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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.

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Theorem 4.2 of On separable algebras over a commutative ring says that the answer is always yes.

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    $\begingroup$ Okay great! I see that Thm 4.2 also has the assumption that A is a faithful R-module. Can that be removed? or is it automatic? $\endgroup$
    – Chris Schommer-Pries
    Commented Oct 23, 2009 at 18:19
  • $\begingroup$ It is not automatic but I imagine will be true in any case that you might be interested in. Otherwise, you could replace R by a suitable quotient ring. $\endgroup$
    – Simon Wadsley
    Commented Oct 24, 2009 at 20:14

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