Throughout is a field and is a Frobenius -algebra. So there exists a bilinear form which is non-degenerate and for all

**Theorem**. There exists a -algebra automorphism of such that for all

*Proof*. Let and define by for all Clearly and so by this lemma, there exists a unique such that for all So for all Define the map by for all So we need to prove that is a -algebra automorphism. First we show that is -linear. If and then

Therefore, since is non-degenerate, and so is -linear. It is easy to see that is injective: if then for all and hence because is non-degenerate (see the Remark in this post). Hence is surjective as well because is finite dimensional. So we only need to prove that is multiplicative. Recall that for all and thus

Therefore because is non-degenerate.

**Definition. **The -algebra automorphism in the above theorem is called the **Nakayama automorphism**.

We proved in here that there exists such that does not contain any non-zero left ideal of In fact we saw that for all So if is the Nakayama automorphism of then for all