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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 $p$?

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up vote 10 down vote accepted

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 $x-n$.

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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.

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This more than answers the question, but at the cost of introducing algebraic number theory and Galois theory where Shreevatsa's answer shows that more modest tools will do. It gives OP something to think about. – Gerry Myerson Jun 8 '11 at 13:22
Must $f$ be monic? – wxu Jun 8 '11 at 15:01

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