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 $R^iF(-)\cong Ext^i(M,-)$?



Equivalent functors
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.


