The Stacks project

Proposition 70.16.1. Let $S$ be a scheme. Let $X$ be a quasi-compact and quasi-separated algebraic space over $S$.

  1. There exists a surjective finite morphism $Y \to X$ of finite presentation where $Y$ is a scheme,

  2. given a surjective ├ętale morphism $U \to X$ we may choose $Y \to X$ such that for every $y \in Y$ there is an open neighbourhood $V \subset Y$ such that $V \to X$ factors through $U$.

Proof. Part (1) is the special case of (2) with $U = X$. Let $Y \to X$ be as in Decent Spaces, Lemma 68.9.2. Choose a finite affine open covering $Y = \bigcup V_ j$ such that $V_ j \to X$ factors through $U$. We can write $Y = \mathop{\mathrm{lim}}\nolimits Y_ i$ with $Y_ i \to X$ finite and of finite presentation, see Lemma 70.11.2. For large enough $i$ the algebraic space $Y_ i$ is a scheme, see Lemma 70.5.11. For large enough $i$ we can find affine opens $V_{i, j} \subset Y_ i$ whose inverse image in $Y$ recovers $V_ j$, see Lemma 70.5.7. For even larger $i$ the morphisms $V_ j \to U$ over $X$ come from morphisms $V_{i, j} \to U$ over $X$, see Proposition 70.3.10. This finishes the proof. $\square$

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Your email address will not be published.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 09YC. Beware of the difference between the letter 'O' and the digit '0'.