Timeline for Are root stacks characterized by their divisor multiplicities?
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9 events
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| Sep 15, 2010 at 3:36 | comment | added | Angelo | Yes, something like this, except that the local rings are not necessarily complete, and might have mixed characteristic anyway, so they won't be power series rings. | |
| Sep 14, 2010 at 1:04 | comment | added | Anton Geraschenko | 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. | |
| Sep 14, 2010 at 1:04 | comment | added | Anton Geraschenko | 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 $n$. | |
| Sep 13, 2010 at 21:44 | comment | added | Anton Geraschenko | 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. | |
| Sep 8, 2010 at 5:24 | history | edited | Angelo | CC BY-SA 2.5 |
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| Sep 8, 2010 at 5:22 | comment | added | Angelo | 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). | |
| Sep 8, 2010 at 5:15 | history | edited | Angelo | CC BY-SA 2.5 |
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| Sep 7, 2010 at 20:04 | comment | added | Anton Geraschenko | 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? | |
| Sep 7, 2010 at 19:21 | history | answered | Angelo | CC BY-SA 2.5 |