# Definition of projective cover

In general we define the projective cover a module $M$ over an arbitrary ring $R$ as a surjective $R$-map $f: P \rightarrow M$ such that $\operatorname{ker}(f)$ is superfluous.

I read (if I recall correctly in Lambek's book) that $f$ is a projective cover if and only if $\operatorname{ker}(f) \subset \operatorname{rad}(P)$. However I don't see why, is this always true or do we need additional assumptions on $P$ or $M$?

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thanks, so is this always true no matter if $M$ is finitely generated or not? i.e no further assumptions on $P$ or $M$? –  user6495 Apr 22 '12 at 19:19
The thing that can go wrong is when $rad(M)=M$. (In that case, $rad(M)$ is obviously not superfluous.) However if memory serves, projective modules always have maximal submodules, so $rad(P)\neq P$. –  rschwieb Apr 23 '12 at 1:48