# isogenous elliptic curves have same rank

This is based on exercise 14.3 from Cassels, Lectures on Elliptic Curves. Let $$E:y^2=x(x^2+ax+b), E':y^2=x(x^2+a_1x+b_1)$$ be two elliptic curves over $$\mathbb{Q}$$, with $$a_1=-2a$$, $$b_1=a^2-4b$$. We know that this means there exists a 2-isogeny from $$E$$ to $$E'$$.

Part (a) of the exercise is proving that the groups $$E(\mathbb{Q})$$ and $$E'(\mathbb{Q})$$ have isomorphic odd-order torsion. I managed to do this, however the second part is: Assuming the Mordell-Weil theorem, show that $$E(\mathbb{Q})$$ and $$E'(\mathbb{Q})$$ have the same rank.

I know that $$E(\mathbb{Q}) = E^{\text{tors}}(\mathbb{Q}) \oplus \mathbb{Z}^r$$, where $$r$$ is the rank. How can I use this to prove that the curves have the same rank?

• Use the induced group homomorphism $E(\mathbb Q)\to E'(\mathbb Q)$, which has finite kernel by definition. Commented Jun 12, 2019 at 8:05
• See this MO-question for an answer. Commented Jun 12, 2019 at 9:00

Consider the degree-$$2$$ isogeny $$\phi:E\to E'$$ and let $$\widehat{\phi}:E'\to E$$ be its dual isogeny. This just means that $$\widehat{\phi}\circ\phi=[2]:E\to E$$, where $$[2]$$ means is multiplication-by-$$2$$ map.

Now, since you assume the Mordell-Weil theorem, you know that $$E(\mathbb{Q})\cong T_{1}\times\Bbb Z^{r_{1}}$$ and $$E'(\Bbb Q)\cong T_{2}\times\Bbb Z^{r_{2}}$$, for some $$r_{1},r_{2}\in\Bbb{Z}_{\geq 0}$$, where $$T_{1},T_{2}$$ are the torsion subgroups of $$E(\Bbb Q),E'(\Bbb Q)$$, respectively. It is easy to see that the map $$\phi:E\to E'$$ restricts to a map

$$\phi:\Bbb Z^{r_{1}}\to\Bbb Z^{r_{2}}$$

since it sends torsion-free elements to torsion-free elements. This can be seen as follows: take a point $$P\in E(\Bbb Q)$$ and suppose that $$\phi(P)$$ it a torsion element in $$E'(\Bbb Q)$$ (i.e., not in $$\Bbb Z^{r_{2}}$$); then there exists $$m\in\Bbb Z$$ such that $$\phi([m]P)=[m]\phi(P)=O_{E'}$$ (here I denote by $$O_{E'}$$ the point at infinity of $$E'$$); hence, composing by $$\widehat{\phi}$$, we derive that $$[2m]P=0$$, which proves that $$P$$ is a torsion point of $$E(\Bbb Q)$$.

We can also observe that the restricted map $$\Bbb Z^{r_{1}}\to\Bbb Z^{r_{2}}$$ is also injective. This can be seen from the fact that isogenies have finite kernels, but in this case it follows also from the fact that $$\phi$$ is a degree-$$2$$ isogeny. Indeed, let $$P\in E(\Bbb Q)$$ be a torsion-free point and suppose that $$\phi(P)=0$$. Then, $$[2]P=\widehat{\phi}(\phi(P))=0$$, a contradiction to the choice of $$P$$.

We conclude (from the injectivity of $$\phi$$ restricted to the torsion-free part of $$E(\Bbb Q)$$) that $$r_{1}\leq r_{2}$$. Applying the same to $$\widehat{\phi}$$ we get the converse inequality and hence the equality $$r_{1}=r_{2}$$.

• Very nice answer! I was wondering: this reasoning is applicable also when we know that $E$ and $E'$ are isogenous but we don't know if $\phi:E\rightarrow E'$ is a 2-isogeny, right? Because you still know that $\phi\circ \hat{\phi}=[\deg\phi]$
– kubo
Commented May 28 at 9:31
• Exactly, the proof does not depend on the particular degree of the isogeny. Commented May 28 at 19:13