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I have been trying to learn the method of Chabauty and Coleman to find rational points on curves; I have been reading an exposition by McCallum and Poonen which was pointed out to me by Emerton in this question.

Let $X$ be a curve of genus $g$ over $\mathbb{Q}$ with jacobian variety $J$, let $p$ be a prime of good reduction, and let $\overline{J(\mathbb{Q})}$ be the $p$-adic closure of the Mordell-Weil group $J(\mathbb{Q})$ in $J(\mathbb{Q}_p)$. Denote by $r'$ the dimension of the $p$-adic manifold $\overline{J(\mathbb{Q})}$.

The main assumption of the approach is that $r' < g$. This is automatic if $r < g$, where $r$ is the rank of $J$, because in general one has $r' \leq r$. This last inequality needn't be equality, "since $\mathbb{Z}$-independent points in log $J(\mathbb{Q})$ need not be $\mathbb{Z}_p$-independent".

How do I compute $r'$?

I wrote down a toy example, that is, $X : y^2 = x^5 + 17$. Here $r = 2$, and the method might work if $r'$ was 0 or 1, but I don't know how to check this.

I suspect that $r' = 2$, in which case the method is not even applicable, and I must think harder, but my question is not about this example, rather the general approach.

Is there an example of a curve $X$ with $r = g = 2$ but with $r' = 0$ or 1?

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  • $\begingroup$ If you know how to compute the $p$-adic logarithm on the formal group for your $J$, then you could just check by computing with sufficient precision if the image of the generators of $J(\mathbb{Q})$ are independent in the Lie algebra. $\endgroup$
    – Chris Wuthrich
    Commented Mar 26, 2011 at 13:03
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    $\begingroup$ There is a conjecture due to Bjorn Poonen (www-math.mit.edu/~poonen/papers/leopoldt.pdf) which implies that if $J$ is simple, then $r'=r$. So, if you believe it you will have to pick $J$ to be a product of two elliptic curves, with ranks 2 and 0 respectively.For any Jacobian of this type $r'$ will be 1. $\endgroup$
    – Tzanko Matev
    Commented Mar 28, 2011 at 16:28
  • $\begingroup$ Thanks for that, it was just what I needed. In the toy example, $J$ is indeed simple, so there (assuming the conjecture) $r' = 2$, and the Chabauty-Coleman method will not work. $\endgroup$
    – Barinder Banwait
    Commented Mar 28, 2011 at 21:31
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    $\begingroup$ A small remark. I just read the paper on Bjorn Poonen his website and in his article and he talks about "Guess" and "Question" and not about conjecture. I have the feeling he does this on purpose and that this is meant to indicate that he is not really convinced yet that the answer to his question will be positive. $\endgroup$
    – Maarten Derickx
    Commented May 28, 2013 at 1:20

1 Answer 1

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If $r=2$ then $r'>0$. For an example where $r'=1$, take a curve such that the jacobian has a nontrivial endomorphism $f$ and such that the group of points in the jacobian is generated by $P,f(P)$ for some point $P$ Now find a prime $p$ splitting in $\mathbb{Q}(f)$ so that $f(P)=\alpha P$ for some $p$-adic number $\alpha$. Then $r'=1$.

Added later: In your toy example, the endomorphism ring contains the fifth roots of unity so it may fall in my example above for those primes that split in that ring. A reasonable conjecture would be that if the endomorphism ring of the jacobian is $\mathbb{Z}$, then $r' = \min \{ g,r \}$. This is likely to be very hard to prove, as it is an abelian variety analogue of Leopoldt's conjecture.

Added much, much later: There are problems with my construction and what I say is probably not right. M.D. had a good reason to be skeptical in the comments. This was pointed out to me by Bjorn Poonen in a recent email exchange. His question is discussed further in a paper of M. Waldschmidt On the p-adic closure of a subgroup of rational points on an Abelian variety, Afr. Mat. 22 (2011), no. 1, 79–89.

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  • $\begingroup$ Thanks for the hints. In your first paragraph, is there always a $P \in J(\overline{\mathbb{Q}})$ such that $J(\mathbb{Q}) \subset \langle P,f(P)\rangle$? $\endgroup$
    – Barinder Banwait
    Commented Mar 26, 2011 at 22:15
  • $\begingroup$ @Barinder. No, not in general. But I expect that a situation in which this happens can be arranged. $\endgroup$
    – Felipe Voloch
    Commented Mar 27, 2011 at 0:28
  • $\begingroup$ 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!) $\endgroup$
    – Barinder Banwait
    Commented Mar 28, 2011 at 21:46
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    $\begingroup$ 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. $\endgroup$
    – Felipe Voloch
    Commented May 27, 2013 at 11:55
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    $\begingroup$ 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 $P=0$. $\endgroup$
    – Maarten Derickx
    Commented May 28, 2013 at 2:10

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