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Jul 3, 2022 at 7:59 history wiki removed Stefan Kohl
May 8, 2010 at 2:40 history edited muad CC BY-SA 2.5
(spelling) changed Noncomutative to Noncommutative
Apr 26, 2010 at 0:18 comment added José Hdz. Stgo. You're right about the existence of that result, Pietro. Yet, according to Herstein: "that theorem as proved has one drawback; true enough, it implies commutativity but only very few commutative rings exist which satisfy its hypothesis."
Apr 24, 2010 at 18:19 comment added Pietro I know of a slightly different theorem by Jacobson. The way you state yours, it sounds like the choice of n has to be uniform across the ring. The one I heard about says that if, for each x, there is an n(x) such that x^n(x) = x, then the ring is commutative. In one aspect this is less general than your version, since we require x^n - x = 0, but on the other hand the absolute freedom in the exponent was very surprising to me.
Apr 24, 2010 at 17:27 history edited José Hdz. Stgo. CC BY-SA 2.5
added 21 characters in body; deleted 8 characters in body; deleted 11 characters in body
Apr 24, 2010 at 17:22 history answered José Hdz. Stgo. CC BY-SA 2.5