Let $X$ be a locally Noetherian scheme defined over an algebraically closed field $k$ and let $Y\subset X$ be a closed subscheme. Suppose there is an algebraic group $G$ acting on $X$ and a subgroup $H\subset G$ acting on $Y$, in such a way that the closed immersion $Y\to X$ descends to a morphism $$ f:[Y/H]\to [X/G]. $$ Let us assume $f$ is representable and $[Y/H]$ is DeligneMumford. I am looking for a condition (maybe on $X$ and $Y$, but preferably on the group actions) ensuring that $f$ is locally of finite type. For instance I am happy with assuming that the action of $H$ on $Y$ is proper, but I am not sure it helps. Thanks!
1 Answer
Yes.
Since the question is local in $X$ we can assume that it is a finite type affine scheme. We always have the induced map of stacks $$f:[Y/H]\rightarrow [X/H]\rightarrow [X/G]$$ given by tensoring a locally trivial $H$ torsor with $G$ over $H$. Now we need to show that the fibre product $[Y/H]\times_{[X/G]} X$ is finite type.
As a stack this is isomorphic to $[(Y\times G)/H]$. The quotient of $Y\times G$ by the action of $H$ always exists in the category of algebraic spaces. This is in Knutson's book, II.6.7. Or see Edidin+Graham's https://arxiv.org/abs/alggeom/9609018, Prop.22: from the proof, the quotient is finite type. By your conditions on $f$, the quotient is then in fact a finite type algebraic scheme.

$\begingroup$ Thanks! So, $[(Y\times G)/H]$ is of finite type because $Y\times G\to [(Y\times G)/H]$ is an atlas and $Y\times G$ is of finite type? Or am I misunderstanding your secondlast sentence? $\endgroup$ Mar 11, 2017 at 13:50

$\begingroup$ I guess above I was using $Y\times_H G$ as an atlas, and reasoning that it was of finite type, but yes, $Y\times G$ also provides a finite type atlas. $\endgroup$ Mar 13, 2017 at 10:37

$\begingroup$ btw you don't need any conditions on $X$ for the statement you want. A morphism $Z\rightarrow [X/G]$ is given by a principal $G$bundle on $Z$ and a $G$equivariant map from the total space to $X$. If you take an etale neighbourhood $U$ on which this bundle is trivial, the map of schemes $U\times_{[X/G]} [X/H]\rightarrow U$ becomes $U\times (G/H)\rightarrow U$, which is finite type. $\endgroup$ Mar 13, 2017 at 10:41