In "Brauer groups and quotient stacks", Edidin et. al prove the following theorem:

Theorem 2.7. Let $\mathcal{X}$ be an algebraic stack over a Noetherian base (of finite type). Then the diagonal $\mathcal{X}\to \mathcal{X}\times \mathcal{X}$ is quasi-finite if and only if there is a finite surjective morphism $X\to \mathcal{X}$ for a scheme $X$.

On the other hand, Kresch in "Cycle groups for Artin stacks" proves the following:

Proposition 3.5.7. Let $\mathcal{X}$ be a stack of finite type over a field. The the following are equivalent: 1) The diagonal is quasi-finite; 2) The stabilizer $\mathcal{X}\times_{\mathcal{X}\times\mathcal{X}}\mathcal{X}\to \mathcal{X}$ is quasi-finite. Further, if $\mathcal{X}$ has quasi-finite diagonal $\mathcal{X}$ admits a stratification by quotient stacks.

Now, suppose that $\mathcal{X}$ is already a quotient stack $[Y/G]$, say with $Y$ an affine scheme and $G$ some group scheme (both of finite type over a field). Then $\mathrm{id}: Y\to\mathcal{X}$ is a finite surjective morphism, so by 2.7 above have quasi-finite diagonal. Then by 3.5.7 the stabilizer is quasi-finite, but this seems false in general. For instance, take $G=GL(n)$ and then you are almost guaranteed to have non-finite stabilizers.

What am I missing here? It's obvious that there's something here that I've gotten wrong.


1 Answer 1


The morphism $Y \to [Y/G]$ is a $G$-torsor, so it is finite only if $G$ is finite.

  • $\begingroup$ Yes, of course! Thanks! I had my suspicions that there was some condition that wasn't satisfied, but I couldn't figure it out. It felt so obvious that the map was finite (easily fooled, it seems). $\endgroup$ Aug 13, 2010 at 12:01

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