The Stacks project

Theorem 68.10.2 (David Rydh). Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X$ is decent, then there exists a dense open subspace of $X$ which is a scheme.

Proof. Assume $X$ is a decent algebraic space for which the theorem is false. By Properties of Spaces, Lemma 66.13.1 there exists a largest open subspace $X' \subset X$ which is a scheme. Since $X'$ is not dense in $X$, there exists an open subspace $X'' \subset X$ such that $|X''| \cap |X'| = \emptyset $. Replacing $X$ by $X''$ we get a nonempty decent algebraic space $X$ which does not contain any open subspace which is a scheme.

Choose a nonempty affine scheme $U$ and an ├ętale morphism $U \to X$. We may and do replace $X$ by the open subscheme corresponding to the image of $|U| \to |X|$. Consider the sequence of open subspaces

\[ X = X_0 \supset X_1 \supset X_2 \ldots \]

constructed in Lemma 68.8.2 for the morphism $U \to X$. Note that $X_0 = X_1$ as $U \to X$ is surjective. Let $U = U_0 = U_1 \supset U_2 \ldots $ be the induced sequence of open subschemes of $U$.

Choose a nonempty open affine $V_1 \subset U_1$ (for example $V_1 = U_1$). By induction we will construct a sequence of nonempty affine opens $V_1 \supset V_2 \supset \ldots $ with $V_ n \subset U_ n$. Namely, having constructed $V_1, \ldots , V_{n - 1}$ we can always choose $V_ n$ unless $V_{n - 1} \cap U_ n = \emptyset $. But if $V_{n - 1} \cap U_ n = \emptyset $, then the open subspace $X' \subset X$ with $|X'| = \mathop{\mathrm{Im}}(|V_{n - 1}| \to |X|)$ is contained in $|X| \setminus |X_ n|$. Hence $V_{n - 1} \to X'$ is an ├ętale morphism whose fibres have degree bounded by $n - 1$. In other words, $X'$ is reasonable (by definition), hence $X'$ contains a nonempty open subscheme by Proposition 68.10.1. This is a contradiction which shows that we can pick $V_ n$.

By Limits, Lemma 32.4.3 the limit $V_\infty = \mathop{\mathrm{lim}}\nolimits V_ n$ is a nonempty scheme. Pick a morphism $\mathop{\mathrm{Spec}}(k) \to V_\infty $. The composition $\mathop{\mathrm{Spec}}(k) \to V_\infty \to U \to X$ has image contained in all $X_ d$ by construction. In other words, the fibred $U \times _ X \mathop{\mathrm{Spec}}(k)$ has infinite degree which contradicts the definition of a decent space. This contradiction finishes the proof of the theorem. $\square$

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Your email address will not be published.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 086U. Beware of the difference between the letter 'O' and the digit '0'.