I'm only ever thinking about finite-dimensional $k$-algebras instead of general rings. Let $A$ be such an algebra. My question is:
- Are full finite abelian subcategories of $A\text{-mod}$ precisely the categories of modules over quotients $A/I$ for ideals $I\leq A$?
I think I can show: $I\leq A$ then $A/I\text{-mod} \subset A\text{-mod}$ is full (of course it's finite abelian).
But if $S \subset A\text{-mod}$, then I don't see why $\operatorname{End}_S(P) \cong A/I$, for a progenerator $P \in S$ and some ideal $I\leq A$. My idea is that the minimal projective generator of $S$ is a quotient of the minimal projective generator of $A\text{-mod}$, namely the quotient by the sum of projective covers of simples in $A\text{-mod}$ that are not in $S$. How to take it from there?
No. Let $M$ be any finitely generated $A$-module with $\operatorname{End}_A(M)=k$. Then the category of finite direct sums of copies of $M$ is a full finite abelian subcategory of $A\text{-mod}$ (it is equivalent to $k\text{-mod}$), but unless $M$ is simple, it won't be of the form $(A/I)\text{-mod}$, as it won't be closed under taking quotients and submodules.