Let K/Q be a field, probably not a finite extension. Is it possible for a polynomial to be irreducible over K but have a root in every completion of K? What about all but finitely many completions?
This question is related to the question "Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?", and should help prove that such a polynomial can not exist for any subfield of the algebraic closure of the rationals.
The idea is that we make the candidate polynomial monic and have algebraic integers for coefficients, then take any maximal ideal in the ring of integers of the candidate field and complete it using the ideal - since the polynomial must have a root in the residue field, it will have a root in the completion. I'm wondering if this forces the polynomial to have a root in the original field - hence the question.
The same question only for function fields is also interesting, in order to prove the above for subfields of the algebraic closure of Fp(t)