Torsion Points

Consider the multiplication-by-$m$ map $[m]$. Then the group of all points $P$ such that $[m] P = O$ is denoted $E[m]$. [My notes are unclear here, but I think we are now working over the algebraic closure of $K$.]

Let $q = \mathrm{char} K$.

Fact: If $m$ is coprime to $q$ then $|E[m]| = m^2$, and furthermore $E[m] \cong \mathbb{Z}_m \times \mathbb{Z}_m$.

Fact: If $E[q] \ne \{O\}$ then $E[q^v] \cong \mathbb{Z}_{q^v}$ for all $v \gt 0$.

We can combine the above results (using the fact that $E[ab] = E[a] \times E[b]$ for coprime $a,b$. Let $m$ be a positive integer. Write $m = q^v n$ where $q$ does not divide $n$.

  • If $E[q] = \{O\}$ we have $E[m] \cong \mathbb{Z}_n \times \mathbb{Z}_n$

  • Otherwise $E[m] \cong \mathbb{Z}_n \times \mathbb{Z}_m$