Timeline for Rational points à la Chabauty-Coleman
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Feb 15, 2019 at 7:01 | history | edited | Felipe Voloch | CC BY-SA 4.0 |
correct erroneous assertion
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| May 28, 2013 at 2:10 | comment | added | Maarten Derickx | I fail to see why $f(P) = aP$ even when $P$ is close enough to $0$ in $J(\mathbb Q_p)$. Assuming your argument works leads to strange things. Now if $p$ splits as $p_1p_2$ in $\mathbb Q(F)$ then applying your argument both to the completion w.r.t. $p_1$ and $p_2$ i get two $p-adic$ integers $a_1$ and $a_2$. By choosing p bigger then the discriminant of the order generated by $f$ one can make sure that $a_1-a_2$ is a unit. Now if I were to believe your argument then $f(P)=a_1P=a_2P$ but then $(a_1-a_2)P=0$ and hence $P=0$. | |
| May 28, 2013 at 2:09 | comment | added | Maarten Derickx | Ok, so my first part in trying to understand your argument is fixed :). Your answer seems valuable since it would give a criterion for when it might be possible to use Chabauty-Coleman in a cases $r \geq g$. I run into trouble in trying to make things concrete however. And I'm starting to suspect that this is because there is an error in your argument, but I'm quite new to this p-Adic analytic stuff so it might also be me. | |
| May 27, 2013 at 11:55 | comment | added | Felipe Voloch | You are correct. To define $\alpha P$ I need some assumption on $P$, e.g. that is zero $\mod p$. But, for the purpose of answering the question, I can always pass to a subgroup of finite index of the Mordell-Weil group where this is satisfied. | |
| May 27, 2013 at 4:01 | comment | added | Maarten Derickx | I'm curious on how you define $\alpha P$ for some $p$-adic number $\alpha$. From your argument its seems that you want to write $f = \sum a_i[p^i]$ in a completion of $\mathbb Q(f)$. But I don't see how $\sum a_i[p^i](P)$ will converge in $J(\mathbb Q_p)$. If $\bar P \in J(\mathbb F_p)$ is not of $p$-power order $[p^i](P)$ will not even become $0$ in $J(\mathbb F_p)$ for large $i$! | |
| Mar 31, 2011 at 13:30 | vote | accept | Barinder Banwait | ||
| Mar 28, 2011 at 21:46 | comment | added | Barinder Banwait | A comment above speaks of a conjecture due to Poonen which implies that if $J$ is simple, then $r'$ = min $g,r$. If true, then the toy example does not fall into your first paragraph scenario (I've inadvertently learned quite a lot about this particular curve!) | |
| Mar 27, 2011 at 0:28 | comment | added | Felipe Voloch | @Barinder. No, not in general. But I expect that a situation in which this happens can be arranged. | |
| Mar 26, 2011 at 22:15 | comment | added | Barinder Banwait | Thanks for the hints. In your first paragraph, is there always a $P \in J(\overline{\mathbb{Q}})$ such that $J(\mathbb{Q}) \subset \langle P,f(P)\rangle$? | |
| Mar 26, 2011 at 7:56 | history | edited | Felipe Voloch | CC BY-SA 2.5 |
added 245 characters in body; added 154 characters in body
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| Mar 26, 2011 at 2:54 | history | answered | Felipe Voloch | CC BY-SA 2.5 |