$C: Y^2=X(X^2+aX+b)$

$D: Y^2=X(X^2+a_1X+b_1)$

where $a,b,\in\mathbb Z a_1=-2a,b_1=a^2-4b,b(a^2-4b)\neq0$

Let $C_{oddtors}(\mathbb Q)$ denote the set of torsion elements of $C(\mathbb Q)$ which have odd order and $D_{oddtors}(\mathbb Q)$ denote the set of torsion elements of $D(\mathbb Q)$ which have odd order. Show that $C_{oddtors}(\mathbb Q)$ and $D_{oddtors}(\mathbb Q)$ are isomorphic.

I don't quite know where to start on this?

I've already done a section on a 2-isogeny on an elliptic curve and I know that this is where I get the two curves from.

I've considered trying to finding the discriminant and perhaps using Nagell-Lutz Theorem to give an idea of what the torsion points could be.

$d_C=b^2(4b-a^2)$ and $d_D=-16b(a^2-4b)^2$ but then how can I purposely restrict to just looking at the odd torsions?

Any hints in the right direction will be appreciated.

Also, does the question implicitly imply that $C_{eventors}(\mathbb Q)$ and $D_{eventors}(\mathbb Q)$ are not necessarily isomorphic?

  • $\begingroup$ Those two are sets, so when you talk about isomorphism between them, what exactly do you mean? That they have the same number of elements? That's the only set-isomorphism I can think of... $\endgroup$
    – DonAntonio
    Feb 28, 2014 at 13:28
  • $\begingroup$ By the group law, they are also groups and hence makes sense to take about isomorphism? $\endgroup$
    – Haikal Yeo
    Feb 28, 2014 at 16:43
  • 1
    $\begingroup$ An isogeny is also a homomorphism of groups (at least if it takes the point at infinity to the point at infinity). It seems to me that everything would follow from the kernel of a $2$-isogeny to be $2$-torsion. Thus the isogeny will map odd torsion to odd torsion. Furthermore, the dual isogeny does the same in the opposite direction, and their composite is the identity multiplied by the degree... $\endgroup$ Feb 28, 2014 at 18:10
  • $\begingroup$ I don't understand why it follows from the kernel of a 2-isogeny to be 2-torsion though. Could you perhaps elaborate it into an answer? $\endgroup$
    – Haikal Yeo
    Feb 28, 2014 at 18:14

1 Answer 1


Let $E$ and $E'$ be elliptic curves, and let $\phi:E\to E'$ be a $p$-isogeny (i.e., $\phi$ is an isogeny of degree $p$), where $p$ is prime. In particular, $\phi$ is a group homomorphism from $E$ to $E'$ and its kernel $\ker(\phi)$ is a group of size $p$.

  • Prove that every $P$ in $\ker(\phi)$ has order dividing $p$.

  • Prove that the prime-to-$p$ torsion subgroup of $E$ injects into $E'$. Hint: if $P$ is a torsion point of order $n$ with $\gcd(n,p)=1$, and $\phi(P)=\mathcal{O}_{E'}$, the zero of $E'$, then $P\in\ker(\phi)$... So what is $n$?

  • Now consider the dual isogeny $\hat{\phi}:E'\to E$ to show that the prime-to-$p$ torsion subgroup of $E'$ injects into $E$.

About your last question, the "even torsion" in two isogenous curves need not be isomorphic. Let $E:y^2=x^3-x$ and $E':y^2=x^3+4x$. Then $E$ and $E'$ are $2$-isogenous, but $E(\mathbb{Q})_\text{tors}\cong \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ and $E'(\mathbb{Q})_\text{tors}\cong \mathbb{Z}/4\mathbb{Z}$.

  • $\begingroup$ In your second bullet point, do you mean that the image of $P$ is the ring of integers of $E'$? What does that mean? $\endgroup$
    – Haikal Yeo
    Mar 2, 2014 at 11:27
  • $\begingroup$ $\mathcal{O}_{E'}$ is the zero element of $E'$. $\endgroup$ Mar 2, 2014 at 13:00
  • $\begingroup$ Might you also be able to answer the last question about even torsion? I'm guessing that they will still be isomorphic... $\endgroup$
    – Haikal Yeo
    Mar 3, 2014 at 15:00
  • $\begingroup$ @HaikalYeo, I updated my answer. $\endgroup$ Mar 3, 2014 at 15:58

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