Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Definitions/Background

Suppose $S$ is a scheme and $D\subseteq S$ is an irreducible effective Cartier divisor. Then $D$ induces a morphism from $S$ to the stack $[\mathbb A^1/\mathbb G_m]$ (a morphism to this stack is the data of a line bundle and a global section of the line bundle, modulo scaling). For a positive integer $k$, the root stack $\sqrt[k]{D/S}$ is defined as the fiber product

$\begin{matrix} \sqrt[k]{D/S} & \longrightarrow & [\mathbb A^1/\mathbb G_m] \\ p\downarrow & & \downarrow \wedge k \\ S & \longrightarrow & [\mathbb A^1/\mathbb G_m] \end{matrix}$

where the map $\wedge k: [\mathbb A^1/\mathbb G_m]\to [\mathbb A^1/\mathbb G_m]$ is induced by the maps $x\mapsto x^k$ (on $\mathbb A^1$) and $t\mapsto t^k$ (on $\mathbb G_m$). The morphism $p:\sqrt[k]{D/S}\to S$ is a coarse moduli space and is an isomorphism over $S\smallsetminus D$. Moreover, there is a divisor $D'$ on $\sqrt[k]{D/S}$ such that $p^*D$ is $kD'$.

The data of a morphism from $T$ to $\sqrt[k]{D/S}$ is equivalent to the data a morphism $f:T\to S$ and a divisor $E$ on $T$ such that $f^*D = kE$.

The question

Suppose $\mathcal X$ is a DM stack, that $f:\mathcal X\to S$ is a coarse moduli space, that $f$ is an isomorphism over $S\smallsetminus D$, and that $f^*D = kE$ for an irreducible Cartier divisor $E$ on $\mathcal X$. Is the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ an isomorphism?

I get the strong impression that the answer should be "yes", at least if additional conditions are placed on $\mathcal X$.

A counterexample

Here's a counterexample to show that some additional condition needs to be put on $\mathcal X$. Take $G$ to be $\mathbb A^1$ with a doubled origin, viewed as a group scheme over $\mathbb A^1$. Then $\mathcal X=[\mathbb A^1/G]\to \mathbb A^1$ is a coarse moduli space ("there's a $B(\mathbb Z/2)$ at the origin"). If we take $D\subseteq \mathbb A^1$ to be the origin, then the pullback to $\mathcal X$ is the closed $B(\mathbb Z/2)$ with multiplicity 1. Yet the induced morphism from $\mathcal X$ to $\sqrt[1]{D/\mathbb A^1}\cong \mathbb A^1$ is not an isomorphism.

In this case, $\mathcal X$ is a smooth DM stack, but has non-separated diagonal.

share|improve this question
    
I should have mentioned that this question arose while trying to understand the proof of Theorem 5.2 of Fantechi-Mann-Nironi's paper "Smooth Toric DM Stacks": arxiv.org/abs/0708.1254. –  Anton Geraschenko Jul 17 '11 at 20:56
add comment

1 Answer 1

What do you mean by an irreducible Cartier divisor? Assume that $S$ and $D$ are regular, that $\mathcal X$ is normal and has finite inertia, and that $f^*D = kE$ for a reduced divisor $E$ on $\mathcal X$. Also assume that $\mathcal X$ is tame in codimension 1. Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. I claim that is representable in codimension 1; this follows from the fact that $\mathcal X$ is ramified of degree $k$ at the generic point of each irreducible component of $D$ (this can be done, for example, by taking the strict henselization of $S$ at the generic point of such a component, thus reducing to the case that $S$ is an henselian trait, which is easy, using the tameness hypothesis). Thus $\mathcal X\to \sqrt[k]{D/S}$ is a proper morphism with finite fibers, $\mathcal X$ is normal, $\sqrt[k]{D/S}$ is regular, and is an isomorphism in codimension 1. By purity of branch locus, it must be étale; and then it must be an isomorphism.

I think that all of the hypotheses are necessary. For example, already when $D$ is a nodal curve on a smooth surface $S$ there are counterexamples: there is a smooth stacks having $S$ as its moduli space, which is ramified of order $k$ along $D$ (this is different from $\sqrt[k]{D/S}$, because the latter is singular). 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. There are also counterexamples when $\mathcal X$ is not normal, or when it is not tame.

share|improve this answer
    
Thanks! By irreducible Cartier divisor, I meant one which is not the sum of two other effective divisors; I'm happy to take "reduced" if there is a problem with this notion of irreducible. I don't see how you get representability. More details would be appreciated, even though that's not usually the nature of an exercise :-). Once you have proper and representable, you combine that with birational (which was essentially given) and quasi-finite (since both are quasi-finite over $S$) and apply Zariski's Main Theorem. Is that what you had in mind, or is there a simpler way? –  Anton Geraschenko Sep 7 '10 at 20:04
    
My previous post was very rash. I edited it, I hope that now it is clearer, and correct (the first version was just plain wrong). –  Angelo Sep 8 '10 at 5:22
    
I'm having some trouble digesting this answer. What does it mean for a morphism to be representable in codimension 1? How does it imply the map is an isomorphism in codimension 1? (This means it's an isomorphism when you pull back to any codimension 1 point, right?) After you apply purity to get that the map is etale, don't you still need representability to conclude that it's an isomorphism? This last part doesn't bother me so much; I know an argument that any etale map of orbifolds is representable. The main thing I don't understand is how to get etaleness. –  Anton Geraschenko Sep 13 '10 at 21:44
1  
Martin Olsson helped clear things up for me at tea today. Here's what I got from our conversation. It's enough to show the map is an isomorphism at the generic point of $D$, so we base change by the strict hensilization of the DVR $\mathcal O_{S,D}$. Then $\mathcal X$ must be of the form $[A/G]$, where $G$ is a finite group and $A$ is a strictly henselian ring (so a DVR). (I'm not completely sure what hypotheses have been used here.) The action of $G$ on the tangent space of the closed point of $A$ must be faithful, so $G\hookrightarrow \mathbb G_m$, so $G$ is $\mu_n$ for some $n$. –  Anton Geraschenko Sep 14 '10 at 1:04
1  
Counting ramification, we get $n=k$. By dancing around a bit, we can show that $A=L[[t]]$ with the action you'd expect, where $L$ is the function field of $D$. This is precisely the $k$-th root stack of $\mathcal O_{X,D}^{sh}$ along the closed point. So the map to the root stack is an isomorphism over the generic point of $D$, so we can apply purity to get etaleness. –  Anton Geraschenko Sep 14 '10 at 1:04
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.