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.


Theorem 4.2 of this paper says that the answer is always yes.

  • $\begingroup$ 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 Oct 23 '09 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 Oct 24 '09 at 20:14

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