Theorem 65.10.5. 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 65.10.3 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 39.3.3. According to Groupoids, Lemma 39.20.6 we see that $U/R \cong U'/R'$. By Lemma 65.10.1 $R'$ 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 $U$. 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 65.10.1 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 65.10.2 each $F_ i \to F$ is representable, and an open immersion. By Lemma 65.10.4 applied to $(U_ i, R_ i)$ we see that $F_ i$ is an algebraic space. Then by Lemma 65.10.3 we see that $U_ i \to F_ i$ is étale and surjective. From Lemma 65.8.4 it follows that $\coprod F_ i$ is an algebraic space. Finally, we have verified all hypotheses of Lemma 65.8.5 and it follows that $F = U/R$ is an algebraic space. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (1)

Comment #216 by David Holmes on

There are also: