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)
Best Answer
For a finite extension $K/\mathbf{Q}$, the answer is no. Suppose that $f$ is an irreducible polynomial with coefficients in $K$ and splitting field $L$. If $G$ is the Galois group of $L/K$, then the polynomial $f$ gives rise to a faithful transitive permutation representation $G \rightarrow S_d$, where $d$ is the degree of $f$. If $P$ is a prime in $O_K$ that is unramified in $L$, then $f$ has a root over $O_{K,P}$ if and only if the corresponding Frobenius element $\sigma_P \in S_d$ has a fixed point. On the other hand, a theorem of Jordan says that every transitive subgroup of $S_d$ contains an element with no fixed points. By the Cebotarev density theorem, it follows that $f$ fails to have a point modulo $P$ for a positive density of primes $P$.
For an infinite extension $K/\mathbf{Q}$, the answer is (often) yes. Let $K$ be the compositum of all cyclotomic extensions. Then $x^5 - x - 1$ is irreducible over $K$, because its splitting field over $\mathbf{Q}$ is $S_5$. On the other hand, it has a root in every completion (easy exercise).
Finally, your claim that "since a the polynomial must have a root in the residue field, it will have a root in the completion" is false. Hensel's lemma comes with hypotheses.