<Mod-three Galois representation of an elliptic curve> is the representation
$$
\rho_3:G_{\mathbb Q}\longrightarrow\operatorname{GL}(E[3])\simeq\operatorname{GL}_2(\mathbb F_3)
$$
in this case. Twisting by the quadratic character $\chi_d$ replaces it by $\chi_d\rho_3$. Thus $E^d$ has a rational point of order three exactly when $E[3]$ contains a nonzero vector $v$ satisfying
$$
\rho_3(\sigma)v=\chi_d(\sigma)v
$$
for every $\sigma\in G_{\mathbb Q}$: the line $\mathbb F_3v$ is a one-dimensional Galois subrepresentation with character $\chi_d$.

The two-dimensional representation $E[3]$ has at most two distinct one-dimensional characters among its <Jordan–Hölder factors>. Therefore at most two quadratic characters $\chi_d$, and hence at most two rational isomorphism classes of twists, can have a rational point of order three.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 1 / d / Solution

{c}
{title2=$\rho_{E,3}:G_K\to\operatorname{GL}_2(\mathbb F_3)$}

The absolute Galois group acts on $E[3]\simeq\mathbb F_3^2$, giving the mod-three Galois representation. Its invariant vectors are precisely the rational points of order dividing three.


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