# Inertia subgroup of finite extension over $\mathbb{Q}_p$ whose ramification index is not divisible by $p$

Let $$K$$ be an extension of $$\mathbb{Q}_p$$ and let $$L/K$$ be a finite extension with $$p \nmid e$$ where $$e = e(L/K)$$ is the ramification index of $$L/K$$. Let $$I=I(L/K)$$ be the intertia subgroup of $$L/K$$.

Question Is there a result showing that $$I$$ must be cyclic?

I still have a vague understanding of things like inertia subgroups, Galois groups over $$\mathbb{Q}_p$$, etc., so I think a reference to obtain the basics to solve the above question would be the best for me. I don't mind an answer to my question though. Thanks!

• Yes this is true. By definition, $I_K$ is the Galois group of $L/K^{un}$, where $K^{un}$ is the maximal unramified subextension of $L$. By assumption, $L/K^{un}$ is totally tamely ramified. But every totally tamely ramified extension of degree $e$ is of the form $K(\pi^{1/e})$ where $\pi$ is a uniformiser. (See Theorem 11.9 of these notes.) Oct 28, 2020 at 15:42
• Sorry, isn't $L=K^{\text{un}}(\pi^{1/e})$?
– Lios
Oct 28, 2020 at 16:08

Yes, this is standard stuff. A finite, normal, totally tamely ramified extension $$L\supset K$$ of local fields has a Galois group that injects into $$\kappa^\times$$, where $$\kappa$$ is the (common) residue field.
If $$\pi$$ is a uniformizer of $$L$$, you send $$\sigma\in\text{Gal}^L_K$$ to the image of $$\frac{\sigma(\pi)}\pi$$ in $$\mathcal O_L/\pi\mathcal O_L=\kappa$$. Show it’s a homomorphism, and injective if the degree is prime to $$p$$. So the Galois group is cyclic.