Does any polynomial with integer coefficients split over some prime field?

Question

Assume that $Q(x)$ is a polynomial with integer coefficients. Is there a prime number $p$ such that the equation $Q(x)=0$ has all its roots in the finite field $\mathbb{Z}/p\mathbb{Z}$?

I asked this question in RG, too.

https://www.researchgate.net/post/Can_a_given_polynomial_with_integer_coefficients_be_splitted_in_some_Z_p

Answer 1

Yes, that's true by Chebotarev's Density Theorem. A more elementary proof works as follows: Let $\beta$ be a primitive element of the splitting field of $Q(x)$, and $F(x)$ be the minimal polynomial of $\beta$. Then every root of $Q(x)$ is a polynomial in $\beta$ with rational coefficients. Thus it suffices to show that there are infinitely many primes $p$ modulo which $F(x)$ factors into linear factors. As $\mathbb Q(\beta)/\mathbb Q$ is Galois, the Galois group of $F(x)$ acts regularly on the roots of $F(x)$. By Dedekind now, whenever $F(X)$ has root modulo $p$, $F(X)$ splits into linear factors modulo $p$, and we are done. (For all of this, one has to exclude finitely many primes dividing the denominators of certain rational coefficients, or modulo which $F(x)$ has multiple roots.)