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!

  • 1
    $\begingroup$ 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.) $\endgroup$
    – Mathmo123
    Oct 28, 2020 at 15:42
  • 2
    $\begingroup$ Sorry, isn't $L=K^{\text{un}}(\pi^{1/e})$? $\endgroup$
    – Lios
    Oct 28, 2020 at 16:08

1 Answer 1


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .