# Cyclic Field Extension of Local Field

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

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.

• Thank you for your answer. One question: Why is the maximal unramified subextension $F/K$ of L/K cyclic? – KarlPeter Dec 19 '18 at 0:09
• 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. – nguyen quang do Dec 19 '18 at 7:33

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 $$\sigma(\pi_K^{1/n}) = \zeta_n \pi_K^{1/n}$$.

• Hi. How do you deduce that $L:K$ is totally ramified from the given condition that it is tamely ramified? – KarlPeter Dec 16 '18 at 21:23
• 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 $K$ $char(K)=p$. – KarlPeter Dec 17 '18 at 1:53
• 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 $l=k$... – KarlPeter Dec 17 '18 at 1:53
• @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 $\kappa_K$). – reuns Dec 17 '18 at 18:35
• 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 $\kappa_K = \mathbb{Q}(\zeta_{n^\infty})$? – KarlPeter Dec 17 '18 at 22:54