Theorem 66.10.2 (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 64.13.1 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 étale 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 66.8.2 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 étale 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 66.10.1. 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. $\square$

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