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.



Separable and finitely generated projective but not Frobenius?
Theorem 4.2 of [On separable algebras over a commutative ring](https://projecteuclid.org/journals/osaka-journal-of-mathematics/volume-4/issue-2/On-separable-algebras-over-a-commutative-ring/ojm/1200691953.full) says that the answer is always yes.


