Show that the subgroup $\{D \in \text{Pic}F: 2D \sim 0\}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

Question

Let $F = y^2z - x^3 - \lambda xz^2 - \mu z^3$ be a smooth elliptic curve. Show that the subgroup $S = \{D \in \text{Pic}F: 2D \sim 0\}$ of $\text{Pic}F$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

Let $\Phi$ be the canonical bijective map between $V(F)$ and $\text{Pic}F$ that maps $P \mapsto P- P_0$ for a fixed base point $P_0$. I have already shown that for any $P \in V(F)$ such that $P- P_0 \in S$, we have

\begin{equation*} P \oplus P = 2\Phi^{-1}(P-P_0) = \Phi^{-1}(2P - 2P_0) =\Phi^{-1}(0) = P_0 \end{equation*}

This means that $P$ is an element of order 2 in the groups of $V(F)$ (if $P \neq P_0$), so the structure of $S$ is like $(\mathbb{Z}/2\mathbb{Z})^n$. How can I show that $S$ has exactly four elements?

Answer 1

Let $E = V(F)$ and suppose $P$ is a $2$-torsion point of $E$. On the affine open where $z \neq 0$, we can write \begin{equation} \label{eqn} \tag{$*$} E: y^2 = \overbrace{x^3 + \lambda x + \mu}^{f(x)} \end{equation} Write $P = (x_0,y_0)$ on this affine open. Since $2P = 0$, then $$ (x_0,y_0) = P = -P = (x_0,-y_0) \, , $$ so $y_0 = 0$. Substituting this into the local equation (\ref{eqn}) for $E$, we find $0 = f(x_0) = x_0^3 + \lambda x_0 + \mu$. Since $E$ is nonsingular, then $f$ has distinct roots, so there are $3$ possible values for $x_0$, and thus $3$ $2$-torsion points. Together with the identity, which we did not yet include, since it lies on the line $z = 0$ at infinity, this yields $4$ points $P$ such that $2P = 0$.

More generally, one can show that the $m$-torsion subgroup $E[m]$ of an elliptic curve $E$ is always isomorphic to $\newcommand{\Z}{\mathbb{Z}} (\Z/m\Z) \times (\Z/m\Z)$ for every $m \geq 1$ not dividing the characteristic of the base field. Over $\mathbb{C}$, this can be easily seen using the complex uniformization of $E$ as a complex torus $\mathbb{C}/\Lambda$, but the general proof is a bit more involved.