Lemma 79.11.7. Let $S$ be a scheme. Let $X \to B$ be a morphism of algebraic spaces over $S$. Let $G$ be a group algebraic space over $B$ and let $a : G \times _ B X \to X$ be an action of $G$ on $X$ over $B$. If

1. $a$ is a free action, and

2. $G \to B$ is flat and locally of finite presentation,

then $X/G$ (see Groupoids in Spaces, Definition 77.19.1) is an algebraic space and $X \to X/G$ is surjective, flat, and locally of finite presentation.

Proof. The fact that $X/G$ is an algebraic space is immediate from Theorem 79.10.1 and the definitions. Namely, $X/G = X/R$ where $R = G \times _ B X$. The morphisms $s, t : G \times _ B X \to X$ are flat and locally of finite presentation (clear for $s$ as a base change of $G \to B$ and by symmetry using the inverse it follows for $t$) and the morphism $j : G \times _ B X \to X \times _ B X$ is a monomorphism by Groupoids in Spaces, Lemma 77.8.3 as the action is free. The assertions about the morphism $X \to X/G$ follow from Lemma 79.11.6. $\square$

Comment #1323 by Kestutis Cesnavicius on

Perhaps it is worth changing "$X/B$" to "$X$ over $B$" because the similar notation $X/G$ used later has a completely different meaning.

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