A RESULT OF COMMUTATIVITY OF RINGS

In this paper we prove the following:THEOREM. Let 1$" id="E1" xmlns:mml="http://www.w3.org/1998/Math/MathML"> n > 1 and m be fixed relatively prime positive integers and k is any non-negative integer. If R is a ring with unity 1 satisfying x k [ x n , y ] = [ x , y m ] for all x , y ∈ R then R is commutative.

(i) (ii) (xy )" x"y" (iii) (xy)* where n > and m are fixed relatively prime positive integers and k is any non-negative integer, then R is commutative.In this paper we prove the theorem stated in the abstract which improve above theorem of Psomopolous [12] where conditions (ii) and (iii) are superfluous.
Throughout, R will denote an associative ring with unit 1.We use the following notations.Z(R ), the center of R.
[x, y xy yx C(R), the commutator ideal of R. N(R), the set of all nilpotent elements of R. D (R), the set of all zero divisors in R.
2. MAIN RFULTS.We state our main result as follows.MAIN THEOREM.Let n > and m be fixed relatively prime positive integers and k is any non- negative integer.If R is a ring with unity satisfying (*) x*[x", y Ix, y'] for all x, y E R then R is commutative.We begin with the following lemmas which will be used in proving our main theorem.
LEMMA 1 ([2], Theorem 1).LetR be a ring satisfying an identityq(X) O, where q(X) is a polynomial identity in non-commuting in-determinates, its coefficient being integers with highest common factor one.
If there exists no prime p for which the ring of 2 x 2 matrices over GF(p) satisfies q(X) 0, then R has a nil commutator ideal and the nilpotent elements of R form an ideal.
LEMMA 3 ([9]).Let R be a ring with unity and let f: R R be a function such that f(x / 1) for all x R. If for some positive integer n, x"f(x) 0 for all x in R, then necessarily f(x) 0. LEMMA 4. If R is a ring satisfying (*) in the hypothesis of the main theorem then PROOF.By Lemma 3 of [12] we have N(R)CZ(R) when R satisfies x[x',y]-[x,y"] for all x, y E R. This is a polynomial identity with coprime integral coefficients.But if we consider (i)x -e, and y e2, if n > 1, m 1 and (ii) x e2 and y ezz if n 1 and m 1, we find that no ring of 2 2 matrices over GF(p), p a prime, satisfies this identity.Hence by Lemma 1, C(R) is a nil ideal and thus C(R C_N(R C_Z(R PROOF OF MAIN THEOREM.By Lemma 4, we have c(R) _V(R)___ Z(R) Thus all commutators are central.Moreover, we know thatR is isomorphic to a subdirect sum of subdirectly irreducible ringsR each of which a homomorphic image ofR satisfies the hypotheses of the theorem.Thus we can assume that R is subdirectly irreducible ring.Hence/, the intersection of all non-zero ideals is non-zero.CASE 1.Let n 1 and m 1.
Replacing y by x, we get m[x,a]-0 for all x R.
(2.23) Butm and n are relatively prime.Hence there exists integers ct and such thatm x+n 1. Multiplying (2.11) by and (2.23) by and adding, we get Ix,a]-0 for all Hence a Z(R), which proves our claim.
We know that y-a and y,a _Z(R).Thus We know that (m, n) I. Hence there exists integers a and 5 such that mct + n I. Multiplying (2.24) by x -x")x and multiplying (2.25) by x -x')x ' and adding, we get (x-x')(x-x')x'/[x,y]=O forail x,y .R V. GUPTA This can be written as (x-x2h(x))x'/'/[x,y]-O forall x,y R (2.26) where h (x) is a polynomial in x with integers coefficients.Suppose R is not cummutative.Then by a well known result of Hetstein [6], there exists x R such that x x2h(x) q Z(R).From this it is clear that x Z(R).Hence x and x x2h(x) is not a zero divisor.Hence (x -x2h(x))x p /q/t is also not a Zero divisor.Thus Ix,y] 0 for all y R (2.27)This gives a contradiction.Hence R is commutative.