# Are root stacks characterized by their divisor multiplicities?

## 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.

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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

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.
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
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
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