MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\cal X$ be a DM stack of finite type over a field (if necessary, I will assume that $k=\mathbb{C}$ and $\cal X$ is a scheme, or even a variety) and $G$ be a finite group. Then we have a quotient stack $[\mathcal{X}/G]$ and in fact we have a natural morphism $\phi:\mathcal{X}\to [\mathcal{X}/G]$.

Is $\phi$ always flat? If not, under what conditions is $\phi$ flat?

share|cite|improve this question
up vote 7 down vote accepted

The map $\phi$ is in fact étale. By definition a map $B \to [\mathcal X/G]$ (say $B$ is a scheme) is a $G$-torsor $E \to B$ and a $G$-equivariant map $E \to \mathcal X$. The base change of $\phi$ is exactly $E \to B$. Since the property of being étale is stable under base change and local, it suffices to show that $E\to B$ is étale. But part of the definition of being a torsor is being locally trivial in whatever topology you are considering, so now it suffices to check that the map from $G$ to a point is étale, which is clear.

The general principle here is that properties of a group scheme $G$ should carry over to properties of the universal map $t \colon \mathrm{pt} \to BG$, e.g. if $G$ is smooth then $t$ is smooth.

share|cite|improve this answer

The consideration of group actions is too special. Note that the two maps $G \times X \rightrightarrows X$ (projection and action) are fppf. Rather generally, if $X$ is a scheme of finite presentation over a base scheme $S$ and if $R \rightarrow X \times_S X$ is a finitely presented monomorphism of schemes that defines an equivalence relation (on the functor $X$) and for which the projections $R \rightrightarrows X$ are flat then the "quotient stack" $[X/R]$ is an Artin stack (by a deep theorem of Artin in case $R$ isn't smooth over $X$) and the natural map $X \rightarrow [X/R]$ is relatively representable and fppf (i.e., for any scheme $T$ and morphism $T \rightarrow [X/R]$ the pullback $X \times_{[X/R]} T$ is a scheme whose projection map to $T$ is fppf). Strictly speaking, if the maps $R \rightrightarrows X$ aren't affine (though they are for the group action case in the question) then this relative representability has to be taken in the sense of algebraic spaces, due to issues with effective fppf descent for schemes away from the setting of affine morphisms.

If you don't understand the reason for the above being true, say at least in the special case when the maps $R \rightrightarrows X$ are smooth (e.g., etale) and $S$ is noetherian, then you'd be well-served by reviewing your study of quotients in the framework of stacks. Of course, to make a link back to the question it is an important exercise to check that in the case of $R = G \times X \rightrightarrows X$ arising from the action of an fppf affine $S$-group $G$ on $X$ (or smooth affine $S$-group if you prefer to avoid invoking deep theorems of Artin), $[X/R]$ equipped with its map from $X$ over $S$ "is" the fibered category of $G$-torsors over $X$ (defined in the evident manner). In particular, it is ad hoc in such situations to say that $[X/G]$ is defined to be the fibered category of $G$-torsors on $X$; it is much better to understand how this identification emerges from the general perspective of $[X/R]$ defined and studied more generally by means of fppf (or etale) descent theory with equivalence relations.

(The hypothesis of $S$-affineness on $G$ is relevant to ensure effectivity of descent for $G$-torsors within the framework of schemes, but in fact if one appeals to some deep results of Artin then this effectivity can be proved to remain valid within the more general setting of algebraic spaces when $G$ is merely fppf over $S$. For example, the case when $G$ is an abelian scheme is interesting.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.