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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 simple $R$-module. If one knows that $H^0(D,M)=0$, is $M=0$? If not, under what further conditions can one show $M=0$?

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

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  • $\begingroup$ If $M=0, M^D=0$, no? I don't understand how one gets the if and only if statement... $\endgroup$ – debanjana May 15 at 3:23
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    $\begingroup$ 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. $\endgroup$ – Alvaro Martinez May 15 at 11:16

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