# Image of a character remains the same when restricting to a totally ramified extension

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.

• A definition of unramified character $\chi$ is that $\chi$ kills the inertia subgroup. Using this definition I think it should be obvious? $L/K$ being cyclic should be irrelevant. I'm also wondering if $F/K$ is necessarily cyclic? I feel you wanted to say that $F = K(\zeta_n)$, which is not necessarily cyclic, isn't it? – dyf Nov 20 '18 at 20:37
• @dyf: I also think that $L/K$ being cyclic is irrelevant, but I just mentioned it because it appears to be cyclic in one of my problems. "I'm also wondering if $F/K$ is necessarily cyclic?" I think this must be true because every finite subgroup of $\mathbb{C}^*$ must be cyclic, otherwise we would not have the correspondence for the characters. – Diglett Nov 20 '18 at 20:45
• For the former, I thought if if $F/L/K$, then $I_{F/L} \subset I_{F/K}$, now now that I think of it, I am not even sure this is right (sorry to disappoint you). For the latter, I see what you mean now, I subtly used Kronecker-Weber to think of $F$ as $K(\zeta_n)$, but indeed I should not have done that. (Thanks!) – dyf Nov 20 '18 at 21:04