## Irreducible polynomial over number field with roots in every completion?

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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)

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## 2 Answers

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.

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I had the cyclotomic thing in my mind a few hours before I posted, but then forgot it... – Dror Speiser Nov 29 2009 at 4:49
As for Hensel, sure, but if I'm not mistaken an irreducible integer polynomial will satisfy Hensel's requirements for all but finitely many places. – Dror Speiser Nov 29 2009 at 5:19
That's interesting, do you know where the argument breaks down if K/Q is infinite? – Rebecca Bellovin Nov 29 2009 at 8:05
Let k be a residue field of O_K. Because K/Q is infinite, it is possible that k is infinite. Then one can't characterize the elements of k as the fixed points of some Frobenius map. In my opinion, this is the largest break in the argument, although probably not the only one. – David Speyer Nov 30 2009 at 0:17

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This has 16 upvotes? What's wrong with you people? – Lavender Honey Jan 22 2010 at 22:44
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Edit: I see I missed the note that $K$ might not be finite over $\mathbb{Q}$. This answer is not correct for $K$ of infinite degree over $\mathbb{Q}$, as in FC's answer above.

No. This is a consequence of the Chebotarev density theorem. To see how it follows, look at exercise 6 at the end of Cassels and Frohlich's "Algebraic Number Theory".

Briefly, the Chebotarev density theorem says that for a Galois extension of global fields $L/K$ and for a finite set $S$ of places of $K$, the proportion of primes of $K$ splitting in $L$ is $1/[L:K]$. If $G=\text{Gal}(L/K)$ and $E$ is the fixed field of some $H\subset G$, it is possible to show that the proportion of places of $K$ with a split factor in $E$ is $|\bigcup_{\rho\in G}\rho H\rho^{-1}|/|G|$, and a lemma on finite groups says that this quotient is not $1$ unless $H=G$.

In your case, take $E=K[x]/(f)$ and $L$ to be a normal extension of $K$ containing $E$. Then "$v$ has a split factor" means "$f$ has a root in the completion $K_v$". If $f$ has a root in each completion (or even a set of completions with density $1$, which includes the case "all but finitely many"), we must have $H=G$ and $E=K$. So $f$ already had a root in $K$.

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