# Unramified prime

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.

## 2 Answers

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.

• Thank you @Ferra. However, according to en.wikipedia.org/wiki/Stickelberger%27s_theorem I don't find any relation with my problem. Do you mean Stickelberger's theorem in this post? math.stackexchange.com/questions/394785/… – Desunkid Dec 3 '19 at 20:27
• I mean this one: link.springer.com/chapter/10.1007/978-1-4684-0065-6_43. – Ferra Dec 3 '19 at 20:50
• Thank you @Ferra. I have managed with case $p$ odd now but can not use this method to case $p=2$. – Desunkid Dec 4 '19 at 19:37
• See Carlitz, L. A theorem of Stickelberger. Math. Scand. 1 (1953), 82--84. – Ferra Dec 4 '19 at 20:15
• Once the correspondence of the primes of $\Bbb{Z}[\alpha]$ above $p$ with the factorization of $f\in \Bbb{F}_p[x]$ is known (Dedekind criterion) and that the conjugates of a root are its $p^l$-th powers then the proof is straightforward – reuns Dec 5 '19 at 0:14

Factoring $$f\in \Bbb{F}_p[x]$$ we have $$f(x)= \prod_{k\le t} g_k(x)=\prod_{k \le t} \prod_{l=1}^{d_k} (x-a_k^{p^l}), \qquad \qquad a_k^{p^{d_k}} =a_k$$ Put an ordering on the set of roots : $$a_k^{p^l} if $$k< k_2$$ or $$k=k_2,1\le l, we obtain $$\Delta(f)^{1/2} = \prod_{a_k^{p^l} Let $$\phi$$ be the Frobenius of $$f$$'s splitting field, since $$\Delta(f)\in\Bbb{F}_p^*$$ then $$\Delta(f)$$ is a square iff $$\frac{\phi(\Delta(f)^{1/2})}{\Delta(f)^{1/2}}=1$$.

Since $$\phi(\Delta(f)^{1/2})=\prod_{a_k^{p^l} then $$\frac{\phi(\Delta(f)^{1/2})}{\Delta(f)^{1/2}}=\prod_{a_k^{p^l} a_{k_2}^{p^{l_2+1}}} (-1)$$

Given $$a_k^{p^l} ,

• If $$k then $$a_k^{p^{l+1}} ,

• otherwise $$k=k_2$$ and $$a_k^{p^{l+1}} >a_k^{p^{l_2+1}}$$ iff $$l_2=d_k$$

which gives $$\left( \dfrac{\Delta(f)}{p} \right) =\frac{\phi(\Delta(f)^{1/2})}{\Delta(f)^{1/2}}= \prod_{k \le t}\prod_{l=1}^{d_k-1} (-1)= (-1)^{n-t}$$