Coefficients in Tate cohomology, mod-p Tate cohomology vs integral Tate cohomology I am new to group cohomology and Tate cohomology.
I have some questions in that regard. I have not yet understood exactly what information we hope to gain from the (Tate) cohomology modules.
i) What is the significance in studying $\hat{H}^{*}(G,\mathbb{Z})$ versus $\hat{H}^{*}(G,M$), where M is some $\mathbb{Z}G$-module?
ii) How is cohomology with integral coefficients related to mod-p cohomology? That is what information does one gain by studying $\hat{H}^*(G,\mathbb{Z})$ versus $\hat{H}^*(G, \mathbb{F}_p)$?
iii) Are there differences/similarities between the rings $\hat{H}^*(G, \mathbb{F}_p)$ and $\hat{H}^*(G, \mathbb{Z}/p\mathbb{Z})$, where the former is computed with $\mathbb{F}_p$ regarded as a trivial $\mathbb{F}_pG$-module, and the latter is computed by viewing $\mathbb{Z}/p\mathbb{Z}$ as a trivial $\mathbb{Z}G$-module? Are these exactly the same?
I would also be interested in examples, need not be too detailed, that would illustrate what changes when one changes coefficients like above.
 A: Of course, I couldn't give a complete answer, but I can give a few remarks :


*

*Well, of course, $\mathbb{Z}$ is a $\mathbb{Z}[G]$-module, so you can say that it makes sense to study first a specific, not too complicated module, before considering general ones. Especially since it is a module that appears a lot in mathematics ; for instance, when you study Pontryagin duality, you use $\hat{H}^{n+1}(G,\mathbb{Z}) = \hat{H}^n(G,\mathbb{Q}/\mathbb{Z})$ a lot. The relation $Hom_G(\mathbb{Z},M) = M^G$ makes $G$-morphisms $\mathbb{Z}\rightarrow M$ very interesting, and they induce maps $\hat{H}^n(G,\mathbb{Z})\rightarrow \hat{H}^n(G,M)$. It also has some interesting properties in itself : it is torsion-free (so for instance $H^1(G,\mathbb{Z})= 0$) ; it satisfies $\mathbb{Z}\otimes M = M$ for any $G$-module $M$ (very useful for cup-products because you get $\hat{H}^p(G,\mathbb{Z})\times \hat{H}^q(G,M)\rightarrow \hat{H}^{p+q}(G,M)$), and in particular $\mathbb{Z}\otimes\mathbb{Z} = \mathbb{Z}$ which implies that $\hat{H}^*(G,\mathbb{Z})$ is a graded ring.

*You mention yourself the second point : it reduces modulo $p$, so understanding $\hat{H}^n(G,\mathbb{Z})$ gives information on all of the $\hat{H}^n(G,\mathbb{F}_p)$, since there is the reduction map $\hat{H}^n(G,\mathbb{Z})\rightarrow \hat{H}^n(G,\mathbb{F}_p)$. You even get from $0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{F}_p\rightarrow 0$ the exact sequence $0\rightarrow \hat{H}^n(G,\mathbb{Z})/p\hat{H}^n(G,\mathbb{Z}) \rightarrow \hat{H}^n(G,\mathbb{F}_p)\rightarrow \hat{H}^{n+1}(G,\mathbb{Z})[p]\rightarrow 0$ which gives lots of information. The point is that $\hat{H}^*(G,\mathbb{Z})$ simultaneously controls (at least partially) all the $\hat{H}^*(G,\mathbb{F}_p)$.

*There is (unless I'm mistaken) no difference between the two objects you describe, because the category of $\mathbb{F}_p[G]$-modules is the same thing as the category of $\mathbb{Z}[G]$-module that are a $p$-torsion groups. So computing an injective resolution of $\mathbb{Z}/p\mathbb{Z}$ in either category makes no difference, and $Ext_{\mathbb{Z}[G]}^n(\mathbb{Z},\mathbb{Z}/p\mathbb{Z})$ is the same as $Ext_{\mathbb{F}_p[G]}^n(\mathbb{F}_p,\mathbb{Z}/p\mathbb{Z})$ (and likewise for $Tor$).
