# Equivalent module categories

Let $A$ and $B$ be rings and let $A\text{-mod}$ and $B\text{-mod}$ be their abelian module categories. Let $F:A\text{-mod}\to M\text{-mod}$ and $F':B\text{-mod}\to A\text{-mod}$ be functors which afford an equivalence between the two module categories (i.e. such that $F\circ F'\simeq \operatorname{id}_{A\text{-mod}}$ and $F'\circ F\simeq \operatorname{id}_{B\text{-mod}}$).

Claim Every finitely-generated $A$-module is a homomorphic image of a direct sum of copies of $P$, where $P$ is the image under $F'$ of the regular representation of $B$.

Attempt at a solution We will need to use the fact that every $A$-module is a homomorphic image of a direct sum of copies of the regular representation of $A$. I can't seem to figure out where the natural isomorphisms come into it.

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Let $M$ be an $A$-module. Then certainly $F(M)$ is a homomorphic image of a direct sum of copies of $_BB$, the regular representation of $B$ -- suppose $F(M)=\varphi(_BB^{\oplus k})$. Then $\varphi\in\mathrm{Hom}(_BB^{\oplus k},F(M))$ which gives $F'(\varphi)\in\mathrm{Hom}(F'(_BB^{\oplus k}),F'\circ F(M))$. Since $F'\circ F(M)\cong M$ by a natural isomorphism afforded by the equivalence, it remains to show that $F'(_BB^{\oplus k})\cong (F'(_BB))^{\oplus k}=P^{\oplus k}$. This is a straightforward but fiddly exercise in manipulating diagrams. Finally, notice that $$B\cong \mathrm{End}_B(_BB)^{op}\cong \mathrm{End}_B(F(P))^{op}\cong \mathrm{End}_A(P)^{op},$$ where the first equality is a general fact about rings, the second comes from the equivalence, and the third comes from the fact that any $A$-homomorphism $\rho:F(P)\to F(P)$ has a corresponding $B$-homomorphism $F'(\rho):P\to P$ and vice versa.
($k$ might be infinite) –  Mariano Suárez-Alvarez Jan 17 '12 at 2:06
Edited the question -- $M$ should be finitely-generated. –  Clinton Boys Jan 17 '12 at 2:22