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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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    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.2012-04-23