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
$X^n- b \in K[X]$.
Additionally, we impose the condition
$$\vert k \vert =1 \operatorname{ mod } n$$
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?
Best Answer
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 $F^\times/F^{\times n}$.