Irreducible polynomial modulo every prime?

There exist irreducible polynomials in $\mathbb{Z}[x]$ (e.g. $x^4-10x^2+1$) which are reducible modulo every prime $p$. (A proof can be found in J.S. Milne's Fields and Galois Theory, page 13.) This kind of polynomial is so "bad". I want to know if there exists some non-trivial "good" polynomials.

State precisely:

Does there exist a polynomial $f(x)\in \mathbb{Z}[x]$ with degree $>1$ such that $f(x)$ is irreducible in $\mathbb{F}_p[x]$ for any prime number $p$?

Solution 1:

No, there is no such polynomial. Any polynomial $f(x) \in \mathbb{Z}[x]$ with degree greater than $1$ is reducible modulo every prime factor of every value it takes.

For, take any value of $n$ for which $f(n) \neq \pm 1$. (There must exist such $n$ because $f$ can take the values $1$ and $-1$ only finitely many times.) Consider any prime factor $p$ of $f(n)$. Then $f(n) \equiv 0 \mod p$, which means that $f$ is reducible in $\mathbb{F}_p[x]$: it is divisible by the polynomial $x-n$.

Solution 2:

The answer is 'no', except for linear polynomials. Indeed, if $f(x)\in\mathbb{Z}[x]$ is irreducible and of degree greater than 1, then its splitting field is a non-trivial extension of $\mathbb{Q}$, and in such an extension, infinitely many primes split, which means that $f$ splits modulo infinitely many primes. An even stronger statement is true: if $G$ is the Galois group of $f$, then the set of primes that split completely, i.e. primes (up to finitely many exceptions) modulo which $f$ splits into linear factors, has density $1/|G|$ by the Chebotarev density theorem.