# Let $f \in \mathbb{Z}[x]$. Does $\bar{f}$ have as many roots in $\mathbb{F}_p$ as $f$ has in $\mathbb{C}$ for infinitely many primes $p$?

Let $f \in \mathbb{Z}[x]$ be a nonconstant polynomial. Consider $\bar{f} \in \mathbb{F}_p[x].$ Let $\rho_p$ be the number of distinct roots of $\bar{f}$ in $\mathbb{F}_p$, and let $\rho$ be the number of distinct roots of $f$ in $\mathbb{C}$. Do you have a reference for showing that $\rho_p = \rho$ for infinitely many primes $p$?

My background to this question: I actually use this result to calculate the precise maximal subgroup growth of groups of the type $\mathbb{Z}^k \rtimes \mathbb{Z}$ (and of similar metabelian groups).

• Do you really mean roots in $\mathbb F_p$ itself, or is it OK if they're in the algebraic closure $\overline{\mathbb F_p}$? – LSpice Apr 25 '18 at 19:54
• You're wanting primes to split completely? – paul garrett Apr 25 '18 at 19:55
• @LSpice I really want them in $\mathbb{F}_p$ itself, not in the algebraic closure. – Andrew James Kelley Apr 25 '18 at 19:58

Yes. Chebotareff (really Frobenius here) density theorem says that each cycle type in the Galois group of $f$ occurs as the splitting type of $f$ modulo $p$ infinitely often. Your case corresponds to the cycle type of the identity element. The Chebotareff theorem says that this happens asymptotically once in $|Gal(f)|$ primes (the statement about infinite number of primes is actually much easier, and follows from the Chinese remainder theorem, but what the heck).
Yes. I will focus on the case that $f$ is irreducible, thus also separable, over $\mathbf Q$. (The general case can be deduced from this, the main case of interest.) For large primes $p$ the reduced polynomial mod $p$ is separable. There is a theorem in algebraic number theory that "in each number field, infinitely many primes split completely". This follows from the zeta-function of each number field having a simple pole at $s=1$. In the application of this to the field $K = \mathbf Q(a)$, where $f(a) = 0$, a large prime $p$ splitting completely in $K$ will be a prime $p$ for which $f \bmod p$ splits into distinct linear factors.