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 standard ring theory textbooks. Here we only discuss a very special case of the theorem, i.e. when for all

**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**. Clearly 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 hence is central. Therefore is central.

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