# The Stacks Project

## Tag 03JH

Lemma 57.12.1. 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 57.4.4 we see that $$\coprod X_i \longrightarrow X'$$ is a surjective map of sheaves on $(\textit{Sch}/S)_{fppf}$. 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 25.15.4 to see that $X'$ is a scheme. $\square$

The code snippet corresponding to this tag is a part of the file spaces-properties.tex and is located in lines 1506–1511 (see updates for more information).

\begin{lemma}
\label{lemma-subscheme}
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.
\end{lemma}

\begin{proof}
Let $U \to X$ be an \'etale 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 \ref{lemma-characterize-surjective}
we see that
$$\coprod X_i \longrightarrow X'$$
is a surjective map of sheaves on $(\Sch/S)_{fppf}$.
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 \ref{schemes-lemma-glue-functors}
to see that $X'$ is a scheme.
\end{proof}

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