Timeline for Reduction of torsion points on Neron Model
Current License: CC BY-SA 3.0
13 events
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| Jun 12, 2017 at 14:36 | comment | added | Jędrzej Garnek | @Lukas - thanks for the discussion. I also thought about it, but unfortunately, I think it's not that easy :( If $\mathcal{A}$ is the Neron model over $K$, then not every point in $A(L)$ corresponds to point in $\mathcal{A}(L)$. This would hold, if $L/K$ would be unramified (by Neron Mapping Property), but we don't know that apriori. | |
| Jun 12, 2017 at 12:18 | comment | added | Lukas | I believe you can actually consider the Néron model over $\mathcal{O}_K$. A posteriori, the Néron model over $\mathcal{O}_L$ will be just a base change of the first. | |
| Jun 12, 2017 at 10:37 | comment | added | Jędrzej Garnek | I'm still not sure if $P$ belongs to $\mathcal{A}_{\sigma}(R')$ - so that the "$P^{\sigma} - P$" makes sense. | |
| Jun 12, 2017 at 10:37 | comment | added | Jędrzej Garnek | Ok, sounds promising. So we can just conclude that $P^{\sigma} - P \in \mathcal{A}_{\sigma}(R')[n]$ and it must be zero, since it reduces to zero? | |
| Jun 12, 2017 at 10:27 | comment | added | Lukas | You are completely right. The point $P^\sigma$ is no longer a point of the same $\mathcal{A}(R')$, but is rather a point of $\mathcal{A}_\sigma(R')$, where $\mathcal{A}_\sigma/\mathcal{O}_L$ is the "same" abelian scheme, except the structural morphism is the twisted $\mathcal{A}\to Spec\,\mathcal{O}_L\overset{\sigma}{\to}Spec\,\mathcal{O}_L$. | |
| Jun 12, 2017 at 10:15 | comment | added | Jędrzej Garnek | @Lukas - how do you know that $Gal(L/K)$ acts on $\mathcal A$? I mean the generic fiber of $\mathcal{A}$ is defined over $L$ and not over $K$. So if $P \in \mathcal{A}(R')$ then we have no longer guarantee that $P^{\sigma}$ is a well defined point in $\mathcal{A}(R')$, right? | |
| Jun 12, 2017 at 10:06 | comment | added | Lukas | I believe your proof of 2. is basically correct. $Gal(L/K)$ acts on $\mathcal{A}$ semilinearly, meaning it does not preserve the base ring $\mathcal{O}_L$ (there's the action on $L$ !). If you restrict the action to the inertia, then you obtain an action on the special fiber by endomorphisms because the inertia preserves the residue field. | |
| Jun 12, 2017 at 8:52 | history | edited | Jędrzej Garnek | CC BY-SA 3.0 |
I found the solution to my problem.
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| Jun 11, 2017 at 22:47 | history | edited | Jędrzej Garnek | CC BY-SA 3.0 |
new thoughts on the problem
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| Jun 10, 2017 at 16:29 | history | edited | Jędrzej Garnek | CC BY-SA 3.0 |
better wording
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| Jun 10, 2017 at 15:19 | comment | added | user19475 | Doesn't 1. follow from the infinitesimal lifting criterion for étaleness? | |
| Jun 10, 2017 at 14:55 | review | First posts | |||
| Jun 10, 2017 at 15:16 | |||||
| Jun 10, 2017 at 14:53 | history | asked | Jędrzej Garnek | CC BY-SA 3.0 |