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.
Best Answer
Yes, this is true and can be proved by applying Chebotarev's density theorem. For a hint, see exercise 7.2 in http://websites.math.leidenuniv.nl/algebra/Lenstra-Chebotarev.pdf
See also [Cassels-Fröhlich], p. 362, Exercise 6.2.