## Tag `02Z6`

Chapter 56: Algebraic Spaces > Section 56.14: Examples of algebraic spaces

Example 56.14.7. 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 56.14.3 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 [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 $\mathop{\mathrm{Spec}}(k) \to X$ 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}
```

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