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Let $F:\mod A\to \mod B$ be a functor from the category of finitely generated $A$-modules to the finitely generated $B$-modules. It is well-known that if $M$ is an $A-B$ bimodule, then $F(M)\cong M\otimes_A F(A)$. However, I am not able to see what would be the image of $m\otimes n\in M\otimes_A F(A)$ under this isomorphism. Any ideas would help. Thanks!

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It's true for finitely presented modules. Are you assuming that $A$ is noetherian? –  Zhen Lin Sep 28 '13 at 10:37
    
Yes, I am assuming it is noetherian and artenian. –  Gary Sep 28 '13 at 18:50

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Let me show that the claim is true when you replace "finitely generated" with "finitely presented". Since you are interested in the case where $A$ is noetherian, this is enough.

To begin, notice that $F(A^{\oplus n}) \cong F(A)^{\oplus n}$, since $F$ preserves direct sums. (In fact, any additive functor preserves direct sums.) Then, if $M$ is finitely presented, there is a right exact sequence of the form $$A^{\oplus m} \longrightarrow A^{\oplus n} \longrightarrow M \longrightarrow 0$$ and so we have $$F(A)^{\oplus m} \longrightarrow F(A)^{\oplus n} \longrightarrow F(M) \longrightarrow 0$$ because $F$ is right-exact. Thus $F$ is entirely determined by $F(A)$. Now, it is straightforward to see that $F(A)$ is an $A$-module, and ${-} \otimes_A F(A)$ is right exact, so this argument implies that $F$ is isomorphic to ${-} \otimes_A F(A)$ as a functor.

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$F (A)$ is an $A$-module because $A$ acts on itself and $F$ is an additive functor. (Consider the maps $A \to A$ defined by $x \mapsto a x$ for a fixed element $a$.) I already explained how $F$ is determined by $F (A)$. –  Zhen Lin Oct 9 '13 at 14:00

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