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

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!

• You can take $L = K$. – Lorem Ipsum Jan 26 at 22:04

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 $$2$$.