Is a polynomial irreducible modulo a silly large prime

Question

Suppose I have a large ($\approx 2^{64}$) prime $p$, and a polynomial $g(x)$ of degree $d$ that irreducible in $\mathbb{Z}[x]$. If $d << p$, is it likely that $g(x)$ is irreducible in $\mathbb{Z}/p\mathbb{Z}[x]$.

Answer 1

Adding to the comment, there are polynomials in $\mathbb{Z}[X]$ which are irreducible in $\mathbb{Q}[X]$ yet reducible in $\mathbb{F}_p[X]$ for every prime $p$.

To construct such a polynomial, choose a (finite) Galois extension $K/\mathbb{Q}$ whose Galois group is not cyclic, and choose an element $\alpha\in K$ such that $K=\mathbb{Q}(\alpha)$ and such that $\alpha$ is integral over $\mathbb{Z}$.

Then if $f(X)$ is the minimal polynomial for $\alpha$, then $f$ is reducible modulo $p$ for every prime $p$. The proof uses local fields and essentially boils down to the fact that Galois groups of extensions of finite fields are always cyclic.

Answer 2

In the other direction, we have for example the Weak Tchebotarev Theorem that says that if $f\in\mathbb Z[X]$ is irreducible and $[K_f:\mathbb Q]=n$, where $K_f$ denotes the splitting field of $f$, then the set $P$ of primes over which $f$ factorizes as a product of linear factors has density $1/n$ in the set of all primes, so in particular it is not bounded. If the Galois group of $f$ is abelian, then the set $P$ is in fact characterized by congruences; cyclotomic polynomials provide examples of this.

Notice this is not quite what you asked, as this set $P$ is the set of worst possible primes, while you are considering primes over which the polynomial simply factors.

There is a very nice article on this subject, [Wyman, B. F.. “What Is a Reciprocity Law?”. The American Mathematical Monthly 79.6 (1972): 571–586]