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Let $F$ be a non-Archidean local field of characteristic 0, and $\zeta_n$ the set of $n$-th roots of unity in the algebraic closure of $F$. Assume $\zeta_n\subseteq F$. Let $\chi:F^\times\to\mathbb{C}^\times$ be a character on $F$.

Is it true that $\chi$ has an $n$-th root (i.e. there exists a character $\mu$ such that $\chi=\mu^n$) if and only if $\chi$ is trivial on $\zeta_n$?

I vaguely remember that this is true when $n=2$. But is it true for general $n$? If so, how does one prove it?

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    $\begingroup$ 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. $\endgroup$
    – sdr
    Commented Jan 4 at 13:49
  • $\begingroup$ Thanks! This makes sense. $\endgroup$
    – Windi
    Commented Jan 5 at 15:53

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