In characteristic zero, the <isogeny of elliptic curves> $\phi:E\to E'$ is finite and separable, so
$$
[K(E):K(E')]=\deg\phi=\#E[\phi].
$$
For every $T\in E[\phi]$, translation $\tau_T(P)=P+T$ satisfies $\phi\tau_T=\phi$. It therefore induces a $K(E')$-automorphism of $K(E)$. These translations are distinct, giving $\#E[\phi]$ automorphisms of an extension of the same degree. The extension is consequently Galois, and
$$
E[\phi]\longrightarrow\operatorname{Gal}(K(E)/K(E')),
\qquad T\longmapsto\tau_T^*,
$$
is an isomorphism.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 125 / 5 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-125/5/a/solution