Remark 59.69.5. Let $k$ be an algebraically closed field. Let $n$ be an integer prime to the characteristic of $k$. Recall that

$\mathbf{G}_{m, k} = \mathbf{A}^1_ k \setminus \{ 0\} = \mathbf{P}^1_ k \setminus \{ 0, \infty \}$

We claim there is a canonical isomorphism

$H^1_{\acute{e}tale}(\mathbf{G}_{m, k}, \mu _ n) = \mathbf{Z}/n\mathbf{Z}$

What does this mean? This means there is an element $1_ k$ in $H^1_{\acute{e}tale}(\mathbf{G}_{m, k}, \mu _ n)$ such that for every morphism $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ the pullback map on étale cohomology for the map $\mathbf{G}_{m, k'} \to \mathbf{G}_{m, k}$ maps $1_ k$ to $1_{k'}$. (In particular this element is fixed under all automorphisms of $k$.) To see this, consider the $\mu _{n, \mathbf{Z}}$-torsor $\mathbf{G}_{m, \mathbf{Z}} \to \mathbf{G}_{m, \mathbf{Z}}$, $x \mapsto x^ n$. By the identification of torsors with first cohomology, this pulls back to give our canonical elements $1_ k$. Twisting back we see that there are canonical identifications

$H^1_{\acute{e}tale}(\mathbf{G}_{m, k}, \mathbf{Z}/n\mathbf{Z}) = \mathop{\mathrm{Hom}}\nolimits (\mu _ n(k), \mathbf{Z}/n\mathbf{Z}),$

i.e., these isomorphisms are compatible with respect to maps of algebraically closed fields, in particular with respect to automorphisms of $k$.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).