I'm searching for a recommendable reference dealing with theory of non-Archimedean local fields where I can find proofs of the following claims about finite extensions $L/K$ of non-Archimedean local fields with finite residue fields $l / k$. I'm pretty sure that this request might not have a research level but up to now I haven't obtained a satisfying answer asking the same at MSE.

Firstly, we use notations: We denote by $q \in O_K$ a uniformizer of $O_K$ and $\pi \in O_L$ a uniformizer of $O_L$. Then $n= [L:K]=e \cdot f$ with $f= [l:k]$ and $(q)O_L= (\pi)^eO_L$.

Now I'm looking for rigorous proofs of the following statements:

(i) Case $L/K$ is Galois & unramified ($\Rightarrow$ $e=1$ & $[L:K]= [l:k]$) and $p$ the characteristic of finite field $k$:

The claim is $G= \operatorname{Gal}(L/K)= \operatorname{Gal}(l/k)=g$. I'm looking for a proof making an explicit construction of the "lift" $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$, i.e. if we have a $k$-automorphism of $l= k(a)= k[X]/ (X^f-1)$, how can it be uniquely lifted to a $K$-automorphism of $L$?

(ii) If $L/K$ is unramified and $K= \mathbb{Q}_p$, then $L$ is cyclotomic: $L= \mathbb{Q}_p(\zeta_n)$, for $\zeta_n$ an appropriate root of unity.

(iii)(1) Case $L/K$ totally ramified with $n =e$ coprime to $p=\operatorname{char}(k)$: there exists $b \in K$ with $L= K(\sqrt[e]{b})$.

How can this $b$ be constructed? Can the $b$ be chosen as a uniformizer: i.e. $b= uq$ with $u \in O_K^\times$?

(2) If moreover $p \mid \lvert k \rvert -1$, why is $L/K$ cyclic & Galois extension?


See Fesenko and Vostokov - Local fields and their extensions. (i) is Proposition 3.3(2). (ii) is Proposition 3.2(1). (iii)(1) is Proposition 3.5(1) (and, yes, $b$ may be chosen as a uniformiser). For (iii)(2), I think you meant $e \mid \lvert k\rvert - 1$, not $p \mid \lvert k\rvert - 1$ (which is impossible). Then ($k$, hence) $K$ contains an $e$th root of unity, so $L = K(\sqrt[e]q)$ is (Galois and) cyclic over $K$.

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  • $\begingroup$ Hi, thank you for your answer. A couple remarks: (you are right, of course, in (iii) I wanted to require $e \mid \lvert k\rvert - 1$). This requirement implies by finiteness of $k$ that $k$ contains an $e$-th root of unity. Why this imply that $K$ has also one? Do you use at this point the Hensel's lemma? Or is it a more more elementary result? Futhermore, you continue, that this imply that $L = K(\sqrt[e]q)$ (<= here we use already (iii)(1) ) is cyclic? How do you deduce this? $\endgroup$ – Rachmaninow98 May 21 at 17:50
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    $\begingroup$ Indeed, I use Hensel's lemma to deduce that $K$ contains a (I forgot to say primitive) $e$th root of unity because $k$ does. It is a general fact in Galois theory (nothing to do with valued fields) that a radical extension is cyclic if and only if the ground field contains an appropriate root of unity: en.wikipedia.org/wiki/Radical_extension#Solvability_by_radicals . In this case, $X^e - q$ is irreducible by Eisenstein's criterion. $\endgroup$ – LSpice May 21 at 18:16
  • $\begingroup$ Great, that was exactly what I was looking for. Thank you a lot! $\endgroup$ – Rachmaninow98 May 21 at 18:28
  • $\begingroup$ A nitpick: The proof in Proposition 3.3(2) that gives an answer two my (i) is indeed quite understandable (verify $Gal(L/K) \to Gal(l/k)$ surjective + $\vert Gal(L/K) \vert =\vert Gal(l/k) \vert$, but I'm quite curious if it also possible to build expliitely a lift $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$, since I want to understand what this lift does "concretly" on $L$. Unfortunately, the proof 3.3(2) from the book not gives this "inside look". Do you know if it also possible to argue via these "lifts"? I mean every $\alpha \in Gal(l/k)$ is a power of $\endgroup$ – Rachmaninow98 May 21 at 20:45
  • $\begingroup$ Frobenius $a \mapsto a^d$ with $d= \vert k \vert$. That is $\operatorname{Gal}(L/K)$ has also to be generated by a lift of this Frobenius. But I have no idea how this lift of the Frobenius in $Gal(L/K)$ shold act. My first idea was to do Teichmüller lifts of $\zeta_{d-1} \in l$ to $L$ and define to lift of the Frobenius on only on these Teichmüller lifts. Problem: do the Teichmüller lifts generate $L$ as $K$-algebra? Do you probably see another possibly more conventional approach to lift the Frobenius and thus to obtain the desired map $\operatorname{Gal}(l/k) \to \operatorname{Gal}(L/K)$? $\endgroup$ – Rachmaninow98 May 21 at 20:45

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