Let $K$ be a local field complete with a discrete valuation. In a book I am reading it states that $\operatorname{Gal}(K^{nr}/K)$ is isomorphic to $\operatorname{Gal}(\bar{k}/k)$, but I wasn't sure how this was true.

How does one see this? I would greatly appreciate any comments and suggestions. Thank you.

  • $\begingroup$ The residue field of $K^{nr}$ is $\overline k$, and $K$-automorphisms of $K^{nr}$ induce $k$-automorphisms of $\overline k$. $\endgroup$ Jan 23, 2018 at 20:03
  • $\begingroup$ There are more precise statements than just the isomorphism that you cited. See math.stackexchange.com/q/2607904/300700 $\endgroup$ Jan 24, 2018 at 10:33
  • $\begingroup$ By Hensel lemma every unramified extension is of the form $K(\zeta_{q^n-1})/K$ and $[K(\zeta_{q^n-1}):K]=[k(\zeta_{q^n-1}):k]$. So in both case the Galois group is (topologically generated by) $\zeta_{q^n-1}\mapsto \zeta_{q^n-1}^q$. $\endgroup$
    – reuns
    Jul 19, 2022 at 2:30

1 Answer 1


There is a unique unramified extension $K_n/K$ of degree $n$ for each $n$ given by adjoining a $q^n-1^{th}$ root of unity to $K$, where $q$ is the size of the residue field $k$. In fact, each of these unramified extensions is a cyclic Galois extension where the reduction map Gal$(K_n/K) \to$ Gal$(k_n/k)$ is an isomorphism ($k_n$ is the residue field of $K_n$). Thus by taking limits, the union of the $k_n$ is $\bar{k}$ and the union of the $K_n$ is $K^{nr}$, so we get an isomorphism Gal$(K^{nr}/K) \cong$ Gal$(\bar{k}/k) \cong \lim_n \mathbb{Z}/n\mathbb{Z} = \hat{\mathbb{Z}}$.

To prove the facts I have mentioned, one can use the Teichmüller character $\omega: k^\times \to \mathcal{O}_K^\times$, where $\mathcal{O}_K$ is the ring of integers. The Teichmüller character is given by $\omega(\bar{x}) = \lim_n x^{q^n}$, where $x$ is any lift of $\bar{x}$ in $\mathcal{O}_K$.

It is easy to see that $x^{q^n}$ is a Cauchy sequence, and the fact that $\omega(\bar{x})$ doesn't depend on $x$ follows from uniqueness in Hensel's lemma on the polynomial $x^{q-1}-1$. One writes $\omega(x)$ for $x \in \mathcal{O}_K$ to mean $\omega$ applied to the reduction of $x$.

Now the important fact about the Teichmüller character that we need is that if $\mu_K'$ denote the roots of unity of $K$ of order coprime to $q$, then $\mu_K' = \omega(k^\times).$ This follows from looking at the formula for $\omega$ and noting that $\omega$ fixes every element of $\mu_K'$.

Theorem There is a unique unramified extension of degree $n$ for each $n$ given by $K_n=K(\mu_{q^n-1})$, which is cyclic Galois. Moreover the natural map Gal$(K_n/K) \to$ Gal$(k_n/k)$ is an isomorphism.

Proof Let $K_n = K(\mu_{q^n-1})$. Then $k_n$ contains $\mathbb{F}_{q^n}$ since by our fact about the Teichmüller character the residue field contains a $q^n-1^{th}$ root of unity. Let $\bar{g}$ be the minimal polynomial of $a$, a primitive $q^n-1^{th}$ root of unity in $k$, and let $g$ be a lift of $\bar{g}$. By Hensel's lemma, there is a unique root of $g$ in $K_n$ reducing to each conjugate of $a$. Letting $\alpha$ be the lift of $a$, we must have $K(\alpha) = K_n$, since $\omega(\alpha)$ is a primitive $q^n-1^{th}$ root of unity. Thus $n = [K_n:K] \geq [k_n:k] \geq n$, so equality must hold. Then if $\pi$ is a uniformizer of $K$, $[k_n:k] = [K_n:K] = [\mathcal{O}_{K_n}/\pi\mathcal{O}_{K_n}:\mathcal{O}_{K_n}/\pi\mathcal{O}_{K_n}]$, so $\pi\mathcal{O}_{K_n}$ is the maximal ideal of $\mathcal{O}_{K_n}$, so $\pi$ stays prime in $K_n$, and the extension is unramified.

Now $K_n$ contains all the conjugates of $\alpha$ so is Galois. Reduction mod the maximal ideal of $\mathcal{O}_{K_n}$ gives a homomorphism from Gal$(K_n/K)$ to Gal$(k_n/k)$, but since $\alpha$'s conjugates reduce to $a$'s conjugates, and Gal$(K_n/K),$ Gal$(k_n/k)$ act faithfully on the conjugates of $\alpha$ and $a$ respectively, the homomorphism is injective and hence an isomorphism. Thus $K_n$ is a cyclic extension.

For uniqueness, if $L$ is unramified degree $n$, let $\pi$ be as before, $n = [L:K] = [\mathcal{O}_{L}/\pi\mathcal{O}_{L}:\mathcal{O}_K/\pi\mathcal{O}_K] = [l:k]$. Write $l = k(a)$, and since $\omega(a)$ is a $q^n-1^{th}$ root of unity, $L \supset K(\mu_{q^n-1})$, but they are the same degree, and hence are equal.

I hope this has made it more clear why unramified extensions of (non-archimedian) local fields are easy to understand. It turns out the totally ramified abelian extensions are also easy to understand, and are constructed via Lubin Tate theory.

  • 1
    $\begingroup$ Great answer! ;) $\endgroup$
    – ArtW
    Jan 23, 2018 at 20:50
  • $\begingroup$ Does something similar work when $k$ is perfect, and not necessarily finite? $\endgroup$
    – Johnny T.
    Jan 24, 2018 at 10:02
  • 1
    $\begingroup$ @JohnnyT. yes. I think this is covered in Serre's local fields. $\endgroup$ Aug 7, 2022 at 20:40

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