Lemma 66.15.3. Let $S$ be a scheme. Let $k$ be a field. Let $X$ be an algebraic space over $S$ and assume that there exists a surjective étale morphism $\mathop{\mathrm{Spec}}(k) \to X$. If $X$ is quasi-separated, then $X \cong \mathop{\mathrm{Spec}}(k')$ where $k/k'$ is a finite separable extension.
Proof. Set $R = \mathop{\mathrm{Spec}}(k) \times _ X \mathop{\mathrm{Spec}}(k)$, so that we have a fibre product diagram
\[ \xymatrix{ R \ar[r]_-s \ar[d]_-t & \mathop{\mathrm{Spec}}(k) \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r] & X } \]
By Spaces, Lemma 65.9.1 we know $X = \mathop{\mathrm{Spec}}(k)/R$ is the quotient sheaf. Because $\mathop{\mathrm{Spec}}(k) \to X$ is étale, the morphisms $s$ and $t$ are étale. Hence $R = \coprod _{i \in I} \mathop{\mathrm{Spec}}(k_ i)$ is a disjoint union of spectra of fields, and both $s$ and $t$ induce finite separable field extensions $s, t : k \subset k_ i$, see Morphisms, Lemma 29.36.7. Because
\[ R = \mathop{\mathrm{Spec}}(k) \times _ X \mathop{\mathrm{Spec}}(k) = (\mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k)) \times _{X \times _ S X, \Delta } X \]
and since $\Delta $ is quasi-compact by assumption we conclude that $R \to \mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k)$ is quasi-compact. Hence $R$ is quasi-compact as $\mathop{\mathrm{Spec}}(k) \times _ S \mathop{\mathrm{Spec}}(k)$ is affine. We conclude that $I$ is finite. This implies that $s$ and $t$ are finite locally free morphisms. Hence by Groupoids, Proposition 39.23.9 we conclude that $\mathop{\mathrm{Spec}}(k)/R$ is represented by $\mathop{\mathrm{Spec}}(k')$, with $k' \subset k$ finite locally free where
\[ k' = \{ x \in k \mid s_ i(x) = t_ i(x)\text{ for all }i \in I\} \]
It is easy to see that $k'$ is a field. $\square$
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