The Stacks project

Lemma 66.34.1. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $G$ be an abstract group with a free action on $X$. Then the quotient sheaf $X/G$ is an algebraic space.

Proof. The statement means that the sheaf $F$ associated to the presheaf

\[ T \longmapsto X(T)/G \]

is an algebraic space. To see this we will construct a presentation. Namely, choose a scheme $U$ and a surjective étale morphism $\varphi : U \to X$. Set $V = \coprod _{g \in G} U$ and set $\psi : V \to X$ equal to $a(g) \circ \varphi $ on the component corresponding to $g \in G$. Let $G$ act on $V$ by permuting the components, i.e., $g_0 \in G$ maps the component corresponding to $g$ to the component corresponding to $g_0g$ via the identity morphism of $U$. Then $\psi $ is a $G$-equivariant morphism, i.e., we reduce to the case dealt with in the next paragraph.

Assume that there exists a $G$-action on $U$ and that $U \to X$ is surjective, étale and $G$-equivariant. In this case there is an induced action of $G$ on $R = U \times _ X U$ compatible with the projection mappings $t, s : R \to U$. Now we claim that

\[ X/G = U/\coprod \nolimits _{g \in G} R \]

where the map

\[ j : \coprod \nolimits _{g \in G} R \longrightarrow U \times _ S U \]

is given by $(r, g) \mapsto (t(r), g(s(r)))$. Note that $j$ is a monomorphism: If $(t(r), g(s(r))) = (t(r'), g'(s(r')))$, then $t(r) = t(r')$, hence $r$ and $r'$ have the same image in $X$ under both $s$ and $t$, hence $g = g'$ (as $G$ acts freely on $X$), hence $s(r) = s(r')$, hence $r = r'$ (as $R$ is an equivalence relation on $U$). Moreover $j$ is an equivalence relation (details omitted). Both projections $\coprod \nolimits _{g \in G} R \to U$ are étale, as $s$ and $t$ are étale. Thus $j$ is an étale equivalence relation and $U/\coprod \nolimits _{g \in G} R$ is an algebraic space by Spaces, Theorem 65.10.5. There is a map

\[ U/\coprod \nolimits _{g \in G} R \longrightarrow X/G \]

induced by the map $U \to X$. We omit the proof that it is an isomorphism of sheaves. $\square$

Comments (0)

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Your email address will not be published.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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 071S. Beware of the difference between the letter 'O' and the digit '0'.