I'd like to know an example of a Frobenius algebra $A$, with a subalgebra $B$ that is itself a Frobenius algebra, such that $A$ is not projective as a left $B$-module. I don't require any compatibility between the Frobenius algebra structures (e.g. trace map) of $A$ and $B$. In all the examples I know of (group algebras of symmetric groups, nilcoxeter algebras, Hecke algebras of type A, etc.), $A$ is even a free $B$-module. But I don't see any reason, a priori, for this to always be the case.

2

$\begingroup$
$\endgroup$

14

$\begingroup$
$\endgroup$

You can take $A={\mathbb C}[x]/x^3$ and $B$ -- subalgebra generated by $x^2$.