# trivial modules of group rings

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

• A simple module is never $0$, so you might have a problem there. – Captain Lama May 2 at 5:27
• What I mean to say is that $M$ does not have any non-trivial sub-modules but might be 0. – debanjana May 2 at 6:06

## 1 Answer

By Maschke's theorem any $$\mathbb{F}_p(D)$$-module $$M$$ is a direct sum of simple modules. The condition $$H^0(D,M)=0$$ only says that $$M^D=0$$, that is, none of these simple summands is trivial. In your case $$M$$ has no nontrivial submodules so the submodule $$H^0(D,M)=M^D$$ is either $$M^D=M$$ or $$M^D=0$$. In the former case $$M$$ is the trivial module or zero. In the latter case (your case) $$M$$ could be zero, but it could also be any nontrivial simple module, so one cannot deduce that $$M=0$$.

• If $M=0, M^D=0$, no? I don't understand how one gets the if and only if statement... – debanjana May 15 at 3:23
• I gave some more details, hope it makes sense to you now. An easy counterexample would be $p=3$, $D=C_2$, and the sign representation $M$, which has $H^0(D,M)=0$ despite $M$ being nonzero. – Alvaro Martinez May 15 at 11:16