Let $K$ be a local field and $G_K$ be the absolute Galois group of $K$.
Definition We call a map $\chi: G_K \to \mathbb{C}^*$ an Artin character if it is a group homomorphism with finite image.
Let $\chi$ be an Artin characeter. Since every finite subgroup in $\mathbb{C}^*$ is cyclic, we know that the image of ${\chi}$ is generated by a primitive $n$-th root of unity $\xi_n$. By the isomorphism theorem, $\chi$ corresponds uniquely to an isomorphism $$ \bar{\chi}: \operatorname{Gal}(F/K) \to \langle \xi_n \rangle $$ for a unique cyclic extension $F/K$ of degree $n$. We say that $F/K$ is cut out by $\chi$.
Question: Let $\chi$ and $\chi'$ be two Artin characters, $F$ and $F'$ be two extensions of $K$ cut out by $\chi$ and $\chi'$, respectively. Let $L/K$ be the extension cut out by the product $\chi \chi'$. Can $L$ be described by $F$ and $F'$?
Approach: As an example, I considered $K = \mathbb{Q}_3$ and $F = \mathbb{Q}_3(\sqrt{3})$ which is a totally ramified extension of $K$ of degree $2$, and $F'=\mathbb{Q}_3(\sqrt{5})$, an unramified extension of degree $2$. Let $\bar{\chi}: \operatorname{Gal}(F/K) \to \langle -1 \rangle$ and $\bar{\chi'}: \operatorname{Gal}(F'/K) \to \langle -1 \rangle$ be the unique isomorphisms, and $\chi,\chi'$ be the Artin characters induced by $\bar{\chi}$ and $\bar{\chi'}$, respectively. Now we would like to consider the product $\phi= \chi \chi'$ which is also an Artin character. My first guess for the extension $L/K$ which is cut out by $\phi$, was $L = FF'$. However, this cannot be true, since $\operatorname{Gal}(FF'/K) \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is not cyclic.
Consider the generator $\bar{\sigma}: \sqrt{3} \mapsto -\sqrt{3}$ of $\operatorname{Gal}(F/K)$, and the $\bar{\mu}: \sqrt{5} \mapsto -\sqrt{5}$ of $\operatorname{Gal}(F'/K)$, and call $\sigma$ and $\mu$ the lifts of $\bar{\sigma}$ and $\bar{\mu}$ to $G_K$, respectively. I know now that $\chi(\sigma) = -1$, but I am not so sure what $\chi(\sigma')$ (and therefore also $\phi(\sigma')$) might be. Same thing for $\mu$. The only thing I know is that the image of $\phi$ must be $\langle -1 \rangle$, so $L$ must be an extension of degree $2$. It might be possible that we can use the properties of local fields (e.g. ramification index) to our advantage, but this is just a vague idea so far.
Does anyone have an idea on how to approach this problem? Thank you!