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Tag 02Z6

Chapter 53: Algebraic Spaces > Section 53.14: Examples of algebraic spaces

Example 53.14.7. 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 53.14.3 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 53.14.6 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}

    Comments (1)

    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 53.14: Algebraic Spaces.

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