Example 65.14.7. 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 65.14.3 holds and we obtain an algebraic space
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 65.14.6 and the theorem just mentioned that either k is \overline{\mathbf{Q}} or isomorphic to \overline{\mathbf{Q}} \cap \mathbf{R}.
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Comment #217 by David Holmes on
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