Lemma 66.14.2. Let $S$ be a scheme. Let $G \to S$ be a group scheme. Let $X \to S$ be a morphism of schemes. Let $a : G \times _ S X \to X$ be an action. Assume that

1. $G \to S$ is finite locally free,

2. the action $a$ is free,

3. $X \to S$ is affine, or quasi-affine, or projective, or quasi-projective, or $X$ is isomorphic to an open subscheme of an affine scheme, or $X$ is isomorphic to an open subscheme of $\text{Proj}(A)$ for some graded ring $A$, or $G \to S$ is radicial.

Then the fppf quotient sheaf $X/G$ is a scheme and $X \to X/G$ is an fppf $G$-torsor.

Proof. We first show that $X/G$ is a scheme. Since the action is free the morphism $j = (a, \text{pr}) : G \times _ S X \to X \times _ S X$ is a monomorphism and hence an equivalence relation, see Groupoids, Lemma 39.10.3. The maps $s, t : G \times _ S X \to X$ are finite locally free as we've assumed that $G \to S$ is finite locally free. To conclude it now suffices to prove the last assumption of Proposition 66.14.1 holds. Since the action of $G$ is over $S$ it suffices to prove that any finite set of points in a fibre of $X \to S$ is contained in an affine open of $X$. If $X$ is isomorphic to an open subscheme of an affine scheme or isomorphic to an open subscheme of $\text{Proj}(A)$ for some graded ring $A$ this follows from Properties, Lemma 28.29.5. If $X \to S$ is affine, or quasi-affine, or projective, or quasi-projective, we may replace $S$ by an affine open and we get back to the case we just dealt with. If $G \to S$ is radicial, then the orbits of points on $X$ under the action of $G$ are singletons and the condition trivially holds. Some details omitted.

To see that $X \to X/G$ is an fppf $G$-torsor (Groupoids, Definition 39.11.3) we have to show that $G \times _ S X \to X \times _{X/G} X$ is an isomorphism and that $X \to X/G$ fppf locally has sections. The second part is clear from the fact that $X \to X/G$ is surjective as a map of fppf sheaves (by construction). The first part follows from the isomorphism $R = U \times _ M U$ in the conclusion of Proposition 66.14.1 (note that $R = G \times _ S X$ in our case). $\square$

Comment #4253 by Remy on

The proof also shows that $X \to X/G$ is a $G$-torsor.

Comment #4254 by Laurent Moret-Bailly on

Another case for (3): $G\to S$ is radicial.

Comment #4424 by on

@Remy: Yes, this is true. I've added it.

@Laurent: Yes, this is another case. I've added it.

See changes here.

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