Reducibility of a polynomial over $\mathbb{Z}_p$

Question

It is well known fact that over the finite field $\mathbb{Z}_p$ of order $p$, there exists irreducible polynomial of every degree.

I came to following natural question from this:

Question: Given any $n\geq 2$ and any polynomial $f(x)=x^n+a_1x^{n-1} + \cdots + a_n$ over $\mathbb{Z}$, there always exists a prime $p$ such that $f(x)$ is reducible over $\mathbb{Z}_p$.

For example, consider $x^2+a$ with $a\in\mathbb{Z}$. Then for infinitely many prime $p$, this polynomial is reducible and for infinitely many primes, it is irreducible also.

Answer 1

Given a polynomial $f=x^{n}+a_{1}x^{n-1}+\cdots+a_{n}\in\mathbb{Z}[X]$, pick $a\in\mathbb{Z}$ such that $f(a)\neq\pm 1$. Consider the prime factorization $$f(a)=\pm p_{1}^{e_{1}}\cdots p_{m}^{e_{m}}.$$

Then $f(a)\equiv 0$ mod $p_{j}$ for $j\in\{1,\ldots,m\}$. Hence in all these $\mathbb{Z}_{p_{j}}$, the polynomial $f$ is reducible, since it has $a$ mod $p_{j}$ as a root.

Answer 2

If $a_n \ne \pm 1$, then let $p$ be a prime factor of $a_n$, so $a_n = k p$ for some $k \in \Bbb Z$. Notice that

$$f(p) = p^n + \sum _{k = 1} ^{n-1} a_k p^k + pk = p \left( p^{n-1} + \sum _{k = 1} ^{n-1} a_k p^{k-1} + p \right) \equiv 0 \pmod p ,$$

and since $f(p) \equiv f(0) \pmod p$ it follows that $x$ divides $f$ over $\Bbb Z/p$, so $f$ is reducible over $\Bbb Z/p$.

For the moment, I do not know what to say about the case $a_n = \pm 1$.