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..)

  • $\begingroup$ 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
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    $\begingroup$ 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

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.


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