Ramification group

Remarks

Each $G_i$ is a normal subgroup of $G$, and $G_{-1} = G \supseteq G_0 \supseteq G_1 \supseteq \cdots$. The group $G_0$ is the inertia group, and $G_i = 1$ for $i$ large. If $p$ is the characteristic of the residue field, then $G_0 / G_1$ is cyclic of order prime to $p$, and $G_i / G_{i+1}$ is an elementary abelian $p$-group for $i \ge 1$. If $\mathcal{O}_L = \mathcal{O}_K[\alpha]$, then $\sigma \in G_i$ exactly when $v_L(\sigma(\alpha) - \alpha) \ge i + 1$. Example: for $L = \Q_2(i)$ and $K = \Q_2$, one has $\mathcal{O}_L = \Z_2[i]$ and $v_L(2) = 2$. For complex conjugation $\sigma$, $v_L(\sigma(i) - i) = v_L(-2i) = 2$, so $\sigma \in G_1$ and $\sigma \notin G_2$. Non-example: if $L / K$ is unramified and $\sigma \neq 1$, then $\sigma$ acts nontrivially on the residue field, so $v_L(\sigma(a) - a) = 0$ for some $a$ and $\sigma \notin G_0$. There is also an upper numbering $G^u$, defined through the Herbrand function, which behaves well under quotients.