# Rank of jacobians of twists of hyperelliptic curves of genus one

For $$a,b \in \mathbb{Z}$$ we define the binary quartic form

$$\displaystyle F_{a,b}(u,v) = a(u^2 - v^2)^2 + 4bu^2 v^2.$$

We shall assume throughout that the discriminant $$\Delta(F_{a,b}) = 4096a^2 b^2 (a-b)^2$$ of $$F_{a,b}$$ is non-zero; that is, the form $$F_{a,b}$$ is non-singular. Consider the twist family of genus one curves given by

$$\displaystyle d z^2 = F_{a,b}(u,v), d \in \mathbb{Z}.$$

Is it known that for infinitely many $$d$$, the curve given above has a rational point and has jacobian whose rank is at least one (so that the curve is isomorphic to its jacobian, which has positive rank)?

The Jacobian of the curve $$z^2 = F_{a,b}(u,v)$$ (i.e., when $$d = 1$$) is given by

$$y^2 = x^3 - \frac{16(a^2 - ab + b^2)}{3} - \frac{64(a+b)(2a-b)(a-2b}{27}$$ $$= \left( x - \frac{4b - 8a}{3}\right)\left( x - \frac{4a - 8b}{3}\right)\left( x - \frac{4b + 4a}{3}\right),$$

which is an elliptic curve with full rational 2-torsion. Therefore it is amenable to the techniques introduced by A. Smith (https://arxiv.org/abs/1702.02325), so we know that for 100% of quadratic twists of this curve the $$2^\infty$$-Selmer rank is at most one. In particular, ordering the twists by $$|d| \leq X$$, that only $$o(X)$$ of such twists will have rank at least two.

This is a stronger conclusion that can be reached for any large family of elliptic curves, since in that case we only have control of the sizes of $$p$$-Selmer groups for $$p = 2,3,5$$ on average due to work of Bhargava and Shankar. However in this setting there is work due to Bhargava and Skinner, and subsequently Bhargava-Skinner-Zhang, which shows that in fact a positive proportion of curves will have positive rank in any large family.

The answer is yes, and it's fairly elementary. By the usual 2-descent, the curve $$C$$ gives a class $$c$$ in $$H^1(\mathbb{Q},E[2])$$, where $$E$$ is the Jacobian you wrote down. As you vary $$d$$, the groups $$H^1(\mathbb{Q},E_d[2])$$ are canonically isomorphic, and $$c$$ is also the class of $$C_d$$. To answer your question, you should condition on the possibility that $$c$$ "comes from" the 2-torsion subgroup of $$E$$ via $$E(\mathbb{Q})/2E(\mathbb{Q}) \to H^1(\mathbb{Q},E[2])$$. I don't know what the conditions on $$a,b$$ for this to happen are, but it won't matter.

If $$C$$ doesn't come from 2-torsion, and if $$C_d$$ has a rational point, then automatically $$E_d$$ has rank at least one (by descent). And as Alex B. says, it's easy to see that there are infinitely many squarefree $$d$$ for which this happens (his argument doesn't quite show this, but it's not hard to fiddle with congruence conditions to produce infinitely many squareclasses).

If $$C$$ does come from 2-torsion, then $$C_d$$ has a rational point for all $$d$$, but there is no guarantee that the rank is at least 1. So your question reduces to asking about the ranks of the twists of $$E$$. But now just apply the argument from the previous paragraph to some other class in $$H^1(\mathbb{Q},E[2])$$ represented by a binary quartic form (there are infinitely many of these).

• Thank you!! This is immensely helpful. – Stanley Yao Xiao Jan 9 at 19:38

If you are willing to assume finiteness of Sha, then the answer is "yes".

The first claim, that for infinitely many $$d$$ (square-free, presumably?) this has a rational point, is easy and does not need any conjectural input: start with any $$u$$, $$v$$ such that $$F_{a,b}(u_0,v_0)$$ is non-zero. Then by writing $$F_{a,b}(u_0,v_0)$$ as a square times the square-free part, you immediately see that there exists a unique square-free $$d\in \mathbb{Z}$$ such that $$F_{a,b}(u_0,v_0)/d$$ is a square, say $$z_0^2$$, so that $$(u_0,v_0,z_0)$$ is a rational point on $$dz^2=F_{a,b}(u,v)$$.

For the second part, observe, e.g. by using the explicit equation for the Jacobian that you have given, that the Jacobian of the twist by $$d$$ is the twist by $$d$$ of the Jacobian. Since these Jacobians have full $$2$$-torsion, it follows from work of Kane that infinitely many of these have $$2$$-Selmer of dimension $$3$$, and if Sha is finite, then $$3$$-dimensional $$2$$-Selmer implies rank $$1$$.