3
$\begingroup$

Let $K=\mathbb{Q}(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(35)$?

Recall that $X_0(35)$ is a hyperelliptic curve of genus 3 and has the simplified affine model: \begin{align*} y^2&=x^8-4x^7-6x^6-4x^5-9x^4+4x^3-6x^2+4x+1\\ &=(x^2+x-1)(x^6-5x^5-9x^3-5x-1) \end{align*} My attempt at finding $K$-rational points on $X_0(35)$ is as follows: First I find a rational map $f$ from $X_0(35)$ to a quotient curve $E$ of $X_0(35)$ with $E$ an elliptic curve (which is induced by the involution $w_5$ by Kubert). Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank 0, $E(K)$ is finite. However, in my case $E(K)$ is of rank 1. As a result it is computationally infeasible to determine the preimages of the infinitely many points of $E(K)$. Is there a way to work-around this issue?

Any help in finding $K$-rational points on $X_0(35)$ would be appreciated.

$\endgroup$
2
  • $\begingroup$ Can you try and brute force enough points and then hope that the Chabauty-Coleman bound is sharp? $\endgroup$
    – Asvin
    Aug 2, 2020 at 0:40
  • $\begingroup$ I'm not familiar with the Chabauty-Coleman technique but I will try to make it work out. Thank you for your input. $\endgroup$
    – 5W1H
    Aug 2, 2020 at 1:13

1 Answer 1

13
$\begingroup$

The group $J_0(35)(\mathbb Q)$ (where $J_0(35)$ is the Jacobian of $X_0(35)$) has rank 0 (as shown for example by a 2-descent computation in Magma); it is isomorphic to ${\mathbb Z}/24{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$, with generators the difference of the two points at infinity on $X_0(35)$ and the 2-torsion point corresponding to the factorization of the polynomial on the right hand side of the hyperelliptic equation. (One can check that the two points generate a group of the given isomorphism type, and the reduction of $J_0(35)$ mod 3 has a group of this type as its group of ${\mathbb F}_3$-points.)

If $P$ is a point in $X_0(35)(K)$ with $x$-coordinate not in $\mathbb Q$, then the sum of $P$ with its Galois conjugate, minus the sum of the two points at infinity, gives rise to a nonzero $\mathbb Q$-rational point on $J_0(35)$. One can check that none of the points is of this form.

This leaves the case when $x(P) \in \mathbb Q$. There are the $\mathbb Q$-rational points (the two points at infinity and the two points with $x(P) = 0$); for all other such points, the Galois conjugate of $P$ must be the image of $P$ under the hyperelliptic involution, so $y(P)$ must be $\sqrt{-2}$ times a rational number. Put differently, $P$ gives rise to a $\mathbb Q$-rational point on the quadratic twist of $X_0(35)$ by $-2$. But this twist does not even have ${\mathbb Q}_2$-rational points, let alone $\mathbb Q$-rational ones.

The conclusion is that $$ X_0(35)(K) = X_0(35)({\mathbb Q}) = \{\infty_+, \infty_-, (0,1), (0,-1)\} . $$

$\endgroup$
2
  • $\begingroup$ I wasn't sure how to deal with the case where x-coordinate is not in $\mathbb{Q}$. Thank you so much for clarifying my confusion. $\endgroup$
    – 5W1H
    Aug 3, 2020 at 20:49
  • $\begingroup$ So, the argument shows that for any quadratic field $K=\mathbb{Q}(\sqrt{d})$, where $d\in \mathbb{Z}$ is square free, $X_0(35)(K)=X_0(35)(\mathbb Q)$ if and only if the quadratic twist of $X_0(35)$ by $d$ does not have any rational point. Even more,$$X_0(35)(K)=X_0(35)(\mathbb Q)\cup X_0^{\langle d\rangle }(35)(\mathbb Q)$$ where the union is disjoint and $X_0^{\langle d\rangle }(35)(\mathbb Q)$ denotes the quadratic twist by $d$ of $X_0(35)(\mathbb Q)$ $\endgroup$ Sep 12, 2020 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy