Ext functor

Remarks

Up to a canonical isomorphism, $\operatorname{Ext}^n_R(M, N)$ does not depend on the choice of the resolution. One has $\operatorname{Ext}^0_R(M, N) \cong \Hom_R(M, N)$, and the same groups are obtained from an injective resolution of $N$. If $R$ is commutative, the Ext groups are $R$-modules. Example: from $0 \to \Z \xrightarrow{\cdot n} \Z \to \Z/n\Z \to 0$ with $n \ge 1$, the complex is $\Z \xrightarrow{\cdot n} \Z$, so $\operatorname{Ext}^0_\Z(\Z/n\Z, \Z) = 0$ and $\operatorname{Ext}^1_\Z(\Z/n\Z, \Z) \cong \Z/n\Z$. Non-example: the sequence $0 \to \Z/n\Z \xrightarrow{\id} \Z/n\Z \to 0$ is exact but $\Z/n\Z$ is not projective over $\Z$ for $n \ge 2$; using it would give $\operatorname{Ext}^1 = 0$, which is wrong. A module $M$ is projective if and only if $\operatorname{Ext}^1_R(M, N) = 0$ for all $R$-modules $N$.