The Stacks project

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 $k$.


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A44. Beware of the difference between the letter 'O' and the digit '0'.