Editing Homological algebra (section)

===Ext functor===
{{Main|Ext functor}}
Let ''R'' be a [[ring (mathematics)|ring]] and let Mod<sub>''R''</sub> be the [[Category (mathematics)|category]] of [[module (mathematics)|modules]] over ''R''. Let ''B'' be in Mod<sub>''R''</sub> and set ''T''(''B'') = Hom<sub>''R''</sub>(''A,B''), for fixed ''A'' in Mod<sub>''R''</sub>. This is a [[left exact functor]] and thus has right [[derived functor]]s ''R<sup>n</sup>T''. The Ext functor is defined by

:<math>\operatorname{Ext}_R^n(A,B)=(R^nT)(B).</math>

This can be calculated by taking any [[injective resolution]]

:<math>0 \rightarrow B \rightarrow I^0 \rightarrow I^1 \rightarrow \cdots, </math>

and computing

:<math>0 \rightarrow \operatorname{Hom}_R(A,I^0) \rightarrow \operatorname{Hom}_R(A,I^1) \rightarrow \cdots.</math>

Then (''R<sup>n</sup>T'')(''B'') is the [[chain complex|cohomology]] of this complex. Note that Hom<sub>''R''</sub>(''A,B'') is excluded from the complex.

An alternative definition is given using the functor ''G''(''A'')=Hom<sub>''R''</sub>(''A,B''). For a fixed module ''B'', this is a [[Covariance and contravariance of functors|contravariant]] [[left exact functor]], and thus we also have right [[derived functor]]s ''R<sup>n</sup>G'', and can define

:<math>\operatorname{Ext}_R^n(A,B)=(R^nG)(A).</math>

This can be calculated by choosing any [[projective resolution]]

:<math>\dots \rightarrow P^1 \rightarrow P^0 \rightarrow A \rightarrow 0, </math>

and proceeding dually by computing

:<math>0\rightarrow\operatorname{Hom}_R(P^0,B)\rightarrow  \operatorname{Hom}_R(P^1,B) \rightarrow \cdots.</math>

Then (''R<sup>n</sup>G'')(''A'') is the cohomology of this complex. Again note that Hom<sub>''R''</sub>(''A,B'') is excluded.

These two constructions turn out to yield [[isomorphic]] results, and so both may be used to calculate the Ext functor.