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:

  1. $E_n$ is a normal subgroup of $G$.
  2. $\cap_nE_n=\lbrace 1\rbrace$
  • 1
    $\begingroup$ Your $E_n$ is presented as the kernel of map from the Galois group to $\text{Aut}(\mathcal{O}_L/Q^{n+1})$ so its normal. One way to show 2. is to note that your group is obviously contained in the decomposition group, so can be computed locally, and then use the fact that $\mathcal{O}_L$ is complete (so that $\varprojlim \mathcal{O}_L/Q^{n+1}=\mathcal{O}_L$). $\endgroup$ – Alex Youcis Oct 4 '16 at 12:22

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 $.

  • $\begingroup$ brilliant solution. $\endgroup$ – learning_math Oct 4 '16 at 18:28

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