# Separable and finitely generated projective but not Frobenius?

Let R be a commutative ring, and $$A$$ an $$R$$-algebra (possibly non-commutative). Then $$A$$ is separable if it is finitely generated (f.g.) projective as an $$(A \otimes_R A^{\mathrm{op}})$$-algebra. Suppose further that $$A$$ is f. g. projective as an $$R$$-module. Does this imply that $$A$$ is a (symmetric) Frobenius algebra?

There are lots of equivalent definitions of a Frobenius algebra. One (assuming $$A$$ is a f.g. projective R-module) is that there exists an $$R$$-linear map $$\mathrm{tr}: A\to R$$, such that $$b(x,y) := \mathrm{tr}(ab)$$ is a non-degenerate.

I know that the answer is yes when $$R$$ is a field. What about other rings?

I am not an expert on algebras, but this question is related to understanding obstructions for extended TQFTs, and so I am very interested in knowing anything I can about it.