Let $L/K$ be a Galois extension of number fields with Galois group $G$. Let $O_K$ and $O_L$ be the ring of algebraic integers of $K$ and $L$ respectively. Let $P\subseteq O_K$ be a prime. Let $Q\subseteq O_L$ be a prime lying over $P$.
The $n$-th ramification group is defined as $$E_n(Q|P)=\lbrace \sigma\in G:\sigma(x)\equiv x\text{ mod } Q^{n+1}\text{ for all } x\in O_L\rbrace$$ In particular, $n=0$ gives the inertia group. How to prove the following:
- $E_n$ is a normal subgroup of $G$.
- $\cap_nE_n=\lbrace 1\rbrace$
The first part is obvious: if $ \sigma \in E_n $ and $ \tau \in G $, then $ \tau \sigma \tau^{-1}(x) \equiv \tau \tau^{-1}(x) \equiv x \pmod{\mathfrak q^{n+1}} $, so $ E_n $ is normal. For the second part, let $ \sigma \in \cap_n E_n $, and pick an arbitrary element $ x \in \mathcal O_L $. The given condition implies that $ x - \sigma(x) $ is divisible by arbitrarily large powers of the prime $ \mathfrak q $, however this is impossible unless $ x = \sigma(x) $ by unique factorization of ideals. Since $ x $ was arbitrary, $ \sigma $ acts trivially on $ \mathcal O_L $, and thus on the fraction field $ L $.