A commutativity theorem for rings

Abstract

Let ft be an associative ring with identity in which every element is either nilpotent or a unit. The following results are established. The set N of nilpotent elements in R is an ideal. If R/N is finite and if x = y (mod N) implies x 2 = J/ 2 or both x and y commute with all elements of N , then R is commutative. Examples are given to show that R need not be commutative if "x 2 = y 2 " is replaced by "x = y "for any integer k > 2 . The case N = (0) yields Wedderburn's Theorem.

Some conditions for finiteness of a ring

“…Now consider N, and let B be any zero ring contained in N. If every non-zero idempotent is regular, there exists a unique non-zero idempotent, necessarily I; and every element is invertible or nilpotent. It follows, again by [7] …”

“…The symbols C and N will be used for the center of R and the set of nilpotent elements of R. The If there exists a non-zero idempotent f # I, the decomposition R fR + (1-f)R shows that R is finite, since both summands are finite by the lemma. Therefore, assume that is the only non-zero idempotent, and use the periodicity of R to obtain the property that every element is either nilpotent or invertible a property that R forces N to be an ideal [7]. The factor ring has ascending chain condition and descending chain condition on subrings, hence is finite by (I).…”

Noncommutativity and noncentral zero divisors

“…Now suppose that D = N, in which case each element of R is either nilpotent or invertible-a condition which implies that N is an ideal [8] Theorem 2.1 is the major step in the proof of the following more general theorem. The remaining part is a simple lemma given in [6], which we need not repeat.…”

A commutativity theorem for power-associative rings

“…Suppose not. Since a + b = b (mod J) and ab t ba , we have by (ii) , (a+b) 2 = b2 and hence a 2 + ab + ba = 0 . Since, moreover, -a + b = b(mod J) , a similar argument shows that a 2 -•ab -ba -0 .…”