# Cyclic totally ramified Galois extension of of non-archimedean fields

Let $$K \subset L := K(a)$$ be a simple totally ramified extension of non-archimedean local fields of degree $$n$$ generated by a $$n$$-th root of $$K$$; ie $$a$$ is a root of irreducible polynomial $$X^n- b \in K[X]$$.

$$\vert k \vert =1 \operatorname{ mod } n$$

for the cardinality of the residue field $$k$$ of $$K$$.

I want to check that $$L/K$$ is Galois and has cyclic Galois group. The Galois problem I was able to solve myself but I don't know how to show that the Galois group is cyclic.

on Galois: $$K$$ has characteristic zero, because it's a local field, therefore the extension is separable. In order to check that it's Galois, we have to check that it's normal. equivalently, $$L$$ contains all roots of $$X^n- b \in K[X]$$.

My key observation was that $$L$$ contains all roots of $$X^n-1$$, because the condition $$\vert k \vert =1 \operatorname{ mod } n$$ is equivalent to that one that $$\vert k \vert-1$$ is divisible by $$n$$ and therefore $$k^{\times}$$ contains all $$n$$-th root. By Hensel's lemma these roots can be lifted to $$n$$-th roots in $$K$$ and obviously the roots of $$X^n- b \in K[X]$$ are $$\zeta_n^m a, m=0,1,..., n-1$$.

Therefore $$L/K$$ is Galois. Why is it cyclic?

• This is just Kummer theory. Note that $|k|\equiv 1\bmod{n}$ implies $K$ contains the $n$th roots of unity (apply Hensel's lemma to $x^n-1\bmod{\mathfrak{p}}$ where $\mathfrak{p}$ is the prime of $\mathcal{O}_K$). Therefore the cyclic extensions of degree $n$ are precisely those generated by $n$th roots as in your example.
– Nico
Feb 24, 2021 at 3:00
• non-archimedean local fields: finite extensions of $\Bbb{Q}_p$ and $\Bbb{F}_p((t))$ (not all of characteristic $0$) Feb 24, 2021 at 3:46

If a primitive root of unity $$\zeta_n\in F$$ and $$a^n\in F$$ then $$F(a)/F$$ is separable because $$a$$ is a root of the separable polynomial $$x^n-a^n$$, which splits completely in $$F(a)$$ so $$F(a)/F$$ is Galois,
It is cyclic because its automorphisms are of the form $$\sigma : a\to \zeta_n^{\phi(\sigma)} a$$ making $$Gal(F(a)/F)$$ a subgroup of $$\Bbb{Z}/n\Bbb{Z}$$.
What Nico said is the converse: every degree $$d|n$$ cyclic extension of $$F$$ is of the form $$F(c)/F$$ with $$c^n\in F$$, one extension per cyclic subgroup of $$F^\times/F^{\times n}$$.