Throughout is a ring.

**Theorem** (Jacobson). If for every there exists some such that then is commutative.

The proof of Jacobson’s theorem can be found in any standard ring theory textbooks. Note that in Jacobson’s theorem, doesn’t have to be fixed, i.e. it could depend on See this post for the proof of the theorem when is fixed. Here we only discuss a very special case of the theorem, i.e. when

**Definitions**. An element is called **idempotent** if The **center** of is

It is easy to see that is a subring of An element is called **central** if Obviously is commutative iff i.e. every element of is central.

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

**Solution**. First note that is reduced, i.e. has no nonzero nilpotent element. For every we have and so is idempotent for all Hence, by Remark 3 in this post, is central for all Now, since

we have and thus is central. Also, since

we have and so is central. Thus is central and so is commutative.

A similar argument shows that if for all then is commutative (see here!).