Question

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 $disc(f)$?

Yildo

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 $p$.