The Stacks project

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). $\square$


Comments (3)

Comment #4253 by Remy on

The proof also shows that is a -torsor.

Comment #4254 by Laurent Moret-Bailly on

Another case for (3): 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.


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 07S7. Beware of the difference between the letter 'O' and the digit '0'.