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Mar 14, 2017 at 12:45 vote accept Andrea Ricolfi
Mar 13, 2017 at 10:41 comment added Ben Davison 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.
Mar 13, 2017 at 10:37 comment added Ben Davison 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.
Mar 11, 2017 at 13:50 comment added Andrea Ricolfi 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 second-last sentence?
Mar 11, 2017 at 13:03 history answered Ben Davison CC BY-SA 3.0