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May 24, 2016 at 2:24 comment added Brett Favre @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 $N(v)$.
Mar 22, 2016 at 14:56 history edited mnr CC BY-SA 3.0
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Mar 18, 2016 at 7:36 comment added David Loeffler 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.
Mar 17, 2016 at 15:03 history edited mnr CC BY-SA 3.0
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Mar 17, 2016 at 14:49 history asked mnr CC BY-SA 3.0