Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

What is the proof that for any integer $n$ and any non-constant, integer coefficient polynomial $P(x)$, there are infinitely many primes congruent to $1 \pmod{n}$ that divide $P(x)$ for some $x$?

share|cite|improve this question
Neglecting the $p \equiv 1 \pmod{n}$ condition, here is a nice explanation of why there are infinitely many primes dividing $P(x)$ for some $x$. – Dan Brumleve Aug 21 '12 at 10:08

1 Answer 1

This is a pretty standard application of Cebatarov density: Let $K$ be the splitting field of $P(x)$. Let $L$ be the composite field of $K$ and $\mathbb{Q}(\zeta_n)$, where $\zeta_n$ is a primitive $n$-th root of unity. A prime $p$ splits in $K$ if and only if1 $P(x)$ splits into distinct linear factors modulo $p$; it splits in $\mathbb{Q}(\zeta_n)$ if and $p \equiv 1 \bmod n$; it splits in $L$ if and only if both are true. By Cebatarov density (or the weaker Theorem of Frobenius) the primes that split in $L$ have density $1/\dim L$.

I can think of ways to make various parts of this argument more elementary, but I don't see how to get away from using algebraic number theory; I'd be curious to see an elementary proof.

1 With finitely many exceptions, related to the fact that the ring of integers in $K$ may not be $\mathbb{Z}[x]/P(x)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.