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. $\square$
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