# If $\zeta$ is an $m$th root of unity, then $1 - \zeta^k \in \mathfrak{q}$ implies $1 -\zeta^k = 0$

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.
• Thanks! Dumb question, but is there an easy way to calculate the norm of $1 - \zeta^k$ when $k$ is a prime power? I want to say that the minimum polynomial of $1 - \zeta^k$ is $\phi_k(1-X)$, and then just look at the constant term. – matt stokes Nov 8 '18 at 20:42
• @mattstokes I think you mean, the norm of $1-\zeta$ where $\zeta$ is a primitive $p^j$-th root of unity. The norm of such a $1-\zeta$ from the field it generates is $p$. There's a simple proof, it is $\Phi_{p^i}(1)$. – Lord Shark the Unknown Nov 8 '18 at 20:50
• Oh yes, I was thinking $\zeta^k$ is a primitive $p^i$th root of unity (got mixed up there). Is this because $\Phi_{p^i}(1)$ is the product of the Galois conjugates of $1-\zeta$? – matt stokes Nov 8 '18 at 21:04