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.


1 Answer 1


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 $a(k)\sim\frac{p^a-1}{p^a}k$)

  • $\begingroup$ 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? $\endgroup$
    – frafour
    Mar 3, 2021 at 17:05
  • 1
    $\begingroup$ @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 $H^i(G,M)=\mathrm{Ext}^i_{\mathbb Z[G]}(\mathbb Z,M)$). $\endgroup$
    – user42024
    Mar 3, 2021 at 20:27

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