For an irreducible monic $f\in\mathbb{Z}[x]$, is it true that $\overline{f}\in\mathbb{F}_p[x]$ will be irreducible for all but finitely many $p$?

Question

Let $f\in\mathbb{Z}[x]$ be an irreducible monic, then it need not be the case that $\overline{f}\in\mathbb{F}_p[x]$ will be irreducible. But will it be irreducible for all but finitely many $p$? My intuition is that this is probably false, but I don't have any idea of how to find a counterexample.

Answer 1

Unfortunately, this is not the case. The prototypical counterexample where this fails most strongly is given by $f(X) = X^{4}+1 \in \mathbb{Z}[X]$. You can check that $f$ is irreducible over $\mathbb{Z}[X]$, but (perhaps) surprisingly, the image of $f$ in $\mathbb{F}_{p}[X]$ is reducible for every prime $p$. You can find more details on this here, for instance.

A more banal counterexample which is somewhat similar in nature is given by $g(X) = X^{2}+1 \in \mathbb{Z}[X]$. Clearly, $g$ is irreducible over $\mathbb{Z}$. However, the image of $g$ in $\mathbb{F}_{p}[X]$ is reducible for any prime $p$ which is congruent to $1$ mod $4$. Indeed, in this case, the group $(\mathbb{F}_{p})^{\times}$ is cyclic of order $p-1$ and so contains an element $\alpha$ of order $4$. Since $-1$ is the unique element of $\mathbb{F}_{p}$ of order $2$, it follows that $\alpha^{2} = -1$, i.e. $\alpha$ is a root of $g$. There are infinitely many primes congruent to $1$ mod $4$, so there are infinitely many primes such that $\overline{g}$ is reducible in $\mathbb{F}_{p}[X]$.

Answer 2

A result in the opposite direction of the provided answer is this: if $ f \in \mathbf Z[X] $ is irreducible monic, then there are infinitely many primes $ p $ such that modulo $ p $, $ f $ splits into linear factors. This follows from the general result that infinitely many primes split completely in any given number field; coupled with the fact that for all but finitely many primes, splitting of $ f $ modulo $ p $ mirrors the splitting of $ p $ in the ring of integers of $ \mathbf Q(a) $, where $ a $ is a root of $ f $. Therefore, there is in fact no $ f $ (of degree greater than $ 1 $) which makes the statement of the question work.