Algebra Colloquium 16 : 3 (2009) 495–502
Algebra Colloquium c 2009 AMSS CAS ° & SUZHOU UNIV
Extensions of McCoy Rings∗ Renyu Zhao College of Economics and Management Northwest Normal University, Lanzhou 730070, China E-mail:
[email protected]
Zhongkui Liu College of Mathematics and Information Science Northwest Normal University, Lanzhou 730070, China E-mail:
[email protected] Received 5 November 2006 Revised 5 May 2007 Communicated by Nanqing Ding Abstract. A ring R is called right McCoy if whenever polynomials f (x), g(x) ∈ R[x] \ {0} satisfy f (x)g(x) = 0, there exists a nonzero r ∈ R such that f (x)r = 0. We continue in this paper the study of right McCoy rings by Nielsen [8]. We first consider properties and basic extensions of right McCoy rings, providing many examples in the process. Next, we show that if R is a right McCoy ring, then R[x] and R[x]/(xn ) are right McCoy rings, where (xn ) is the ideal generated by xn and n is a positive integer; and that for a right Ore ring R with Q its classical right quotient ring, R is right McCoy if and only if Q is right McCoy. 2000 Mathematics Subject Classification: 16W60 Keywords: right McCoy ring, classical right quotient ring
1 Introduction Throughout this paper, all rings are associative with identity. Given a ring R, the polynomial ring over R is denoted by R[x] with x its indeterminate. According to Nielsen [8], a ring R is called right McCoy whenever polynomials f (x), g(x) ∈ R[x] \ {0} satisfy f (x)g(x) = 0, there exists a nonzero r ∈ R such that f (x)r = 0. We define left McCoy rings similarly. If a ring is both left and right McCoy, we say that the ring is a McCoy ring. It is well known that commutative rings are ∗
The second named author was supported by the National Natural Science Foundation of China (10171082), TRAPOYT and the Cultivation Fund of the Key Scientific and Technical Innovation Project, Ministry of Education of China, and NWNU-KJCXGC-0318.
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always McCoy rings [7, Theorem 2], but it is not true for noncommutative rings (see [11]). According Chhawchharia [9], a ring R is called Armendariz Pm to Rege and P n if given f (x) = i=0 ai xi , g(x) = j=0 bj xj ∈ R[x] with f (x)g(x) = 0, this implies that ai bj = 0 for all i, j. Armendariz rings are clearly McCoy rings. According to Cohn [2], a ring R is called reversible if ab = 0 implies that ba = 0 for a, b ∈ R. Reversible rings are McCoy by [8, Theorem 2]. But these implications are not reversible. Indeed, there exists a commutative ring which is not Armendariz (see [9]), and there is a non-reversible (right) McCoy ring (see [8]). In this paper, we continue the study of right McCoy rings by Nielsen [8]. First, we consider properties and basic extensions of right McCoy rings, providing many examples in the process. Next, we show that if R is a right McCoy ring, then R[x] and R[x]/(xn ) are right McCoy rings, where (xn ) is the ideal generated by xn and n is a positive integer; and that for a right Ore ring R with Q its classical right quotient ring, R is right McCoy if and only if Q is right McCoy. 2 Main Results For a ring R and an (R, R)-bimodule M , let T (R, M ) = {(a, x) | a ∈ R, x ∈ M } with the addition componentwise and the multiplication defined by (a1 , x1 )(a2 , x2 ) = (a1 a2 , a1 x2 + x1 a2 ). Then T (R, M ) is a ring which is called the trivial extension of ³ ´ a x R by M . Note that T (R, M ) is isomorphic to the ring consisting of matrices 0 a , where a ∈ R and x ∈ M . We will give more examples of McCoy rings, which are generalizations of trivial extensions. Let S be a ring and define a 0 0 Rn = . .. 0
a12 a13 a a23 0 a .. .. . . 0 0
· · · a1n · · · a2n ¯¯ ¯ · · · a3n ¯ a, aij ∈ S , . ¯ .. . .. ··· a
where n ≥ 2 is a positive integer. Proposition 2.1. Let S be a ring and let Rn (n ≥ 2) be the ring defined as above. Then S is a right McCoy ring if and only if Rn is a right McCoy ring. Proof. (⇒) Without loss of generality, we can assume S 6= 0. Let f (x), g(x) ∈ Rn [x] \ {0} with f (x)g(x) = 0. We will show that there exists 0 6= C ∈ Rn such that f (x)C = 0. In fact, we will show something slightly stronger. Let Eij be the usual matrix unit with 1 in the (i, j)-coordinate and zero elsewhere. We will show that there exists some C = rE1n 6= 0 such that f (x)C = 0, and this will imply that Rn is right McCoy. We proceed with the following two cases. Case 1: Suppose f (x)E11 = 0. In this case, take C = rE1n = E11 rE1n with r ∈ S \ {0}. Case 2: Suppose f (x)E11 6= 0. Since g(x) 6= 0, there exists j ≥ i ≥ 1 such that
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Eii g(x)Ejj 6= 0 with i maximal. One computes X¡ ¡ ¢ ¢ 0 = Eii f (x)g(x) Ejj = Eii f (x) Ekk g(x)Ejj =
X¡
k≤i
¢¡ ¢ ¢ ¡ Eii f (x)Ekk Ekk g(x)Ejj = Eii f (x)Eii Eii g(x)Ejj , ¢¡
k≤i
where the maximality of i is used in the second equality, and¢¡the upper-triangularity ¡ ¢ is used in the last equality. But the equation Eii f (x)Eii Eii g(x)Ejj = 0 is an equation over S[x] (and both polynomials are non-zero). ¡ Since S ¢is a right McCoy ring, there exists a non-zero element r ∈ S such that Eii f (x)Eii r = 0. Then one can easily check that C = rE1n annihilates f (x) on the right. Pq Pp i j (⇐) Suppose f (x) = j=0 bj x ∈ S[x] \ {0} such that i=0 ai x , g(x) = f (x)g(x) = 0. Let F (x) = f (x)In and G(x) = g(x)In , where In is the n × n identity matrix. Then F (x), G(x) ∈ Rn [x] \ {0} and F (x)G(x) = 0. Since Rn is right McCoy, there exists 0 6= C ∈ Rn such that F (x)C = 0. Now it is easy to see that there exists 0 6= r ∈ S such that f (x)r = 0, which implies that S is right McCoy. 2 In [8, Theorem 2], it is shown that reversible rings are McCoy rings. In general, this implication is not reversible. For example, there is a non-reversible ring which is McCoy (see [8]); here we also obtain a counterexample independently by Proposition 2.1. Example 2.2. Let S be a reduced ring. Then S is a McCoy ring. Thus, a b c ¯¯ R = 0 a d ¯¯ a, b, c, d ∈ S 0 0 a is a right McCoy ring by Proposition 2.1, but R is not reversible by [5, Example 1.5]. We know that Armendariz rings are McCoy rings, but the converse is not true in general. For example, there exists a commutative ring which is not Armendariz (see [9]). Also, we have the following counterexample for this situation by Proposition 2.1. Example 2.3. Let S be a right McCoy ring and a a12 a13 a14 ¯ ¯ 0 a a a 23 24 ¯ a, aij ∈ S . R4 = ¯ 0 0 a a34 0 0 0 a Then R4 is right McCoy by Proposition 2.1. But R4 is not Armendariz by [4, Example 3].
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There is also an example in [8] to show that semicommutative rings do not have to be right McCoy rings. In [5, Example 1.3], it is shown that R4 is not semicommutative. Thus, Example 2.3 also implies that right McCoy rings are not necessarily semicommutative. Based on Proposition 2.1, one may suspect that if R is right McCoy, then every n × n upper triangular matrix ring Tn (R) and every n × n full matrix ring Mn (R) are right McCoy, where n ≥ 2. But the following example erases the possibility. Example 2.4. Let R be a ring and let S = Tn (R) (resp., Mn (R)). Let f (x) = (In − Enn ) + E1n x and g(x) = Enn − E1n x. Then f (x), g(x) ∈ S[x] \ {0} with f (x)g(x) = 0. If (aij ) ∈ S such that f (x)(aij ) = 0, then aij = 0 for all i, j. Thus, S is not right McCoy. Let R be a right McCoy ring, then the trivial extension T = T (R, R) is a³ right ´ 0 R McCoy ring by Proposition 2.1. Notice that the prime radical P (T ) of T is 0 0 (hence it is a right McCoy ring by applying the definition of right McCoy rings to rings without identity) and that T /P (T ) ' R is right McCoy. So one may suspect that a ring R is a right McCoy ring if R/P (R) and P (R) are right McCoy rings. However, the following example erases the possibility. Example 2.5. Use the ring in [4, Example 13]. Let Z be the ring of integers and let ½µ ¶¯ ¾ a c ¯ R= ¯ a, b, c ∈ Z, a − b ≡ c ≡ 0 (mod 2) . 0 b Then by [4], P (R) and R/P (R) are right McCoy. Now we claim that R is not right McCoy. Let µ ¶ µ ¶ µ ¶ µ ¶ 2 2 0 2 0 2 0 2 f (x) = + x, g(x) = + x. 0 0 0 0 0 −2 0 0 ³ ´ a c Then f (x), g(x) ∈ R[x] \ {0} with f (x)g(x) = 0. Suppose f (x) 0 b = 0 for some ³ ´ a c 0 b . Then a = b = c = 0, and so R is not right McCoy. Moreover, one may conjecture that R is right McCoy if for any nonzero proper ideal I of R, R/I and I are right McCoy, where I is considered as a right McCoy ring without identity. However, we also have a counterexample to this situation as in the following. ³ ´ F F Example 2.6. Let F be a field and consider the ring R = 0 F . Then R is not right McCoy by Example 2.4. By [4, Example 14], for any nonzero ideal I of R, R/I and I are Armendariz rings, so they are right McCoy. Let A be a ring, B be a subring of A, {Ai }∞ i=1 be a countable set of copies of A, let D be the direct product of all rings A , and R = R(A, B) be the subring of D i L∞ generated by the ideal i=1 Ai and the subring {(b, b, . . . ) | b ∈ B} (see [10]). Then we have the following result.
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Proposition 2.7. If A is a right McCoy ring, then R = R(A, B) is a right McCoy ring. Proof. Let f, g ∈ R[x] \ {0} with f g = 0. We write f = (f1 , f2 , . . . ) and g similarly. For any x = (xi )∞ i=1 ∈ R, we define the support of x as supp(x) = {i | xi 6= 0}. Next, we divide the proof into two cases. Case 1: Suppose there exists some k ∈ / supp(f ). If we fix x ∈ A \ {0}, then the sequence with x in the k th spot and zeros elsewhere annihilates f on the right. Case 2: Suppose every component of f is nonzero. In particular, there exists some k ∈ supp(f ) ∩ supp(g). From fk gk = 0 and the fact that A is right McCoy, there exists some x ∈ A \ {0} such that fk x = 0. Then the same sequence as in Case 1 suffices. 2 For a ring R, Let Y ¯ © ª S = (an )∞ R ¯ an is eventually constant , n=1 ∈ which is a subring of the countable direct product following result.
Q∞ n=1
R. Then we have the
Corollary 2.8. The ring S is right McCoy if and only if R is right McCoy. Pp Proof. 2.7, it suffices to establish the necessity. Let f = i=0 ai xi , PqBy Proposition g = j=0 bj xj ∈ R[x] \ {0} be such that f g = 0. For any i and j, set αi = (ai )∞ n=1 Pp Pq i j and βj = (bj )∞ . Then α , β ∈ S. Let F (x) = α x and G(x) = β i j n=1 i=0 i j=0 j x . ∞ Then F (x), G(x) ∈ S[x] \ {0} and F (x)G(x) = 0. Thus, there exists r = (rn )n=1 ∈ S \ {0} such¡ P that F (x)r ¢ = 0. Since r 6= 0, there exists k ∈ {1, 2, . . . } such that p i rk 6= 0 and a x rk = 0. This means that R is right McCoy. 2 i=0 i Proposition 2.9. The class of right McCoy rings is closed under direct products and direct sums. Proof. This proposition follows by the same method used in the proof of Proposition 2.7. 2 In [1], it is shown that R[x] is an Armendariz ring if and only if R is an Armendariz ring. In [5], it is shown that there exists a reversible ring R with R[x] not reversible. For right McCoy rings, we have the following result. Theorem 2.10. A ring R is a right McCoy ring if and only if R[x] is a right McCoy ring. Proof. (⇒) Suppose that R is a right McCoy ring and let f (T ), g(T ) ∈ R[x][T ] \ {0} with f (T )g(T ) = 0. Let k = max{deg(fi )} + 1. Then the set of coefficients (over R) of the polynomial f (xk ) equals the set of coefficients (over R) of f (T ). Since f (T )g(T ) = 0 and x commutes with elements of R, f (xk )g(xk ) = 0. Since R is right McCoy, there exists 0 6= r ∈ R ⊆ R[x] such that f (xk )r = 0. Thus, f (T )r = 0, which implies that R[x] is right McCoy.
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Pm Pn (⇐) Suppose f (y) = i=0 ai y i , g(y) = j=0 bj y j ∈ R[y] \ {0} with f (y)g(y) = 0. Then there exists 0 6= r(x) ∈ R[x] such that f (y)r(x) = 0. Thus, ai r(x) = 0 for all i. Since r(x) 6= 0, there exists 0 6= rj ∈ R such that ai rj = 0 for all i. Hence, f (y)rj = 0, which implies that R is right McCoy. 2 Corollary 2.11. Let R be a right McCoy ring and let {xα } be any set of commuting indeterminates over R. Then R[{xα }] is right McCoy. Proof. Let f, g ∈ R[{xα }][T ] \ {0} with f g = 0. Then f, g ∈ R[xα1 , . . . , xαn ][T ] for some finite subset {xα1 , . . . , xαn } ⊆ {xα }. By induction, the ring R[xα1 , . . . , xαn ] is right McCoy, so there exists 0 6= r ∈ R[xα1 , . . . , xαn ] such that f r = 0. Hence, R[{xα }] is right McCoy. 2 The following result, similar to [1, Theorem 5] and [5, Theorem 2.5], extends to the class of McCoy rings. Theorem 2.12. Let R be a ring and n any positive integer. If R is right McCoy, then R[x]/(xn ) is a right McCoy ring, where (xn ) is the ideal generated by xn . Proof. Denote x in R[x]/(xn ) by u, so R[x]/(xn ) = R[u] = R + Ru + · · · + Run−1 , where u commutes with elements of R and un = 0. Let f, g ∈ R[u][y] \ {0} be such Pp Pq Pn−1 that f g = 0. Suppose f = i=0 fi y i and g = j=0 gj y j . Let fi = s=0 ais us and ¢ ¡ Pn−1 j t Pn−1 ¡ Pp P n−1 Pq j j¢ t s i i gj = t=0 bt u . Then f = s=0 i=0 as y u and g = t=0 j=0 bt y u . From f g = 0, we have the following equations: p X ³X s+t=k
If
Pp i=0
ais y i
q ´³ X
i=0
´ bjt y j = 0,
k = 0, 1, . . . , n − 1.
(∗)k
j=0
ai0 y i = 0, take r = un−1 . Then 0 6= r ∈ R[u] and fr =
p ³ n−1 X³X s=0
p ´ ´ ³X ´ ais y i us un−1 = ai0 y i un−1 = 0.
i=0
i=0
Pp Pq If i=0 ai0 y i 6= 0, since g 6= 0, there exists l ∈ {0, 1, . . . , n − 1} such that j=0 bjl y j Pq 6 0. Let l0 = min{l | j=0 bjl y j 6= 0}. Then from (∗)l0 , it follows that = 0=
p X ³X s+t=l0
i=0
ais y i
q ´³ X j=0
p q ´ ³X ´³ X ´ ai0 y i bjt y j = bjl0 y j . i=0
j=0
¡ Pp ¢ i i Since R is right McCoy, there exists 0 6= c ∈ R such that c = 0. i=0¢a0 y ¢ ¡ ¡ P P n−1 p i i s n−1 n−1 a y u cu = Let r = cu . Then 0 = 6 r ∈ R[u] and f r = s=0 i=0 s ¢ n−1 ¡ Pp i i a y cu = 0. Therefore, R[u] is a right McCoy ring. 2 i=0 0 We end this paper by generalizing [3, Theorem 12] for Armendariz rings and [5, Theorem 2.6] for reversible rings to McCoy rings. A ring R is called right Ore if given a, b ∈ R with b regular, there exist a1 , b1 ∈ R with b1 regular such that
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ab1 = ba1 . It is a well-known fact that R is a right Ore ring if and only if there exists the classical right quotient ring of R. Let F be a field and R = F {x, y} be the free algebra in two indeterminates over F . For x and y, there do not exist a, b ∈ R such that y −1 x = ab−1 (xy −1 = b−1 a). So the domain R cannot have its classical right (left) quotient ring; hence the hypothesis of the following result is necessary. Theorem 2.13. Let R be a right Ore ring and Q be the classical right quotient ring of R. Then R is right McCoy if and only if Q is right McCoy. Pn Pm Proof. Suppose that R is right McCoy. Let F (x) = i=0 αi xi , G(x) = j=0 βj xj ∈ Q[x] \ {0} be such that F (x)G(x) = 0. By [6, Proposition 2.1.16], we may assume αi = ai u−1 and βj = bj v −1 with ai , bj ∈ R for all i, j and some regular u, v ∈ R. Also, by [6, Proposition 2.1.16], for each j, there Pn regular Pmexist cj ∈ R and some w ∈ R such that u−1 bj = cj w−1 . Put f (x) = i=0 ai xi and g(x) = j=0 cj xj . Then f (x), g(x) ∈ R[x] \ {0}, and 0 = F (x)G(x) =
m+n X³ k=0
=
m+n X³ k=0
X
m+n ´ ´ X³ X αi βj xk = ai (u−1 bj )v −1 xk
X i+j=k
´
ai cj w−1 v −1 xk =
i+j=k
k=0 m+n X³ k=0
X
i+j=k
´ ai cj xk w−1 v −1
i+j=k
= f (x)g(x)(vw)−1 , hence f (x)g(x) = 0 in R[x]. Since R is right McCoy, there exists 0 6= r ∈ R such that f (x)r = 0. By [6, Proposition 2.1.16], there exist r0 ∈ R and some regular ¡ Pm ¢ i 0 tP∈ R such that u−1P r0 = rt−1 , and r0 6= 0 since r 6= 0. So F (x)r0 = i=0 αi x r = m m −1 0 i r x = i=0 ai rt−1 xi P = f (x)rt−1 = 0. Therefore, Q is right McCoy. i=0 ai u Pn m i j Conversely, suppose f (x) = a x , g(x) = b x ∈ R[x] \ {0} with i=0 i j=0 j −1 f (x)g(x) = 0. Then there exists 0 6= ru ∈ Q such that f (x)ru−1 = 0. Thus, f (x)r = 0 with r ∈ R. Hence, R is right McCoy. 2 Acknowledgement. The authors are very much indebted to the referee for his/her careful reading of the paper, and many valuable comments, which have greatly improved the paper, especially simplified the proof of Propositions 2.1 and 2.7, and led to the new version of Proposition 2.7.
References [1] D.D. Anderson, V. Camillo, Armendariz rings and Gaussian rings, Comm. Algebra 26 (1998) 2265–2272. [2] P.M. Cohn, Reversible rings, Bull. London Math. Soc. 31 (1999) 641–648. [3] C. Huh, Y. Lee, A. Smoktunowicz, Armendariz rings and semicommutative rings, Comm. Algebra 30 (2002) 751–761. [4] N.K. Kim, Y. Lee, Armendariz rings and reduced rings, J. Algebra 223 (2000) 477– 488. [5] N.K. Kim, Y. Lee, Extensions of reversible rings, J. Pure Appl. Algebra 185 (2003) 207–223.
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[6] J.C. McConnell, J.C. Robson, Noncommutative Noetherian Rings, John Wiley & Sons, Ltd., Chichester, 1987. [7] N.H. McCoy, Remarks on divisors of zero, Amer. Math. Monthly 49 (1942) 286–295. [8] P.P. Nielsen, Semi-commutativity and the McCoy condition, J. Algebra 298 (2006) 134–141. [9] M.B. Rege, S. Chhawchharia, Armendariz rings, Proc. Japan Acad. (Ser. A Math. Sci.) 73 (1997) 14–17. [10] A.A. Tuganbaev, Rings Close to Regular, Kluwer Academic Publishers, DordrechtBoston-London, 2002. [11] L. Weiner, Concerning a theorem of McCoy, Amer. Math. Monthly 59 (5) (1952) 336–337.
