1
$\begingroup$

Let $E$ be an elliptic curve defined over $\mathbb{Q}$ with equation $y^2=x^3+(ax+b)^2$, where $a,b \in \mathbb{Q}$.
I have to prove that $0$ and $(0, \pm b)$ are the rational $3$-torsion points of $E$. I know that if $Q$ is a $3$-torsion point then $[2]Q=-Q$ and that the $x$-coordinate of a $3$-torsion point satisfies $x(3x^3+4a^2x^2+12abx+12b^2)=0$
Now, clearly $0$ is a $3$-torsion point and if I choose $x=0$ and I plug it into the equation of $E$ I get $y^2=b^2$, thus $y= \pm b$.
But now I don't know to continue and prove that there are no other rational $3$-torsion points (for example if $a,b \in \mathbb{Z}$ I would have used the rational root theorem, but I don't know in this case..)

$\endgroup$
  • $\begingroup$ Over the algebraic closure $\overline{\Bbb{Q}}$ the curve has the full 3-torsion of eight points ($E[3]\simeq C_3\times C_3$). If you are familiar with the Weil pairing then the claim follows easily. For if $E[3]$ is rational over a field $K$, then $K$ must contain the values of the Weil pairing, that is, the third roots of unity. This rules out the possibility that $E(\Bbb{Q})$ could ever contain the full 3-torsion subgroup. $\endgroup$ – Jyrki Lahtonen Feb 23 at 20:37
  • $\begingroup$ I know that on $\overline{\mathbb{Q}}$ there are nine $3$-torsion points (the roots of $\psi_3^2$ and $0$), but I don't know Weil pairing, and I haven't proved anything yet about the group structure of $E[m]$ $\endgroup$ – user289143 Feb 23 at 20:39
  • $\begingroup$ Mind you, that curve has a singularity at $(-4,0)$ if $a=3,b=4$. In that case (probably also for other pairs $(a,b)$) you don¨t have an elliptic curve, This may be needed to rule out eventual other zeros of $\psi_3$. $\endgroup$ – Jyrki Lahtonen Feb 23 at 20:46
  • 1
    $\begingroup$ That's maybe helpful.. since $3x^3+4a^2x^2+12abx+12b^2=3(x^3+a^2x^2+2abx+b^2)+a^2x^2+6axb+9b^2=3y^2+(ax+3b)^2$ and this is $0$ iff $y=0$ and $x=-\frac{3b}{a}$ $\endgroup$ – user289143 Feb 23 at 20:50
1
$\begingroup$

Now $\Delta_E=16b(4a^3b^2-27b^3)$.
We want to find the solution for $3x^3+4a^2x^2+12abx+12b^2=0$ and we can rewrite it as
$3(x^3+a^2x^2+2abx+b^2)+a^2x^2+6abx+9b^2=3y^2+(ax+b)^2$ and this is equal to $0$ iff $y=0$ and $x=-\frac{3b}{a}$.
Thus, if we plug those values to the equation of $E$ we get $0=-\frac{27b^3}{a^3}+4b^2 \Leftrightarrow 4a^3b^2-27b^3=0 \Leftrightarrow \Delta_E=0 $ but then we get a singular curve, hence it cannot be.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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