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Suppose I have a number field $F$, and a finite place $v$ of $F$. Let $E$ be finite extension of $F_v$. I start with a continuous morphism $$ \chi \colon O_{F_v}^\times \to E^\times. $$ where $O_{F_v}$ is the ring of integers of the completion $F_v$ of $F$ at $v$.

My question is if $\chi$ can be extended to a continuous morphism $$ \epsilon \colon {\mathbb A}_F^\times/ \overline{F^\times (F \otimes {\mathbb R})^{\times, 0}} \to E^\times $$ where $\mathbb A_F$ are the adeles of $F$, and $(F \otimes {\mathbb R})^{\times, 0}$ is the identity component of the group $(F \otimes {\mathbb R})^{\times}$.

I am mostly interested in the case where $F$ is totally real.

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    $\begingroup$ Not in general: there's an obstruction coming from the global units -- for an extension to exist, $\chi$ must be trivial on a finite-index subgroup of $\mathcal{O}_F^\times \subset \mathcal{O}_{F_v}^\times$, and if $F$ is not $\mathbf{Q}$ or an imaginary quadratic field, then not all $\chi$'s have this property. $\endgroup$
    – David Loeffler
    Mar 18, 2016 at 7:36
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    $\begingroup$ @DavidLoeffler: your objection is roughly correct but literally false; one can take $v$ to split completely in $F$, take $E$ to be $F_v$, and take $\chi$ to be the obvious inclusion. Then $\chi$ is injective and so not trivial on any finite index subgroup of $\mathcal{O}^{\times}_F$ (if this is infinite), and yet the cyclotomic character exists. In general, one has to take into account all primes above $N(v)$. $\endgroup$
    – Brett Favre
    May 24, 2016 at 2:24

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