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.
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.