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 $x-n$.

  • $\begingroup$ 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 $p$? $\endgroup$ – Silent Apr 27 at 4:22
  • 1
    $\begingroup$ @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]. $\endgroup$ – ShreevatsaR Apr 27 at 5:25
  • $\begingroup$ Thank you very much, Shreevatsa $\endgroup$ – Silent Apr 27 at 6:07

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.

  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Jun 8 '11 at 13:22
  • $\begingroup$ Must $f$ be monic? $\endgroup$ – wxu Jun 8 '11 at 15:01

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