Recall that a ring extension $S \subset R$ (i.e. fancy words for saying that $S$ is a subring of $R$) is called a Frobenius extension if the restriction functor $\operatorname{Res}:R\text{-mod} \to S\text{-mod}$ is Frobenius, i.e. has isomorphic left and right adjoints.
Let now $H$ be a finite-dimensional Hopf algebra over a field, and $A \subset H$ be a sub-Hopf algebra. By this I mean that $A$ is a Hopf algebra by restricting the Hopf algebra structure of $H$ to $A$. The question in the title is:
Question: Is $A \subset H$ a Frobenius extension? Or what is an example where it isn't? $\newcommand{\Res}{\operatorname{Res}} \newcommand{\Coind}{\operatorname{Coind}} \newcommand{\Hom}{\operatorname{Hom}}$
It is motivated by the following chain of facts
- Nichols-Zoeller theorem states that $\Res(H)$ is free.
- Thus, the right adjoint of $\Res$, called coinduction, is exact: it is given by $\Coind(N) = \Hom_S(\Res(H), N)$, and thus exact since $\Res(H)$ is projective.
- As an exact functor, $\Coind$ also possesses a right adjoint (we are dealing with abelian categories here), call that $\Coind^{r.a.}$
Therefore my question is equivalent to
Question': Is $\Coind^{r.a.}$ automatically isomorphic to $\Res$? Or are there examples where it isn't?
For group Hopf algebras it should be true, what about the general case?