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Let $R=\mathbb{F}_p[D]$ where $D$ is a finite group of order prime to $p$. Let $M$ be any finitely generated (left) $R$-module. If one knows that $\textrm{Hom}_R(\mathbb{F}_p,M)=0$, can one show $M=0$? If not, under what further conditions can one show $M=0$?

$R$ is semi simple by Maschke's theorem and breaks up as

$$R= \oplus_i R_i ,$$ where each $R_i=e_i R$ is a simple ring. Here each $e_i$ is an idempotent.

Similarly, any finitely generated (left) $R$-module $M$ breaks up as

$$M= \oplus_i M_i , $$ where each $M_i= e_i M$ is a simple $R$-module.

We are then reduced to the case where we can look at each component individually.

Suppose we further know that $H^0(D,M)=0$. Therefore $H^0(D, M_1) := \textrm {Hom}_R(\mathbb{F}_p, M_1) = 0$.

We have in this case that the action of $R$ is through $R_1$.

Question: I was hoping there might be a way to understand the condition $H^0(D,M)=0$ in terms of the idempotents?

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closed as off-topic by Jeremy Rickard, Andreas Blass, abx, user44191, LSpice May 14 at 11:30

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    $\begingroup$ By Schur’s Lemma, if $M$ is simple but not isomorphic to $\mathbb{F}_p$ then $\text{Hom}_R(\mathbb{F}_p,M)=0$. $\endgroup$ – Jeremy Rickard May 13 at 12:53
  • $\begingroup$ @JeremyRickard Right, but I want to understand the other way implication. $\endgroup$ – debanjana May 13 at 15:27
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    $\begingroup$ If $M$ is a semisimple module, yes $\endgroup$ – YCor May 13 at 15:37
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    $\begingroup$ You have the hom of $M$ (f.g.) equals zero iff none of its direct summands is isomorphic to $\mathbb{F}_p$. I think that is what Jeremy Rickard said. $\endgroup$ – user66288 May 13 at 16:08
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    $\begingroup$ @debanjana I think my comment addresses the implication you want to understand: no, if one knows that $\text{Hom}_R(\mathbb{F}_p,M)=0$, one can’t show $M=0$. $\endgroup$ – Jeremy Rickard May 13 at 17:26