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In http://en.wikipedia.org/wiki/Ext_functor, under "Interesting examples", the second sentence says:

For $\mathbb{F}_p$ the finite field on $p$ elements, we also have that $H^*(G,M) = \text{Ext}^*_{\mathbb{F}_p[G]}(\mathbb{F}_p, M)$, and it turns out that the group cohomology doesn't depend on the base ring chosen.

It comes right after a sentence that says that $H^*(G,M) = \text{Ext}^*_{\mathbb{Z}[G]}(\mathbb{Z}, M)$.

I'm trying to understand the quoted sentence. The module $M$ must be an $\mathbb{F}_p[G]$-module and not just a $\mathbb{Z}[G]$-module, right? And the resulting cohomology is going to be an $\mathbb{F}_p$-module and not just a $\mathbb{Z}$-module, right? Is it computed by taking an $\mathbb{F}_p[G]$-projective resolution of $\mathbb{F}_p$ (instead of a $\mathbb{Z}[G]$-projective resolution of $\mathbb{Z}$?). What exactly is claimed not to depend on the base ring?

I'd be glad to understand the general situation here.

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up vote 2 down vote accepted

The point is that if $M$ is an $R[G]$-module for some ring $R$, then $H^i(G,M)$, which a priori is equal to $Ext^i(\mathbb Z, M)$ in the category of $\mathbb Z[G]$-modules, is also equal to $Ext^i(R,M)$ in the category of $R[G]$-modules. (In your example $R = \mathbb F_p$.)

This is often useful for converting between interesting contexts where some natural coefficient rings arise, and the general group cohomology machine.

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If $M$ is an $R[G]$-module, does $\text{Ext}^i_{\mathbb{Z}[G]}(\mathbb{Z},M)$ have an $R[G]$-module structure as well? –  user3533 May 9 '13 at 19:22
    
@user3533: Dear user, No, no $G$-action, just an $R$-action (coming from the $R$-action on $M$). Why do you think it would? (After all, cohomology has no $G$-action --- unless you want to give it the trivial action; this could be a helpful way to think in some contexts.) Regards, –  Matt E May 9 '13 at 19:26
    
I typed $R[G]$ instead of $R$ by mistake. I need to think for a minute of how this action comes from the $R$-action on $M$. –  user3533 May 9 '13 at 19:28
    
I see it now. Thank you very much. –  user3533 May 9 '13 at 19:29
    
How does one prove that $\text{Ext}^i(\mathbb{Z},M)$ and $\text{Ext}^i(R,M)$ coincide in the way you mentioned? –  user3533 May 9 '13 at 19:33

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