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assume we have an algebraic stack $A$ over $Sch(\mathbb{Z})$ which is quasi-compact and with separated diagonal. Assume that I have a stack $B$ which is obtained by rigidifying $A$ by a subgroup of the inertia group. Assume moreover that $A\rightarrow B$ is faithfully-flat and smooth over $\mathbb{Z}[\frac{1}{d}]$ and that the diagonal of $B$ is finite. Question : is $B$ proper over $\mathbb{Z}$, or at least over $\mathbb{Z}[\frac{1}{d}]$ ?? If so is it trivial to prove it?

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out of curiosity: what do you mean by rigidifying? – Yosemite Sam Oct 4 '11 at 22:38
Is $A$ assumed to be proper? – Keerthi Madapusi Pera Oct 4 '11 at 22:57
@knob call this subgroup $G$ then $B$ is the stack with the same objects of $A$ but with $Hom_B(x,y):=Hom_A(x,y)/y^{*} G$ where the action is the composition action @Keerthi Madapusi Pera I only have these hypoteses and $A$ is not proper – uuuuuuuuuuuu Oct 5 '11 at 10:20
I think $A = B = \mathbb{A}^1$ is a counterexample. – S. Carnahan Oct 5 '11 at 10:25

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