Lemma 66.13.2. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If there exists a finite, étale, surjective morphism $U \to X$ where $U$ is a quasi-separated scheme, then there exists a dense open subspace $X'$ of $X$ which is a scheme. More precisely, every point $x \in |X|$ of codimension $0$ in $X$ is contained in $X'$.
Proof. Let $X' \subset X$ be the maximal open subspace which is a scheme (Lemma 66.13.1). Let $x \in |X|$ be a point of codimension $0$ on $X$. By Lemma 66.11.2 it suffices to show $x \in X'$. Let $U \to X$ be as in the statement of the lemma. Write $R = U \times _ X U$ and denote $s, t : R \to U$ the projections as usual. Note that $s, t$ are surjective, finite and étale. By Lemma 66.6.7 the fibre of $|U| \to |X|$ over $x$ is finite, say $\{ \eta _1, \ldots , \eta _ n\} $. By Lemma 66.11.1 each $\eta _ i$ is the generic point of an irreducible component of $U$. By Properties, Lemma 28.29.1 we can find an affine open $W \subset U$ containing $\{ \eta _1, \ldots , \eta _ n\} $ (this is where we use that $U$ is quasi-separated). By Groupoids, Lemma 39.24.1 we may assume that $W$ is $R$-invariant. Since $W \subset U$ is an $R$-invariant affine open, the restriction $R_ W$ of $R$ to $W$ equals $R_ W = s^{-1}(W) = t^{-1}(W)$ (see Groupoids, Definition 39.19.1 and discussion following it). In particular the maps $R_ W \to W$ are finite étale also. It follows that $R_ W$ is affine. Thus we see that $W/R_ W$ is a scheme, by Groupoids, Proposition 39.23.9. On the other hand, $W/R_ W$ is an open subspace of $X$ by Spaces, Lemma 65.10.2 and it contains $x$ by construction. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: