Why does a prime ideal in a cyclotomic number field factor in this way?

Question

Let $m$ be some natural number and $q$ a prime such that $q\equiv 1\mod m$. Let $\alpha$ be a root of the $m$th cyclotomic polynomial and $\mathbb{Z}[\alpha]$ be the field extension of the integers by adding $\alpha$, and let $\langle q\rangle\subseteq \mathbb{Z}[\alpha]$ be the ideal generated by $q$. Then there will be some $\omega\in\mathbb{Z}_q$ such that the order of $\omega$ is $m$. Then a paper I was reading stated that the ideal $\langle q\rangle$ can be decomposed as: $$\langle q\rangle=\prod_{i\in\mathbb{Z}_m^*}(\langle q\rangle+\langle\alpha-\omega^i\rangle)$$ Where each ideal $\langle q\rangle+\langle\alpha-\omega^i\rangle$ is coprime. They gave no proof of this decomposition, nor that these ideals are coprime. Any ideas on how the proof would go, or what I would look for to find a proof?

Answer 1

After studying a bit more field theory I think I have an answer.

If I take $\omega\in\mathbb{Z}_q$ with order $m$, then if $q$ is large enough that the coefficients of $\phi_m(x)$ are unchanged my modding out by $q$ (which should be easy, since the coefficients are typically small) then $\phi_m(x)$ is unchanged in $\mathbb{Z}_q[x]$ and thus must contain all the primitive $m$th roots of unity as roots, which are all just powers of $\omega$. Thus: $$\phi_m(x)=\prod_{k\in\mathbb{Z}_m^*}(x-\omega^k)\in\mathbb{Z}_q[x]$$

Then let $x\in \prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle$. Since it's a product of elements of each of these ideals, it will have a factor of $q$ (so $x\in\langle q\rangle$) unless $x=\prod_{k\in\mathbb{Z}_m^*}(\zeta-\omega^k)$. But then if I project $x$ into $\mathbb{Z}_q[\zeta]$, then it becomes $\overline{x}=\prod_{k\in\mathbb{Z}_m^*}(\zeta-\omega^k)=\phi_m(\zeta)=0$ because $\zeta$ is defined as a root of this polynomial.

This gives two things: First, that $x\in\langle q\rangle$, so $\prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle\subseteq\langle q\rangle$. Second, define $p(x)=\prod_{k\in\mathbb{Z}_m^*}(x-\omega^k)\in\mathbb{Z}[x]$, and then $p(\zeta)=qf(\zeta)$ for some $\zeta$.

Then for any $j\in\mathbb{Z}_m^*$, I can take $q\prod_{k\in\mathbb{Z}_m^*\setminus\{j\}}(\zeta-\omega^k) \in\prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle$. If I take one element like this for each $j\in\mathbb{Z}_m^*$ and add them together, it's still in the product ideal: $$g(\zeta)=\sum_{j\in\mathbb{Z}_m^*}q\prod_{k\in\mathbb{Z}_m^*\setminus\{j\}}(\zeta-\omega^k) =q\sum_{j\in\mathbb{Z}_m^*}\prod_{k\in\mathbb{Z}_m^*\setminus\{j\}}(\zeta-\omega^k). $$ Consider just $\tilde{g}(x)=\sum_{j\in\mathbb{Z}_m^*}\prod_{k\in\mathbb{Z}_m^*\setminus\{j\}}(x-\omega^k)$, so that $g(x)=q\tilde{g}(x)$. Note that $\tilde{g}(x)$ is just the formal derivative of $p(x)$, and since the roots of $p(x)$ are all distinct, then $gcd(\tilde{g}(x),p(x))=1$. Since $\mathbb{Z}[x]$ is a Euclidean domain, there exists $a(x),b(x)$ such that $a(x)\tilde{g}(x)+b(x)p(x)=1$. Evaluated at $\zeta$ gives $a(\zeta)\tilde{g}(\zeta)+b(\zeta)p(\zeta)=1$, and multiplied by $q$: $$qa(\zeta)\tilde{g}(\zeta)+qb(\zeta)p(\zeta)=q$$ $$a(\zeta)g(\zeta)+(qb(\zeta))p(\zeta)=q$$ But since $g(\zeta),p(\zeta)\in\prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle$, which is an ideal, then multiplying them by $a(\zeta)$ and $qb(\zeta)$, then adding the products together, will still be in this ideal. So $q\in \prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle$, and thus $\langle q\rangle\subseteq \prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle$, so $$\langle q\rangle=\prod_{k\in\mathbb{Z}_m^*}\langle q\rangle + \langle \zeta-\omega^k\rangle.$$

Showing that they are coprime is actually straightforward: For any $j,k\in \mathbb{Z}_m^*$, let $$(\zeta-\omega^j)-(\zeta-\omega^k),q\in\langle q\rangle + \langle \zeta-\omega^k\rangle+\langle q\rangle + \langle \zeta-\omega^j\rangle$$ Then $(\zeta-\omega^j)-(\zeta-\omega^k)=\omega^k-\omega^j\in\mathbb{Z}$. Since $q$ is prime, then $gcd(\omega^k-\omega^j,q)=1$ so there are integers $a,b$ such that $$a(\omega^k-\omega^j)+bq=1$$ And thus $1$ is also in the ideal, so $\mathbb{Z}=\langle q\rangle + \langle \zeta-\omega^k\rangle+\langle q\rangle + \langle \zeta-\omega^j\rangle$