I'm reading "On Ranks of Twists of Elliptic Curves and Power-Free Values of Binary Forms" by Stewart and Top, and struggling to understand the argument on pg 962 which shows that the rank of a particular elliptic curve $E_{D(t)}/\mathbb{Q}(t)$ is exactly 2.

Here are the relevant details:

Start with the elliptic curve $$E/\mathbb{Q}: y^2 = x^3 + 1$$ and the polynomial $$D(t) = 2t(t - 1)(t + 1)(2t + 1)(t + 2) \in \mathbb{Z}[t].$$ Let $C/\mathbb{Q}$ be the curve given by $s^3 = D(t)$ and let $$E_D/\mathbb{Q}(t): y^2 = x^3 + D(t)^2.$$ For each point $P = (x(t), y(t))$ in $E_D(\mathbb{Q}(t))$, we define an element $\phi_P$ of $\text{Mor}_\mathbb{Q}(C, E)$ by $$\phi_P(t, s) = (x(t)/s^2, y(t)/s^3).$$ Then we have a map $$\lambda: E_D(\mathbb{Q}(t)) \to H^0(C, \Omega^1_{C/\mathbb{Q}})$$ given by $$\lambda(P) = \phi_P^\ast \omega_E$$ which is shown to be a homomorphism with finite kernel.

We want to use this homomorphism to show that the rank of $E_D/\mathbb{Q}(t)$ is exactly two.

First, we can find two points, and show they are independent by looking at their images under $\lambda$. This is fine, and shows that the rank is at least 2.

Next, we want to show that the rank is at most 2. We know that the image lands in $H^0(C, \Omega^1_{C/\mathbb{Q}}(\zeta_3))$, the eigenspace on which the automorphism of $C$ given by $\zeta(t, s) = (t, \zeta_3 s)$ acts on differentials as multiplication by $\zeta_3$. This constrains the image to the 3-dimensional space, say spanned by $\omega_1, \omega_2, \omega_3$ (this numbering is different from the paper).

All of this makes sense to me. What doesn't make sense is how the authors constrain the image to a 2-dimensional subspace.

They define three involutions on $C$, called $\sigma_1, \sigma_2, \sigma_3$, and show that the space of $\sigma_i^\ast$-invariant holomorphic differentials is generated by $\omega_i$. Hence the quotient of $C$ by $\sigma_i$ is an elliptic curve. In two cases, the curve is isogenous to $E$ over $\mathbb{Q}$, and in the third case it is not. From this, they somehow infer that the rank is 2. I'm very confused about this, and would appreciate some more details or a reference.


  • $\begingroup$ I think you answered your own question... the image must land in the space corresponding to $\omega_1, \omega_2$ (or $\omega_3, \omega_4$ as labelled in the original paper), since the third case yields a curve which is not isogenous, which is the natural notion of 'isomorphic' for elliptic curves. $\endgroup$ Jul 10, 2015 at 2:08
  • $\begingroup$ That's the part I don't understand... why the image must land in a space which yields an isogenous curve in this way. $\endgroup$
    – stl
    Jul 10, 2015 at 12:45

1 Answer 1


Suppose that $\omega$ is the invariant differential of an elliptic quotient $C/\sigma$ of $C$. In particular, $\omega$ is non-zero. If $\omega$ lies in the image of $\lambda$, then there exists a point $P = (x(t), y(t))$ on $E_D(\mathbb{Q}(t))$ such that $\lambda(P) = \omega$. Let $\rho_1(P)$ denote the morphism in $\operatorname{Mor}_{\mathbb{Q}}(C/\sigma, E)$ corresponding to $\lambda_1(P)$. Then, as a morphism of curves, it must be either finite or constant. Since $\omega$ is non-constant, it must be the former; so $\rho_1(P)$ is an isogeny between $C/\sigma$ and $E$. Hence, this shows that any elliptic quotient corresponding to an element in the image of $\lambda$ must be isogenous to $E$.

Since, as you noted in the question, that the quotients $C/\sigma_1, C/\sigma_2$ are isogenous but $C/\sigma_3$ is not isogenous to the first two, $E$ can be isogenous to at most two of them. Since we have the lower bound of two, the rank is exactly two.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.