Some Polynomial Identities that Imply Commutativity of Rings

Size: px

Start display at page:

Download "Some Polynomial Identities that Imply Commutativity of Rings"

Joanna Lewis
5 years ago
Views:

1 International Journal of Algebra, Vol. 4, 2010, no. 27, Some Polynomial Identities that Imply Commutativity of Rings M. S. Khan Department of Mathematics and Statistics College of Science, P.O. Box: 36 Postal Code 123, Al Khodh Muscat, Sultanate of Oman Abstract In this paper, we establish some commutativity theorems for certain rings with polynomial constraints as follows: Let R be an associative ring, and for all x, y R, and fixed non-negative integers m>1,n 0,r > 0,s 0,t 0,p 0,q 0 such that P (x, y) =±Q(x, y), where P (x, y) =y s [x, y]y t and Q(x, y) =x p [x m,y n ] r y q. First,it is shown that a semiprime ring R is commutative if and only if R satisfies the above conditions together with some constraints. Secondly, we investigate the commutativity of rings with identity if it satisfies some related polynomial identities, and then these results are extended to one sided s-unital rings.finally, we give some examples and open problems that appreciate our results. 1. Introduction Throughout this paper, R will denote an associative ring (may be without identity 1), Z(R) the centre of R, N(R) the set of nilpotent elements of R, C(R) the commutator ideal of R, and B a nonempty subset of R. As usual, for any x, y R, the symbol [x, y] will stand for the commutator xy yx. ByGF (q)) we mean the Galois field (finite field) with q elements,and by (GF (q)) 2 the ring of all 2 2 matrices over GF (q)). We denote e 11,e 12,e 21,e 22 in (GF (p)) 2, for a prime p. The symbol, Z[X], is the totality of polynomials in X with coefficients in Z, the ring of integers. A ring R is said to be left(resp. right)sunital if x Rx (resp.x xr) for all x in R. Further, R is called s-unital if x Rx xr for all x in R. If R is s-unital(resp. left or right s-unital), then for any finite subset F of R there exists an element e R such that ex = xe = x( resp.ex = x or xe = x) for all x F. Such an element e will be called a pseudo-identity (resp.pseudo-left identity or pseudo-right identity) of F in R (see [1]). Now,we define the following ring properties :

2 1308 M. S. Khan (P 1 ) P (x, y) =±Q(x, y), where P (x, y) =y s [x, y]y t and Q(x, y) =x p [x m,y n ] r y q, with t =0, for all x, y R, m>1,r > 0,n 0,s 0,p 0,q 0 are fixed integers. (P 2 ) P (x, y) =±Q(x, y), where P (x, y) =y s [x, y]y t and Q(x, y) =x p [x m,y n ] r y q, with s =0, for all x, y R, m>1,r > 0,n 0,s 0,p 0,q 0 are fixed integers. (P 3 ) P (x, y) =±Q(x, y), where P (x, y) =y s [x, y]y t and Q(x, y) =x p [x m,y n ] r y q, with s =0, & t = 0 for all x, y R, m>1,r > 0,n 0,p 0,q 0 are fixed integers. (P 4 ) P (x, y) =±Q(x, y), for all x, y R, where m = m(x, y) > 1,n = n(x, y) 0, r = r(x, y) > 0,s = s(x, y) 0,t = t(x, y) 0,p = p(x, y) 0,q = q(x, y) 0 are integers. (P 5 ) For each x R, there exits a polynomial f(x) Z[X] such that x x 2 f(x) B, where B R. A well- know theorem of Jacobson [8] assert that if R is a ring such that every element of R is equal to some power of itself ( i.e., R satisfies the relation x n(x) = x, x R, where n(x) > 1 is a positive integer) then R is commutative. Jacobson s proof involves the axiom of choice while Herstein [3] gave a proof of the theorem not involving axiom of choice. These results at the same time extend theorem of Wedderburn which states that every finite division ring is a field. A celebrated theorem of Herstein [5; Theorem 2] generalized Jacobson result as follow: Let R be a ring in which for every x, y R, there exits a positive integer m = m(x, y) > 1 such that [x, y] = [x m,y], then R is commutative. A well- know result of Herstein [4] asserts that a ring satisfying the polynimial identity (x + y) m = x m + y m for some m>1 must have nil commutator ideal. Recently, second author[13] has shown that a ring with unity 1 is commutative if, for every x, y R, R satisfies one of the polynomial identities x s [x, y]x t = ±y p [x m,y m ] r y q, where m>1,r >0 and n,s,p,q are fixed non-negative integers. The objective of this paper is to extend above results. In Section 2, we prove commutativity of semiprime rings satisfying (P 1 )or(p 2 ). In Section 3 we establish commutativity of rings with unity that satisfy the property (P 1 )or(p 2 ) and also prove commutativity of rings (may be without unity) satisfying the property (P 3). In Section 4, attempts are made to prove

3 Polynomial identities 1309 commutativity of s-unital rings satisfying the property (P 1 )or(p 2 ). Finally, we conclude our discussion with some open problems in Section Commutativity of Semiprime Rings Theorem 2.1. Let R be a semiprime ring. Then the following statements are equivalent: (a) R satisfies the polynomial identity (P 1 ). (b) R satisfies the polynomial identity(p 2 ). (c) R is commutative. Before proving this theorem, we state the following lemma. Lemma 2.2([10, Theorem 1]). Let f be a polynomial in n non-commuting indeterminates x 1,x 2,x 3,...,x n with integer coefficients. Then the following statements are equivalent: (i) For any ring R satisfying the polynomial identity f = 0, C(R) is nil. (ii) For every prime p, (G(F (p)) 2 fails to satisfy f =0. (iii) Every semiprime ring satisfying f = 0 is commutative. Next, we establish the following result. Lemma 2.3. Let R satisfy (P 1 )or(p 2 ). Then C(R) N(R). Proof. Suppose that R satisfy (P 1 ).One observe that x = e 11 +e 12, y = e 12 fail to satisfy this equality in (GF (p)) 2, p a prime. Using Lemma 2.2, we obtain the required result. On the other hand, if R satisfies (P 2 ) then use the similar arguments with the choice of x = e 11, y = e 12 to obtain C(R) N(R). Now, we are in a position to suppy the proof of the main theorem of this section. Proof of theorem 2.1. Clearly, every commutative ring R satisfies the one of the conditions (P 1 )or(p 2 ). Conversely, if a semiprime ring R satisfies (P 1 )or (P 2 ), then Lemma 2.3 together with Lemma 2.2 implies that R is commutative. Remark 2.4. Since there are non-commutative rings with R 2 Z(R) neither of the conditions (P 1 ) nor (P 2 ) guarantees the commutativity in arbitrary rings.

4 1310 M. S. Khan One might ask a natural question: What additional conditions are needed to ensure the commutativity for arbitrary rings which satisfies (P 1 )or(p 2 )?. 3. Commutativity of Rings with 1 Theorem 3.1. Let R be a ring with 1. Then the following statements are equivalent: (a) R is commutative. (b) R satisfies (P 1 ). (c) R satisfies (P 2 ). In the sequel, we state the following lemmas. Lemma 3.2([9,p.221]). If [x, y] commutes with x then [x n,y]=nx n 1 [x, y] for all n 1. Lemma 3.3([13]). Let R be a ring with unity land x, y R. If kx m [x, y] =0 and k(x +1) m [x, y] = 0 for some integer m 1 and k 1 then k[x, y] =0. Lemma 3.4. Let R be a ring with unity satisfy (P 1 )or(p 2 ). Then N(R) Z(R). Proof. Let R satisfy (P 1 ) and let a N(R). Then there exists an integer k l, l 1 such that a l Z(R) k l, l minimal. (3.1) Take l =1. Then for each such a, the result is trivial. Suppose that l>1. Next, choose integer m 1 = m 1 (a l 1,y) 1,r 1 = r 1 (a l 1,y) 0,n 1 = n 1 (a l 1,y) 0,s 1 = s 1 (a l 1,y) 0,p 1 = p 1 (a l 1,y) 0,q 1 = q 1 (a l 1,y) 0. Replacing x by a l 1 in (P 1 ), we get y s 1 [a l 1,y]=±(a l 1 ) p 1 [(a l 1 ) m 1,yn 1 ]r 1 (al 1 ) q 1 =0. In view of (3.1) and the fact that (l 1)n 1 l for n 1 1, we see that y1 s[al 1,y] = 0. An application of Lemma 3.3, this yields [a l 1,y]=0, which implies that a l 1 Z(R) which contradicts the minimality of l in (3.1) and N(R) Z(R). Similarly, we can prove the result if R satisfies (P 2 ).

5 Polynomial identities 1311 Particularly, the following result is derived from Lemma 3.4. Lemma 3.5. Let R be a ring with unity satisfy (P 3 ). Then N(R) Z(R). Proof of Theorem 3.1. Clearly, (a) (b) and (c). Next, we show that (b) (a). Let R satisfy the property (P 1 ). Here, we prove the result for the case (b) (a) (i.e., y s [x, y] =±x p [x m,y n ] r y q, where m 1,r 0,n 0,s 0,p 0,q 0). If n = 0 then we have y s [x, y] = 0, for all x, y R, and hence by Lemma 3.3, we obtain the result. Now, suppose that n>0. Combining Lemma 3.4, one get C(R) N(R) Z(R) (3.2) Now, we break the proof in two parts: Case(i). Let r = 1 in the hypothesis of (P 1 ). Then we have y s [x, y] =±x p [x m,y n ]y q, x, y R. (3.3) Replacing x by 2x in the relation (3.3) and using (3.3), we get ±(2x) p [(2x) m,y n ]y q y s [2x, y] =0 ±2 p+m (x p [x m,y n ]y q ) 2(y s [x, y]) =0 (2 p+m 2)y s [x, y] =0. In view of Lemma 3.3, this gives β[x, y] =0, where β =2 p+m 2 0. Using (3.2) and Lemma 3.2, one can obtain [x, y β ]=βy β 1 [x, y β ], forall x, y R, that is y β Z(R), x, y R where β =2 p+m 2. (3.4) Combining Lemma 3.2 with the relation (3.2) and (3.3) repeatedly, we get (1 x (m 1)(p+m 1) )[x m,y]y s = y s [x m,y] x (m 1)(p+m 1) mx m 1 [x, y]y s = y s [x m,y] x (m 1)(p+m 1) mx m 1 (±x p [x m,y n ]y q ) = y s [x m,y] x mp [x m2,y n ]y q = 0. Using Lemma 3.2, we get (1 x (m 1)(p+m 1) )[x m,y]=0.

6 1312 M. S. Khan Replace y by y n in the last equation to get (1 x (m 1)(p+m 1) )[x m,y n ]=0. This implies that (1 x (m 1)(p+m 1) )[x m,y]x p x q = 0 or (1 x (m 1)(p+m 1) )y s [x, y] =0. By Lemma 3.2, this gives (1 x (m 1)(p+m 1) )[x, y] =0. So (1 x β(m 1)(p+m 1) )[x m,y]=0 (β =2 ( p + m) 2 0) Using (3.4) in the last identity, we have [y, x x β(m 1)(p+m 1) ]=0, x, y R. An application of Herstein Theorem [5] gives the required result. Case(ii). Let r>1 Then using (3.2) in the relation (3.3) repeatedly, for any positive integer α, we obtain y s+rns+...+rα 1 n (α 1)s [x, y] = y s+rns+...+rα 1 n (α 1)s (±[x m,y n ] r x p y q ) = (±) r y s+rns+...+rα 1 n (α 1)s [x m2,y n2 ] r2 (x mp,y nq ) r x p,y q =... = (±) rα [x mα,y nalpha ](x mp,y nq ) r2 (x mα 1p,y nα 1q ) r...(x mp,y nq ) r x p,y q. In view of (3.2) commutators are nilpotent, this implies that y s+rns+...+rα 1 n (α 1)s [x, y] =0, x, y R. Hence the proof is complete by using Lemma 3.2. Similar arguments with necessary variations would show that (c) (a). Remarks 3.6. The proof of above theorem (i.e., (b) (a)) could have been obtained by [12,Theorem]. Here we give a direct proof with view to preparing some ground work for the Theorem 3.7 that proves commutativity of arbitrary

7 Polynomial identities 1313 rings (see [2],for details). Theorems 3.7. Let R be a ring (may be without unity 1). Then the following statement are equivalent: (a) R is commutative. (b) R satisfy (P 3 ). Proof. It is easy to check that every commutative ring satisfies (P 3 ). That is (a) implies (b). Now, it is enough to show that (b) implies (a). Trivially, for n = 0 in (P 3 ), yields the required result. Now, let n>0 and R satisfy (P 3 ), that is [x, y] =±x p [x m,y n ] r y q, where m>1,r >0,n 0,p 0,q 0. (3.5) From Lemma 2.3 and Lemms 3.5 one can observe that they are still valid in the present situation, and hence we have C(R) N(R) Z(R) (3.6) Let r =1. Then (3.5) can be written as [x, y] =±x p [x m,y n ] r y q, x, y R (3.7) Apply similar techniques as used to get (3.4), we write y β Z(R), x, y R and β =2 p+m 2. (3.8) Using Lemma 3.2 and (3.7)repeatedly, we get [x m,y] x β(m 1)(p+m 1) [x m,y]=0. Hence, a technique similar to used in Theorem 3.1 gives that [y, x x β(m 1)(p+m 1) ]=0 (β =2 p+m 2 0), x, y R. Thus, R is commutative by Herstein [5,Theorem].

8 1314 M. S. Khan For r>1. Then for an arbitrary integer α, applying similar techniques as used in the proof of Theorem 3.1, we write [x, y] =(±) rα (m α n α x mα 1 y nα 1 ) rα [x, y] rα (x mα 1p y nα 1q ) rα 1...(x mp y nq ) r x p y q But, since the commutators are nilpotent, this completes the proof. Remark 3.8. The above theorem reduces to Theorem 2 of [3] if p = 0 and q =0. Example 3.9. The ring of 3 3, without unit, of strictly upper triangular matrices over Q, the ring of rational numbers shows that the hypotheses of Theorem 3.1 alone without additional condition 1, does not guarantee commutativity. Remark 2.4 indicates that Theorem 3.1 cannot generalized to arbitrary rings. It is natural to look the extension of this theorem. In this context, we discuss some results in the next section. 4. Commutativity of s-unital Rings A ring R is called left(resp. right) s-unital if x Rx(resp.x xr) for all x R. A ring R is called s-unital if and only if x Rx xr, x R. If R is s-unital(resp. left or right s-unital), then for any finite subset F of R there exist an element e R such that ex = xe = x(resp.ex = x or xe = x), x F. Such an element e is called pseudo identity of F. Now, we prove the following theorem Theorem 4.1. Let R be a left(resp. right) s-unital ring. If R satisfy (P 1 )(resp. (P 2 )). Then R is commutative(and conversely). Proof. Clearly, every commutative left(resp. right) s-unital ring satisfy (P 1 )(resp. (P 2 )). Conversely, suppose that R is a left(resp. right) s-unital ring satisfying (P 1 )(resp. (P 2 )), and x an arbitrary element of R. Set an element e R such that ex = x (resp.xe = x). Replacing y by e in (P 1 )(resp. (P 2 )), we get e s [x, e] =±x p [x m,e n ] r e q (resp. [x, e]e s = ±x p [x m,e n ] r e q ) Let r = 1. Then this gives x = x(e x p+m 1 e x p+m 1 e) xr (resp. x =

9 Polynomial identities 1315 (e ± x p+m 1 e x p e n x m 1 )x Rx), for m>1. Take r>1. Then we get e s [x, e] = 0 (resp. [x, e]e s = 0). This implies that x xr (resp.x Rx) for all x R. Hence, R is right(resp. left) s-unital. So, R is commutative by Theorem 3.1. this completes the proof. Remark 4.2. The following example demonstrates that there are noncommutative left(resp. right) s-unital ring satisfying (P 1 )(resp. (P 2 )). Example 4.4. Let R 1 = 0 0, 1 0, , (resp. R 2 = , , , be subrings of 2 2 matrices over GF(2). It is obivious that (R 1 ) (resp. (R 2 )) is a left (resp. right) s-unital ring, and for any fixed integers m>1,r >0,s> 0,n 0,p 0,q 0, (R 1 ) (resp. (R 2 )) satisfies (P 1 )(resp. (P 2 )). Remark 4.3. One can observed in the above theorems, the exponent in the underlying conditions are assumed to be global. It would be interesting to further generalized these results for the case when they are assumed to be local (local means they depend on the pair of elements x, y for their values) ) 5. Some Open Problems In retrospect,one can ask the following questions : 5.1 A left(resp. right) s-unital ring R is commutative if and only if R satisfies (P 4 ). 5.2 Let R be a left(resp. right) s-unital ring. If there exist a nil subset B of R for which R satisfies (P 5 ) and the condition (P 4 ), then R is commutative. Acknowledgments A part of the present research work was carried out while Professor M.A.Khan was a visiting consultant for the project ( Account No. IG/SCI/DOMS/09/14) of the first author. References [1] H.A.S. Abujabal and M.A. Khan, Commutativity of certain class of rings, Georgian Math.J., 5 (1998), [2] M. Ashraf, Certain polynomial identities and commutativity of rings,math. Slovaca, 50 (2000),

10 1316 M. S. Khan [3] H.E. Bell, On some commutativity theorems of Herstein, Arch.Math.(Besel), 24 (1955), [4] I.N. Herstein, An elementary proof of a theorem of Jacobson, Duke Math. J., 21 (1954), [5] I.N. Herstein, A note on rings with central nilpotent elements, Proc. Amer. Math. Soc., 5 (1954), 620. [6] I.N. Herstein, Two remarks on commutativity of rings, cand. J. Math.,7 (1955), [7] Y. Hirano, M. Hongon and H. Tominaga, Some polynomial identities and commutativity of s-unital rings, Math. J. Okayama Univ., 29 (1982), [8] M. Hongan and H. Tominaga, Some commutativity theorems for semiprime rings, Hokkaido Math. J., 10 (1981), [9] N. Jacobson, Structure theory for algebraic algebras of bounded degree, Ann. of Math., 46 (1945), [10] N. Jacobson, Structure of Rings, Amer. Math. Soc. Colloq. Publ. Providence, [11] T. P. Kezlan, A note on commutativity of semiprime PI-rings, Math. Japon., 27 (1982), [12] Komatsu, H., Commutativity theorems for rings-ii, Hiroshima Math.J., 22 (1985), [13] Moharram. A. Khan, A commutativity study for certain rings, (In press). [14] Moharram. A. Khan, Commutativity of rings with variable constraints, Publ. Math.Debercen, 58 (2001), [15] W.K. Nicholson and A. Yaqub, A commutativity theorem for rings and groups, Canad. Math. Bull., 22 (1979), Received: April, 2010

ON STRUCTURE AND COMMUTATIVITY OF NEAR - RINGS

Proyecciones Vol. 19, N o 2, pp. 113-124, August 2000 Universidad Católica del Norte Antofagasta - Chile ON STRUCTURE AND COMMUTATIVITY OF NEAR - RINGS H. A. S. ABUJABAL, M. A. OBAID and M. A. KHAN King