The Stacks project

Lemma 65.16.8. 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 65.16.2 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 64.10.2 it is also an open subspace of $X$. By Lemma 65.13.1 we conclude that $X$ is a scheme. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03KX. Beware of the difference between the letter 'O' and the digit '0'.