Let $E$ be a quadratic extension of a local nonarchimedean field $F$ of characteristic zero (and odd residual characteristic). Let $\sigma$ be a generator of the Galois group $G = Gal(E/F)$. I'm looking for an elementary proof that there are infinitely many distinct (unitary) characters $\chi$ of $E^\times$ such that ${}^\sigma \chi = \chi \circ \sigma \neq \chi$.


Here's a sketch: the character $\chi$ is Galois invariant, i.e., ${}^\sigma \chi = \chi$ if and only if $\chi$ is trivial on the kernel of the norm map $N\_{E/F}: E^\times \rightarrow F^\times$. Let $K = \ker N\_{E/F}$. The group $K$ is a closed subgroup of the compact group $U\_E$ of units in $E^\times$. We can extend any non-trivial character $\tilde \chi$ of $K$ to $E^\times$ to obtain a nontrivial character $\chi$ of $E^\times$ such that ${}^\sigma \chi \neq \chi$.


The part of the argument that is missing is to show that either:


(a) the character group $\widehat K = Hom(K,S^1)$ of $K$ is infinite, 


or, if (a) is false (?),


(b) if $\widehat K$ is finite, then we need to show that there are infinitely many distinct extensions of at least one $\tilde \chi \in \widehat K$ to $E^\times$


I expect that (a) is true. Any suggestions to address this fact (or a reference) would be greatly appreciated.



Characters of the kernel of the norm map of an extension of local fields
The sequences


$ 1 \rightarrow K \rightarrow E^\times \rightarrow \operatorname{Im} N\_{E/F} \rightarrow 1$


and


$ 1 \rightarrow \operatorname{Im} N\_{E/F} \rightarrow F^\times \rightarrow G \rightarrow 1$


are exact (the second by Local Class Field Theory). Taking Pontryagin duals we obtain exact sequences:


$ 1 \rightarrow \widehat{\operatorname{Im} N\_{E/F}} \rightarrow \widehat{E^\times} \rightarrow \widehat{K} \rightarrow 1$


and


$ 1 \rightarrow \widehat{G} \rightarrow \widehat{F^\times} \rightarrow \widehat{\operatorname{Im} N} \rightarrow 1$


It follows that $\widehat{\operatorname{Im} N\_{E/F}} \cong \widehat{F^\times} / \widehat{G}$ is infinite; indeed, $G$ and $\widehat G$ are finite groups.
Thus, we have a short exact sequence 


$ 1 \rightarrow \widehat{F^\times} / \widehat{G} \rightarrow \widehat{E^\times} \rightarrow \widehat{K} \rightarrow 1$


and the fibres of the restriction map $\widehat{E^\times} \rightarrow \widehat{K}$ are infinite.


