McCoy Rings Relative to a Monoid

International Journal of Algebra, Vol. 4, 2010, no. 10, 469-476 McCoy Rings Relative to a Monoid M. Khoramdel Department of Azad University, Boushehr, Iran M khoramdel@sina.kntu.ac.ir Mehdikhoramdel@gmail.com S. Dolati Pishhesari Department of Azad University, Boushehr, Iran saboura dolati@yahoo.com Abstract For a monoid M, we introduce M-McCoy rings, which are a generalization of McCoy rings, and investigate their properties. Every reversible ring is M-McCoy for any unique product monoid M. It is also shown that a finitely generated Abelian group M is torsion free if and only if there exists a ring R such that R is M-McCoy. Mathematics Subject Classification: Primary 16U99; Secondary 16S15 Keywords: Reversible ring; Monoid rings; M-McCoy condition 1 Introduction Throughout in this paper, all rings are associative with identity. A ring R is right McCoy ring if the equation fxgx = 0 where fx,gx R[x]\{0}, implies that there exists s R\{0} such that fxs = 0. Left McCoy rings are defined similarly[5]. McCoy rings were chosen because McCoy [4] had noted that every commutative rings satisfy this condition. Let M be a monoid. In the following, e will always stand for the identity of M. A ring R is called a right M-McCoy ringa McCoy ring relative to M, if whenever elements α = a 1 g 1 +... + a n g n, β = b 1 h 1 +... + b m h m R[M] satisfy αβ = 0, then αs = 0 for some s R\{0}. Left M-McCoy rings are defined similarly. If M = {e} then every ring is M-McCoy. Let M =N {0}, +. Then a ring R is M-McCoy if and only if R is McCoy ring. The following results will give more examples of M-McCoy rings.

470 M. Khoramdel and S. Dolati Pishhesari Recall that a monoid M is called an u.p.- monoid unique product monoid if for any two nonempty finite subsets A, B M there exist an element g M, uniquely presented in the form ab where a A and b B. The class of u.p. -monoids is quite large and important see [1]. For example, this class includes the right or left ordered monoids, submonoids of a free groups, and torsion free nilpotent group. Every u.p.- monoid M has no nonunity of finite order. For notation M n R,T n R, I n and e ij, 1 i, j n denote the n n matrix ring over R, the upper triangular matrix ring over R, the identity matrix and the matrix with 1 at i, j- entry and 0 elsewhere, respectively. 2 Main results Definition 2.1 Let M be a monoid. A ring R is called right M-McCoy ring if whenever elements α = a 1 g 1 +... + a n g n, β = b 1 h 1 +... + b m h m R[M] satisfy αβ =0, then αs =0for some s R\{0}. Left M-McCoy rings are defined analogously. Recall that a ring R is called reversible if ab = 0 then ba = 0 for all a, b R. Proposition 2.2 Let M be an up-.monoid and R a reversible ring. Then R is M-McCoy. proof: Letα = a 1 g 1 +... + a n g n and β = b 1 h 1 +... + b m h m R[M] be such that αβ =0. We claim that there exists s R\{0} such that αs =0. We proceed by induction on n. Let n =1. Then α = a 1 g 1. By [1, Lemma 1.1] g 1 h i g 1 h j for i j. Thus a 1 b j =0for all j. Let n 2. Since M is u.p.-monoid, there exist i,j with 1 i n and 1 j m such that g i h j is uniquely presented by considering two subset A = {g 1,g 2,..., g n } and B = {h 1,h 2,..., h m } of M. We may assume, without loss of generality, that i =1,j =1. Thus a 1 b 1 g 1 h 1 =0and hence a 1 b 1 =0. Since R is reversible, b 1 a 1 =0follows. Thus b 1 αβ =0and so n m b 1 a i g i b j h j =0. i=2 j=1 By induction, we have n i=2 b 1 a i g i s =0for some s R\{0}. Thus αsb 1 = 0. This prove the results. Similarly, by induction on m we prove that R is left M-McCoy.

McCoy rings relative to a monoid 471 Let M, be an ordered monoid. If for any g, ǵ, h M, g < ǵ implies gh < ǵh and hg < hǵ, then M, is called a strictly ordered monoid. Corollary 2.3 Let M be a strictly totally ordered monoid and R a reversible ring. Then R is M-McCoy. Corollary 2.4 Let R be a reversible ring. Then R is Z-McCoy, that is, for any α = a n g n +... + a q g q, β = b m h m +... + b p h p R[x, x 1 ],ifαβ =0, then there exist t, s R\{0} such that αs =0and tβ =0. Taking M =N {0}, + in Corollary 2.3, it follows that every reversible ring is McCoy; [5,Theorem 2]. Proposition 2.5 Let M be a monoid with M 2. Then a ring R is the right resp.,left M-McCoy if and only if the ring a a 12 a 1n 0 a a 2n R n :=..... a, a ij R. a is right resp.,left M-McCoy for any n 1. proof:let R be an M-McCoy ring and α = A 1 g 1 +... + A n g n, β = B 1 h 1 +... + B m h m be elements of R n [M]. Assume that αβ =0.Let a i a i 12 a i 1n b j b j 0 a i a i 12 b j 1n A i = 2n.....,B 0 b j b j j = 2n........ a i b j It is easy to see that there exists an isomorphism R n [M] = R[M] n defined by s k=1 a k a k 12 a k 1n 0 a k a k 2n...... a k g k sk=1 a k g sk=1 k a k 12 g k sk=1 a k 1n g k 0 sk=1 a k g k sk=1 a k 2n g k....... sk=1 a k g k

472 M. Khoramdel and S. Dolati Pishhesari Thus α 1 α 12 α 1n 0 α 1 α 2n α =.....,β =. α 1 are elements of R[M] n such that αβ =0. β 1 β 12 β 1n 0 β 1 β 2n...... β 1 Thus α 1 β 1 =0. Since R is right M-McCoy, there exists s R\{0} such that α 1 s =0. Therefore, αse 1n =0and it follows that R n is M-McCoy. Conversely, let R n be M-McCoy ring and α = a 1 g 1 +... + a n g n, β = b 1 h 1 +... + b m h m satisfy αβ =0in R[M]. From the isomorphism we have, α 0 0 β 0 0 0 α 0 0 β 0........... =0.. α β Then as in proof of another side there exists s R\{0} such that αs =0. This proves the results. A ring R is called M-Armendariz ring, if whenever elements α = a 1 g 1 +... + a n g n, β = b 1 h 1 +... + b m h m R[M] satisfy αβ = 0, then a i b j = 0 for each i, j. M-Armendariz rings are M-McCoy by definition. However there exists an M-McCoy ring which is not M-Armendariz. If R is an M-McCoy ring, then R 4 is M-McCoy too, by Proposition 2.5. However R 4 is not M-Armendariz by [7, Remark 1.8]. Based on Proposition 2.5, we may suspect M n R ort n R over an M- McCoy ring is still M-McCoy. But the following example erases the possibility. Example 2.6 Let M be a monoid with M 2 and R a ring. Take e g M. LetA = C = e 12,B = e 11,D = e 22, A 0 B 0 α 1 = e + g β 1 = C 0 A 0 e + 0 I n 2 g β 2 = C 0 e + D 0 g

McCoy rings relative to a monoid 473 α 2 = A 0 0 I n 2 e + B 0 g are elements of M n R[M] so α 1 β 1 =0and α 2 β 2 =0. But if Sβ 1 =0or α 2 T =0for some S, T M n R[M]\{0}, then S = T =0, thus M n R is neither left nor right M-McCoy. It is natural to ask whether R is an M-McCoy ring for a monoid M, if for any nonzero proper ideal I of R, R/I, I are M-McCoy, where I is considered as a M-McCoy ring without identity. However, we have a negative answer to this question by the following example. Example 2.7 Let F be a field and consider R = T 2 F, which is not M- McCoy, for u.p.- monoid M by Example 2.6. Next we show that R/I and I are M-McCoy for some nonzero proper ideal I of R. Note that the only nonzero proper ideal of R are F F F F, 0 F 0 F and 0 F Let I =. Then R/I = F and so R/I is M-McCoy. Now let α = n ai b i i=1 g i and β = m cj d j j=1 h j be nonzero elements of I[M] such that αβ =0. From the isomorphism T 2 F [M] = T 2 F [M] defined by:. n ai b ni=1 i a g 0 c i i g ni=1 i b i g i i=1 i 0 ni=1 c i g i we have α 1 β 1 = α 1 β 2 =0, where α 1 = m i=1 a i g i, β 1 = n j=1 c j h j and β 2 = n j=1 d j h j F [M]. If α 1 =0, then αe 11 =0. Suppose α 1 0. Since β 0, then β 1 0or β 2 0. From the equation and the condition that F is right M-McCoy, we have α 1 s =0for some nonzero s F, therefore αse 11 =0. Thus I is right M-McCoy, and I is left M-McCoy since e 12 β =0. Proposition 2.8 Let M be a cancelative monoid and N is an ideal of M. If R is right N-McCoy, then R is right M-McCoy. proof: Suppose that α = a 1 g 1 +... + a n g n, β = b 1 h 1 +... + b m h m R[M] satisfy αβ =0. Take g N, then gg 1,..., gg n,h 1 g,..., h m g N and gg i gg j

474 M. Khoramdel and S. Dolati Pishhesari and h i g h j g when i j. Now from n m a i gg i b j h j g=0 i=1 j=1 and from the hypothesis, there exists s R\{0} such that n i=1 a i gg i s =0. Thus αs =0, it follows that R is M-McCoy. Lemma 2.9 Let M be a cyclic group of order n 2 and R a ring with 0 1, then R is not M-McCoy. proof: Suppose M = {e, g, g 2,..., g n 1 }.Letα =1e +1g +1g 2 +... +1g n 1 and β =1e + 1g. Then αβ =0, but if αs =0then s =0. Thus R is not M-McCoy. Lemma 2.10 Let M be a monoid and N a submonoid of M. If R is M- McCoy then R is N-McCoy. Let T G be the set of elements of finite order in an ablian group G. Then G is said to be torsion free if T G ={e}. Theorem 2.11 Let G be a finitely generated Abelian group. Then the following conditions on G are equivalent. 1G is torsion free. 2There exists a ring R with R 2 such that R is G-McCoy. proof:2 1 If g T G and g e, then N =< g>is cyclic group of finite order. If a ring R {0} is G-McCoy, then by Lemma 2.10 R is N-McCoy, a contradiction with Lemma 2.9. Thus every ring R {0} is not G-McCoy. 1 2. If G is finitely generated Ablian group with T G ={e}, Then G = Z Z... Z, a finite direct product group Z, by [7,Lemma1.13], G is u.p.-monoid. Let R be a commutative ring then by Proposition 2.2, R is G-McCoy. A classical right quotient ring for a ring R is a ring Q which contains R as a subring in such a way that every regular element i.e., non-zero-divisor of R is invertible in Q and Q = {ab 1 : a, b R, b is regular }. A ring R is called right Ore if given a, b R with b regular there exist a 1,b 1 R with b 1 regular

McCoy rings relative to a monoid 475 such that ab 1 = ba 1. Classical left quotient rings and left Ore rings are defined similarly. It is a well-known fact that R is a right resp., left Ore ring if and only if the classical right resp., left quotient ring of R exists. Theorem 2.12 Suppose that there exists the classical right quotient ring Q of a ring R. Then R is right M-McCoy if and only if Q is right M-McCoy. proof:. Let A = n i=1 α i g i and B = m j=1 β j h j be nonzero elements of R[M] such that AB =0. Since Q is the classical right quotient ring, we may assume that α i = a i u 1,β j = b j v 1 with a i,b j R for all i, j and regular elements u, v R. For each j, there exists c j R and a regular element w R such that u 1 b j = c j w 1. Denote A 1 = n i=1 a i g i and B 1 = m j=1 c j h j. Then the equation A 1 B 1 vw 1 =0, thus there exist a nonzero element s R\{0} such that As =0, then α i us =0for every i. It implies that Aus =0and us is nonzero element of Q. Hence Q is right M-McCoy.. Letα = n i=1 a i g i, β = m j=1 b j h j R[M] such that αβ =0. Then there exists a nonzero element K Q such that αk =0since Q is right M- McCoy. Because Q is a classical right quotient ring, we can assume K = au 1 for some a R\{0} and regular element u. Then αau 1 = αk =0implies that αa =0. Therefore, R is a right M-McCoy ring. By the Goldie Theorem, if R is semiprime left and right Goldie ring, then R has the classical left and right quotient ring. Hence there exists a class of rings satisfying the following hypothesis. Corollary 2.13 Suppose that there exists the classical left and right quotient ring Q of a ring R. Then R is M-McCoy if and only if Q is M-McCoy. In following proposition we consider the case of direct limit of direct system of M-McCoy rings. Proposition 2.14 The direct limit of a direct system of M-McCoy rings is also M-McCoy. proof: Let D = {R i,α ij } be a direct system of M-McCoy rings R i for i I and ring homomorphisms α ij : R i R j for each i j satisfying α ij 1 = 1, where I is directed partially ordered set. Set R = limr i be direct

476 M. Khoramdel and S. Dolati Pishhesari limit of D with l i : R i R and l j α ij = l i. We will prove that R is M-McCoy ring. Take x, y R, then x = l i x i,y = l j y j for some i, j I and there is k I such that i k, j k. Define x + y = l k α ik x i +α jk y j and xy = l k α ik x i α jk y j, where α ik x i and α jk y j are in R k. Then R forms a ring with l i 0 = 0 and l i 1 = 1. Now suppose AB =0for A =Σ m s=1a s g s, B =Σ n t=1b t h t in R[M]\{0}. There exist i s,j t,k I such that a s = l is a is, b t = l jt b jt, i s k, j t k. So a s b t = l k α iska is α jtkb jt. Thus AB =Σ m s=1 l kα iska is g s Σ n t=1 l kα jtkb jt h t =0. But R k is M-McCoy ring and so there exist 0 d R k such that Al k d =0. Thus R is M-McCoy ring. References [1] G.F. Birkenmeier, J. K. Park, Triangular matrix representation of ring extensions, J Algebra, 265 2003, 457-477. [2] C. Huh, Y. Lee, A. Smoktunowicz, Armendariz rings and semicommutative rings, Comm Algebra, 30 2 2002, 751-761 [3] N. Kim, Y. Lee, Extension of reversible rings, J. Pure Appl. Algebra, 185 2003, 207-223. [4] N. McCoy, Remarks on divisors of zero, Amer. Math. Monthly, 49 1942, 286-295. [5] P.P. Nielsen, semi-commutativity and the McCoy condition, J. Algebra, 298 2006, 134-141. [6] G. Shin, Prime ideals and sheaf representation of pseudo symmetric rings, Trans Amer math. Soc., 184 1973, 43-60. [7] L.Zhongkui, Armendariz rings relative to monoid, Comm. Algebra, 33 3 2005, 649-661. [8] R. Zhao, Z. Liu, Extension of McCoy rings,manuscript 2007. Received: April, 2008