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Jan 5 at 15:53 comment added Windi Thanks! This makes sense.
Jan 4 at 13:49 comment added sdr Suppose $A \subset B$ are abelian groups, and $\Gamma$ is a divisible group. Then it's standard that any homomorphism $A \rightarrow \Gamma$ extends to $B$ -- this is Baer's criterion for $\mathbb{Z}$. Now take $B = F^{\times}$, $A$ the subgroup of $n$th powers, and $\Gamma = \mathbb{C}^{\times}$ to get the claim you want. If $\chi$ is trivial on $\mu_n(F) :=$ the $n$th roots of unity of $F^{\times}$ (without assumptions -- there may not exist any), then it factors through $F^{\times}/\mu_n(F) = A$, which by the above extends to $B$, which was your question.
Jan 4 at 9:13 history asked Windi CC BY-SA 4.0