Let be a field and let be a -algebra. Let be the Jacobson radical of

**Theorem**. (Amitsur) If as cardinal numbers, then is nil.

*Proof***. **If is finite, then and so is Artinian. Thus is nilpotent and therefore nil. So we may assume that is infinite. Let So is invertible in for all In particular, if then is invertible and thus is invertible. Now consider the set

Clearly for any non-zero elements we have if and only if Thus, since is infinite, we have Therefore the elements of cannot be -linearly independent. So there exist an integer and non-zero elements such that

Obviously all commute with each other and with all because and are in So multilpying through by will clear denominators and we will get

Define the polynomial Then by Besides is not identically zero because for example So we have proved that is algebraic over Thus

for some integer and some Now, is not invertible because Thus On the other hand, if for all then becomes and we are done. So we may assume that there exists such that and

If we let then we can re-write as

Since the element is invertible and thus gives us

An interesting case of the theorem is when is finitely generated and is uncountable. Then clearly is countable, because is finitely generated, and thus So is nil.