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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