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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.

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    Well, in fact there are two statements which are true: *(i)* the result you claimed originally without the finite generation hypothesis and *(ii)* if a module is finitely generated, then its image under an equivalence like the one you are considering is also finitely generated. It is best if you prove the two things separately :)2012-01-17