# Irreducible polynomials with a root modulo almost all primes

Let $f \in \mathbb{Z}[x]$ be a non-zero polynomial which is irreducible over $\mathbb{Q}$. Suppose that $f$ has a root in $\mathbb{F}_p$ for almost all primes $p$. Must $f$ be linear?

Here by almost all, I mean all but finitely many. Here is an equivalent way to state this question in terms of number fields.

Let $\mathbb{Q} \subset k$ be a number field. Suppose that for almost all primes $p$, the ring of integers of $k$ contains a prime ideal of norm $p$. Must we have $k = \mathbb{Q}$?

If $k$ is as above and is also Galois, then the Chebotarev density theorem implies that we indeed have $k = \mathbb{Q}$. The Galois case is quite special however, as here if a prime is split then it is also completely split. It is the non-Galois case that I am really interested in.

• In fact even more is true. One can prove that if $f$ is irreducible, then $f$ has roots for roughly $1/deg f$ primes. – Alvin Dec 4 '13 at 15:44
• See Serre's On a theorem of Jordan ams.org/journals/bull/2003-40-04/S0273-0979-03-00992-3/… for a typically elegant treatment of this question. – Lucia Dec 4 '13 at 18:19
• @Alvin: In general the proportion of primes for which there is a root is strictly larger than $1/\mathrm{deg}(f)$. One gets $1/\mathrm{deg}(f)$ in the Galois case, which as I say I already know about. – Daniel Loughran Dec 5 '13 at 10:00
• @Lucia: As far as I can tell, Serre does not address my question in his paper. He shows that there is a positive density $c \geq 1/\mathrm{deg}(f)$ for the collection of primes for which $f$ has a root. My question is whether $c = 1$ if and only if $f$ is linear. – Daniel Loughran Dec 5 '13 at 10:04
• @Lucia: Actually I see now that remarks on p433 that one can use the Chebotarev density theorem to give me what I want. Thanks! – Daniel Loughran Dec 5 '13 at 10:14

It seems this can be done "by hand", using much less than what goes into the full proof of Chebotarev. (But it's the same circle of ideas.) It is known that for any number field $K/\mathbf{Q}$, we have $\zeta_K(s) \sim \frac{\kappa}{s-1}$ as $s\downarrow 1$, for some $\kappa > 0$. Hence, $\log \zeta_K(s) = \log \frac{1}{s-1} + O(1)$, as $s\downarrow 1$. On the other hand, starting from the Euler product representation $\zeta_K(s) = \prod_{\mathfrak{p}} (1-|\mathfrak{p}|^{-s})^{-1}$, we also see that $\log\zeta_K(s) = \sum_{\mathfrak{p}} \frac{1}{|\mathfrak{p}|^s} + O(1) = \sum_{p} \frac{n(p)}{p^s} + O(1)$, where $n(p)$ is the number of degree one prime ideals $\mathfrak{p}$ of $\mathcal{O}_K$ that lie above the rational prime $p$. (Note that I'm using the somewhat nonstandard notation $|\cdot|$ for the norm map on ideals.)
In particular, when $K=\mathbf{Q}$, we get that $\sum_{p} \frac{1}{p^s} = \log\frac{1}{s-1} + O(1)$ as $s\downarrow 1$.
Now suppose that $k$ is a number field with the property that all but finitely many rational primes $p$ have a degree $1$ prime $\mathfrak{p}$ of $\mathcal{O}_k$ lying above them. Applying the result of the first paragraph to this $k$, we see that $$\log \frac{1}{s-1} \sim \log \zeta_k(s) = \sum_{p} \frac{n(p)}{p^s} + O(1)\geq \sum_{p} \frac{1}{p^s} + O(1) \sim \log \frac{1}{s-1}.$$ Hence, $$\sum_{p:~n(p)>1}\frac{n(p)-1}{p^s} = o\left(\log\frac{1}{s-1}\right),$$ as $s\downarrow 1$. Consequently, $$\sum_{p:~n(p)>1}\frac{1}{p^s} = o\left(\log\frac{1}{s-1}\right).$$ That is, the (Dirichlet) density of primes $p$ having more than one degree one prime ideal above them must be zero. But unless $[k:\mathbf{Q}] = 1$, this contradicts that a positive proportion of rational primes split completely in $k$. (It is in this final step that $n > 1$ is important.)
It might seem that I smuggled in Chebotarev to say that a positive proportion of rational primes split in $k$. But no, this follows again from the result in the first paragraph, now with $K$ taken as the Galois closure of $k/\mathbf{Q}$.