Example 63.14.7. 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 63.14.3 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 63.14.6 and the theorem just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to $\overline{\mathbf{Q}} \cap \mathbf{R}$.

Comment #217 by David Holmes on

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

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