Let $C/K$ be a curve of genus > 2 over a number field $K$ and suppose there exists a $p \in C(K)$. Then a recurring theme in studying $C(K)$ is using the map $C \to J(C)$ normalized by sending $p$ to 0, for example Faltings proof that $C(K)$ is finite, Chabauty-Coleman, Mazurs determination of all $\mathbb Q$ rational points on $X_0(N)$.

I want to know if there are also techniques for studying $C(K)$ that don't use the Jacobian.

My main motivation is that I later want to explicitly apply these techniques to a curve $C/\mathbb Q$ for which I can prove that the closure of $J(\mathbb Q)$ is of finite index in $J(\mathbb Q_p)$ for all primes p so that Coleman-Chabauty does not work. The techniques don't necessarily need to fit in a nice theoretical framework, examples in the literature where rational point questions on specific curves are solved without using their Jacobian are welcome to!

One example that I know of is Runge's method which was recently succesfully used to study rational points on certain modular curves by Bilu and Parent http://arxiv.org/abs/0907.3306

  • $\begingroup$ The very conjectural "effective Mordell conjecture" (see Astérisque 183) would provide such a method. The analog of this conjecture is proven over function fields (proofs by Arakelov and Szpiro) and such a method is thus available over function fields. Over number fields, I don't know any method that does not use the Jacobian. $\endgroup$ Aug 24, 2014 at 20:54
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    $\begingroup$ I am wondering whether $J(\mathbb{Q})$ can indeed be dense in $J(\mathbb{Q}_p)$ for all $p$, as you say is true for your $C$. I actually think this can't ever happen, using an argument very much analogous to the one presented in this question: mathoverflow.net/questions/113968/… (to which you supplied the winning answer, incidentally). I am of course also curious what your $C$ is, if you'd care to disclose such information. :) $\endgroup$
    – R.P.
    Aug 25, 2014 at 22:42
  • $\begingroup$ You are right, there was a mistake, I indeed did can not proof density, only that the closure was of the same dimension, which still ruins Coleman-Chabauty. $\endgroup$ Aug 26, 2014 at 16:14
  • $\begingroup$ The $C$ I have in mind are the modular curves $X^+_{ns}(p)$. If you can prove that these don't have rational points except the ones coming from CM for big enough $p$ you will have answered Serre's question whether for all non CM $E/\mathbb Q$ and all $l>37$ the $l$-adic galois representation associated to $E$ will be surjective. $\endgroup$ Aug 26, 2014 at 16:16

2 Answers 2


There is a way of shifting the problem to other curves: start with an unramified covering $\pi \colon D \to C$ that is Galois over $\bar{\mathbb Q}$ and compute the associated Selmer set $\operatorname{Sel}^\pi(C)$. The set is finite and computable, at least in principle; its elements $\xi$ correspond to twists $\pi_\xi \colon D_\xi \to C$ of $\pi$ such that $$C(K) = \bigcup_\xi \pi_\xi(D_\xi(K)),$$ which "reduces" the problem to that of determining the sets of rational points on the curves $D_\xi$ (whose genus is larger than that of $C$, so they tend to be more difficult to deal with). One can then hope that other methods can be applied to these curves. In some cases, for example, these curves map to other curves of lower genus over $K$ or over larger fields, and one might be able to find their $K$-points or the images of the $K$-points on $D_\xi$.

The problems with this approach in practice are:

  • The computation of the Selmer set may be infeasible (it works reasonably well for 2-coverings of hyperelliptic curves, but not in many other cases);
  • As already mentioned, the covering curves are "worse" than $C$, and one needs some luck to reduce the problem of determining $D_\xi(K)$ to something manageable (again chances are best in the hyperelliptic case);
  • It is likely to be hard to get it to work for a family of curves like the $X^+_{\text{ns}}(N)$ that you mention in a comment above. The determination of $X^+_{\text{ns}}(13)({\mathbb Q})$ is a notorious open problem, even though the curve has genus only 3!
    EDIT: This problem is no longer open, see this paper by Balakrishnan, Dogra, Müller, Tuitman and Vonk.
That said, I'd like to see you make progress on this question — this would also be helpful for a project of mine.


Minhyong Kim uses Selmer varieties (which are the "unipotent" analogues of the Jacobian) parametrizing certain $\pi_1^{unip}$-torsors on curves to prove Siegel's finiteness theorem by relating the set of integral points on $\mathbb P^1_{\mathbb Z} - \{0,1,\infty\}$ to integral points on the corresponding Selmer variety.

The first paper in this direction appeared in 2005: http://people.maths.ox.ac.uk/kimm/papers/siegelinv.pdf His method is a refinement (in some sense) of the method of Chabauty-Coleman.

Many other papers are now written by M. Kim and others (including Hadian-Jazi); see for instance http://people.maths.ox.ac.uk/kimm/

There are also lecture videos that you can easily find online given at the Newton institute and the IHES. In these M. Kim explains his work.

Finally, and most importantly, there is an MO post with an explanation of the main ideas here: Why should I believe the Mordell Conjecture?

  • $\begingroup$ The method of Kim is very interesting but he can only prove Siegel's theorem for the projective line, as far I can see. In that case, (much) more elementary proofs can be made quite effective (see Silverman's book on elliptic curves) but even this does not give a method to find all the rational points on a curve. $\endgroup$ Aug 25, 2014 at 21:25
  • $\begingroup$ @DamianRössler Yes, you are right. Note that there are some other papers on his website in which he and others apply these methods to other curves. I didn't realize the OP was asking for effective techniques and probably only read his "I want to know if there are also techniques for studying $C(K)$ that don't use the Jacobian". (Quite interestingly: somethin special to Kim's approach is that it shows that these classical finiteness statements (proven by Siegel) are related to other conjectures like the Fontaine-Mazur conjecture.) $\endgroup$ Aug 25, 2014 at 22:17

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