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I will use as a reference Hartshorne Prop. II.7.14, the universal property of blow-up. $\tilde{X}$ is the blow up of $X$ along a sheaf of ideals. $Z\to X$ is the morphism that is to be lifted to $\tilde{X}$ (if the inverse image sheaf of ideals is invertible).

Does the same universal property hold if $Z$ is not a scheme but an algebraic (Artin) stack?

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1 Answer 1

Yes, this can, for example, be checked using fppf-descent.

Pick a presentation $p:U\to X$ and pull-back everything along $p$. Since blowing-up commutes with flat base change, you may then use the universal property for $Z\times_X U\to U$ and $\tilde{X}\times_X U\to U$ to deduce that if the inverse image of the ideal sheaf to $Z$ is invertible, then there is a unique lifting $Z\times_X U\to \tilde{X}\times_X U$. Similarly, there is a unique lifting $Z\times_X R\to \tilde{X}\times_X R$ where $R=U\times_X U$. Then you conclude that there is a unique lift $Z\to \tilde{X}$ by fppf descent.

[Edit: I misread the question as asking for $X$ to be an algebraic stack. If $Z$ is an algebraic stack (and $X$ is a scheme or an algebraic stack), then you can similarly pick a presentation $q:V\to Z$ and deduce that there are unique morphisms $V\to \tilde{X}$ and $V\times_X V\to \tilde{X}$ and hence, by fppf-descent, a unique morphism $Z\to \tilde{X}$.]

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Looks great! You mean $p$ a presentation of the morphism, right? –  The Prester Feb 6 '13 at 19:17
$p$ is a presentation of the algebraic stack $X$, i.e. a smooth surjective morphism from a scheme. –  Martin Brandenburg Feb 6 '13 at 21:06
It seems to me that the question was what happens when $Z$ (rather than $X$) is an Artin stack.. –  Mattia Talpo Feb 6 '13 at 22:53
Yes, Mattia is right. In the question $Z$ is supposed to be an algebraic stack, not $X$. –  IMeasy Feb 7 '13 at 7:42
If a morphism of sheaves $\mathsf{Stacks} \to \mathsf{Cat}$ is an isomorphism on all schemes, it is an isomorphism. Therefore $Z$ can also be an algebraic stack. –  Martin Brandenburg Feb 7 '13 at 11:00

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