0
$\begingroup$

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?

$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$
    – Nico
    Feb 24, 2021 at 3:00
  • 2
    $\begingroup$ non-archimedean local fields: finite extensions of $\Bbb{Q}_p$ and $\Bbb{F}_p((t))$ (not all of characteristic $0$) $\endgroup$
    – reuns
    Feb 24, 2021 at 3:46

1 Answer 1

2
$\begingroup$

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 $F^\times/F^{\times n}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .