$2$-torsion points of an elliptic curve over $\mathbb{F}_p$

Question

Let $E : y^2=x^3+b$ be an elliptic curve over $\mathbb{F}_p$ ($p$ is a prime) with $b \in \mathbb{F}_p$ and $p=2 \mod 3$.

How can I show that $E[2]$ (the $2$-torsion points of $E$ on a algebraic closure of $\mathbb{F}_p$) is not included in $E(\mathbb{F}_p)$, the points of $\mathbb{F}_p$ in $E$.

Thank you.

Answer 1

I assume that $p>2$ and $b\neq0$ as otherwise the curve is not elliptic.

For curves in this form their $2$-torsion points are the affine points with $y=0$. If $y=0$, then also $x^3+b=0$. The equation $x^3+b=0$ has three zeros in the algebraic closure $\overline{\Bbb{F}_p}$, and your task is to prove that not all of the three solutions are in $\Bbb{F}_p$.

This is where the congruence condition gets used:

• Show that $p-1$ is not divisible by three.
• Because the group $\Bbb{F}_p^*$ is cyclic of order $p-1$ this implies that cubing is an injective (hence also bijective) mapping from $\Bbb{F}_p$ to itself. Do you see why?
• This implies that every non-zero element of $\Bbb{F}_p$ has a single cube root in $\Bbb{F}_p$ (and two other cube roots in the quadratic extension field $\Bbb{F}_{p^2}$.

Anyway, the conclusion is that $E[2]$ has one finite point in $E(\Bbb{F}_p)$ and the two remaining points are in $E(\Bbb{F}_{p^2})\setminus E(\Bbb{F}_p)$.