We have a smash product of Hopf algebras if one acts on other (namely making it module algebra, coalgebra and Hopf algebra) with a compatibility condition (Theorem 17).

Now I ask the same question myself for the case of Frobenius algebras.

I already had a look at the paper entitled Note on Smash Products over Frobenius algebras. I wonder whether we modify the first theorem (smash product) of this paper for two Frobenius algebras with an action of one on another or not. To handle this, of course we need a proper definition of an action of Frobenius algebras and maybe it already exist somewhere (don't know)... However if not, it seems it will not be much difficult with a slight modification on the compatibility condition, analogous to Hopf algebra case; or having the exact definition of the action in a categorical way via short exact sequences.

PS: My keypoint is; since I study Hopf algebras in a diagrammatical way instead of calculations, the only difference between Hopf algebras and Frobenius algebras appears when product meets coproduct in diagrammatical definition. That is why this problem does not seem much complicated for me, after dealing the same issues for the case of Hopf algebras.

I really appreciate if someone can give me some lights for these issues.

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    $\begingroup$ Did you tried to see if A#H is Frobenius if A is a Frobenius algebra and H a finite dimensional Hopf algebra? (Say A is an H-module algebra) $\endgroup$ – Marco Farinati May 18 at 3:06

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