 Solution

Let $G=\mathrm{Gal}(L/K)$. The lower ramification groups are $G_{-1}=G$ and, for $i\ge 0$,

If $L/K$ is totally ramified and ${\pi }_L$ is a uniformizer, then ${\mathcal{O}}_L={\mathcal{O}}_K[{\pi }_L]$. Factoring $F(\sigma {\pi }_L)-F({\pi }_L)$ by $\sigma {\pi }_L-{\pi }_L$ for $F\in {\mathcal{O}}_K[X]$ proves the uniformizer criterion for lower ramification groups

For $\sigma \in G_0$, defineThe inertia group $G_0$ acts trivially on $k_L$, so $\theta (\sigma \tau )=\theta (\sigma )\theta (\tau )$. Its kernel consists exactly of those $\sigma$ for which $\sigma ({\pi }_L)/{\pi }_L\equiv 1$ modulo the maximal ideal, namely $G_1$. The first isomorphism theorem therefore gives an injection