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



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> > 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 $k = \mathbb{Q}$?
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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.



Irreducible polynomials with a root modulo almost all primes
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.


