Let be a ring, which may or may not have We proved in here that if for all then is commutative. A similar approach shows that if for all then is commutative.

**Problem**. Prove that if for all then is commutative.

**Solution**. Clearly is reduced, i.e. has no nonzero nilpotent element. Note that for all because Hence is an idempotent for every because

Thus is central for all by Remark 3 in this post. Therefore is central for all But

and hence is central. Therefore which gives us

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