# rational points of a hyperelliptic curve of genus 3

Let $$K=\mathbb{Q}(\sqrt{-1}).$$ I have the following hyperelliptic curve of genus 3: $$C : y^2 = (x^2-x+1)(x^6+x^5-6x^4 -3x^3+14x^2-7x+1)$$ I want to find $$C(K)$$. My first attempt was to compute the Jacobian of the curve $$J(\mathbb{Q})$$. By Magma, $$J(\mathbb{Q})$$ has rank 0 and the torsion subgroup is isomorphic to $$\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/52\mathbb{Z}.$$ (generated by the root of one of the factors on the right side of the hyperelliptic curve equation and the difference of two points at infinity.) One can see that the reduction of $$J$$ modulo 5 has this group type.

In order to find the rank of $$J(K)$$, I need to compute the rank of the quadratic twist $$J^{-1}(\mathbb{Q})$$. According to Magma, it is impossible to obtain an upper bound for the rank so I got unlucky. When the rank is positive, I don't think it is computationally feasible to extract $$K$$-points on the curve. Is there another method for finding rational points generally on a hyperelliptic curves of even degree and odd genus $$3?$$ Any help would be appreciated.

• You don't need the Jacobian $J(K)$ to compute the points over $K$: as $C(K)$ can be put inside the symmetric product $C(\mathbb{Q})^{(2)}$, which can be send to $J(\mathbb{Q})$ injectively outside the Weierstrass points, and this last group is finite, one should be able to compute all the points but just computing the points of $J(\mathbb{Q})$ explicitly. – Xarles Dec 22 '20 at 11:32

It turns out that $$C(K) = C(\mathbb Q) = \{\infty_+, \infty_-, (0,1), (0,-1), (1,1), (1,-1)\}$$.

To see this, consider a point $$P \in C(K)$$ and write $$\bar{P}$$ for its image under the nontrivial automorphism of $$K$$. Then $$P + \bar{P}$$ is an effective divisor of degree 2 on $$C$$, which is defined over $$\mathbb Q$$ (this is what Xarles is alluding to in his comment). So for any effective rational divisor $$D$$ of degree 2 on $$C$$, the linear equivalence class of $$P + \bar{P} - D$$ is a rational point on $$J$$. As you state correctly, $$J(\mathbb Q)$$ is finite and is isomorphic to $${\mathbb Z}/2 \times {\mathbb Z}/52$$; it is generated by differences of rational points on $$C$$ together with the point of order 2 corresponding to the given factorization of the polynomial in the equation, which makes it easy to enumerate all these 104 points (in Magma, say). Magma represents these points as effective divisors in Mumford representation (minus a multiple of one of the points at infinity). So we look at all these points and check if these divisors are of degree 2 with points in the support defined over $$K$$. The only such divisors we find are sums of two rational points. This shows that there are no exceptional'' $$K$$-points, i.e., no $$K$$-points with $$x$$-coordinate not in $$\mathbb Q$$. If $$x(P) \in \mathbb Q$$ and $$P \notin C(\mathbb Q)$$, then $$P + \bar{P}$$ is in the class of pull-backs of a point on $${\mathbb P}^1$$; such points all give the same point on $$J$$. To deal with these, we observe that such points correspond bijectively to rational points on the quadratic twist $$C^{(-1)}$$. However, there are no such points, since the polynomial on the right hand side of the equation of $$C$$ is always positive. This finishes the proof.

The business with the exceptional points is related to the fact that the map from the symmetric square $$C^{(2)}$$ (there is an unfortunate clash of notations with the quadratic twist) to $$J$$ is not injective: there is a $$\mathbb P^1$$ contained in $$C^{(2)}$$ coming from pulling back points on $$\mathbb P^1$$ under the hyperelliptic double cover, and this $$\mathbb P^1$$ gets contracted to one point, whereas the map is injective otherwise.

Since every quadratic point on $$C$$ gives rise to a rational point on $$C^{(2)}$$, the computation above actually shows that there are the following exceptional quadratic points on $$C$$. $$(\tfrac{1\pm\sqrt{-3}}{2}, 0), (\pm\sqrt{2}, \pm 4\sqrt{2}-5), (\pm\sqrt{2}, \mp 4\sqrt{2}+5), (\tfrac{2\pm\sqrt{2}}{2}, \tfrac{-5\mp4\sqrt{2}}{4}), (\tfrac{2\pm\sqrt{2}}{2}, \tfrac{5\pm4\sqrt{2}}{4}), (-1\pm\sqrt{2}, -11\pm8\sqrt{2}), (-1\pm\sqrt{2}, 11\mp8\sqrt{2})$$ (The signs correspond in each pair of coordinates.)

• Just add that the group of points $J(\mathbb{Q})$ is generated by $\infty^+-\infty^-$ (of order 52) and a point of order 2 coming from the factor of degree 2 of the Weierstrass polynomial, a fact that can be shown easily once one know that the bound for the order of the torsion subgroup is 104. – Xarles Dec 22 '20 at 13:24
• @Xarles You are right, one needs the point of order 2 together with the differences of rational points to generate $J(\mathbb Q)$. I have edited my answer accordingly. – Michael Stoll Dec 22 '20 at 14:51
• I just realized that I didn't check the degree 2 divisors carefully. There are in fact some execptional quadratic points (but none over $K$), which I have now listed in the answer. – Michael Stoll Dec 22 '20 at 15:55
• Thank you for the rigorous answer. I have a simple question. In the non-exceptional case, my understanding is $P + \bar{P}$ is linearly equivalent to $\infty_{+} + \infty_{-}$. Thus $P = (x,\pm \sqrt{f(x)}$ with $x \in \mathbb{Q}$, $f$ is the polynomial on the right-hand side of the equation. So the non-exceptional $K-$points would have rational $x$ if $f(x) = -y_{1}^2$ for some rational $y_{1}.$ But $f$ happens to be positive for all $x$, so it doesn't have real roots and it cannot equal to $-y_{1}^2$ for any rational $y_1.$ This is equivalent to your argument, right? – bijection123 Dec 25 '20 at 5:00
• @bijection123 Yes, this is what I was trying to say. Put differently, there is a local obstruction against such points, since $C^{(-1)}(\mathbb R)$ is empty. – Michael Stoll Dec 25 '20 at 12:18