I am trying to figure out how prime ideals of $\mathbb{Z}$ decompose in the Galois extension $\mathbb{Q}(\zeta_{7})/\mathbb{Q}$.
And so, for this, I've picked the prime ideals $5\mathbb{Z}$, $7\mathbb{Z}$ and $29\mathbb{Z}$ to learn with an example.
From what I understand, the first step is to factor the minimal polinomial of $\zeta_{7}$ (over $\mathbb{Q}$)
\begin{equation} p(x)=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1 \end{equation}
modulo 5 (resp. 7 and 29), so I know the "form" of the ideals above $5\mathbb{Z}$ (resp. $7\mathbb{Z}$ and $29\mathbb{Z}$).
(I can do this for every prime in this example because $\mathbb{Z}[\zeta_{7}]$ is exactly the ring of integers of $\mathbb{Q}(\zeta_{7})$, right? I know this can be more carefully stated in terms of something called "conductor".)
We get
\begin{equation} p(x)\equiv x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\mod 5 \end{equation} \begin{equation} p(x)\equiv (x-1)^{6}\mod 7 \end{equation} \begin{equation} p(x)\equiv (x-7)(x-16)(x-20)(x-23)(x-24)(x-25)\mod 29 \end{equation}
This means that we can write
\begin{equation} 5\mathbb{Z}[\zeta_{7}]\:\text{is a prime ideal} \end{equation} \begin{equation} 7\mathbb{Z}[\zeta_{7}]=(\zeta_{7}-1)^{6}\mathbb{Z}[\zeta_{7}] \end{equation} \begin{equation} 29\mathbb{Z}[\zeta_{7}]=(\zeta_{7}-6)(\zeta_{7}-16)(\zeta_{7}-20)(\zeta_{7}-23)(\zeta_{7}-24)(\zeta_{7}-25)\mathbb{Z}[\zeta_{7}] \end{equation}
And so, 5 is inert, 7 ramifies and 29 splits completely.
Is this correct?
The easiest way to find this decomposition for p is by reducing the polynomial mod p? It took me a while to do this. In this particular extension, can I completely characterize which primes behave like 5, 7 and 29? Also, is there another type of behaviour that my example didn't show?
Thank you!
For this particular extension, because it’s cyclotomic, your task is considerably simplified.
No matter over what field $\Bbb F_p$ you consider your polynomial $P(x)=(x^7-1)/(x-1)$, it’s always the polynomial whose roots are the seventh roots of unity. Now, how big a field $\Bbb F_{p^m}$ do you need to go to to get the seventh roots of unity? In the case $p=5$, you see that $5^6-1$ is the first such that $5^m-1$ is divisible by $7$. Oho.
Oho! You’re asking for the smallest $m$ such that $7|(p^m-1)$, i.e. such that $7k=p^m-1$, i.e. such that $p^m\equiv1\pmod7$. Oho. You’re asking for the period of $p$ in $\Bbb F_7^*\>$(!). No matter what this $m$ is that you find, that’s the degree of the irreducible factors of $P$ over $\Bbb F_p$. For $p=5$, $P$ is still irreducible, for $p=11$, it factors into two cubics, and for $p=29$, $P$ is the product of six linears.
Oh yes, and for $p=7$, you should write out $P(x+1)$ as a $\Bbb Z$-polynomial.