Generalized Periodic and Generalized Boolean Rings

We prove that a generalized periodic, as well as a generalized Boolean, ring is either commutative or periodic. We also prove that a generalized Boolean ring with central idempotents must be nil or commutative. We further consider conditions which imply the commutativity of a generalized periodic, or a generalized Boolean, ring. 2000 Mathematics Subject Classification. 16D70, 16U80. Throughout, R denotes a ring, N the set of nilpotents, C the center, and E the set of idempotents of R. A ring R is called periodic if for every x in R, there exist distinct positive integers m,n such that xm = xn. We now formally state the definitions of a generalized periodic ring and a generalized Boolean ring. Definition 1. A ring R is called generalized periodic if for every x in R such that x ∉ (N∪C), we have xn−xm ∈ (N∩C), for some positive integers m,n of opposite parity. Definition 2. A ring R is called generalized Boolean if for every x in R such that x ∉ (N∪C), there exists an even positive integer n such that x−xn ∈ (N∩C). Theorem 3. If R is a generalized periodic ring, then R is either commutative or periodic. Proof. Let N and C denote the set of nilpotents and the center of R, respectively. We distinguish three cases. Case 1 (N ⊆ C). Then x ∉ C implies x ∉ (N ∪C), and hence there exist distinct positive integers m,n such that xm−xn ∈ N , with n > m. Suppose (xm−xn)k = 0. Then, as is readily verified, ( x−xn−m+1)kxk(m−1) = 0, (1) which, in turn, implies that ( x−xn−m+1)km = (x−xn−m+1)kxk(m−1)g(x) = 0, (2) where g(λ)∈ Z[λ]. (3) We have thus shown that x−xn−m+1 ∈N, ∀x ∉ C, (n−m+1> 1). (4) 458 H. E. BELL AND A. YAQUB Recall that, in our present case, we assumed that N ⊆ C , and hence by (4), x−xn−m+1 ∈ C, ∀x ∉ C, (n−m+1> 1). (5) Since (5) is trivially satisfied if x ∈ C , we see that x−xn(x) ∈ C, for some n(x) > 1, where x ∈ R (arbitrary). (6) Therefore, R is commutative, by a well-known theorem of Herstein [3]. Case 2 (C ⊆ N). Then x ∉ N implies x ∉ (N ∪C), and hence there exist distinct positive integers m,n such that xn−xm ∈ N , with n > m. Repeating the argument used to prove (4), we see that x−xn−m+1 ∈N, ∀x ∉N, (n−m+1> 1). (7) Since (7) is trivially satisfied for all x ∈N , we conclude that x−xk(x) ∈N, for some k(x) > 1, where x ∈ R (arbitrary). (8) By a well-known theorem of Chacron [2], equation (8) implies that R is periodic. Case 3 (C ⊆N and N ⊆ C). In this case, let z ∈ C\N, u∈N\C. (9) Equation (9) readily implies that z+u ∉ C and z+u ∉N , and hence (see Definition 1) (z+u)n−(z+u)m ∈N, for some integers n>m≥ 1. (10) Since z commutes with the nilpotent element u, (10) implies that zn−zm+u′ ∈N, where u′ ∈N, u′commutes with z. (11) Hence zn −zm ∈ N , with n > m ≥ 1. Now, a repetition of the argument used in the proof of (4) shows that z−zn−m+1 ∈N, ∀z ∈ C\N, (n−m+1> 1). (12) Trivially, x−xk ∈N, ∀x ∈N, ∀k∈ Z+. (13) Finally, if x ∉ (N∪C), then xn−xm ∈N, for some integers n>m≥ 1. (14) Again, repeating the argument used in the proof of (4), we see that x−xn−m+1 ∈N, ∀x ∉ (N∪C), (n−m+1> 1). (15) Combining (12), (13), and (15), we conclude that x−xk(x) ∈N, for some k(x) > 1, where x ∈ R (arbitrary). (16) Thus, by Chacron’s theorem [2], R is periodic. This completes the proof. GENERALIZED PERIODIC AND GENERALIZED BOOLEAN RINGS 459 Corollary 4. If R is a generalized Boolean ring, then R is either commutative or periodic. This follows at once, since a generalized Boolean ring is necessarily a generalized periodic ring (see Definitions 1 and 2). Before proving the next theorem, we prove the following lemma. Lemma 5. Let R be a generalized periodic ring. If e is any nonzero central idempotent in R and a∈N , then ea∈ C . Proof. The proof is by contradiction. Suppose the lemma is false, and let η0 ∈N, eη0 ∉ C. (17) Since e∈ C and η0 ∈N , therefore eη0 is nilpotent. Let ( eη0 )α ∈ C, ∀α≥α0, α0 minimal. (18) Since eη0 ∉ C (see (17)), therefore α0 > 1. Let η= (eη0)α0−1. Then, η= (eη0)α0−1 ∈N, η ∉ C (by the minimality of α0), ηk ∈ C, ∀k≥ 2, e∈ C, e = e = 0, e ∉N. (19) Equation (19) implies that e+η ∉ C and e+η ∉N , and hence (see Definition 1) ( e+η)m′ −(e+η)n′ ∈ C, (20) wherem′, n′ are of opposite parity. Combining (20) and (19), we see that (keep in mind that eη= η; see (19)) ( m′ −n′)eη∈ C, (21) where m′ −n′ is an odd integer. Equation (19) also implies that (−e+η) is not in (N∪C), so (−e+η)m′′ −(−e+η)n′′ ∈N, (22) where m′′, n′′ are of opposite parity. Combining (19) and (22), we see that (−e)m −(−e)n ∈N, (23) and hence 2e∈N , sincem′′ andn′′ are of opposite parity. Therefore, (2e)γ = 0, γ ∈ Z+, and thus 2γe= 0, which implies that 2γeη∈ C ; γ ∈ Z+. (24) Now, combining (21) and (24), keeping in mind that (2γ,m′ −n′) = 1, we see that eη ∈ C , and hence, by (19), η = eη ∈ C , which contradicts (19). This contradiction proves the lemma. As usual, [x,y]= xy−yx denotes the commutator of x and y . We are now in a position to prove the following theorem. 460 H. E. BELL AND A. YAQUB Theorem 6. Suppose R is a generalized periodic ring, and suppose that there exists an element c in C , with c = 0, such that c [ x,y ]= 0 implies [x,y]= 0, ∀x,y ∈ R. (25) Then R is commutative. Proof. We distinguish two cases. Case 1 (c ∈N). In this case, ck = 0 for some positive integer k, and hence ck [ x,y ]= 0, ∀x,y ∈ R. (26) Combining (25) and (26), we see that ck [ x,y ]= 0 ⇒ c[ck−1x,y]= 0 ⇒ [ck−1x,y]= 0 ⇒ ck−1[x,y]= 0 ⇒ ··· ⇒ c[x,y]= 0 ⇒ [x,y]= 0. (27) Thus, ck[x,y]= 0 implies [x,y]= 0, and hence R is commutative. Case 2 (c ∉ N). In view of Theorem 3, we may assume that R is periodic. This implies, in particular, that cm is idempotent for some positive integerm. Furthermore, cm = 0 (since c ∉N in our present case). The net result is (since c ∈ C also) cm = e is a nonzero central idempotent in R. (28) Let a ∈ N . By Lemma 5 and equation (28), we have ea ∈ C , and hence [ea,x] = 0 for all x ∈ R, which implies [ cma,x ]= cm[a,x]= 0, ∀x ∈ R. (29) The argument used in Case 1 of Theorem 6 shows that cm[a,x]= 0 implies [a,x]= 0, (30) and hence (see (29)) [a,x]= 0 ∀x ∈ R, ∀a∈N. (31) Thus, R is a periodic ring with the property that N ⊆ C . By a well-known theorem of Herstein [4], it follows that R is commutative, and the theorem is proved. Corollary 7. Suppose that R is a generalized periodic ring with identity 1. Then, R is commutative. Corollary 7 follows at once by taking c = 1 in Theorem 6. Since a generalized Boolean ring is also a generalized periodic ring, therefore we have the following corollary. Corollary 8. Ageneralized Boolean ringwith identity 1 is necessarily commutative. Another corollary is the following result, proved by the authors in [1]. GENERALIZED PERIODIC AND GENERALIZED BOOLEAN RINGS 461 Corollary 9. Suppose that R is a generalized periodic ring containing a central element which is not a zero divisor. Then R is commutative. This follows at once, since the hypotheses of this corollary imply the hypotheses of Theorem 6. Theorem 10. Suppose R is a generalized periodic ring. Suppose, further, that there exists a nonzero central element c such that ca= 0 implies a= 0, ∀a∈N. (32) Then R is commutative. Proof. In [1], the authors proved the following result: IfR is a generalized periodic ring, then the nilpotents N form an ideal and R/N is commutative. (33) Let x,y ∈ R. By (33), for all x̄, ȳ in R/N , x̄ȳ = ȳx̄, and hence [x,y] ∈ N . Taking a= [x,y]∈N in (32), we see that (32) yields c [ x,y ]= 0 implies [x,y]= 0, ∀x,y ∈ R. (34) The theorem now follows at once from Theorem 6. Theorem 11. A generalized Boolean ring R with central idempotents is necessarily nil (R =N) or commutative (R = C). Proof. Since R is also a generalized periodic ring, therefore by Theorem 3, R is commutative or periodic. If R is commutative, there is nothing to prove. So we may assume that R is periodic. We now distinguish two cases. Case 1 (C ⊆N). Recall that, by hypothesis, the set E of idempotents is central, and hence E ⊆ C ⊆N (in the present case). Thus, E ⊆N , and hence E = {0}. Therefore, zero is the only idempotent of R. (35) Let x ∈ R. Since R is periodic, therefore xk is idempotent for some positive integer k, and hence by (35), xk = 0, which proves that R is nil. Case 2 (C ⊆N). Then, for some c ∈ R, we have c ∈ C, c ∈N. (36) Again, since R is periodic, cm is idempotent for some positive integer m. Moreover, cm = 0 (since c ∈N). The net result is (see (36)) e= cm is a nonzero central idempotent of R. (37) Now, suppose a∈N . Since 0 = e∈ C and a∈N , therefore e+a ∈N . Suppose, for the moment, that a ∈ C . Then e+a ∈ C (since e∈ C), and hence e+a ∈ (N∪C). Therefore, by Definition 2, (e+a)−(e+a)n ∈ (N∩C), for some even integer n≥ 2. (38) 462 H. E. BELL AND A. YAQUB Since R is also a generalized periodic ring, therefore by Lemma 5 (see (37)) eai ∈ C, ∀i∈ {1, . . . ,n−1}, (0 = e= e, e∈ C, a∈N). (39) Combining (38) and (39), we see that a−an ∈ C, ∀a∈N\C. (40) Since (40) is trivially satisfied for a∈ (N∩C), therefore a−an ∈ C, ∀a∈N, n≥ 2. (41) We claim that N ⊆ C. (42) The proof is by contradiction. Suppose (42) is false. Then, for some a∈ R, we have a∈N, a ∈ C. (43) Since a∈N , there exists a positive integer σ0 such that aσ ∈ C, ∀σ ≥ σ0, σ0 minimal. (44) Moreover, since a ∉ C (see (43)), therefore σ0 > 1. Now, applying (41) to the nilpotent element aσ0−1, we see that aσ0−1−(aσ0−1)n ∈ C, for some n=n(aσ0−1)≥ 2. (45) Furthermore, since (σ0−1)n≥ (σ0−1)2≥ σ0 (since σ0 ≥ 2), (44) implies that ( aσ0−1 )n = a(σ0−1)n ∈ C. (46) Combining (45) and (46), we conclude thataσ0−1 ∈ C , which contradicts theminimality of σ0 in (44). This contradiction proves (42). Since R is a periodic ring satisfying (42), therefore, by a well-known theorem of Herstein [4], R is commutative. This completes the proof. Corollary 12. A generalized Boolean ring with central idempotents and commuting nilpotents is commutative. This corollary recovers a result proved by the authors in [1]. Corollary 13. If R is a generalized Boolean ring, and if R is 2-torsion-free, then R is nil or commutative. Proof. We claim that all idempotents of R are central. Suppose not, and suppose e is a noncentral idempotent in R. Then −e ∉ (N∪C), and hence (see Definition 2) (−e)−(−e)n ∈ C, n even. (47) Thus, 2e∈ C , and hence [2e,x]= 0 for all x in R. Since R is 2-torsion-free, 2[e,x]= 0 implies [e,x] = 0, and thus e ∈ C , a contradiction. This contradiction proves that all idempotents of R are central, and hence R is nil or commutative, by Theorem 11. GENERALIZED PERIODIC AND GENERALIZED BOOLEAN RINGS 463 Theorem 14. Let R be a generalized Boolean ring in which every finite subring is either commutative or nil. Then R is either commutative or nil. Proof. By contradiction. Thus, suppose R is a generalized Boolean ring such that every finite subring of R is either commutative or nil. Suppose, further, that R is not commutative and not nil either. By Theorem 11, there must exist a noncentral idempotent element e in R, and hence e ∉ (C∪N). Thus (see Definition 2), since −e ∉ (C∪N), (−e)−(−e)n ∈ (N∩C), n even. (48) This implies that 2e ∈ (N ∩C), and hence (2e)k = 2ke = 0, for some k ∈ Z+. Since e ∈ C , we must have the following: Either ex−exe = 0 for some x ∈ R, or x′e−ex′e = 0 for some x′ ∈ R. (49) Suppose u= ex−exe = 0. Then, eu=u = 0=ue=u2, (u= ex−exe = 0). (50) Moreover, 2u= [2e,ex]= 0 (since 2e∈ C). (51) Furthermore, the subring generated by e and u is 〈e,u〉 = {re+su | r , s ∈ Z}. (52) Since 2ke= 0 and 2u= 0, the subring 〈e,u〉 is finite. Indeed, 〈e,u〉 = {re+su | 1≤ r ≤ 2k, 1≤ s ≤ 2}. (53) On the other hand, if x′e−ex′e = 0 for some x′ ∈ R (the only other possibility), then the subring, 〈e,v〉, generated by e and v = x′e−ex′e is (as is readily verified) 〈e,v〉 = {re+sv | 1≤ r ≤ 2k, 1≤ s ≤ 2}. (54) Again, 〈e,v〉 is a finite subring of R. Hence, in either case, we found a finite subring of R, which is neither commutative (since e ∉ C), nor nil (since e ∉N), contradicting our hypothesis. This contradiction proves the theorem. Remark 15. A careful examination of the proof of Theorem 14 shows that we only need to assume that “every subring S, with |S| = 2m for some positive integer m, is commutative or nil” in order for the ground generalized Boolean ring R to be commutative or nil. Indeed, |〈e,u〉| = 2k ·2 = 2k+1, since the representation of any x in this subring in the form x = re+ su; r ,s ∈ Z, is unique. For, suppose x = re+ su and x = r ′e+ s′u. Then, (r −r ′)e = (s′ − s)u. Recall that 2u = 0, and ue = 0. Thus, if s′ −s is even, then (r −r ′)e = 0, and hence re = r ′e, su = s′u. On the other hand, if s′ − s is odd, then (r − r ′)e = u, and hence (r − r ′)ee = ue = 0. Again, we obtain re= r ′e, su= s′u. We conclude with the following examples. 464 H. E. BELL AND A. YAQUB