Projective resolution

Remarks

Every module has a free resolution: take a free module $F_0$ mapping onto $M$, then a free module mapping onto the kernel, and repeat. Example: over $\Z$, the sequence $0 \to \Z \xrightarrow{\cdot n} \Z \to \Z/n\Z \to 0$ is a free resolution of length $1$, since multiplication by $n \ne 0$ is injective and its image $n\Z$ is the kernel of the reduction map. Over $R = \Z/4\Z$, the module $\Z/2\Z$ has the free resolution $\cdots \to R \xrightarrow{\cdot 2} R \xrightarrow{\cdot 2} R \to \Z/2\Z \to 0$, which never stops. Non-example: $0 \to \Z \xrightarrow{\cdot 4} \Z \to \Z/2\Z \to 0$ is not a resolution, because it is not exact at the middle $\Z$. The kernel of the reduction is $2\Z$, but the image of multiplication by $4$ is $4\Z$. One often writes the resolution as the complex $P_\bullet$ together with the map $\varepsilon \colon P_0 \to M$; then $H_0(P_\bullet) \cong M$ and $H_i(P_\bullet) = 0$ for $i > 0$.