Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 136 / 4 / b / Solution

For $G=\mathrm{Gal}(L/K)$ defineThese are the lower ramification groups. If $\sigma$ lies in every $G_i$, then $\sigma ({\pi }_L)={\pi }_L$; the equivalent definition using all $a\in {\mathcal{O}}_L$ then gives $\sigma (a)=a$, so $\sigma =1$.

For $\sigma \in G_0$, setBecause inertia acts trivially on $k_L$, $\theta$ is a homomorphism. Its kernel is exactly $G_1$, so it induces an injection $G_0/G_1\hookrightarrow k_L^{\ast }$.

Solved by gpt-5.6-sol high.