 0 0 Theorem. Let be a monomorphism of a reduced ring R and R be -skew Armendariz. Then R is -rigid. Proof. Assume the conclusion is false. Then there exists an element a R such that a 0 and a a = 0. We certainly have a a = a a = 0 = 0. Also, we have a a = 0. In fact, a a = a a a a = 0. Since R is reduced, so a a = 0. Since a 0, is a monomorphism, and R is reduced it follows that a 0 and a 0. For any p x = a 0 + a x, and q x = b 0 + b x in R x, we have p x q x = a 0 b 0 + a 0 b + a b 0 x + a b x. Especially take a 0 = a a = a b 0 = a b = a. Then p x q x = a a + a a + a a x + a a x = 0inR x. But a 0 b = a 0. This shows that R is not -skew Armendariz, a contradiction. Hence R is -rigid. It is known that, for an endomorphism of a ring R R x is reduced if and only if R is -rigid Hong et al. (00, Proposition ), and that R is -rigid implies R is -skew Armendariz Hong et al. (00, Corollary ). Combining Proposition of Hong et al. (000) with Theorem,we have the following corollaries. Corollary. Let R be a ring, and be an endomorphism of R. Then R is -rigid if and only if is a monomorphism, R being reduced and -skew Armendariz. Corollary. Let R be a reduced ring and be a monomorphism of R. The following statements are equivalent. () R x is reduced. () R is -skew Armendariz. () R is -rigid. In addition, we can restate Proposition and Proposition of Hong et al. (00) in another form. Corollary. Let be an endomorphism of a ring R. If is a monomorphism and R a reduced and -skew Armendariz ring, then the trivial extension T R R of R is -skew Armendariz, where T R R T R R is defined by a b = a b. Corollary. Let be an endomorphism of a ring R. If is a monomorphism and R a reduced and -skew Armendariz ring, then S = {( a b c ) } 0 a d a b c d R 0 0 a is -skew Armendariz, where S S is defined by a ij = a ij. Recall that if is an endomorphism of a ring R, then the map R x R x defined by m i=0 a ix i m i=0 a i x i is an endomorphism of the polynomial ring R x. And clearly this map extends. Usually this extended map R x R x is denoted by and the image of f x in R x by f x. It is an interesting question whether R is -skew Armendariz if and only if R x is -skew Armendariz for any endomorphism of a ring R. Hong et al.
A NOTE ON SKEW ARMENDARIZ RINGS 0 0 (00, Theorem ) give a partial positive answer to this question, which generalized the result of Anderson and Camillo (). Our next proposition is also connected with this question. Proposition. Let R be a reduced ring and be a monomorphism of R. Then R is -skew Armendariz if and only if R x is -skew Armendariz. Proof. Assume that R is -skew Armendariz. Then R is -rigid by Theorem and the hypothesis. We claim that R x is also -rigid. In fact, for any f x = a 0 + a x + +a n x n in R x, where a 0 a a n R, if f x f x = 0, then a 0 + a x + +a n x n a + a x + + a n x n = 0. Comparing the constant term we have a 0 a 0 = 0, so a 0 = 0 since R is -rigid. Now we have a x + +a n x n a x + + a n x n = 0. It gives that a a = 0, and so a = 0. Continuing this process, at last we have a 0 = a = =a n = 0. Hence f x = 0. By Corollary, R x is -skew Armendariz. Conversely, assume that R x is -skew Armendariz. Then R is -skew Armendariz since R is a subring of R x. Remark. By Hong et al. (00, Theorem ), we know that R is -skew Armendariz if and only if R x is -skew Armendariz provided t = I R for some positive integer t. But the proof of Theorem had a gap. To be specific, the gap lies in the third-last line of p. 0, where the authors claimed that p x tk q x tk = 0inR x. We conclude this note by giving a new proof of Theorem in Hong et al. (00). Proposition. Let be an endomorphism of a ring R and t = I R for some positive integer t. Then R is -skew Armendariz if and only if R x is -skew Armendariz. Proof. Assume that R is -skew Armendariz. Suppose that p y = f 0 x + f x y + +f m x y m q y = g 0 x + g x y + +g n x y n in R x y and p y q y = 0. We also let f i x = a i0 + a i x + +a isi x s i gj x = b j0 + b j x + +b jwj x w j for each 0 i m, and 0 j n, where a i0 a isi b j0 b jwj R. We need to prove f i x i g j x = 0inR x for all 0 i m, and 0 j n. Take a positive integer k such that k>deg f 0 x + deg f x + +deg f m x + deg g 0 x + deg g x + +deg g n x, where the degree is as a polynomial in R x and the degree of the zero polynomial is taken to be 0. Since p y q y = 0inR x y, we have f 0 x g 0 x = 0 f 0 x g x + f x g 0 x = 0 ( ) f m x m g n x = 0 in R x. Now put f x = f 0 x t + f x t x tk+ + f x t x tk+ + +f m x t x mtk+m g x = g 0 x t + g x t x tk+ + g x t x tk+ + +g n x t x ntk+n ( )
CHEN AND TONG 0 0 Note that t = I R. Then f x g x = f 0 x t g 0 x t + f 0 x t g x t + f x t g 0 x t x tk+ + +f m x t m g n x t x m+n tk+ in R x. Using and t = I R, we have f x g x = 0inR x. On the other hand, from we have f x g x = a 00 + a 0 x t + +a 0s0 x s 0t + a 0 x tk+ + a x tk+t+ + +a s x tk+s t+ + +a m0 x mtk+m + a m x mtk+t+m + +a msm x mtk+s mt+m b 00 + b 0 x t + + b 0w0 x w 0t + b 0 x tk+ + b x tk+t+ + + b w x tk+w t+ + + b n0 x ntk+n + b n x ntk+t+n + +b nwn x ntk+w nt+n = 0 in R x. Since R is -skew Armendariz and t = I R, so a iu i b jv = a iu itk+ut+i b jv = 0 for all 0 i m 0 j n u 0 s 0 s m v 0 w 0 w n. So, we have f i x t i g j x t = 0 for all 0 i m 0 j n in R x. Now it is easy to see that f i x i g j x = 0inR x for all 0 i m, and 0 j n. Hence R x is -skew Armendariz. Obviously, if R x is -skew Armendariz, then R is -skew Armendariz. ACKNOWLEDGMENTS The authors express their gratitude to the referee for his or her careful reading and helpful comments, which improved the preparation of this paper. REFERENCES Anderson, D. D., Camillo, V. (). Armendariz rings and Gaussian rings. Comm. Alg. ():. MR e:0. Hong, C. Y., Kim, N. K., Kwak, T. K. (000). Ore extensions of Baer and p.p.-rings. J. Pure Appl. Alg. ():-. MR 00f:0. Hong, C. Y., Kim, N. K., Kwak, T. K. (00). On skew Armendariz rings. Comm. Alg. ():0. Krempa, J. (). Some examples of reduced rings. Alg. Colloq. (): 0. CMP No..