Projective cover

Remarks

If $p \colon P \to M$ and $p' \colon P' \to M$ are projective covers, there is an isomorphism $f \colon P \to P'$ with $p' \circ f = p$. Example: let $R = \Z/4$ and $p \colon R \to \Z/2$ the reduction map. The module $R$ is free, hence projective, and $\ker p = 2R$. The submodules of $R$ are $0$, $2R$ and $R$, and $0 + 2R$ and $2R + 2R$ are not $R$, so $2R$ is superfluous and $p$ is a projective cover. Non-example: over $R = \Z$, the reduction map $\Z \to \Z/2$ has kernel $2\Z$, and $2\Z + 3\Z = \Z$ with $3\Z \ne \Z$, so the kernel is not superfluous. In fact the $\Z$-module $\Z/2$ has no projective cover at all. Over a local ring $R$ with maximal ideal $\mathfrak m$, the map $R \to R / \mathfrak m$ is a projective cover, by Nakayama's lemma. Bass proved two characterizations. A ring is semiperfect if and only if every finitely generated left module has a projective cover. A ring is left perfect if and only if every left module has a projective cover.