Let R be a commutative ring, and A an R-algebra (possibly non-commutative). Then A is separable if it is (fin. gen.) projective as an (A tensor_R A^op)-algebra. Suppose further that A is fin. gen. 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 tr: A --> R, such that b(x,y) := 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.