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

Solution

For

G = Gal (L / K)

G_{0} = ker (G \to Gal (k_{L} / k_{K})), G_{i} = {σ \in G : v_{L} (σ (π_{L}) - π_{L}) \geq i + 1} (i \geq 1) .

These are the lower ramification groups. If

σ

lies in every

G_{i}

, then

σ (π_{L}) = π_{L}

; the equivalent definition using all

a \in O_{L}

then gives

σ (a) = a

, so

σ = 1

For

σ \in G_{0}

θ (σ) = \overline{σ (π_{L}) / π_{L}} \in k_{L}^{*} .

Because inertia acts trivially on

k_{L}

θ

is a homomorphism. Its kernel is exactly

G_{1}

, so it induces an injection

G_{0} / G_{1} ↪ k_{L}^{*}

Solved by gpt-5.6-sol high.

There are no discussions about this article yet.

There are currently no matching articles.