Let $f \in \mathbb{Z}[X]$ be a monic irreducible polynomial, n its degree, $\alpha$ a zero of $f$ in some extension field of $\mathbb{Q}$, and $p$ a prime number not dividing the discriminant $\Delta(f)$ of $f$. Denote by $t$ the number of prime ideals $\mathfrak{p}$ of $\mathbb{Z}[\alpha]$ with $p \in \mathfrak{p}$. Prove that $\left( \dfrac{\Delta(f)}{p} \right) = (-1)^{n-t}$.
It is a generalization of a result in quadratic number field
Proposition: Let $d \neq 1$ be squarefree and $p$ an odd prime. Then $p$ is split in $\mathbb{Z}[\sqrt{d}]$ for $\left(\dfrac{d}{p}\right)=1$, inert for $\left(\dfrac{d}{p}\right)=-1$ and ramified for $\left(\dfrac{d}{p}\right)=0$.
The proposition can be deduced from Kummer-Dedekind theorem and explicit description of $\mathcal{O}_{\mathbb{Z}[\sqrt{d}]}$ that $\mathcal{O}_{\mathbb{Z}[\sqrt{d}]}=\mathbb{Z}[\sqrt{d}]$ for $d \equiv 2,3 \;(mod \;4)$ and $\mathcal{O}_{\mathbb{Z}[\sqrt{d}]}=\mathbb{Z}\left[\dfrac{1+\sqrt{d}}{2} \right]$ for $d \equiv 1 \; (mod \;4)$. However I have stuck since we don't have explicit description of ring of integer of $\mathbb{Q}[\alpha]$ in general.
Best Answer
This is known as Stickelberger's Theorem, and the proof is not exactly straightforward. Here is how it goes: by Dedekind criterion the number of primes above $p$ is exactly the number of irreducible factors of $f$ modulo $p$, because $p$ does not divide $\Delta(f)$. So write $f=g_1\ldots g_r$ in $\mathbb F_p[x]$. If $r=1$, the claim holds because on the one hand the Galois group of $g_1$ over $\mathbb F_p$ is $C_{\deg g_1}$, and on the other hand the Galois group of an irreducible polynomial of degree $n$ is contained in $A_n$ if and only if the discriminant is a square. To get the general case, just use the fact that $\Delta(f)$ is $\prod \Delta(g_i)$ up to a square.