# 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 $p$?

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 $x-n$.
• Your proof says that $f(n) \equiv 0 \mod p$ only for primes p which divide $f(n)$ for some $n$. How do we know that $f(n) \equiv 0 \mod p$ for any prime $p$? – Silent Apr 27 at 4:22
• @Silent The only number that is $\equiv0\pmod p$ for every $p$ is $0$. So that's not the question. The question is whether there always exists some prime $p$ for which the polynomial $f$ factors modulo $p$. Example: Take the polynomial $f(x)=x^4+5$. The question asks, could this (or some other polynomial) be irreducible in every $\mathbb{F}_p[x]$? The answer is no: for this example, consider $f(2)=2^4+5=21$, and its factor $p=3$. Then as $f(2)\equiv0\pmod3$, we know that $f$ is not irreducible in $\mathbb{F}_3[x]$: it is divisible by $(x-2)$: indeed, $f(x)=(x-2)(x+2)(x^2+1)$ in F_3[x]. – ShreevatsaR Apr 27 at 5:25
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.
• Must $f$ be monic? – wxu Jun 8 '11 at 15:01