# Equivalent functors

Let $R$ be a commutative Noetherian ring, $M$ is a finitely generated $R$-module. If $F: Mod \to Mod$ is a left exact functor and $R^iF(E)=0$ where $E$ is injective module. Assume that $F(-) \cong Hom(M,-)$, can we infer the $i-th$ right devired functors $R^iF(-)\cong Ext^i(M,-)$?

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You can define $Ext$ by taking derived functors of $Hom$ in either variable. If you are using the second variable, it's tautologically true (see Andreas Blass' answer). So I suspect you might be defining it via a projective resolution of $M$ (or via Yoneda or...) Then there is something to prove, but this can be found in many homological algebra texts. I'm sure someone can elaborate, but it would be good to clarify this. –  Donu Arapura Sep 2 '11 at 17:17

Yes. For example, if you compute right derived functors by injective resolutions, then naturality of the isomorphism between $F$ and $\text{Hom}(M,-)$ will ensure that you have an isomorphism between the two complexes whose cohomology groups give you the two derived functors.