3
$\begingroup$

Hi,

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?

Improve this question
$\endgroup$
4
  • $\begingroup$ out of curiosity: what do you mean by rigidifying? $\endgroup$
    – Yosemite Sam
    Commented Oct 4, 2011 at 22:38
  • 1
    $\begingroup$ Is $A$ assumed to be proper? $\endgroup$
    – Keerthi Madapusi
    Commented Oct 4, 2011 at 22:57
  • $\begingroup$ @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 $\endgroup$
    – uuuuuuuuuuuu
    Commented Oct 5, 2011 at 10:20
  • 1
    $\begingroup$ I think $A = B = \mathbb{A}^1$ is a counterexample. $\endgroup$
    – S. Carnahan ♦
    Commented Oct 5, 2011 at 10:25

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .