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Let $K$ be a local field (therefore complete, discrete non-archimedian valuation field) with perfect residual field $\kappa_K:= \mathcal{O}_K/\pi_K$.

Assume that $L/K$ is a field extension of $K$ of degree $n= [L:K]$ and $L$ contains all $n$-th roots and is tamely ramified. How to show that then $L/K$ is a cyclic extension (Therefore that $Gal(L/K)$ is a cyclic group)?

It looks (taking into account the condition that it contains the $n$-th roots) that possibly I could use the Kummer theory to get the claim but I don’t see how to apply it here. Especially I’m not really familar with local fields. Could anybody help?

Btw: I know that the main result of Kummer theory provides a bijective correspondence between abelian extensions of $K$ and subgroups $W$ of multiplicative group: $(K^*)^n \subset W \subset K^*$. Can group this correspondence be derived a correspondence to cyclic extensions?- as in my case

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I keep your notation $L/K$ of degree $n$, and write as usual $n=ef$, where $e$ (resp. $f$) is the ramification (resp. inertia) index. The maximal unramified subextension $F/K$ (= inertia subfield) of $L/K$ is classically known to be cyclic of degree $f$.

1) Let us first study $L/F$, which, by hypothesis, is tamely ramified (the residual characteristic $p$ of $K$ does not divide $e$) and contains the $n$-th (hence the $e$-th) roots of unity. Denote by $\pi$ (resp. $\Pi$) a uniformizer of $F$ (resp. $L$), and by $U_i , i\ge 0$, the subgroup of units $U_i=1+ (\pi^i)$ of $F$. It is known that $U_0/U_1$ is canonically isomorphic to the residual multiplicative group ${\kappa}^*$ of $F$, and for $i \ge 1, U_i/ U_{i+1}$ is non canonically isomorphic to $(\kappa,+)$, see e.g. Cassels-Fröhlich, chap. 1. By definition $\Pi^e /\pi =u\in U_0$, and $u=u_1w$, with $u_1 \in U_1$ and $w$ representing a class of $U_0/U_1$. But, because $p \nmid e$, multiplication by $e$ in $U_1/U_2$ is an isomorphism, hence we can repeat the approximation process to get $u_1=u_2{w_1}^e$, with $u_2 \in U_2$ and $w_1$ representing a class of $U_1/U_2$, etc. On taking the projective limit, we obtain (with a change of notations) $\Pi^e=\pi$. This shows that $L=F(\sqrt [e]\pi)$ is an Eisenstein extension, thus tamely totally ramified with ramification index $e$. Since $F$ contains a primitive $e$-th root of unity, Kummer theory tells us that $L/F$ is a cyclic Galois extension of degree $e$.

2) We can't say much more about $L/K$ since we don't even know if it is Galois. However, even in the Galois situation, I think that your proposition (the cyclicity of $L/K$) does'nt hold in general. For example, fix an uniformizer $\pi$ of $K$ and take $L=E.F$, with $E=K(\sqrt [e]\pi)$. For ramification reasons, $E, F$ are linearly disjoint, hence $L/K$ is an abelian extension, with Galois group $G \cong Z/eZ \times Z/fZ$. If $e, f$ are coprime, $G$ is cyclic. But if for instance $e=f=$ a prime $q$, $G$ is not cyclic.

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  • $\begingroup$ Thank you for your answer. One question: Why is the maximal unramified subextension $F/K$ of L/K cyclic? $\endgroup$ – KarlPeter Dec 19 '18 at 0:09
  • $\begingroup$ This is a classical result : an unramified extension is cyclic, obtained by adding a primitive root of 1 of order not divisible by the residual characteristic. See e.g. Cassels-Fröhlich, chap. 1. $\endgroup$ – nguyen quang do Dec 19 '18 at 7:33
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If $\kappa_K$ is contained in $\overline{\mathbb{F}_p}$, or if $L/K$ is totally ramified (see comments below)

The valuation gives an absolute value $|x| = q^{-v(x)} $.

  • If $x \in L, |x| < 1$ then $(1+x)^{1/n} = \sum_{m=0}^\infty {1/n \choose k} x^k $ converges and is $\in L$.

    Proof : (In characteristic $p \nmid n$ then ${1/n \choose k} \in \mathbb{Z}_p$ then reduce it modulo $p$)

    So $|{1/n \choose k}| \le 1$ and the series converges, and since $L$ is complete the limit is in $L$.

  • Since $L/K$ is totally ramified of degree $n$ then $\pi_L^n = u^{-1} \pi_K$ where $|u| = 1$. So there is a root of unity $\zeta \equiv u \bmod (\pi_L)$ such that $u \zeta^{-1} = 1+x, |x| < 1$, so that $\varpi_L = \zeta^{1/n} (1+x)^{1/n} \pi_L \in L$ satisfies $ \varpi_L^n = \pi_K$.

Whence $L = K(\pi_K^{1/n})$ and $Gal(L/K) = \langle \sigma \rangle$ with $\sigma(\pi_K^{1/n}) = \zeta_n \pi_K^{1/n}$.

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  • $\begingroup$ Hi. How do you deduce that $L:K$ is totally ramified from the given condition that it is tamely ramified? $\endgroup$ – KarlPeter Dec 16 '18 at 21:23
  • $\begingroup$ I’m using the definitions for totally/tamely ramification from math.uga.edu/~pete/8410Chapter4.pdf It says especially that if we consider a finite field extension $L/K$. Denote by $k := \kappa_K$ (resp l:= \kappa_L) is the residual fields of $K$ (resp L) then: $L/K$ totally ramified iff e(L/K) =[L:K] (or as you used $\pi_K \mathcal{O}_L = \pi_L ^{e(L/K)} \mathcal{O}_L$. This is equivalently to $l=k$ and $L/K$ is tamely ramified iff $e(L/K)$ is prime to the characteristic of $K$ $char(K)=p$. $\endgroup$ – KarlPeter Dec 17 '18 at 1:53
  • $\begingroup$ We know that $k$ is perfect. I don't see how do you conclude that $L/K$ totally ramified. What can we say about the resudual field $l$ of $L$. We need an argument that $l=k$... $\endgroup$ – KarlPeter Dec 17 '18 at 1:53
  • $\begingroup$ @KarlPeter Let $\kappa_K = \mathbb{Q}(\zeta_{n^\infty})$ then it has non-cyclic Galois extensions $\kappa_L/\kappa_K$ of degree $n$ so $K = \kappa_K((t)),L = \kappa_L((t))$ is a non-cyclic unramified extension of degree $n$. That's why your statement meant that $L/K$ is totally ramified (and doesn't have wild ramification, since all the $\zeta_{n^r}$ are then in $\kappa_K$). $\endgroup$ – reuns Dec 17 '18 at 18:35
  • $\begingroup$ Two points aren't clear to me: Firstly, why we can assume wlog that the residual field $\kappa_K$ is already the perfect closure $\mathbb{Q}(\zeta_{n^\infty})$ Naively - since by assumption it is perfect - it is just a subfield of $\mathbb{Q}(\zeta_{n\infty})$ in case of characteristic zero. Or did you implicitely used an argument that allows to reduce this to the case $\kappa_K = \mathbb{Q}(\zeta_{n^\infty})$? $\endgroup$ – KarlPeter Dec 17 '18 at 22:54

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