Let $K$ be a local field. Let $n$ be an arbitrary natural number and $e$ be any divisor of $n$.
Question Does there exist an extension $L/K$ with the following properties?
- $L/K$ is a cyclic extension, i.e. it is Galois with cyclic Galois group,
- $L/K$ has degree $n$ and
- $L/K$ has ramification index $e$.
My actual problem is it to find any cyclic extension $L/K$ with a fixed degree which is not totally ramified to construct so-called unramified characters of this fixed degree, but this question seems to be more interesting. I though about taking some cyclic extension $L'/K$ (if that's possible) and then take an intermediate subextension which must be cyclic due to the Fundamental Theorem of Galois Theory. But this approach does not resolve the problem with the degree resp. ramification indices.
Could you please help me with that problem? Thank you!
This is not always possible. Let's stick to the totally ramified case $n=e$ to see this.
The theory of higher ramification groups poses some restrictions on the degree of totally ramified Galois extensions (cyclic or not). Let $q=p^n$ be the cardinality of the residue field $\kappa$ of $K$. Then for any Galois extension $L/K$ we have a filtration of $G=\mathrm{Gal}(L/K)$ given by the (higher) ramification groups $G \supset G_0 \supset G_1\supset \dots $. In the case that $L/K$ is totally ramified, we have $G=G_0$. One can show that $G_0/G_1$ injects into $\kappa^\times$, hence the order divides $q-1$ and for $i \geq 1$ $G_i/G_{i+1}$ injects into $\kappa$, hence the order is a power of $p$. Thus in this case the order of $\mathrm{Ga}(L/K)$ is of the form $u \cdot p^k$ where $u$ divides $q-1$. Thus it is not possible to have a totally ramified Galois extension of $K$ when the order has prime divisor that is not $p$ and doesn't divide $q-1$. For details regarding ramification groups, see these notes on algebraic number theory, especially lemma 9.1.5 (p. 193) and lemma 9.3.7 (p. 200).