International Journal of Algebra, Vol. 7, 2013, no. 9, 429-438 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ija.2013.2894 Generalized Boolean and Boolean-Like Rings Hazar Abu Khuzam Department of Mathematics American University of Beirut Beirut, Lebanon hazar@aub.edu.lb Adil Yaqub Department of Mathematics University of California Santa Barbara, CA. 93106, USA yaqub@math.ucsb.edu Copyright 2013 Hazar Abu Khuzam and Adil Yaqub. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract A generalized Boolean ring is a ring R such that, for all x, y in R\(N C), x n y xy n (N C), where n is a fixed even integer, and N, C are the set of nilpotents and center of R, respectively. A Boolean-like ring is a ring R such that, for all x,y R\(N C), x n y xy n, where again n is a fixed even integer. The commutativity behavior of these rings is considered. In particular, it is proved that a Boolean-like ring with identity is commutative.
430 Hazar Abu Khuzam and Adil Yaqub Mathematics Subject Classification: 16U80, 16D70 Keywords: Generalized Boolean ring, Boolean-like ring, Jacobson radical, commutator ideal, weakly periodic-like ring Throughout R is a ring, N is the set of nilpotents, J is the Jacobson radical, and C is the center of R. As usual, [x,y] denotes the commutator xy yx. Definition 1. Let n be a fixed positive even integer. A ring R is called a generalized Boolean ring if (1) x n y xy n (N C) for all x, y in R\(N C), (n even). R is called a Boolean - like ring if (2) x n y xy n = 0 for all x, y R\(N C), (n even). In preparation for the proofs of the main theorems, we state the following lemmas. Lemma 1. ([1]) Suppose R is a ring in which each element x is central or potent in the sense that x k = x for some k > 1. Then R is commutative. Lemma 2. Suppose R is a ring and x R. Suppose that (3) x m x m+1 f (x) N, m a positive integer and f ( λ ) Z [ λ ]. Then x x 2 f(x) N. Proof. Note that (x x 2 f(x)) m = (x x 2 f(x) ) x m-1 g(x), for some g ( λ ) Z [ λ ]. = ( x m x m+1 f (x) ) g(x) N, and hence x x 2 f(x) N, proving the lemma. Lemma 3. Suppose R is a generalized Boolean ring. Then, (4) For all x R, either x C or 2x N.
Generalized Boolean and Boolean-like rings 431 Proof. By contradiction. Suppose that x R, x C and 2x N. Then x C and x N, and hence by (1), x n ( x) x( x) n N, n even. Hence 2x n+1 N, which implies that 2x N, contradiction, which proves the lemma. Lemma 4. Suppose R is a generalized Boolean ring, and suppose all idempotents are central. Then, N J, (J is the Jacobson radical of R). Proof. Let a N, x R. If ax N, then ax is right quasi regular (r.q.r.). Also, since ax C implies ax N, it follows that ax C implies ax is r.q.r. A similar argument shows that (ax) 2 N implies ax is r.q.r. Moreover, (ax) 2 C implies ((ax) 2 ) k = (ax) 2 (ax) 2 (ax) 2 = a k y for some y in R, k any positive integer. Since a N, a k = 0 for some k, and thus ((ax) 2 ) k = 0, which implies ax N. So ax is r.q.r. again. The only case left to consider is: ax (N C) and (ax) 2 (N C). By (1), (ax) n (ax) 2 (ax) ((ax) 2 ) n N and thus (ax) n+2 (ax) 2n+1 N. Since n+2 2n+1, we conclude that (ax) q = (ax) q+1 g(ax), g ( λ ) Z [ λ ], and hence (ax) q = (ax) q (ax g(ax)) =. = (ax) q (ax g(ax)) q. Therefore, (5) (ax) q = (ax) q e, e 2 =e ar, e = (ax g(ax)) q. Hence, for some r R, e = e.e = e(ar) = aer ( since e is central). By re-iterating, we see that e = aer = a 2 er 2 = = a k er k for all positive integer k. Since a N, a k = 0 for some k, and thus e = 0, which implies by (5) that (ax) q =0. Thus, ax is r.q.r. This shows that ax is r.q.r. for all a N, x R, and hence a J, which proves the lemma. We are in a position to prove the main theorems Theorem 1. Any subring and any homomorphic image of a generalized Boolean ring is also a generalized Boolean ring. This follows readily from Definition 1.
432 Hazar Abu Khuzam and Adil Yaqub Theorem 2. If R is a generalized Boolean ring with central nilpotents, then R is commutative. Proof. Suppose x R, x C. We now distinguish two cases. Case 1. x 2 C. Since N C, (1) implies that (6) x n y x y n N for all x, y R\C. Also, since x C and x 2 C, it follows by (6) that x n (x 2 ) x(x 2 ) n N and hence x n+2 x 2n+1 N, which implies by Lemma 2 that x x n N C. Thus, x x n C for all x in R, and hence by a well known theorem of Herstein [2], R is commutative. Case 2. x 2 C. In this case, x x 2 C (since x C), and hence by (6) x n (x x 2 ) x( x x 2 ) n N, which implies that x n+1 x n+2 x(x n nx n+1 +x n+2 g(x)) N, for some g ( λ ) Z [ λ ]. Therefore, (7) (n 1)x n+2 x n+3 g(x) N. Since x C, 2x N (by Lemma 3), and hence nx N ( since n is even), which implies that (8) nx n+2 N. Thus, (n-1)x n+2 -x n+3 g(x) and nx n+2 are two commuting nilpotents, and hence their difference is in N; that is, x n+2 x n+3 g(x) N. Therefore, by the proof of Lemma 2, x x 2 g(x) N, which implies that x x 2 f(x) N, where f ( λ ) = g ( λ ) Z [ λ ]. Thus x x 2 f(x) C ( since N C ) for all x R, which implies that R is commutative, by [2]. This proves the theorem. Theorem 3. Suppose R is a generalized Boolean ring with identity. Then all idempotents of R are central. Proof. Let e 2 =e R, x R, and let a = xe exe, suppose a 0. Since a N, 1+ a N. Moreover, 1+a C, since 1+ a C implies that a C, and hence ae = ea; that is, a = 0, contradiction. Thus (1+ a) (N C). Moreover, a 0, e (N C). So, by (1), (1+a) n e (1+a) e n C, which implies that nae ae C; which implies (n 1)a = 0 since ae=a and ea=0. Moreover, since a 0, e 0, and hence -e 0, which implies that -e (N C), and e (N C). Therefore, by (1),
Generalized Boolean and Boolean-like rings 433 (-e) n e-(-e)e n C, and hence, since n is even, 2e C. Thus, a(2e) = (2e)a, which implies that 2a = 0. The net result is: (9) (n 1) a = 0 and 2a = 0. Since n is even, (9) implies that a=0, contradiction. Hence, a = 0, and thus xe = exe. Similarly, ex = exe, and the Theorem is proved. Theorem 4. Suppose R is a generalized Boolean ring with identity. Then N is an ideal and R/N is commutative (and thus the commutator ideal of R is nil). Proof. By Theorem 3, all idempotents of R are central, and hence by Lemma 4, N J, where J is the Jacobson radical of R. We claim that (10) J N C. Suppose not. Let j J, j (N C). Then, 1+j (N C), and hence by (1), (1+j) n j (1+j) j n N, which implies that j j 2 f(j) N, for some f ( λ ) Z [ λ ]. Therefore, j(1 j f(j)) N, and, moreover, 1 j f(j) is a unit in R. Thus, 1 1, and hence j N, contradiction. This contradiction proves (10). Hence (11) N J N C. We claim that (12) N is an ideal of R. To prove this, suppose a N, x R. Then, by (11), a J, x R and hence ax J N C (by (11)). Hence ax N or ax C (which also implies that ax N), which proves that ax N. Similarly xa N. Next, suppose a N, b N, Then a J, b J (by (11)), and hence a b J N C (by (11)). So, a b N or a b C. If a b C, then a commutes with b, and hence a b N again, which proves (12). Let S = R / N. Then, by (1), (13) x n y = xy n for all noncentral elements of S. Moreover, S has an identity. Now suppose x is a noncentral element of S. Then, 1+x is also a noncentral element of S, and hence by (13), x n (1+x) = x(1+x) n. Therefore, for all noncentral elements x in S, x x 2 f(x) = 0 for some f ( λ ) Z [ λ ], which implies x x 2 f(x) is in the center of S, for all elements x of S. Hence, by Herstein s theorem [2], S is commutative; that is, R/N is commutative, which proves the theorem. Theorem 5. Suppose R is a generalized Boolean ring with identity, and suppose (14) N J is commutative.
434 Hazar Abu Khuzam and Adil Yaqub Then R is commutative. Proof. By Theorem 4, N is an ideal, and hence N J, which implies by (14), that (15) N is a commutative ideal of R. We claim (16) [a, x]x = 0 for all a N, x R. The proof is by contradiction. Suppose that for some a N, x R, [a, x] x 0, which implies x N (by (15)) and, of course, x C. So x (N C). Moreover, 1 1+ a C (since [a, x] x 0) and 1 + a N, which implies that 1+a (N C). Hence, by (1), (17) (1+a) n x (1+a) x n N C. Now, since N is an ideal (see (15)) and a N, (17) implies that (18) x x n N. Moreover, since N is a commutative ideal (see (15)), N 2 C. Also, (17) clearly implies that (19) [(1+a) n x (1 +a) x n, x] = 0 (a N, N 2 C). Thus, by (19), (20) [nax ax n, x] = 0. Let x x n = a o. By (18), a o N, and thus (21) x n = x a o, (a o N). Combining (20) and (21), we get [nax a(x a o ), x ] = 0, and hence [nax ax, x ] = 0 since aa o N 2 C; that is, [(n 1)ax, x ] = 0. Therefore, (22) (n 1) [ax, x ] = 0, (a N, x R). Recall that, by Lemma 3, x C or 2x N. If x C, then clearly [ax, 2x] = 0. Moreover, if 2x N, then [ax, 2x] = 0, by (15) which implies (23) 2 [ax, x] = [ax, 2x] = 0 (a N, x R). Combining (22) and (23), we see that (n 1) [ax, x] = 0 and 2 [ax, x] = 0, and n is even. So (24) [a, x] x = [ax, x] = 0, contradiction. This contradiction proves (16).
Generalized Boolean and Boolean-like rings 435 To complete the proof, since (16) is true for all x in R, we may replace x by x+1 in the above argument to obtain (see (24)) (25) [a, x+1] (x+1) = 0. In view of (24) and (25), we see that [a, x] = 0 for all a N, x R, which proves that (26) N C The theorem now follows from Theorem 2 and (26). In order to deal with the case where R does not have an identity in Theorem 5, we introduce the following Definition 2. A ring R is called weakly periodic-like if every element x in R\C is of the form (27) x = a + b, a N, b potent in the sense that b k = b for some k>1, x C. Theorem 6. Suppose R is a generalized Boolean ring, not necessarily with identity, and suppose, further, R is a weakly periodic- like ring and all the idempotents of R are central. Then N is an ideal and R/N is commutative (and thus the commutator ideal of R is nil). Proof. By Lemma 4, N C. We now prove that (28) J N C. Suppose not, let j J, j N, j C. Then, by (27), (29) j = a +b, a N, b k = b for some k>1, (j C). By (29), we see that (30) j a = ( j a) k, and hence = for all positive integers m. Since a N, = 0 for some m 1, and hence by (30), j a J; that is b J (see (29)). Hence, by (29), b k 1 is an idempotent element in J, and thus b k 1 = 0. So, b = b k = 0, and hence by (29), j = a N, contradiction. This proves (28). A combination of N J and (28) yields (31) N J NU C In the proof of Theorem 4, it was shown that in any ring R, (31) implies that (32) N is an ideal. Next we show that (33) for any x R, either x C or x - x k N for some k>1. To see this, note that by Definition 2, (34) x = a + b, a N, b k = b for some k>1, x R\C,
436 Hazar Abu Khuzam and Adil Yaqub and hence (35) x a = ( x a) k, k>1, (a N, x R\C). Combining (32) and (35), we obtain (33). Therefore, by (32) and (33), we see that every element of R/N is central or potent (satisfying x k = x, x R/N, k>1), and hence by a Theorem of Bell (Lemma 1), R/N is commutative, which proves the theorem. Theorem 7. Suppose R is a generalized Boolean ring and, moreover, R is weakly periodic-like. Suppose, further that all the idempotents of R are central and N J is commutative. Then R is commutative. Proof. By Lemma 4, N J, and hence N = N J is commutative. Thus, (36) N is commutative. Next we show that (37) All potent elements of R are central. Suppose that b is a potent nonzero element in R, and suppose b k = b, k>1. Let e = b k 1. Then, e is a nonzero idempotent of R and, moreover, e is central (by hypothesis). Hence, er is a ring with identity e. It is readily verified that er satisfies all the hypotheses imposed on R in Theorem 5. (To verify that (14) holds in er, keep in mind, that J(eR) = ej(r) J(R)). Thus, by Theorem 5, (38) er is commutative. Let y N. Then, by (38), e [b, y] = [eb, ey] = 0. Recalling that e 2 = e = b k-1 C and b k = b, we see that 0 = e [b, y] = b k-1 [b, y] = b k y b k 1 yb = b k y yb k = by yb, and hence by = yb for all y R, which proves (37). To complete the proof, suppose x R\C, y R\C. Then, by (27), (39) x = a + b, a N, b potent, and also y = a' + b', a' N, b' potent, and hence [x, y] = [a + b, a' + b']=[a, a'], by (37), and [a, a'] = 0, by (36). Hence [x, y] = 0, and thus R is commutative, which proves the Theorem. We now focus our attention on Boolean- like rings. Theorem 8. A Boolean-like ring with identity is commutative. Proof. First, we prove that
Generalized Boolean and Boolean-like rings 437 (40) for all a N and all units u R, [a, u] = 0. To prove this, suppose that (since a N) (41) [a σ, u] = 0 for all σ σ o, σ o minimal. We claim that (42) σ o = 1. Suppose not. Then, by the minimality of σ o in (41), 1 + a σo-1 C, and of course 1 + a σo-1 N (since a N). If u C, then (40) is trivially satisfied. So we may assume that u C, and of course u N. Hence, by (2), (1 + a σo-1 ) n u = ( 1 + a σo-1 ) u n, n even, and thus (since 1 + a σo-1 is a unit) ( 1 + a σo-1 ) n-1 = u n 1 which implies that (43) [( 1 + a σo-1 ) n-1, u] =0. By (43) and (41), we see that (44) (n 1) [a σo-1, u] =0, n even. Since u C, 2u N ( by Lemma 3), and hence (2u)u -1 N; that is, 2 N, which implies that R is of characteristic 2 k for some k 1, and thus (45) 2 k [a σo-1, u] =0. Since n is even, (44) and (45) imply that [a σo-1, u] =0, which contradicts the minimality of σ o in (41). This contradiction proves that σ o = 1, and hence by (41), [a, u] = 0, which proves (40). Next, we prove that (46) N is commutative. Let a N, b N. Then 1 + b is a unit in R, and hence by (40), [a, 1+b] =0, which implies [a, b] = 0, and (46) is proved. Hence by Theorem 5 and (46), R is commutative. We conclude with the following: Remark 1. Theorem 5 and 7 are not true if the hypothesis n is even is deleted in Definition 1, as can be seen by taking R = a b c 0 0 a 2 0 a 0 : a,b,c GF(4) ; n = 7.
438 Hazar Abu Khuzam and Adil Yaqub Indeed all the hypotheses of Theorem 5 and 7 hold except that n isn t even. But R is not commutative. Remark 2. Theorem 8 is not true if R does not have an identity, as can be seen by taking R = 0 0 0 1 1 0 1 1,,, : 0,1 GF(2), n = 2. 0 0 0 1 1 0 1 1 Related work appears in [3]. References [1] H.E. Bell, A near-commutativity property for rings, Result. Math. 42 (2002), 28-31. [2] I.N. Herstein, A gemeralization of a theorem of Jacobson III, Amer. J. Math. 75 (1953), 105-111. [3] A. Yaqub, A generalization of Boolean rings, Intern. J. Algebra, 1 (2007), 353-362. Received: August 5, 2012