Let $p$ be a prime and let $K$ be a finite extension of $\mathbb{Q}_p$. Suppose $L/K$ is a tamely ramified Galois extension. I want to show that if $\sigma$ is a lift of the Frobenius element of the Galois group of the residue field extension and $\tau$ is a generator of $\textrm{Gal}(L/E)$ where $E$ is the maximal unramified subextension of $L$, then $\sigma\tau\sigma^{-1} = \tau^q$ where $q$ is the order of the residue field of $K$.

I know that $L = E(\lambda^{1/e}$) where $\lambda$ is some uniformizer for $E$ and $e$ is the ramification index of $L/K$. If $\sigma(\lambda^{1/e})$ is another $e$-th root of $\lambda$, then I can show the desired relation holds. However, it seems to me that we can't be sure this is the case, because $\lambda$ may not be in $K$.

  • $\begingroup$ The title doesn't correspond to the question. (To answer the implied question in the title, the Galois group is not always a semidirect product of the inertia group with a lift of Frobenius.) $\endgroup$
    – user760870
    Apr 29, 2020 at 11:42
  • $\begingroup$ @user760870 I think it is a semidirect product provided it's a tamely ramified extension a $p$-adic field? Using OP's notation, Galois group is generated by $\sigma, \tau$, subjected to $\sigma\tau\sigma^{-1} = \tau^q$. Am I missing something? $\endgroup$
    – pisco
    Apr 29, 2020 at 12:18
  • $\begingroup$ @pisco Yes you are missing something. I suggest thinking about Galois extensions $K/\mathbf{Q}_p$ of degree $4$ (when $p$ is odd) . $\endgroup$
    – user760870
    Apr 29, 2020 at 16:54
  • $\begingroup$ @user760870 Yes, you're right. I had the wrong impression that $\tau$ and $\sigma$ generates disjoint subgroups, which is not always the case. Thank you very much for pointing this out. $\endgroup$
    – pisco
    Apr 29, 2020 at 18:48
  • $\begingroup$ @user760870 So are you saying this relation may not always hold, or just that it may not be enough to determine the Galois group? $\endgroup$ Apr 29, 2020 at 23:42

1 Answer 1


Let $\pi$ be a uniformizer of the totally ramified extension $L/E$. We have a map: $$\theta: \text{Gal}(L/E) \to U_L/U^{(1)}_L\qquad \tau_0\mapsto \tau_0(\pi)/\pi$$ where $U_L$ is the unit group of $\mathcal{O}_L$, and $U^{(1)}_L = 1+\mathfrak{p}_L$. This map is independent of the unifromizer $\pi$ chosen. Its kernel is the wild ramification group, which is trivial since $L/K$ is assumed to be tamely ramified. Hence $\theta$ is injective.

Denote $\tau(\pi)\equiv a\pi \pmod{\mathfrak{p}_L^2}$. Then $$\tag{1}\sigma\tau(\pi) \equiv \sigma(a)\sigma(\pi) \equiv a^q\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$ because $\sigma$ acts like Frobenius. Moreover, since $\theta$ is independent of unifromizer, we have $\tau(\sigma(\pi))\equiv a\sigma(\pi) \pmod{\mathfrak{p}_L^2}$, so $$\tau^2(\sigma(\pi))\equiv \tau(a)\tau(\sigma(\pi))\equiv \tau(a)a\sigma(\pi) \equiv a^2\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$ the last equality follows from the fact that $\tau$ acts on residual field trivially. Induction shows $$\tau^q(\sigma(\pi))\equiv a^q\sigma(\pi) \pmod{\mathfrak{p}_L^2}$$

Comparing with $(1)$ shows that $\sigma\tau$ and $\tau^q\sigma$ have same image under $\theta$. Injectivity of $\theta$ proves your claim.


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