Timeline for Does Chabauty-Coleman method give an algorithm for finding rational points?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 12, 2016 at 20:51 | vote | accept | SashaP | ||
| Sep 12, 2016 at 20:26 | comment | added | Michael Stoll | Sha: Finding points gives you successively better lower bounds on the rank, computing Selmer group gives you upper bounds. When Sha is finite, the bounds will eventually agree, and then you know that $n$ is sufficiently large. Zeros: The functions are $p$-adic analytic functions, which you can only know to finite precision. How do you want to tell two simple, but very close zeros from a double zero? You would have to distinguish between (say) $t^2$ and $t^2+at^n$ without knowing a bound for $n$. Send me an email if you have further questions. | |
| Sep 12, 2016 at 19:10 | comment | added | SashaP | To be clear, surely your algorithm is more effective in practice, I just want to understand if one can prove effective Mordell conjecture for cruves with $r<g$ by Coleman's methods. | |
| Sep 12, 2016 at 19:08 | comment | added | SashaP | Say, if equation of my curve has integral coefficients, and after computing $n$ digits and substituting, I get a result with valuation less than $n$, I know that this is not a root.Then I divide my series by $(x-root)$ and repeat until there are slopes $<-1$. After (order of zero of the form in the reduction)+1 steps I will be done by Coleman's bound. Where the problem with multiple roots occur? | |
| Sep 12, 2016 at 19:03 | comment | added | SashaP | Sorry, I do not understand what is the problem with multiple and collateral zeroes. If I know that the Newton polygon starts with a slope $\lambda$, I can iteratively calculate the digits of the root with valuation $-\lambda$. This procedure has to terminate because the root I am looking for is actually rational, and I can detect then it terminates substituting the root in the equation of my curve and checking if I got zero. If this is not a collateral root, I will eventually get it. But if it is not, I will notice this because I would get results that are too far from zero[tbc] | |
| Sep 12, 2016 at 18:44 | comment | added | SashaP | Sure, I understand that if we have a bound for $\# Sha$ then we can compute the rank. But in principle, how can I be sure that I picked sufficiently large $n$? Sha may contain $n$-torsion and I cannot detect which torsion in Selmer group came from $J(\mathbb{Q})$ and which from Sha. This is probably not a problem if I am for some reason sure that Chabauty method is applicable for my curve, beause I only need a bound for rank which I will eventually obtain for a large $n$. But what if I want to check whether method is applicable or not(i.e. to avoid false negative result)? | |
| Sep 12, 2016 at 15:14 | comment | added | Michael Stoll | ... In practice, this is feasible for hyperelliptic curves with $n = 2$ and in a few other rather special cases. (For elliptic curves, it looks a bit better; if the coefficients of the defining equation are not too large, one can do $n = 2,3,4,8,9$.) | |
| Sep 12, 2016 at 15:11 | comment | added | Michael Stoll | 2) That finiteness of Sha implies computability of $J(\mathbb Q)$ is a standard fact. At least in principle, one can compute the $n$-Selmer group $S_n(J)$ of $J$ for any $n \ge 2$. It contains an isomorphic image of $J(\mathbb Q)/n J(\mathbb Q)$ and the quotient is the $n$-torsion of Sha. So when Sha is finite, then for all sufficiently large prime $n$, $\#S_n(J) = n^r$, where $r$ is the rank of $J(\mathbb Q)$. On the other hand, one will find $r$ independent points in $J(\mathbb Q)$ by systematic search. ... | |
| Sep 12, 2016 at 15:07 | comment | added | Michael Stoll | @Sasha 1) The problem with zeros of analytic functions is that you may not be able to detect multiple zeros with a finite amout of computation; this is why we pick a special kind of prime, for which there are only simple zeros. Also, you usually have zeros that do not come from rational points, so you need some additional method to show this (this is the role of the Mordell-Weil Sieve). [to be continued] | |
| Sep 12, 2016 at 13:18 | comment | added | SashaP | Thank you! So, the method with zeroes of analytic function can not be made into an algorithm? Also, could you please give a reference on how finiteness of Sha implies computability of $J(\mathbb{Q})$, or I overlooked it in your paper? | |
| Sep 11, 2016 at 18:23 | history | answered | Michael Stoll | CC BY-SA 3.0 |