Lemma 70.17.3 (0B80)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 70: Limits of Algebraic Spaces
Section 70.17: Obtaining schemes
Lemma 70.17.3 (cite)

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Lemma 70.17.3. Let $f: X \to S$ be a quasi-compact and quasi-separated morphism from an algebraic space to a scheme $S$ . If for every $x \in |X|$ with image $s = f(x) \in S$ the algebraic space $X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S,s})$ is a scheme, then $X$ is a scheme.

Proof. Let $x \in |X|$ . It suffices to find an open neighbourhood $U$ of $s = f(x)$ such that $X \times _ S U$ is a scheme. As $X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s})$ is a scheme, then, since $\mathcal{O}_{S, s} = \mathop{\mathrm{colim}}\nolimits \mathcal{O}_ S(U)$ where the colimit is over affine open neighbourhoods of $s$ in $S$ we see that

$X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}_{S, s}) = \mathop{\mathrm{lim}}\nolimits X \times _ S U$

By Lemma 70.5.11 we see that $X \times _ S U$ is a scheme for some $U$ . $\square$

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