The Stacks project

Example 64.14.7. Let $S = \mathop{\mathrm{Spec}}(\mathbf{Q})$. Let $U = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})$. Let $G = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with obvious action on $U$. Then by construction property $(*)$ of Lemma 64.14.3 holds and we obtain an algebraic space

\[ X = \mathop{\mathrm{Spec}}(\overline{\mathbf{Q}})/G \longrightarrow S = \mathop{\mathrm{Spec}}(\mathbf{Q}). \]

Of course this is totally ridiculous as an approximation of $S$! Namely, by the Artin-Schreier theorem, see [Theorem 17, page 316, JacobsonIII], the only finite subgroups of $\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ are $\{ 1\} $ and the conjugates of the order two group $\text{Gal}(\overline{\mathbf{Q}}/\overline{\mathbf{Q}} \cap \mathbf{R})$. Hence, if $\mathop{\mathrm{Spec}}(k) \to X$ is a morphism with $k$ algebraic over $\mathbf{Q}$, then it follows from Lemma 64.14.6 and the theorem just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to $\overline{\mathbf{Q}} \cap \mathbf{R}$.

Comments (1)

Comment #217 by David Holmes on

Typo: on the first line of the final paragraph, should read 'ridiculous'.

There are also:

  • 9 comment(s) on Section 64.14: Examples of algebraic spaces

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Your email address will not be published.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 02Z6. Beware of the difference between the letter 'O' and the digit '0'.