Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 2 down vote accepted

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 $\Lambda$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.