TITLE: Primes for which a polynomial splits completely
QUESTION [8 upvotes]: Suppose that $f(x) \in \mathbb{Z}[x]$ is an irreducible polynomial over $\mathbb{Q}$. Nevertheless, it may be the case that $f(x)$ is reducible modulo $p$ for some prime $p$. What is the density of primes $p$ for which $f(x)$ splits completely into linear factors over $\mathbb{F}_p$?

REPLY [6 votes]: The density is $\frac{1}{|G|}$ where $G$ is the Galois group of $f$. In particular it is at least $\frac{1}{(\deg f)!}$. This is a corollary of the Frobenius density theorem, a slightly weaker but (apparently; I don't know this first-hand) considerably easier version of the Chebotarev density theorem. 
For example, let $f(x) = \Phi_n(x)$ be the $n^{th}$ cyclotomic polynomial. The primes for which $f(x)$ splits are precisely the primes congruent to $1 \bmod n$; the density of these is $\frac{1}{\varphi(n)}$ by Dirichlet's theorem on arithmetic progressions. And indeed the Galois group is $(\mathbb{Z}/n\mathbb{Z})^{\times}$, which has size $\varphi(n)$.