## The Nakayama automorphism of a Frobenius algebra

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
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Throughout $k$ is a field and $A$ is a Frobenius $k$-algebra. So there exists a bilinear form $B : A \times A \longrightarrow k$ which is non-degenerate and $B(xy,z)=B(x,yz),$ for all $x,y,z \in A.$

Theorem. There exists a $k$-algebra automorphism $\sigma$ of $A$ such that $B(x,y)=B(y,\sigma(x)),$ for all $x,y \in A.$

Proof.  Let $x \in A$ and define $g : A \longrightarrow k$ by $g(y)=B(x,y),$ for all $y \in A.$ Clearly $g \in A^*$ and so by this lemma, there exists a unique $\theta_x \in A$ such that $g(y)=B(y,\theta_x),$ for all $y \in A.$ So $B(x,y)=B(y,\theta_x),$ for all $y \in A.$ Define the map $\sigma : A \longrightarrow A$ by $\sigma(x)=\theta_x,$ for all $x \in A.$ So we need to prove that $\sigma$ is a $k$-algebra automorphism. First we show that $\sigma$ is $k$-linear. If $\alpha \in k$ and $x,y,z \in A,$ then

$B(z, \alpha \sigma(x) + \sigma(y))=\alpha B(z, \sigma(x)) + B(z, \sigma(y))=\alpha B(x,z)+B(y,z)$

$=B(\alpha x + y,z)=B(z, \sigma(\alpha x + y)).$

Therefore, since $B$ is non-degenerate, $\sigma(\alpha x + y)=\alpha \sigma(x) + \sigma(y)$ and so $\sigma$ is $k$-linear. It is easy to see that $\sigma$ is injective: if $\sigma(x)=0,$ then $B(x,y)=B(y,\sigma(x))=B(y,0)=0,$ for all $y \in A$ and hence $x = 0$ because $B$ is non-degenerate (see the Remark in this post). Hence $\sigma$ is surjective as well because $A$ is finite dimensional. So we only need to prove that $\sigma$ is multiplicative. Recall that $B(xy,z)=B(x,yz),$ for all $x,y,z \in A$ and thus

$B(z, \sigma(xy))=B(xy, z)=B(x,yz)=B(yz, \sigma(x))=B(y, z \sigma(x))=B(z \sigma(x), \sigma(y))$

$=B(z, \sigma(x)\sigma(y)).$

Therefore $\sigma(xy)=\sigma(x)\sigma(y)$ because $B$ is non-degenerate. $\Box$

Definition. The $k$-algebra automorphism $\sigma$ in the above theorem is called the Nakayama automorphism.

We proved in here that there exists $f \in A^*$ such that $\ker f$ does not contain any non-zero left ideal of $A.$ In fact we saw that $B(x,y)=f(xy),$ for all $x,y \in A.$ So if $\sigma$ is the Nakayama automorphism of $A,$ then $f(xy)=f(y \sigma(x)),$ for all $x,y \in A.$

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