[3.1.12, CLO]

Lemma 66.15.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If $X_{red}$ is a scheme, then $X$ is a scheme.

Proof. Let $U' \subset X_{red}$ be an open affine subscheme. Let $U \subset X$ be the open subspace corresponding to the open $|U'| \subset |X_{red}| = |X|$. Then $U' \to U$ is surjective and integral. Hence $U$ is affine by Proposition 66.15.2. Thus every point is contained in an open subscheme of $X$, i.e., $X$ is a scheme. $\square$

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