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I'm reading Benson's Representations and Cohomology I, Section 1.9.

Could someone please clarify to me the following sentence at page 18 lines -7,-6,-5:

"So given a simple $\bar\Lambda$-module $S_j$, it has a projective cover $P_j=\bar Q_j$ for some projective indecomposable $\Lambda$-module $Q_j$ unique up to isomorphism.''

$\hskip250pt$ Thanks in advance!

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1 Answer 1

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I don't have the book available, and the link didn't work for me, so I can't pin down the precise context, but I think I can answer your question:

First let me suppose for a moment that $\Lambda = \overline{\Lambda}$.

Then the statement just becomes the following:

If $S$ is a simple $\Lambda$-module, then it admits a projective cover which is indecomposable.

Why is this? Well, any module admits a projective cover $P \to S$ which is unique up to isomorphism, characterized by the property that if $M$ is any non-zero submodule of $P$, then the induced map $M \to S$ is non-zero. Suppose now that $M$ is a non-zero direct summand of $P$: then $M \to S$ is non-zero, and hence surjective (because $S$ is simple). Also, $M$ is projective, being a summand of a projective module. Thus $M$ is also a projective cover of $S$, and slightly more argument (essentially, the same arguments that show uniqueness of projective covers) shows that in fact $M = P.$ This proves the indecomposability of $P$.

Returing to your context: since $S$ is a simple $\overline{\Lambda}$-module, it is also a simple $\Lambda$-module. If we let $Q$ denote its projective cover as a $\Lambda$-module, then I think that $\overline{Q}:= \overline{\Lambda}\otimes_{\Lambda} Q$ will be its projective cover as an $\overline{\Lambda}$-module. (This just uses the surjectivity of $\Lambda \to \overline{\Lambda}$.) The preceding argument shows that $Q$ is indecomposable, and $Q$ is unique up to isomorphism, since it is the projective cover of $S$ over $\Lambda$.

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Thanks for your clarifying answer. One little thing: I think the property characterizing projective covers is that the only submodule mapping surjectively is the whole projective module. This simplifies your argument. Thanks again! –  Fred Aug 18 '12 at 4:07

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