# Finding $K$-rational points on $X_0(35)$

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.

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

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 $$X_0(35)(K) = X_0(35)({\mathbb Q}) = \{\infty_+, \infty_-, (0,1), (0,-1)\} .$$
• 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.
• 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 $X_0(35)(\mathbb Q)$ Sep 12, 2020 at 10:44