Timeline for A good reduction property
Current License: CC BY-SA 3.0
6 events
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| Dec 2, 2011 at 14:19 | comment | added | Sebastian Petersen | Dear Ariyan, thanks for the helpful references! There is also the paper of Pop "Henselian implies large" which contains useful information in the first section after the intro. My impression is that the answer is yes. Maybe I write a brief note on this for internal use (i.e. of course not to be published). I this sees the day, I send you a copy. Keep me informed if you have any news. Best, Sebastian | |
| Nov 25, 2011 at 19:32 | comment | added | Ariyan Javanpeykar | @Sebastian: Maybe the following notes of M. Romagny on the construction of the Neron model math.jussieu.fr/~romagny/exposes/Neron_models.pdf might come in handy. I would actually like to see how this works myself.... | |
| Nov 25, 2011 at 13:10 | comment | added | Ariyan Javanpeykar |
...Corollary 6.2.12 in Q. Liu's book. Now, you can take $U$ to be the inverse image of $\mathrm{Spec} \ R \backslash \pi(\mathcal{X} \backslash \mathcal{X}_{sm} )$ under $\pi$. This is non-empty because $X$ is smooth over $K$. Note that $U$ is the "biggest" choice possible.) It is clear that $x$ extends to a section of $U$ over $\mathrm{Spec} \ R \backslash \pi(\mathcal{X} \backslash \mathcal{X}_{sm} )$. I'm not sure if one can extend $x$ to a section of $U$ over $\mathrm{Spec} \ R$.
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| Nov 25, 2011 at 13:10 | comment | added | Ariyan Javanpeykar | Assume that $X$ is projective over $K$. Then there is a closed immersion $X\to \mathbf{P}^n_K$ for some $n\geq 0$. Let $\mathcal{X}$ be the Zariski closure of $X$ in $\mathbf{P}^n_{R}$. Note that $\mathcal{X}\times_R K =X$. By the valuative criterion of properness, there exists a unique $s \in \mathcal{X}(R)$ corresponding to $x$ on the generic fibre. Let $U\subset \mathcal{X}$ be any non-empty open subscheme such that $U\to \mathrm{Spec} \ R$ is smooth. (The set $\mathcal{X}_{sm}$ of points where $\pi:\mathcal{X}\to \mathrm{Spec} \ R$ is smooth is open in $\mathcal{X}$. See, for example.... | |
| Nov 25, 2011 at 12:18 | comment | added | Martin Bright | When R is a Dedekind domain, I believe this follows from the "smoothening process" which is the first step in constructing Néron models. Chapters 1 & 3 of Bosch, Lütkebohmert & Raynaud, "Néron models", contain all you could want to know about this case. | |
| Nov 25, 2011 at 11:45 | history | asked | Sebastian Petersen | CC BY-SA 3.0 |