Short exact sequence of modules and elliptic curves

Question

Let $E/\mathbb{Q}$ be an elliptic curve with a 3-torsion point $P$. Let $E_{d}$ denote the quadratic twist of $E$ by $d$. Then the action of $\sigma \in \operatorname{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on $E_{d}[3]$ can be written as the matrix $$\begin{pmatrix} \chi_{d}(\sigma) & \ast\\0 &\chi_{d}(\sigma)\omega(\sigma)\end{pmatrix}$$ where $\chi_{d}$ is the quadratic character associated to the quadratic twist and $\omega$ is the mod 3 cyclotomic character. Why does this yield the short exact sequence of modules: $$0 \rightarrow M_{1} \rightarrow E_{d}[3] \rightarrow M_{2} \rightarrow 0$$ where $M_{1}$ is the module $\mathbb{F}_{3}$ endowed with the Galois action through $\chi_{d}$ and $M_{2}$ is the module $\mathbb{F}_{3}$ endowed with the Galois action through $\chi_{d}\omega$?

Answer 1

This fact is true in general, elliptic curves play no particular role here. Suppose you have a prime $p$ and a $2$-dimensional $\mathbb{F}_p$-vector space $V$, with an action of a group $G$. Suppose that $P$ and $Q$ form a basis of $V$, and suppose that the action of $\sigma\in G$ on $V$ is given by a matrix: $$\begin{pmatrix} \phi(\sigma) & \tau(\sigma)\\0 &\psi(\sigma) \end{pmatrix}.$$ It follows that $\phi$ and $\psi$ are $\mathbb{F}_p^\times$-valued characters on $G$. Then, $V_1=\langle P \rangle$ is an invariant subspace under the action of $G$, and clearly $G$ acts on $V_1$ via $\phi$. Hence, the sequence $$0 \to V_1 \to V \to V/V_0 \to 0$$ is an exact sequence of $\mathbb{F}_p[G]$-modules.

What is the action of $G$ on $V_2=V/V_0$? Let $[v]$ be a class in $V_2=V/V_0$, where $v\in V$. Since $P$ and $Q$ form a basis, we have $v=\lambda P + \mu Q$ for some $\lambda,\mu\in \mathbb{F}_p$. Thus, if $\sigma\in G$, we have $$\sigma(v) = \lambda \sigma(P)+\mu \sigma(Q) = (\lambda \phi(\sigma) + \mu \tau(\sigma)) P + \mu \psi(\sigma) Q.$$ Therefore, $\sigma([v]) = [\sigma(v)]\in V_2 = V/V_1 = V/\langle P \rangle$ is given by $$\sigma([v]) = [\sigma(v)]=[(\lambda \phi(\sigma) + \mu \tau(\sigma)) P + \mu\psi(\sigma) Q] = [\mu\psi(\sigma)Q]=\psi(\sigma)[\mu Q] = \psi(\sigma)[v].$$ Hence, $G$ acts on $V_2$ via $\psi$.