Isomorphism of algebras vs isomorphism of categories of representations

Question

Let $A, B$ be finite-dimensional algebras over a field, so that their categories of modules $A\text{-mod}$ and $B\text{-mod}$ are finite abelian in the sense of EGNO Tensor categories.

Given an isomorphism $f:A \to B$ of algebras, the pullback $f^* : B\text{-mod} \to A\text{-mod}$ is an isomorphism of finite abelian categories. Furthermore, it commutes with the forgetful functors, in the sense that \begin{align} \hom_A(A, f^*V) \cong \hom_B(B, V), \quad g \mapsto (b \mapsto b.g(1)) \end{align} natural in $B$-modules $V$. Here, $A$ and $B$ are the regular modules. (actually, this isomorphism is

I was wondering if we can reverse this statement. However, I don't see it, and I don't understand where isomorphism vs equivalence plays a role.

So far I did

• take $F: B\text{-mod} \to A\text{-mod}$ an equivalence of finite abelian categories
• such that there exists $\nu : \hom_A(A, F-) \overset{\sim}{\Rightarrow} \hom_B(B,-)$

We have the vector space isomorphism $FB \cong \hom_A(A,FB) \xrightarrow{\nu_B} \hom_B(B,B) \cong B$. Note also that $F$ is an equivalence so $A \cong FGA$, where $G$ is a quasi-inverse of $F$. In particular, then, \begin{align} \hom_B(B,V) \cong \hom_A(A,FV) \cong \hom_A(FGA,FV) \cong \hom_B(GA,V) \end{align} whence by Yoneda $A \cong FB$ in $A\text{-mod}$. In particular, $A \cong B$ as vector spaces.

I feel I'm overlooking something very obvious here. There at least should be some condition on $\nu$; and I guess $F$ should actually do nothing on morphisms?

Any hints would be appreciated.

Answer 1

$\newcommand{\id}{\operatorname{id}} \newcommand{\tensor}{\otimes}$

This follows from Lemma 1.6 in Hopf Monads (Bruguieres, Virelizier). Indeed, if $A$ is an algebra, then $A\otimes -$ is a monad on $\text{Vect}$, and the category of modules over the monad is precisely the category of modules over $A$.

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EDIT: I will now be more explicit. Without loss we can actually assume that the forgetful functor sends a module to its underlying vector space (this functor is isomorphic to the one I wrote above). Let's at first not restrict to equivalences/isomorphisms of categories. So our initial set-up is the following:

• $F : B\text{-mod} \to A\text{-mod}$ a functor (+ appropriate adjectives) which commutes with the forgetful functors to $\text{Vect}$.

This already implies two things:

1. $Ff = f$, i.e. $F$ doesn't do anything on intertwiners.
2. $F(V,\rho) = (V,\hat{\rho})$ with $\hat{\rho} : A\otimes V\to V$, for a $B$-module $(V,\rho)$. [Note that we denote a module by a pair consisting of a vector space and an action]

Define for any vector space the linear map \begin{align} f_{V} = \widehat{m_B \tensor \id_V} \circ \id_A \tensor 1_B \tensor \id_V : A \tensor V \to B \tensor V \ , \end{align} where $1_B$ is the unit of $B$. Note that this is natural in $V$ and thus in fact determined by its value $f=f_k$ on the field, i.e. $f_V = f \tensor \id_V.$

Let $(V,\rho) \in B\text{-mod}$. Consider the free $B$-module $(B \tensor V, m_B \tensor \id_V)$, where $m_B$ is the multiplication of $B$. We can consider $\rho$ as a morphism in $B\text{-mod}$, from the free module above to $(V,\rho)$. Then in particular $F\rho = \rho$ is an $A$-module map, which means \begin{align} F\rho \circ \widehat{m_B \tensor \id_V} = \widehat{\rho} \circ \id_A \tensor F\rho \ . \end{align} Thus we find for the interplay between $\rho$ and $f_{V}$ \begin{align} F\rho \circ f_{V} &= F\rho \circ \widehat{m_B \tensor \id_V} \circ \id_A \tensor 1_B \tensor \id_V \\ &= \widehat{\rho} \circ \id_A \tensor F\rho \circ \id_A \tensor 1_B \tensor \id_V \\ &= \widehat{\rho} \ . \end{align} That is, we have that $\widehat{\rho} = \rho \circ f_{V} = \rho \circ f \tensor \id_V$ as linear maps.

Finally, the claim that $F = f^*$ follows once one has proved that the following are equivalent for a linear map $f:A \to B$ between algebras:

• $f$ is an algebra map
• If $(V,\rho)$ is a $B$-module, then $(V, \rho \circ f \tensor \id_V)$ is an $A$-module.

I will leave this as a good exercise (if done correctly, the proof takes less than one A6-sized page)