Let $m \in \mathbb{Z}$ such that $m$ is not a prime power, and suppose that $\zeta$ is a primitive $m$th root of unity. Let $q$ be a prime number such that $q$ doesn't divide $m$, and suppose that $\mathfrak{q}$ is a prime ideal of $\mathbb{Z}[\zeta]$ such that $q \in \mathfrak{q}$. I want to show that $1 - \zeta^k \in \mathfrak{q}$ implies $1 - \zeta^k = 0$.
I've been stuck on this problem for a while, and I feel like it shouldn't be too hard to prove.
I know that if $\gcd(k,m) = 1$, then $\zeta^k$ is an $m$th root of unity and $1- \zeta^k$ is a unit in $\mathbb{Z}[\zeta]$. So we may assume that $\gcd(k,m) = d > 1$. Now, $1 - \zeta^k = (1-\zeta)(1 +\zeta + \dots + \zeta^{k-1})$, which implies that $1 +\zeta + \dots + \zeta^{k-1} \in \mathfrak{q}$, since $\mathfrak{q}$ is prime and $1-\zeta$ is a unit in $\mathbb{Z}[\zeta]$.
I think this is the right direction to go in, but I just cant seem to get anywhere with this. I know that somehow I would like to get that $m \mid k$, or that $1 +\zeta + \dots + \zeta^{k-1} \in \mathfrak{q}$ somehow implies that $1 +\zeta + \dots + \zeta^{k-1} = 0$. I'm also having trouble seeing how $q \nmid m$ comes into play. Does it have something to do with $\mathfrak{q}$ being unramified? What am I missing?
Of course, $\zeta^k$ is a primitive $r$-th root of unity for some $r\mid m$, and unless $r$ is a prime power $1-\zeta^k$ is a unit in the ring of algebraic integers.
So we may assume $r$ is a prime power; if $r=1$ we get the conclusion $1-\zeta^k=0$. Otherwise, the norm of $1-\zeta^k$ in any field containing it is a power of $p$, the prime in question. But $p\ne q$ as $q\nmid m$. Then $1-\zeta^k$ can only lie in prime ideals dividing $p$, and $\mathfrak q$ is not such an ideal.