Problem I want to prove: Let $\chi: G_K \to \mathbb{C}^*$ be an unramified character and let $L/K$ be a cyclic totally ramified extension. Then $\chi(G_K)=\chi(G_L)$.
All I managed to do was considering all definitions and characterizations (without further success):
Definition 1: Let $G_K$ be the absolute Galois group of a local field $K$. We will call a group homomorphism $\chi: G_K \to \mathbb{C}^*$ with finite image a character on $K$.
Remark: Since every finite subgroup of $\mathbb{C}^*$ is cyclic, it is generated by a primitive root of unity. So in our case, every character $\chi$ corresponds to a unique cyclic Galois extension $F/K$ of degree $n$, the cardinality of the image of $\chi$, and an isomorphism $$\bar{\chi}: \operatorname{Gal}(F/K) \xrightarrow{\sim} \langle \xi_n \rangle \subseteq \mathbb{C}^*$$ where $\xi_n$ is a primitive $n$-th root of unity. We also say that $\chi$ cuts out the extension $F/K$.
Definition 2: We call a character $\chi: G_K \to \mathbb{C}^*$
- unramified if the restriction of $\chi$ to $F$ is the trivial map, i.e. $\chi|_F(\sigma)=1$ for all $\sigma \in G_F$, and
- totally ramified if $\chi(I_K) = \chi(G_K)$ where $I_K$ denotes the inertia subgroup of $K$.
Remark A character $\chi$ which cuts out an extension $F/K$ is unramified (resp. totally ramified) if and only if $F/K$ is unramified (resp. totally ramified). Another characterization for $\chi$ being unramified (resp. totally ramified) is that the order of $\chi(I_K)$ (also called the ramification index of $\chi$) is equal to $1$ (resp. $[F:K]$) where $I_K$ denotes the inertia subgroup of $K$.
The intuitive approach for the Problem should somehow deal with the fact that the residue fields (resp. the inertia subgroups) remain the same when going from $K$ to $L$. But I am not able to proceed with the technical proof.
Could you please help me with my problem? Thank you in advance!