# Cohomology of finite $p$-groups over integers in local fields

Let $$p$$ be a prime, $$G$$ be a finite group of order $$p^a$$. Let $$M$$ be a $$\mathbb{Z}[G]$$-module. Then $$H^n(G, M)$$ is annihilated by $$p^a$$ for all $$n \geq 1$$ (see e.g. Brown, Corollary III.10.2).

In particular this is true for $$\mathbb{Z}_p[G]$$-modules, and I was wondering if this version of the result can be generalized to local fields other than $$\mathbb{Q}_p$$. More precisely, let $$\mathbb{K}$$ be a non-Archimedean local field with ring of integers $$\mathfrak{o}$$, uniformizer $$\omega$$, maximal ideal $$\mathfrak{p}$$, residue field $$\mathfrak{k}$$ of characteristic $$p$$. Let $$G$$ be a finite $$p$$-group and $$M$$ an $$\mathfrak{o}[G]$$-module. What is the least $$a$$ such that $$H^n(G, M)$$ is annihilated by $$\omega^a$$?

It should be easy to use the statement at the beginning of this post to generalize this to any local field of characteristic $$0$$, and $$a$$ should still be uniformly bounded, with the bound depending only on the order of $$\mathfrak{k}$$ and the order of $$G$$. But what about characteristic $$p$$? This case seems to be substantially different.

I think this question is interesting in general, but there is probably no chance to get a general answer without assuming anything extra. So let me point out the specific setting that I am interested in:

1. $$n = 1$$ and especially $$n = 2$$: this comes up in an extension problem so I only really care about these degrees.

2. $$M$$ is a free $$\mathfrak{o}/\mathfrak{p}^k$$-module. So I already know that $$H^n(G, M)$$ is annihilated by $$\omega^k$$, that is $$a \leq k$$. But I would like to get better estimates, ideally I would like to find a bound of the form $$a \leq a(k)$$ such that $$(k - a(k)) \to \infty$$ as $$k \to \infty$$. In characteristic $$0$$ I can get $$a(k)$$ to be a constant so this is a significant weakening, but still somewhat strong.

3. Even though I am interested in the general case, I have trouble understanding even the first example: $$G = \mathbb{Z}/2\mathbb{Z}$$ and $$\mathbb{K} = \mathbb{F}_2((X))$$, so $$\omega = X$$ and $$\mathfrak{o} = \mathbb{F}_2[[X]]$$. Explicitly: if $$M$$ is a free $$\mathbb{F}_2[X]/X^k$$-module with an action of $$G$$, what is the least $$a$$ such that $$X^a H^n(G, M) = 0$$ for $$n = 1, 2$$? After trying a bit I think that one cannot do better than $$a(k) = k/2$$ (see the previous point), but I am not able to prove that this works.

If you take $$G=\mathbb Z/p^a\mathbb Z$$, then the cohomology of any $$M$$ are computed by the Tate complex $$M\xrightarrow{\sigma-\mathrm{id}} M \xrightarrow{1+\sigma+\ldots+ \sigma^{p^a-1} }M\xrightarrow{\sigma-\mathrm{id}} M \xrightarrow{1+\sigma+\ldots+ \sigma^{p^a-1} }M\xrightarrow{\sigma-\mathrm{id}}\ldots$$. In particular if the action on $$M$$ is trivial then the norm map $$1+\sigma+\ldots+ \sigma^{p^a-1}$$ is given by multiplication by $$p^a$$ and $$\sigma-\mathrm{id}=0$$. If $$M$$ is of characteristic $$p$$ then $$H^{i}(G,M)\simeq M$$ for all $$i$$. Note that if $$M$$ had a structure of an $$\mathfrak o$$-module this is an isomorphism of $$\mathfrak o$$-modules (I assume here that $$G$$ acts $$\mathfrak o$$-linearly): this shows that in general $$a(k)=k$$. This is also true for any $$G$$: you can argue inductively by considering lower central series and the Hochschield-Serre spectral sequence to reduce to the case of products of cyclic groups (where you can use Kunneth formula). In more detail the $$E^{0,1}$$ term there is never hit by a differential and thus by induction $$H^1(G, M)$$ naturally contains $$M$$.
Allowing $$M$$ (and $$k$$) vary for a fixed $$G$$ you can always make $$a(k)$$=0. Indeed in the case of a cyclic group $$\mathbb Z/p^a\mathbb Z$$ we can take $$M=\mathfrak o/\omega^{i\cdot p^a}$$ and the action of $$\sigma$$ by $$1+\omega^i$$. Then $$\sigma^{p^a}=1$$, so we have an action and $$1+\sigma+\ldots \sigma^{p^a-1}=\frac{(1+\omega^i)^{p^a}-1}{\omega^i}=\omega^{i\cdot (p^a-1)}$$, while $$1-\sigma=\omega^i$$. Looking at the Tate complex we get that $$H^{>0}(G,M)=0$$. Then for any $$G$$ you can take the a cyclic subgroup $$H$$ in the center $$Z(G)\subset G$$ and induce the above representation to $$G$$. In fact you could also induce from the trivial subgroup right away (in other words just take the regular module $$M=\mathfrak o[G]$$). The above construction might be slightly better because there the annulating power of $$\omega$$ for $$H^0$$ also decreases (so if you also include $$H^0$$ in the definition of $$a(k)$$ you still get $$a(k)\sim\frac{p^a-1}{p^a}k$$)
• Thank you very much for the answer! To be sure: $\sigma$ is the automorphism given by the action of the generator of $G$? And $1$ is the identity map? Also could you give a reference for the first statement, the fact that cohomology is computed by this resolution? Mar 3, 2021 at 17:05
• @frafour Yep that's right $\sigma$ is the action of a chosen generator and $1=\mathrm{id}$ is the identity map. I am not sure about the precise reference (probably anything on Tate cohomology) but this just follows from the fact that the analogous complex $\ldots \xrightarrow{1+\sigma +\ldots } \mathbb Z[G] \xrightarrow{\sigma - \mathrm{id}} \mathbb Z[G]$ is a free resolution of trivial module $\mathbb Z$ as a $\mathbb Z[G]$-module (plus the formula $H^i(G,M)=\mathrm{Ext}^i_{\mathbb Z[G]}(\mathbb Z,M)$). Mar 3, 2021 at 20:27