A celebrated theorem of Jacobson states that

**Theorem. **Let be a ring, not necessarily containing . If, for each there exists a positive integer such that , then is commutative.

This is a very strong and difficult result (although not very useful in practice). However, we can obtain some special cases via elementary means.

**Proposition 1. **Let be a ring such that for each we have . Then is commutative.

*Proof.* Let . Then , i.e., . Again, , i.e., . Thus , i.e., for each . Thus , as desired.

The next case is already considerably harder.

**Proposition 2. **Let be a ring such that for each we have . Then is commutative.

*Proof.* Let . Then shows that

,

and shows that

.

Hence

for all .

Plugging into gives , i.e., for each .

Plugging into gives , i.e., for each . Replacing by gives , i.e., .

Also, multiplying by first on the left and then on the right and then subtracting the two gives .

From the last two paragraphs we conclude that for all .

**Corollary. **Let be a ring such that for each we have for some . Then is commutative.

*Proof.* Note that if for some then . Hence the result follows by **Proposition 2**.