For $G=\operatorname{Gal}(L/K)$ define
$$
G_0=\ker(G\to\operatorname{Gal}(k_L/k_K)),\qquad
G_i=\{\sigma\in G:v_L(\sigma(\pi_L)-\pi_L)\geq i+1\}\quad(i\geq1).
$$
These are the lower <ramification group>[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 $\sigma\in G_0$, set
$$
\theta(\sigma)=\overline{\sigma(\pi_L)/\pi_L}\in k_L^*.
$$
Because inertia acts trivially on $k_L$, $\theta$ is a homomorphism. Its kernel is exactly $G_1$, so it induces an injection $G_0/G_1\hookrightarrow k_L^*$.

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-136/4/b/solution