Lemma 66.13.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. There exists a largest open subspace $X' \subset X$ which is a scheme.

Proof. Let $U \to X$ be an étale surjective morphism, where $U$ is a scheme. Let $R = U \times _ X U$. The open subspaces of $X$ correspond $1 - 1$ with open subschemes of $U$ which are $R$-invariant. Hence there is a set of them. Let $X_ i$, $i \in I$ be the set of open subspaces of $X$ which are schemes, i.e., are representable. Consider the open subspace $X' \subset X$ whose underlying set of points is the open $\bigcup |X_ i|$ of $|X|$. By Lemma 66.4.4 we see that

$\coprod X_ i \longrightarrow X'$

is a surjective map of sheaves on $(\mathit{Sch}/S)_{fppf}$. But since each $X_ i \to X'$ is representable by open immersions we see that in fact the map is surjective in the Zariski topology. Namely, if $T \to X'$ is a morphism from a scheme into $X'$, then $X_ i \times _{X'} T$ is an open subscheme of $T$. Hence we can apply Schemes, Lemma 26.15.4 to see that $X'$ is a scheme. $\square$

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