Hi there,

Here is a part of the book of Murty "an introduction to Sieve methods and applications" page 36

"At this point we invoke some algebraic number theory let $K=Q(\theta)$ where $\theta$ is the solution of the polynomial $f(x)$. The ring of integers $O_{k}$ of K is a Dedekind Domain. it's a classical theorem of Dedekind that for all but finitely many primes $\delta_{f}(p)$ is the number of prime ideals $p$ of $O_{k}$ such that the norm $N_{K/Q}(p)=p$"

Note: $\delta_{f}(p)$:= number of solutions of $f(x)$ modulo $p$

Question: What can be said about the size of the set of primes breaking this rule? In my opinion, might they be those dividing $disc(f)$?


  • $\begingroup$ I believe you can even narrow it further to those primes dividing $disc(f) / disc(\mathcal{O})$. $\endgroup$
    – user13113
    May 9, 2011 at 8:47

1 Answer 1


This statement is true for all $p$ not dividing $disc(f)$, as you say. Moreover you can write down exactly what the primes dividing $p$ are: they're the ideals $(p, g(\theta))$ for each irreducible factor $g$ of $f$. The norm of such a prime is the degree of $g$, so the number of degree 1 primes is the number of roots of $f$ mod $p$.

  • $\begingroup$ I don't think so, the formulation gives the number of degree $1$ primes over $p$ which correspond exactly to the roots. $\endgroup$ May 9, 2011 at 7:42
  • 1
    $\begingroup$ Sorry, yes, I was too hasty and misread the question. Thanks Torsten for the correction. $\endgroup$ May 9, 2011 at 7:53

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