EXTENSIONS OF EXTENDED SYMMETRIC RINGS

Transcription

1 Bull Korean Math Soc , No 4, pp EXTENSIONS OF EXTENDED SYMMETRIC RINGS Tai Keun Kwak Reprinted from the Bulletin of the Korean Mathematical Society Vol 44, No 4, November 2007 c 2007 The Korean Mathematical Society

2 Bull Korean Math Soc , No 4, pp EXTENSIONS OF EXTENDED SYMMETRIC RINGS Tai Keun Kwak Abstract An endomorphism α of a ring R is called right left symmetric if whenever abc = 0 for a, b, c R, acαb = 0 αbac = 0 A ring R is called right left α-symmetric if there exists a right left symmetric endomorphism α of R The notion of an α-symmetric ring is a generalization of α-rigid rings as well as an extension of symmetric rings We study characterizations of α-symmetric rings and their related properties including extensions The relationship between α-symmetric rings and extended Armendariz rings is also investigated, consequently several known results relating to α-rigid and symmetric rings can be obtained as corollaries of our results 1 Introduction Recall that a ring is reduced if it has no nonzero nilpotent elements Lambek called a ring R symmetric [13] provided abc = 0 implies acb = 0 for a, b, c R Every reduced ring is symmetric [16, Lemma 11] but the converse does not hold by [2, Example II5] Cohn called a ring R reversible [5] if ab = 0 implies ba = 0 for a, b R Historically, some of the earliest known results about reversible rings although not so called at the time were due to Habeb [6] It is obvious that commutative rings are symmetric and symmetric rings are reversible; but the converses do not hold by [2, Examples I5 and II5] and [14, Examples 5 and 7] Another generalization of a reduced ring is an Armendariz ring Rege and Chhawchharia called a ring R Armendariz [15] if whenever any polynomials fx = a 0 + a 1 x + + a m x m, gx = b 0 + b 1 x + + b n x n R[x] satisfy fxgx = 0, then a i b j = 0 for each i and j This nomenclature was used by them since it was Armendariz who initially showed that a reduced ring always satisfies this condition [3, Lemma 1] For a ring R with a ring endomorphism α : R R, a skew polynomial ring also called an Ore extension of endomorphism type R[x; α] of R is the ring obtained by giving the polynomial ring over R with the new multiplication xr = αrx for all r R Received March 6, Mathematics Subject Classification Primary 16W20, 16U80; Secondary 16S36 Key words and phrases reduced rings, symmetric rings, extended Armendariz rings This work was supported by the Daejin University Research Grants in c 2007 The Korean Mathematical Society

3 778 TAI KEUN KWAK The Armendariz property of a ring was extended to skew polynomial rings but with skewed scalar multiplication in [8, 9]: For an endomorphism α of a ring R, R is called α-armendariz resp α-skew Armendariz if for p = m a i x i and q = n b j x j in R[x; α], pq = 0 implies a i b j = 0 resp a i α i b j = 0 for all j=0 0 i m and 0 j n On the other hand, an endomorphism α of a ring R is called rigid [12] if aαa = 0 implies a = 0 for a R, and R is an α-rigid ring [7] if there exists a rigid endomorphism α of R Note that any rigid endomorphism of a ring is a monomorphism, and α-rigid rings are reduced rings by [7, Proposition 5] Any α-rigid ring is α-armendariz [7, Proposition 6], but the converse is not true, in general; every α-armendariz ring is α-skew Armendariz, but the converse does not hold by [9, Theorem 17 and Example 18] In [8, Proposition 3], R is an α-rigid ring if and only if R[x; α] is reduced Motivated by the above, in this paper we introduce the notion of an α- symmetric ring for an endomorphism α of a ring R, as a generalization of α- rigid rings and an extension of symmetric rings, and study α-symmetric rings and their related properties The relationship between α-symmetric rings and extended Armendariz rings is also investigated Consequently, several known results are obtained as corollaries of our results Throughout this paper R denotes an associative ring with identity and α denotes a nonzero and non identity endomorphism, unless specified otherwise 2 Properties of α-symmetric rings We begin with the following definition Definition 21 An endomorphism α of a ring R is called right left symmetric if whenever abc = 0 for a, b, c R, acαb = 0 αbac = 0 A ring R is called right left α-symmetric if there exists a right left symmetric endomorphism α of R R is α-symmetric if it is both right and left α-symmetric Observe that every subring S with αs S of a right α-symmetric ring is also right α-symmetric; and any domain R is α-symmetric for any endomorphism α of R, but the converse does not hold see Example 271 below The next example shows that the concept of α-symmetric is not left-right symmetric Example 22 Let Z be the ring of integers Consider a ring { } R = a, b, c Z 0 c 1 1 Note that for A = and B = R, we have AB = O but BA O Thus R is not reversible, and so R is not symmetric i Let α : R R be an endomorphism defined by i=0

4 EXTENSIONS OF EXTENDED SYMMETRIC RINGS 779 a 0 α = 0 c a b, B = 0 c 0 c a b If ABC = O for A = and C = 0 c R, then we get aa a = nd so aa a = 0 Thus this yields ACαB = O, and 1 1 hence R is right α-symmetric However, for A =, B = 1 0 and C = R with ABC = O, we have αbac O, and thus R is not left α-symmetric ii Let β : R R be an endomorphism defined by β = 0 c 0 c By the similar method to i, we can show that R is left β-symmetric However, for A =, B = and C = R with ABC = O, we have ACβB O, and thus R is not right β-symmetric Proposition 23 1 For a ring R, R is right α-symmetric if and only if ABC = 0 implies ACαB = 0 for any three nonempty subsets A, B and C of R 2 Let R be a reversible ring R is right α-symmetric if and only if R is left α-symmetric Proof 1 It suffices to show that ABC = 0 for any three nonempty subsets A, B and C of R implies ACαB = 0, when R is right α-symmetric Let ABC = 0 Then abc = 0 for a A, b B and c C, and hence acαb = 0 by the condition Thus ACαB = a A,b B,c C acαb = 0 2 Let abc = 0 for a, b, c R If R is right α-symmetric, then acαb = 0 Since R is reversible, we have αbac = nd hence R is left α-symmetric The converse is similar Example 22 shows that the condition R is reversible in Proposition 232 cannot be dropped as well as there exists a right symmetric endomorphism α of a ring R such that R is not symmetric The next example provides that there exists a commutative reduced ring R which is not α-symmetric for some endomorphism α of R Example 24 Let Z 2 be the ring of integers modulo 2 and consider a ring R = Z 2 Z 2 with the usual addition and multiplication Then R is a commutative reduced ring, and so R is symmetric Now, let α : R R be defined by αa, b = b, a Then α is an automorphism of R For a = 1, 0, b = 0, 1, c = 1, 1 R, abc = 0 but acαb = 1, 0 0, and thus R is not right α-symmetric

5 780 TAI KEUN KWAK Recently, the reversible property of a ring is extended to a ring endomorphism in [4] as follows: An endomorphism α of a ring R is called right reversible if whenever ab = 0 for a, b R, bαa = 0 A ring R is called right α-reversible if there exists a right reversible endomorphism α of R The notion of an α- reversible ring is a generalization of α-rigid rings as well as an extension of reversible rings Theorem 25 Let R be a right α-symmetric ring Then we have the following 1 For a, b, c R, abc = 0 implies acα n b = 0, bcα n a = 0, and abα n c = 0 for any positive integer n, especially, R is a right α-reversible ring 2 Let α be a monomorphism of R Then we have the following i R is a symmetric ring ii For a, b, c R abc = 0 implies α n abc = nd aα n bc = 0 for any positive integer n Conversely, if α m abc = 0, aα m bc = 0, or abα m c = 0 for some positive integer m, then abc = 0 Proof 1 Let a, b, c R with abc = 0 Since R is right α-symmetric, acαb = 0 Then 0 = acαb = acαb 1 implies acα 2 b = 0 Continuing this process, we have acα n b = 0 for any positive integer n Similarly, 1 abc = 0 implies bcαa = 0 By the same method as above, we obtain bcα n a = 0 for any positive integer n Finally, 0 = abc = abc 1 implies abαc = 0, and thus abα n c = 0 for any positive integer n 2 Suppose that α is a monomorphism i: Let a, b, c R with abc = 0 Then acαb = 0, and so αbαac = 0 by 1 Since α is a monomorphism, bac = nd acb = 0 Thus R is symmetric ii: Note that R is symmetric and so reversible Let abc = 0 Then bcα n a = 0 by 1 Since R is reversible, α n abc = 0 Next, from abc = 0 we have acα n b = 0 by 1 Since R is symmetric, aα n bc = 0 Conversely, if α m abc = 0 for some positive integer m then α m aα m bc = α m abc = 0 by i, and thus abc = 0, since α is a monomorphism Similarly, if aα m bc = 0 then acα m b = 0, since R is symmetric Hence α m acα m b = 0 by i, and acb = nd so abc = 0 By the same method as above, we can obtain that abα m c = 0 implies abc = 0 Corollary 26 Every symmetric ring is reversible Notice that for any positive integer n, aα n b = 0 is equivalent to arα n b = 0, when R is a right α-symmetric ring with ab = 0 for a, b R: For, abr = 0 implies arαb = 0 for any r R This shows that arα n b = 0 for any positive integer n and any r R from Theorem 251, and thus arα n b = 0 We remark that the converse of Theorem 251 does not hold For example, the ring R with an endomorphism α in Example 221 is right α-symmetric However, for A = = B R, we have Aα n B = O = Bα n A for any positive integer n but AB O

6 EXTENSIONS OF EXTENDED SYMMETRIC RINGS 781 In the next example, part 1 shows that there exists a right α-symmetric ring R for an automorphism α, but R is not semiprime and so not α-rigid, and part 2 illuminates that there exists a commutative domain and an α- symmetric ring R, but R is not α-rigid where α is not a monomorphism Example 27 1 Consider a ring { R = Let α : R R be an endomorphism defined by α = } a, b Z a b Clearly, R is not semiprime and hence R is not α-rigid Note that α is an automorphism Moreover, R is right α-symmetric: Indeed, let ABC = O for a b A =, B = a b and C = R, then we get aa a = nd aa b + ab a + ba a = 0 If a = 0 then ba a = 0, if a = 0 then ab a = 0, and a = 0 then aa b = 0 These imply that aa a = 0 and aa b + ab a + ba a = 0 Thus ACαB = O, and hence R is right α-symmetric 2 Let R = F [x] be the polynomial ring over a field F Define α : R R by αfx = f0 where fx R Then R is a commutative domain and so reduced, but α is not a monomorphism Since R is a domain, R is right α-symmetric for any endomorphism α of R However, R is not α-rigid by [8, Example 52] The class of semiprime rings and the class of right α-symmetric rings do not depend on each other by Example 24 and Example 271 There exists a skew polynomial ring R[x; α] over a symmetric ring R which is not a symmetric ring For example, consider the commutative ring R = Z 2 Z 2 and the automorphism α of R defined by αa, b = b, a, as in Example 24 Then R is a symmetric ring, but R[x; α] is not reversible hence not symmetric: Indeed, for p = 1, 0, q = 0, 1x R[x; α], we get pq = 0 but 0 0, 1x = qp However, we have the following theorem Theorem 28 1 For a ring R, R is α-rigid if and only if R is semiprime and right α-symmetric and α is a monomorphism 2 If the skew polynomial ring R[x; α] of a ring R is a symmetric ring, then R is α-symmetric Proof 1 Let R be α-rigid Note that any α-rigid ring is reduced and α is a monomorphism by [7, p 218] We show that R is right α-symmetric Assume that abc = 0 for a, b, c R Then we obtain bac = 0, since R is reduced and so symmetric Thus acαbαacαb = acαbacα 2 b = 0 Since R is α-rigid, acαb = nd thus R is right α-symmetric The converse follows from [4, Proposition 253] and Theorem 251

7 782 TAI KEUN KWAK 2 Suppose that abc = 0 for a, b, c R Let p = a, q = b and h = cx in R[x; α] Then pqh = abcx = 0 R[x; α] Since R[x; α] is symmetric, we get 0 = phq = acxb = acαbx, and so acαb = 0 Thus R is right α-symmetric and therefore R is α-symmetric by Proposition 232 Corollary 29 [10, Proposition 271] A ring R is reduced if and only if R is a semiprime and symmetric ring Observe that the class of right α-symmetric rings and the class of α-armendariz rings do not depend on each other by Example 272 and [11, Example 14] Theorem 210 Let R be an α-armendariz ring The following statements are equivalent: 1 R[x; α] is symmetric 2 R is α-symmetric 3 R is right α-symmetric 4 R is symmetric Proof 1 4 by [9, Theorem 36 1] and 1 2 by Theorem is trivial Now we show 3 4 Suppose abc = 0 for a, b, c R Then acαb = 0, and so acb = 0 by [9, Proposition 13 2] Thus R is symmetric The next result is a direct consequence of Theorem 210 Corollary 211 [10, Proposition 34] Let R be an Armendariz ring R is symmetric if and only if R[x] is symmetric Notice that the converse of Theorem 282 does not hold and the condition R is an α-armendariz ring in Theorem 21re not superfluous by Example 272: Indeed, consider A = R[y; α] = F [x][y; α] Now, let p = 1, q = xy and h = x A Then pqh = 0, but phq = x 2 y 0 Hence A is not symmetric Note that R is not α-armendariz by [9, Example 19] 3 Extensions of α-symmetric rings Given a ring R and an R, R-bimodule M, the trivial extension of R by M is the ring T R, M = R M with the usual addition and the following multiplication: r 1, m 1 r 2, m 2 = r 1 r 2, r 1 m 2 + m 1 r 2 r m This is isomorphic to the ring of all matrices, where r R and 0 r m M and the usual matrix operations are used For an endomorphism α of a ring R and the trivial extension T R, R of R, ᾱ : T R, R T R, R defined by ᾱ = αa αb 0 αa

8 EXTENSIONS OF EXTENDED SYMMETRIC RINGS 783 is an endomorphism of T R, R Since T R, 0 is isomorphic to R, we can identify the restriction of ᾱ on T R, 0 to α Note that the trivial extension of a reduced ring is symmetric by [10, Corollary 24] For a right α-symmetric ring R, T R, R needs not to be an ᾱ- symmetric ring by the next example Example 31 Consider the right α-symmetric ring { } R = a, b Z in Example 271 where α is defined by a b α = For and A = C =, B = T R, R, ABC = O but ACᾱB O Thus R is not ᾱ-symmetric Recall that another generalization of a symmetric ring is a semicommutative ring A ring R is semicommutative if ab = 0 implies arb = 0 for a, b R Historically, some of the earliest results known to us about semicommutative rings although not so called at the time was due to Shin [16] He proved that any symmetric ring is semicommutative [16, Proposition 14] but the converse does not hold [16, Example 54a] Semicommutative rings were also studied under the name zero insertive by Habeb [6] Proposition 32 Let R be a reduced ring If R is an α-symmetric ring, then T R, R is an ᾱ-symmetric ring Proof Let ABC = O for a b A =, B = a b and C = T R, R Then we have 1 aa a = 0; and 2 aa b + ab a + ba a = 0 In the following, we freely use the fact that R is a reduced ring if and only if for any a, b R, ab 2 = 0 or, a 2 b = 0 implies ab = 0; and every reduced ring

9 784 TAI KEUN KWAK is semicommutative From Eq1, we get aa a = nd ara Ra = 0 From Eq2 a, we have aa b a + ab a 2 + ba a 2 = nd so ab a + ba a = 0 Then 0 = aab a + ba a = a 2 b a, and hence ab a = 0 So Eq2 becomes 3 aa b + ba a = 0 If we multiply Eq3 on the left side by a, then 0 = a 2 a b = aa b and so ba a = 0 Then aa a = 0, ab a = 0, aa b = nd ba a = 0, and hence we obtain aa αa = 0, aa αb = 0, ab αa = nd ba αa = 0, since R is α-symmetric Thus ACᾱB = O and therefore T R, R is ᾱ-symmetric Corollary 33 [10, Corollary 24] Let R be a reduced ring, then T R, R is a symmetric ring The trivial extension T R, R of a ring R is extended to a ring a a 12 a 13 a 1n a 23 a 2n T n = a a 3n a, a ij R 0 a for any n 3 and an endomorphism α of a ring R is also extended to the endomorphism ᾱ : T n T n defined by ᾱa ij = αa ij The following example shows that T n cannot be ᾱ-symmetric for any n 3, even if R is an α-rigid ring Example 34 Let α be an endomorphism of an α-rigid ring R Note that if R is an α-rigid ring, then αe = e for e 2 = e R by [7, Proposition 5] In particular α1 = 1 First, we show that T 3 is not ᾱ-symmetric For A = 1 0 1, B = 0 1 0, C = T 3, ABC = O But we have ACᾱB = CB O T 3 In case of n 4, we can also prove that T n is not ᾱ-symmetric by the same method as the above Recall that if α is an endomorphism of a ring R, then the map ᾱ : R[x] R[x] defined by ᾱ m i= ix i = m i=0 αa ix i is an endomorphism of the polynomial ring R[x] and clearly this map extends α The Laurent polynomial ring R[x, x 1 ] with an indeterminate x, consists of all formal sums n i=k a ix i, where a i R and k, n are possibly negative integers The map ᾱ : R[x, x 1 ] R[x, x 1 ] defined by ᾱ n i=k a ix i = n i=k αa ix i extends α and also is an endomorphism of R[x, x 1 ] Multiplication is subject to xr = αrx and rx 1 = x 1 αr The following results extend the class of right α-symmetric rings Theorem 35 Let R be a ring 1 R[x] is right ᾱ-symmetric if and only if R[x; x 1 ] is right ᾱ-symmetric

10 EXTENSIONS OF EXTENDED SYMMETRIC RINGS If R is an Armendariz ring, then R is right α-symmetric if and only if R[x] is right ᾱ-symmetric Proof 1 It is sufficient to show necessity Let fx, gx and hx R[x; x 1 ] with fxgxhx = 0 Then there exists a positive integer n such that f 1 x = fxx n, g 1 x = gxx n and h 1 x = hxx n R[x], and so f 1 xg 1 xh 1 x = 0 Since R[x] is right ᾱ-symmetric, we obtain f 1 xh 1 xᾱg 1 x = 0 Hence fxhxᾱgx = x 3n f 1 xh 1 xᾱg 1 x = 0 Thus R[x; x 1 ] is right ᾱ- symmetric 2 It also suffices to establish necessity Let fx = m i= ix i, gx = n j=0 b jx j and hx = l k=0 c ix i R[x] with fxgxhx = 0 By [1, Proposition 1], a i b j c k = 0 for all i, j and k, and so a i c k αb j = 0 since R is Armendariz and right α-symmetric This yields fxhxᾱgx = 0, and thus R[x] is right ᾱ-symmetric Corollary 36 1 [10, Lemma 322] For a ring R, R[x] is symmetric if and only if so is R[x; x 1 ] 2 [10, Proposition 34] Let R be an Armendariz ring R is symmetric if and only if R[x] is symmetric Note that Example 22i and Example 24 show that Armendariz rings and right α-symmetric rings do not depend on each other For an ideal I of R, if αi I then ᾱ : R/I R/I defined by ᾱa + I = αa + I is an endomorphism of a factor ring R/I The homomorphic image of a symmetric ring may not necessarily be symmetric by [10, p163] One may conjecture that R is α-symmetric if for any right α-symmetric nonzero proper ideal I of R, R/I is ᾱ-symmetric, where I is considered as a ring without identity However the next example erases the possibility F F Example 37 For a field F, consider a ring R = and an endomorphism α of R defined by 0 F a b α = 0 c 0 c 0 F For a right ideal I = of R, it can be easily checked that I is right α-symmetric and the factor ring { } a 0 R/I = + I a, c F 0 c is reduced Observe that R/I is ᾱ-symmetric, where ᾱ is an identity map on R/I However, for A =, B = and C = R, we have ACαB O and ABC = O Thus R is not right α-symmetric

11 786 TAI KEUN KWAK Theorem 38 Let R be a reduced ring and n be any positive integer If R is right α-symmetric with α1 = 1, then R[x]/ x n is a right ᾱ-symmetric ring, where x n is the ideal generated by x n Proof Let S = R[x]/ x n If n = 1, then S = R If n = 2, then S is ᾱ- symmetric by Proposition 32, since S = T R, R Now, we assume n 3 Let f = a 0 + a 1 x + + a n 1 x n 1, g = b 0 + b 1 x + + b n 1 x n 1 and h = c 0 + c 1 x + + c n 1 x n 1 S with fgh = 0, where x = x + x n Note that a i b j c k x i+j+k = 0 for all i, j and k with i + j + k n Hence it suffices to show the cases i + j + k n 1 Since fgh = 0, we have the following equations: 1 a 0 b 0 c 0 = 0 2 a 0 b 0 c 1 + a 0 b 1 c 0 + a 1 b 0 c 0 = 0 3 a 0 b 0 c 2 + a 0 b 1 c 1 + a 0 b 2 c 0 + a 1 b 0 c 1 + a 1 b 1 c 0 + a 2 b 0 c 0 = 0 n-2 a 0 b 0 c n 2 + a 0 b 1 c n a n 3 b 1 c 0 + a n 2 b 0 c 0 = 0 n-1 a 0 b 0 c n 1 + a 0 b 1 c n a n 2 b 0 c 1 + a n 2 b 1 c 0 + a n 1 b 0 c 0 = 0 Recall that R is a reduced ring if and only if for any a, b R, ab 2 = 0 implies ab = 0, and every reduced ring is semicommutative We use these facts in the following Eq1 and Eq2 b 0 c 0 give a 1 b 0 c 0 2 = 0, and so a 1 b 0 c 0 = nd a 0 b 0 c 1 + a 0 b 1 c 0 = 0; multiplying b 1 c 0 gives 0 = a 0 b 1 c 0 2 = a 0 b 1 c 0, so we have 2 a 0 b 0 c 1 = 0, a 0 b 1 c 0 = nd a 1 b 0 c 0 = 0 From Eqs1, 2 and 3 b 0 c 0, we get a 2 b 0 c 0 = nd 3 a 0 b 0 c 2 + a 0 b 1 c 1 + a 0 b 2 c 0 + a 1 b 0 c 1 + a 1 b 1 c 0 = 0, in a similar way If we multiply Eq3 on the right side by b 1 c 0, b 0 c 1, b 2 c nd b 1 c 1 respectively, then we obtain a 1 b 1 c 0 = 0, a 1 b 0 c 1 = 0, a 0 b 2 c 0 = 0, a 0 b 1 c 1 = 0, and a 0 b 0 c 2 = 0 in turn Inductively we assume that a i b j c k = 0 for i + j + k = 0, 1,, n 2 We apply the above method to Eqn-1 First, the induction hypotheses and Eqn-1 b 0 c 0 give a n 1 b 0 c 0 = nd n-1 a 0 b 0 c n 1 + a 0 b 1 c n a n 2 b 0 c 1 + a n 2 b 1 c 0 = 0 If we multiply Eqn-1 on the right side by b 1 c 0, b 0 c 1,, and b 1 c n 2 respectively, then we obtain a n 2 b 1 c 0 = 0, a n 2 b 0 c 1 = 0,, a 0 b 1 c n 2 = nd so a 0 b 0 c n 1 = 0, in turn This shows that a i b j c k = 0 for all i, j and k with i + j + k = n 1 Consequently, a i b j c k = 0 for all i, j and k with i + j n 1, and thus a i c k α t b j = 0 for any positive integer t by Theorem 251 This yields fhᾱg = 0, and therefore S is right ᾱ-symmetric Corollary 39 [10, Theorem 23] If R is a reduced ring, then R[x]/ x n is a symmetric ring for any positive integer n Let R be an algebra over a commutative ring S Recall that the Dorroh extension of R by S is the ring D = R S with operations r 1, s 1 + r 2, s 2 = r 1 + r 2, s 1 + s 2 and r 1, s 1 r 2, s 2 = r 1 r 2 + s 1 r 2 + s 2 r 1, s 1 s 2, where r i R

12 EXTENSIONS OF EXTENDED SYMMETRIC RINGS 787 and s i S For an endomorphism α of R and the Dorroh extension D of R by S, ᾱ : D D defined by ᾱr, s = αr, s is an S-algebra homomorphism In the following, we give some other example of right α-symmetric rings Proposition If e is a central idempotent of a ring R with αe = e and α1 e = 1 e, then er and 1 er are right α-symmetric if and only if R is right α-symmetric 2 If R is a right α-symmetric ring with α1 = 1 and S is a domain, then the Dorroh extension D of R by S is ᾱ-symmetric Proof 1 It is enough to show the necessity Suppose that er and 1 er are right α-symmetric Let abc = 0 for a, b, c R Then 0 = eabc = aebc and 0 = 1 eabc = a1 ebc By hypothesis, we get 0 = acαeb = aceαa = eacαb and 0 = acα1 eb = ac1 eαb = 1 eacαb Thus acαb = eacαb + 1 eacαb = 0, and therefore R is right α-symmetric 2 Let r 1, s 1, r 2, s 2, r 3, s 3 D with r 1, s 1 r 2, s 2 r 3, s 3 = 0 Then r 1 r 2 r 3 + s 1 r 2 r 3 + s 2 r 1 r 3 + s 3 r 1 r 2 + s 1 s 2 r 3 + s 1 s 3 r 2 + s 2 s 3 r 1 = nd s 1 s 2 s 3 = 0 Since S is a domain, we get s 1 = 0, s 2 = 0 or s 3 = 0 In the following computations, we freely use the assumption that R is right α-symmetric with α1 = 1 If s 1 = 0, then 0 = r 1 r 2 r 3 + s 2 r 1 r 3 + s 3 r 1 r 2 + s 2 s 3 r 1 and so 0 = r 1 r 3 +s 3 αr 2 +s 2 = r 1 r 3 αr 2 +r 1 s 3 αr 2 +r 1 r 3 s 2 +r 1 s 3 s 2 This yields r 1, s 1 r 3, s 3 ᾱr 2, s 2 = 0 Similarly, let s 2 = 0 Then r 1 + s 1 r 2 r 3 + s 3 = 0, and so r 1 + s 1 r 3 + s 3 αr 2 = 0, and hence r 1 r 3 αr 2 + s 1 s 2 αr 2 + s 3 r 1 αr 2 +s 1 s 3 αr 2 = 0 Thus we have r 1, s 1 r 3, s 3 ᾱr 2, s 2 = 0 Finally, let s 3 = 0 Then r 1 + s 1 r 2 + s 2 r 3 = 0, and so 0 = r 1 + s 1 r 3 αr 2 + s 2 = r 1 r 3 + s 1 s 3 αr 2 + s 2 r 1 r 3 + s 1 r 3 This imply r 1, s 1 r 3, s 3 ᾱr 2, s 2 = 0 Therefore the Dorroh extension D is ᾱ-symmetric Corollary [10, Proposition 362] For an abelian ring R, R is symmetric if and only if er and 1 er are symmetric for every idempotent e of R if and only if er and 1 er are symmetric for some idempotent e of R 2 [10, Proposition 421] Let R be an algebra over a commutative ring S, and D be the Dorroh extension of R by S If R is symmetric and S is a domain, then D is symmetric Note that the condition α1 = 1 in Proposition 3102 cannot be dropped by the next example Example 312 Let R = Z 2 Z 2 and α : R R be defined by αa, b = 0, b Consider the Dorroh extension D of R by the ring of integers Z Then we have 1, 0, 01, 0, 11, 0, 0 = 0 in D, but in D 1, 0, 01, 0, 0ᾱ1, 0, 1 = 1, 0, 0 0

13 788 TAI KEUN KWAK Acknowledgments The authors would like to thank the referee for helpful comments and suggestions References [1] D D Anderson and V Camillo, Armendariz rings and Gaussian rings, Comm Algebra , no 7, [2], Semigroups and rings whose zero products commute, Comm Algebra , no 6, [3] E P Armendariz, A note on extensions of Baer and PP-rings, J Austral Math Soc , [4] M Başer, C Y Hong, and T K Kwak, On extended reversible rings, Algebra Colloq to appear [5] P M Cohn, Reversible rings, Bull London Math Soc , no 6, [6] J M Habeb, A note on zero commutative and duo rings, Math J Okayama Univ , [7] C Y Hong, N K Kim, and T K Kwak, Ore extensions of Baer and pp-rings, J Pure Appl Algebra , no 3, [8], On skew Armendariz rings, Comm Algebra , no 1, [9] C Y Hong, T K Kwak, and S T Rizvi, Extensions of generalized Armendariz rings, Algebra Colloq , no 2, [10] C Huh, H K Kim, N K Kim, and Y Lee, Basic examples and extensions of symmetric rings, J Pure Appl Algebra , no 1-3, [11] C Huh, Y Lee, and A Smoktunowicz, Armendariz rings and semicommutative rings, Comm Algebra , no 2, [12] J Krempa, Some examples of reduced rings, Algebra Colloq , no 4, [13] J Lambek, On the representation of modules by sheaves of factor modules, Canad Math Bull , [14] G Marks, Reversible and symmetric rings, J Pure Appl Algebra , no 3, [15] M B Rege and S Chhawchharia, Armendariz rings, Proc Japan Acad Ser A Math Sci , no 1, [16] G Y Shin, Prime ideals and sheaf representation of a pseudo symmetric ring, Trans Amer Math Soc , Department of Mathematics Daejin University Pocheon , Korea address: tkkwak@daejinackr