MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $P(x)$ be an irreducible polynomial in $\mathbb{Z}[x]$ of degree $n.$ By $N(k,x)$ we denote the number of primes up to $x,$ such that $P(x)$ has exactly $k$ solutions in $\mathbb{Z}_p.$ Is it possible to get the asymptotic for $N(k,x)?$ I am mostly interested in the case when $k=1$ and Galois group of $P(x)$ is not $S_n.$

share|cite|improve this question
Sorry, if I sound dumb. What is the meaning of "up to $x$"? As $x$ is a symbolic variable, and this upto $x$ sounds like bounded above by $x$, I am confused. – P Vanchinathan Apr 4 '14 at 2:49
For that matter, what does it mean for a polynomial to have a solution? – Gerry Myerson Apr 4 '14 at 4:08

Yes, this is possible. The more natural method is by using Chebotarev's density theorem. If $G$ is the Galois group of $P(x)$, and $p$ is a prime not dividing the discriminant of $P$, the number of roots $k$ of $P(x)$ in $\mathbb Z/p \mathbb Z$ is the number of fixed points of the Frobenius element $\sigma_p$ of $G$. So let $D_k$ be the set of elements of $G$ which have exactly $k$-fixed prime. Cheobotarev density theorem tells you that the number $N(k,x)$ of primes up to $x$ with $\sigma_p$ in $G$ is $\sim \frac{|D_k|}{|G|} Li(x) \sim \frac{|D_k|x}{|G|\log x}$.If you want more precise result with an error term, you need to apply an effective version of Chebotarev's density theorem. Serre's paper "Quelques applications du théorème de Chebotarev" is a good place to start, if you read French.

share|cite|improve this answer
Thanks for the reference. So in fact, |D_1| can be $0.$ Would it be possible to estimate the number of the primes from below then? – Bob Apr 4 '14 at 1:01
"$k$-fixed prime" supposed to be "$k$ fixed points"? – Gerry Myerson Apr 4 '14 at 4:11

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.