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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.

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2 Answers 2

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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.

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  • $\begingroup$ 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/… $\endgroup$
    – Desunkid
    Dec 3, 2019 at 20:27
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    $\begingroup$ I mean this one: link.springer.com/chapter/10.1007/978-1-4684-0065-6_43. $\endgroup$
    – Ferra
    Dec 3, 2019 at 20:50
  • $\begingroup$ Thank you @Ferra. I have managed with case $p$ odd now but can not use this method to case $p=2$. $\endgroup$
    – Desunkid
    Dec 4, 2019 at 19:37
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    $\begingroup$ See Carlitz, L. A theorem of Stickelberger. Math. Scand. 1 (1953), 82--84. $\endgroup$
    – Ferra
    Dec 4, 2019 at 20:15
  • $\begingroup$ 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 $\endgroup$
    – reuns
    Dec 5, 2019 at 0:14
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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 : $a_k^{p^l} <a_{k_2}^{p^{l_2}}$ if $k< k_2$ or $k=k_2,1\le l<l_2\le d_k$, we obtain $$\Delta(f)^{1/2} = \prod_{a_k^{p^l} <a_{k_2}^{p^{l_2}}} (a_k^{p^l} -a_{k_2}^{p^{l_2}})$$ 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 $\phi(\Delta(f)^{1/2})=\prod_{a_k^{p^l} <a_{k_2}^{p^{l_2}}}(a_k^{p^{l+1}} -a_{k_2}^{p^{l_2+1}})$ then $$\frac{\phi(\Delta(f)^{1/2})}{\Delta(f)^{1/2}}=\prod_{a_k^{p^l} <a_{k_2}^{p^{l_2}},a_k^{p^{l+1}} >a_{k_2}^{p^{l_2+1}}} (-1)$$

Given $a_k^{p^l} <a_{k_2}^{p^{l_2}}$,

  • If $k<k_2$ then $a_k^{p^{l+1}} <a_{k_2}^{p^{l_2+1}}$,

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

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

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