Lemma 69.17.3. 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 69.5.11 we see that $X \times _ S U$ is a scheme for some $U$. $\square$

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).