most recent 30 from http://mathoverflow.net 2013-05-21T01:05:14Z http://mathoverflow.net/feeds/question/74371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74371/equivalent-functors minhtringuyen 2011-09-02T16:01:37Z 2011-09-02T16:48:46Z

<p>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 $R^iF(-)\cong Ext^i(M,-)$?</p>

http://mathoverflow.net/questions/74371/equivalent-functors/74378#74378 Andreas Blass 2011-09-02T16:48:46Z 2011-09-02T16:48:46Z

<p>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.</p>