# Trivial modules of group rings [closed]

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?

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

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• By Schur’s Lemma, if $M$ is simple but not isomorphic to $\mathbb{F}_p$ then $\text{Hom}_R(\mathbb{F}_p,M)=0$. – Jeremy Rickard May 13 at 12:53
• @JeremyRickard Right, but I want to understand the other way implication. – debanjana May 13 at 15:27
• If $M$ is a semisimple module, yes – YCor May 13 at 15:37
• 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. – user66288 May 13 at 16:08
• @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 $M=0$. – Jeremy Rickard May 13 at 17:26