I have a basic question about the meaning of Tate twists in étale cohomology.
I want the understand a statement of the form $$H^1(U,\Lambda) \cong \Lambda(-1)$$ in which $U$ is the spectrum of a localization of a regular, strictly henselian local ring $A$ – I don't think giving more details would be useful – and $\Lambda = \mathbb{Z}/\ell^n \mathbb{Z}$ for some prime number $\ell$ which is invertible in $A$ (and some positive integer $n$).
I don't know how to "read" this statement. I'm well aware of the existence of the étale sheaves $\Lambda(1)$, $\Lambda(-1)$, … and I know how they are defined. But $H^1(U,\Lambda)$ is a cohomology group (or $\Lambda$-module), not a sheaf. So, what is $\Lambda(-1)$ in the above equality? At first I thought that this would mean that the underlying group is precisely $\Lambda$ equipped with some Galois action – but of which Galois group? there is no obvious one – which explains the "(-1)". But I guess this is wrong…
Also, how is the above equality related to the statement $$H^1(U,\mu_{\ell^n}) = H^1(U,\Lambda(1)) \cong \Lambda?$$ This is a statement which I think I understand; $\Lambda(1)$ is a perfectly honest étale sheaf, you can consider étale cohomology with coefficients in this sheaf and get the group $\Lambda$ as your output. However, there must be some subtle differences between these two isomorphisms…?
Any help would be greatly appreciated.
Best Answer
You are right that $H^1(U, \Lambda)$ should be interpreted as a group plus a Galois action.
Notation: $U$ is the spectrum of the henselian local ring $(A, \mathfrak{m}, k)$.
Lemma: The category of étale covers of $U$ is naturally equivalent to the category of étale covers of $Spec(k)$.
The only place I've seen this spelled out in detail is in Lemma 10.143.8 of the Stacks project. The functor is the natural one given by moding out $S\mapsto S/\mathfrak{m}S$.
This induces an isomorphism in étale cohomology $$H^i(U,\Lambda)\stackrel{\sim}{\to}H^i(Spec(k), \Lambda).$$ I think this is the key confusing abuse of notation in the question.
Now we have a canonical identification by taking the stalk of $\Lambda$ at the geometric point with standard Galois cohomology $$H^i(Spec(k),\Lambda)\stackrel{\sim}{\to}H^i(Gal(k^{sep}/k), \Lambda).$$
Since this latter group is Galois cohomology, it has a natural Galois action on it (in fact it is the one I wrote in the comment and can already be read off without identifying the étale cohomology of $Spec(k)$ with Galois cohomology).