# The Stacks Project

## Tag 02Z6

Example 56.14.7. Let $S = \mathop{\rm Spec}(\mathbf{Q})$. Let $U = \mathop{\rm Spec}(\overline{\mathbf{Q}})$. Let $G = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with obvious action on $U$. Then by construction property $(*)$ of Lemma 56.14.3 holds and we obtain an algebraic space $$X = \mathop{\rm Spec}(\overline{\mathbf{Q}})/G \longrightarrow S = \mathop{\rm Spec}(\mathbf{Q}).$$ Of course this is totally ridiculous as an approximation of $S$! Namely, by the Artin-Schreier theorem, see [JacobsonIII, Theorem 17, page 316], 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 $\mathop{\rm Spec}(k) \to X$ is a morphism with $k$ algebraic over $\mathbf{Q}$, then it follows from Lemma 56.14.6 and the theorem just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to $\overline{\mathbf{Q}} \cap \mathbf{R}$.

The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 2378–2401 (see updates for more information).

\begin{example}
\label{example-Qbar}
Let $S = \Spec(\mathbf{Q})$.
Let $U = \Spec(\overline{\mathbf{Q}})$.
Let $G = \text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})$ with obvious
action on $U$. Then by construction property $(*)$ of
Lemma \ref{lemma-quotient} holds and we obtain an algebraic space
$$X = \Spec(\overline{\mathbf{Q}})/G \longrightarrow S = \Spec(\mathbf{Q}).$$
Of course this is totally ridiculous as an approximation of $S$!
Namely, by the Artin-Schreier theorem,
see \cite[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
$\Spec(k) \to X$ is a morphism with $k$ algebraic over $\mathbf{Q}$,
then it follows from Lemma \ref{lemma-quotient-field-map} and the theorem
just mentioned that either $k$ is $\overline{\mathbf{Q}}$ or isomorphic to
$\overline{\mathbf{Q}} \cap \mathbf{R}$.
\end{example}

Comment #217 by David Holmes on May 17, 2013 a 4:41 pm UTC

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

There are also 7 comments on Section 56.14: Algebraic Spaces.

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