Timeline for Smoothen a nodal curve
Current License: CC BY-SA 4.0
20 events
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| Aug 8, 2018 at 23:20 | comment | added | gdb | ... this descent datum. If you are not familiar with general theory of descent, I recommend to look into Chapter 6 of the book "Neron Models", especially ch 6.1 and 6.2 (or anything else). | |
| Aug 8, 2018 at 23:08 | comment | added | gdb | > But I am not quite sure why the descent work, (Sorry I didn't learn this well). It works essentially because your deformation problem is really defined over $k$ and also any semi-stable $n$-pointed curve has a canonical ample line bundle. The first condition gives you a descent datum, but it is possible that this descent datum is not effective. Here an ample line bundle comes into play, it is a general theorem that any descent datum for a pair $(X,L)$, where $L$ is an ample line bundle, is effective. This is the main idea, what is written in the ans is a rigorous way to construct ... | |
| Aug 8, 2018 at 23:03 | comment | added | gdb | According to infinitesimal criterion of unramifiedness, this precisely means that the automorphism scheme of a pair $(X_0', D_0')$ is an unramified scheme. I should say that though this argument essentially gives you a prove of pro-representability of the deformation problem you should use Schlessinger's criterion (or something else) to make it rigorous. | |
| Aug 8, 2018 at 23:01 | comment | added | gdb | @Qixiao That's not quite correct. The main reason is that in the definition of the deformation functor you fix an isomorphism of the special fibre of a deformation $X' \to \operatorname{Spec} A$ and $X_0'$ (compatible with sections since we deform a curve with given sections). I mean that this isomorphism is a part of data and not just a condition. Therefore, what one really want to check is that for any artinian $k'[[t]]$-algebra $R$ and a deformation $(X', D') \to \operatorname{Spec} A$ a natural map $Aut(X',D') \to Aut(X_0', D_0')$ is bijective. | |
| Aug 8, 2018 at 21:23 | comment | added | user39380 | (I think the deformation functor being pro-representable at a point means the corresponding point in the moduli stack has no automorphism, if the central curve has automorphism, then I think the algebriaized family over $\mathrm{Spec}(k'[[t]])$ don't have universal property and not clear why the covering data glues..) | |
| Aug 8, 2018 at 21:22 | comment | added | user39380 | Thanks a lot! I thought the problem with non-rational nodes was, one need to find Galois-equivariant deformation, (suppose we know the rational node case). I think your argument by directly descending the universal family would avoid the question. But I am not quite sure why the descent work, (Sorry I didn't learn this well). | |
| Aug 8, 2018 at 11:10 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 8, 2018 at 8:03 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 7, 2018 at 22:04 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 7, 2018 at 21:23 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 7, 2018 at 20:49 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 7, 2018 at 20:44 | comment | added | gdb | @Qixiao I updated the answer to address the non-rationality issue (didn't have good access to the Internet to do it earlier) | |
| Aug 7, 2018 at 20:37 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 3, 2018 at 17:41 | comment | added | user39380 | Thanks a lot! (it"s really not in a hurry) | |
| Aug 3, 2018 at 15:33 | comment | added | gdb | However, I need to check that everything indeed works fine before posting this as an answer (but I am 95% sure that basically the same proof should work). I will come back to this issue later (hopefully, I will say whether it works in 24 hours). | |
| Aug 3, 2018 at 15:28 | comment | added | gdb | @Qixiao Fair point. Certainly, Talpo and Vistoli assume that singularities are rational. One reason for it is that they use a different definition of a double point. Probably this simplifies the proof and they don’t really need a classification result from Freitag and Kiehl’a book (just usual Artin approximation is sufficient for their purposes). So, in my recollection exactly the same proof as one presented in these notes should work in general, if one uses this classification result. | |
| Aug 3, 2018 at 14:55 | comment | added | user39380 | Thanks a lot! But does the result there applies if the nodes are not rational? | |
| Aug 3, 2018 at 13:42 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 3, 2018 at 13:35 | history | edited | gdb | CC BY-SA 4.0 |
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| Aug 3, 2018 at 12:50 | history | answered | gdb | CC BY-SA 4.0 |