Injective resolution

Remarks

Example: over $\Z$, the sequence $0 \to \Z \to \Q \to \Q/\Z \to 0$ is an injective resolution of $\Z$. It is exact, and $\Q$ and $\Q/\Z$ are divisible groups, which are injective modules over $\Z$. Non-example: $0 \to \Z \xrightarrow{\cdot 2} \Z \to \Z/2\Z \to 0$ is exact, but $\Z$ is not an injective $\Z$-module, since it is not divisible, so clause 1 fails. One writes the resolution as the cochain complex $I^\bullet$ with the map $\eta$; then $H^0(I^\bullet) \cong X$ and $H^n(I^\bullet) = 0$ for $n > 0$. If $\mathcal{A}$ has enough injectives, every object has an injective resolution; this holds for modules over any ring, since every module embeds into an injective module.