1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $K$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.