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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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You need to switch "ker(f) essential" with "ker(f) superfluous".

The radical of a projective module is its largest superfluous submodule.

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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
    
Ah sorry if you meant is the original post true, then sure. The superfluous kernel will have to be contained in the largest superfluous submodule rad(P), and any submodule of rad(P) is superfluous. –  rschwieb Apr 23 '12 at 2:41

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