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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!

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  • $\begingroup$ 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? $\endgroup$ – dyf Nov 20 '18 at 20:37
  • $\begingroup$ @dyf: Thank you for your response? "Using this definition (of yours) I think it should be obvious?" It might be but I do not see it immediately. Could you please elaborate this point? $\endgroup$ – Diglett Nov 20 '18 at 20:44
  • $\begingroup$ @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. $\endgroup$ – Diglett Nov 20 '18 at 20:45
  • $\begingroup$ 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!) $\endgroup$ – dyf Nov 20 '18 at 21:04

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