ON COMMUTATIVE SUBRINGS OF INFINITE RINGS THOMAS J. LAFFEY In this note we prove THEOREM 1. Let R be any (associative) infinite ring. Then R has an infinite commutative subring. Proof. Assume the result is false and let R be a counter-example. We obtain several properties of R which we eventually show are contradictory. LetxeR. (J) If {reR\xr = 0} is infinite, then Ann (x) — {reR\xr = rx = 0} is infinite. For let H = {reR\xr = 0}. H is a subring of R and thus Hx = {hx\heH} is a commutative subring of R. ThusHx is finite and thus {heH\hx = 0} is infinite. By symmetry we have (2) If{r e R\rx = 0} is infinite then Ann (x) is infinite. We now prove: (3) If x is nilpotent, then Ann (x) is infinite. If x = 0, this is trivial. Suppose x # 0 but that x" = 0 for some integer n > 1. First A-"" 1 R is a subring of R, so, since x(x"~ l R) = 0, (1) implies that Ann (x) is infinite if x"' 1 R is infinite. If x"~ l R is finite, then (1) implies that Ann (x"' 1 ) is infinite. Suppose now that n— 1 > k ^ 1 and that we have shown that Ann (x"~ k ) is infinite. If x(Ann (x n ~ k )) isfinite,then {re Ann (x"~ k )\xr = 0} is infinite. In this case (1) implies that Ann (x) is infinite. If x(Ann (x"~ k )) is infinite, then, since (1) implies that Ann (x"~ (fc+1) ) is infinite. Thus in any case, Ann (y~ (fc + 1) ) is infinite. Induction on k completes the proof of (3). (4) If x, yeR are such that xy = 0, then either Ann (x) is infinite or Ann (y) is infinite. For xyR = 0, so, by (1), Ann (x) is infinite if yR is infinite. If yR is finite, then (1) implies that Ann (y) is infinite. Let X(R) = {re R\Ann (r) is infinite}. (5) X(R) * {0}. Suppose conversely that X(R) = {0}. Let xeR, x # 0. Since the ring generated by x is commutative, it must be finite. By (3), R has no nonzero nilpotent elements. Received 7 June, 1971. [BULL. LONDON MATH. SOC, 4 (1972), 3-5]