Timeline for Can an abelian variety/Q have no non-trivial points over Q_sol?
Current License: CC BY-SA 3.0
16 events
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| S Mar 25, 2015 at 0:26 | history | suggested | R.P. | CC BY-SA 3.0 |
title asked for something different than the main body
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| Mar 24, 2015 at 23:59 | review | Suggested edits | |||
| S Mar 25, 2015 at 0:26 | |||||
| Mar 1, 2015 at 14:18 | comment | added | Tim Dokchitser | @Pablo That would be still only a conjectural answer and not for all abelian varieties. Probably someone will come along with something much better. But if not, I'll certainly post it. | |
| Mar 1, 2015 at 14:14 | comment | added | Pablo | @TimDokchitser could you please explain how is the field $F$ chosen? It would be appropriate to post it as an answer (if you wish, of course). | |
| Mar 1, 2015 at 9:29 | comment | added | Tim Dokchitser | ...Maria Sabitova has several papers on root numbers of abelian varieties and their twists. I think she reviews root numbers of abelian varieties and their connection to the parity of ranks. | |
| Mar 1, 2015 at 9:26 | comment | added | Tim Dokchitser | ... If now $K/F$ is quadratic in which all bad primes of $A$ and $\infty$ split, except for one prime above $p$ which ramifies, then the root number of $A/K$ is $-1$ (this is like the `Heegner condition' for elliptic curves), so the parity conjecture will imply that $A(K)$ has odd rank, in particular infinite... | |
| Mar 1, 2015 at 9:23 | comment | added | Tim Dokchitser | @Pablo I think it would be something like this. Suppose $A$ has at least one prime $p$ of bad reduction where the potentially toric part has odd dimension. (Generically, you would expect this to happen, even if "odd" is replaced by 1 and without "potentially", e.g. a generic curve would have a prime $p$ that divides the discriminant exactly once.) By assumption, over a (soluble) extension $F/{\mathbb Q}$ the variety $A/F$ has split semistable reduction at $p$, and this is all that is needed for parity... | |
| Mar 1, 2015 at 9:07 | comment | added | Pablo | @TimDokchitser I was familiar with the work of Fehm and Petersen but the connection with the parity conjecture is new to me. Could you sketch the argument for the existence of such $K$ assuming the parity conjecture/ Birch and Swinnerton-Dyer conjecture? | |
| Mar 1, 2015 at 8:59 | comment | added | Tim Dokchitser | I believe this is not known, though it is even conjectured that $A(F)\ne\{0\}$ over the maximal abelian extension of $F$ for any number field $F$ (Fehm, Petersen, On the rank of abelian varieties over ample fields, IJNT 6 (2010), no. 3, 579–586). And for "generic" abelian varieties the existence of such a $K$ would follow from the parity conjecture. But, unconditionally, I am not sure you can prove this except for "easy" families, e.g. Jacobians of hyperelliptic curves. | |
| Mar 1, 2015 at 8:52 | comment | added | Laurent Berger | @Pablo sorry, my mistake! | |
| Mar 1, 2015 at 8:50 | comment | added | Pablo | @LaurentBerger I explicitly wrote "$A(K)$ is nontrivial" and not "$A(K)$ is nonempty" in the body of the question in grey. That is, I ask whether the group $A(K)$ may be trivial for every finite solvable extension $K/\mathbb{Q}$. | |
| Mar 1, 2015 at 8:47 | comment | added | Laurent Berger | By definition, an abelian variety over a field K has a rational point over K, so in your question, you presumably mean a homogenous space for your abelian variety. | |
| Mar 1, 2015 at 7:54 | comment | added | Pablo | @VesselinDimitrov you are right! So, maybe I should try to look for curves on my abelian variety since there are some works about solvable points on curves? | |
| Mar 1, 2015 at 7:51 | comment | added | Vesselin Dimitrov | See Theorem 0.0.2 in their paper "Solvable points on genus one curves" in the Duke Math. Journal, 2008. While the statement is very likely to be true without the condition of existence of points in all $\mathbb{Q}_p$, the latter is quite essential to their proof. | |
| Mar 1, 2015 at 1:45 | comment | added | Vesselin Dimitrov | Rather than elliptic curves (where you could just adjoin the points of order two), much more interesting is the case of smooth genus one curves over $\mathbb{Q}$. For these, as you probably know, there is the Ciperiani-Wiles theorem proving the existence of a $\mathbb{Q}^{\mathrm{sol}}$ point under some (fairly mild) technical conditions. (There should be a $\mathbb{Q}_p$-point for all $p$, and the Jacobian, an elliptic curve over $\mathbb{Q}$, should be semistable.) | |
| Mar 1, 2015 at 0:28 | history | asked | Pablo | CC BY-SA 3.0 |