The Stacks Project

Tag 02WW

Theorem 56.10.5. Let $S$ be a scheme. Let $U$ be a scheme over $S$. Let $j = (s, t) : R \to U \times_S U$ be an étale equivalence relation on $U$ over $S$. Then the quotient $U/R$ is an algebraic space, and $U \to U/R$ is étale and surjective, in other words $(U, R, U \to U/R)$ is a presentation of $U/R$.

Proof. By Lemma 56.10.3 it suffices to prove that $U/R$ is an algebraic space. Let $U' \to U$ be a surjective, étale morphism. Then $\{U' \to U\}$ is in particular an fppf covering. Let $R'$ be the restriction of $R$ to $U'$, see Groupoids, Definition 38.3.3. According to Groupoids, Lemma 38.20.6 we see that $U/R \cong U'/R'$. By Lemma 56.10.1 $R'$ is an étale equivalence relation on $U'$. Thus we may replace $U$ by $U'$.

We apply the previous remark to $U' = \coprod U_i$, where $U = \bigcup U_i$ is an affine open covering of $S$. Hence we may and do assume that $U = \coprod U_i$ where each $U_i$ is an affine scheme.

Consider the restriction $R_i$ of $R$ to $U_i$. By Lemma 56.10.1 this is an étale equivalence relation. Set $F_i = U_i/R_i$ and $F = U/R$. It is clear that $\coprod F_i \to F$ is surjective. By Lemma 56.10.2 each $F_i \to F$ is representable, and an open immersion. By Lemma 56.10.4 applied to $(U_i, R_i)$ we see that $F_i$ is an algebraic space. Then by Lemma 56.10.3 we see that $U_i \to F_i$ is étale and surjective. From Lemma 56.8.3 it follows that $\coprod F_i$ is an algebraic space. Finally, we have verified all hypotheses of Lemma 56.8.4 and it follows that $F = U/R$ is an algebraic space. $\square$

The code snippet corresponding to this tag is a part of the file spaces.tex and is located in lines 1571–1579 (see updates for more information).

\begin{theorem}
\label{theorem-presentation}
Let $S$ be a scheme. Let $U$ be a scheme over $S$.
Let $j = (s, t) : R \to U \times_S U$
be an \'etale equivalence relation on $U$ over $S$.
Then the quotient $U/R$ is an algebraic space,
and $U \to U/R$ is \'etale and surjective, in other words
$(U, R, U \to U/R)$ is a presentation of $U/R$.
\end{theorem}

\begin{proof}
By Lemma \ref{lemma-when-it-works-it-works}
it suffices to prove that $U/R$ is an algebraic space.
Let $U' \to U$ be a surjective, \'etale morphism.
Then $\{U' \to U\}$ is in particular an fppf covering.
Let $R'$ be the restriction of $R$ to $U'$, see
Groupoids, Definition \ref{groupoids-definition-restrict-relation}.
According to
Groupoids, Lemma \ref{groupoids-lemma-quotient-groupoid-restrict}
we see that $U/R \cong U'/R'$.
By Lemma \ref{lemma-pullback-etale-equivalence-relation} $R'$ is an
\'etale equivalence relation on $U'$. Thus we may replace $U$ by $U'$.

\medskip\noindent
We apply the previous remark to $U' = \coprod U_i$, where
$U = \bigcup U_i$ is an affine open covering of $S$. Hence we
may and do assume that $U = \coprod U_i$ where
each $U_i$ is an affine scheme.

\medskip\noindent
Consider the restriction $R_i$ of $R$ to $U_i$.
By Lemma \ref{lemma-pullback-etale-equivalence-relation}
this is an \'etale equivalence relation.
Set $F_i = U_i/R_i$ and $F = U/R$.
It is clear that $\coprod F_i \to F$ is surjective.
By Lemma \ref{lemma-finding-opens} each $F_i \to F$
is representable, and an open immersion.
By Lemma \ref{lemma-presentation-quasi-compact}
applied to $(U_i, R_i)$ we see that $F_i$ is an algebraic space.
Then by Lemma \ref{lemma-when-it-works-it-works} we see that
$U_i \to F_i$ is \'etale and surjective.
From Lemma \ref{lemma-coproduct-algebraic-spaces}
it follows that $\coprod F_i$ is an algebraic space.
Finally, we have verified all
hypotheses of Lemma \ref{lemma-glueing-algebraic-spaces}
and it follows that $F = U/R$ is an algebraic space.
\end{proof}

Comment #216 by David Holmes on May 17, 2013 a 4:39 pm UTC

Typo: in the first line of the proof, 'suffice' should read 'suffices'.

There are also 6 comments on Section 56.10: Algebraic Spaces.

Add a comment on tag 02WW

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 lower-right corner).