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

In the smooth case, I think that the answer is positive over an arbitrary regular excellent base. The argument was in my PhD thesis; it was done over a field, but I think it adapts to this case. Cove …

There is a paper by Matthieu Romagny, "Group actions on stacks", in the Michigan Journal of Mathematics http://www.math.jussieu.fr/~romagny/articles/group.pdf, where the first two points are discussed …

Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. … For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. …

Yes. The criterion for an Artin stack to be Deligne-Mumford is that it should have unramified diagonal (this is somewhere in Laumon Moret Bailly, I don't have it here). If the stack is fibered in sets …

I think that what they might have in mind is that for non-quasi-separated Deligne-Mumford algebraic stacks one should not assume that the diagonal is represented by schemes, but by algebraic spaces. … For quasi-separated Deligne-Mumford stacks this implies representability by schemes, but this is not true in general. …

If $X \to Y$ is a proper morphism of DM stacks, where $X$ has finite inertia (the hypotheses in my paper are more stringent, but the theory has been refined since then), there exists a finite map $V \to …

They only work with stacks that are locally finitely presented. …

Take the complement of a non-empty open subscheme, and use noetherian induction.

By a family of gerbes you mean, I suppose, a gerbe over $X \times \mathbb A^{1}$. In any case, it has a class in $\mathrm H^0(\mathbb A^1, \mathrm R^2 \mathrm{pr}_{2*}\mathbb G_{\rm m})$. Since $\math …

Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $S \times_{ …

It is certainly not true that $\mathcal X \to \mathcal X^H$ is a good moduli morphism, unless $H$ is linearly reductive, because when you push forward the cohomology of $H$ will come into play. On th …

I have heard this result attributed to Kai Behrend, who apparently came up with it while writing his part of the now defunct many-authored stack book. However it is certainly possible that someone els …

For the examples you mention, this boils down to representability of Hom sheaves of flat finitely presented proper schemes, which is due to Grothendieck. These Hom sheaves are not of finite type, thou …

From the definition it follows easily that tame stacks with affine moduli spaces have the property you require. … There are several different characterizations of tame stacks; see the paper "Tame stacks in positive characteristic" by Dan Abramovich, Martin Olsson and myself. …