Lemma 66.16.8. Let $S$ be a scheme. Let $X \to \mathop{\mathrm{Spec}}(k)$ be étale morphism over $S$, where $k$ is a field. Then $X$ is a scheme.

Proof. Let $U$ be an affine scheme, and let $U \to X$ be an étale morphism. By Definition 66.16.2 we see that $U \to \mathop{\mathrm{Spec}}(k)$ is an étale morphism. Hence $U = \coprod _{i = 1, \ldots , n} \mathop{\mathrm{Spec}}(k_ i)$ is a finite disjoint union of spectra of finite separable extensions $k_ i$ of $k$, see Morphisms, Lemma 29.36.7. The $R = U \times _ X U \to U \times _{\mathop{\mathrm{Spec}}(k)} U$ is a monomorphism and $U \times _{\mathop{\mathrm{Spec}}(k)} U$ is also a finite disjoint union of spectra of finite separable extensions of $k$. Hence by Schemes, Lemma 26.23.11 we see that $R$ is similarly a finite disjoint union of spectra of finite separable extensions of $k$. This $U$ and $R$ are affine and both projections $R \to U$ are finite locally free. Hence $U/R$ is a scheme by Groupoids, Proposition 39.23.9. By Spaces, Lemma 65.10.2 it is also an open subspace of $X$. By Lemma 66.13.1 we conclude that $X$ is a scheme. $\square$

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