Let $K$ be a local field and $G_K$ be the absolute Galois group of $K$. An Artin character is any continuous group homomorphism $G_K \to \mathbb{C}^*$ whose image is finite.
Correspondence: Now I would like to understand why each Artin character $\chi$ corresponds to a unique pair $(F/K, \bar{\chi})$ where $F/K$ is a cyclic Galois extension (i.e. its Galois group is cyclic) and $\bar{\chi}: \operatorname{Gal}(F/K) \to \langle\xi_n\rangle \subset \mathbb{C}^*$ is an isomorphism where $\xi_n$ is an $n$-th primitive root of unity.
I know that every finite subgroup of $\mathbb{C}^*$ is cyclic. Therefore the image of an Artin character is generated by a primitive root of unity, i.e. $\rho(G_K) = \langle \xi_n \rangle$. To get an isomorphism, I assume that we must take the preimage of $\langle \xi_n \rangle$ under $\rho$, and since $\langle \xi_n \rangle$ is finite, the preimage is a finite cyclic subset of $G_K$. Why does it has to be a finite Galois extension then? And why is this extension unique?
For instance, if $K=\mathbb{Q}_3$ we can find two different extensions $F = \mathbb{Q}_3(\sqrt{3})$ and $F' = \mathbb{Q}_3(\sqrt{2})$ (they are different because $F/K$ is totally ramified whereas $F'/K$ is unramified). Say we have two isomorphisms $\bar{\chi}: \operatorname{Gal}(F/K) \to \{\pm 1\}$ and $\bar{\chi}': \operatorname{Gal}(F'/K) \to \{\pm 1\}$. How would $\chi: G_K \to \mathbb{C}^*$ and $\chi': G_K \to \mathbb{C}^*$ look like?
It would be nice if you could answer my question and verify my reasoning before. Thank you!