Let $R$ and $S$ be rings. I'm trying to prove that if the categories $\operatorname{mod}R$ and $\operatorname{mod}S$ of finitely generated (left) modules are equivalent, then so are the categories $\operatorname{Mod}R$ and $\operatorname{Mod}S$. (That is, $R$ and $S$ are Morita equivalent.) Rings and Categories of Modules (second edition) by Anderson and Fuller provides a hint (page 266, exercise 4), which I've paraphrased here:
Let $F:\operatorname{mod}R \to \operatorname{mod}S$ be an equivalence with inverse equivalence $G$. If $R^n \to G(S)$ is a split epimorphism, then so is $F(R)^n \to FG(S) \cong S$.
I can see that, since $G(S)$ is a finitely generated $R$-module (by assumption), there exists an epimorphism $R^n \to G(S)$ for some $n$. Moreover, this epimorphism splits because $G(S)$ is projective. Hence there is a split epimorphism $F(R)^n \to S$. However, I'm not sure how to use the hint at this point. I suppose we could observe that $F(R)^n \cong S \oplus X$ for some (finitely generated) $S$-module $X$, but I don't know how this would help us construct an equivalence between $\operatorname{Mod}R$ and $\operatorname{Mod}S$.
How can I use the hint to prove this fact? Alternatively, does anyone have a different approach to proving it?
Best Answer
You want to see that $F(R)$ is a progenerator, that is, a finitely generated projective generator (in the full category of modules) with $\operatorname{End}_S(F(R))=R$.
The $S$-module $F(R)$ is finitely generated by assumption; its endomorphism ring is $R$ because $F$ is an (additive) equivalence.
The key is that epimorphisms in $\operatorname{mod}R$ are surjective (prove it).
In order to see that $F(R)$ is projective, consider an epimorphism $S^n\to F(R)$, that becomes an epimorphism $G(S^n)\to R$, which therefore splits. Hence also $S^n\to F(R)$ splits and so $F(R)$ is projective.
In order to see that $F(R)$ is a generator, you need to find an epimorphism $F(R)^m\to S$, for some $m$. This is easy, because $G(S)$ is finitely generated, so there is an epimorphism $R^m\to G(S)$ and you can apply $F$.