Is the Inertia quotiented by the first group of ramification $I/G_1\simeq (\mathcal{O}_L/P)^*$?

Question

Consider $K$ number field, and $L/K$ a finite Galois extension with $G$ its Galois group and $P$ a prime of $\mathcal{O}_L$ above $p$ prime in $K$. Moreover suppose that the characteristic of the residue field $\mathcal{O}_K /p $ is a prime, say $q$. I recall that the $m$th group of ramification is defined as follows: $$G_m := \{ \sigma \in G : \sigma(x)\equiv x \bmod P^{m+1}, \forall x\in \mathcal{O}_L \}.$$This is clearly normal in the decomposition group $D$ and I have showed that, for $\pi \in P\setminus P^2$ uniformizer and for $m\ge 1$ and $\sigma \in G_0=I$ (the inertia group), $$\sigma \in G_m \iff \sigma(\pi)\equiv \pi \bmod P^{m+1}. $$ Now, I have showed that the map $\phi \colon G_0 \rightarrow (\mathcal{O}_L /P)^* , \sigma \mapsto \dfrac{\sigma(\pi)}{\pi}\bmod P $ has $G_1$ as kernel. So clearly it follows that $$G_0/G_1 \simeq \phi(G_0)\subseteq (\mathcal{O}_L /P)^*$$ Is the morphism $\phi$ surjective? If not, can someone find a counterexample? Thanks in advance!

Answer 1

Take $K={\bf Q}$, $L={\bf Q}(\sqrt{5})$ and $p=5$. Then $P=(\sqrt{5})$ and $O_L/P={\bf F}_5$, while $G_0/G_1$ has order $2$.