I'm trying to solve the following:
Let $p$ be a prime number and $\zeta \in \mathbb{C}^* $ such that $ord(\zeta) = p$. Show that $(1-\zeta)$ is a prime ideal in $\mathbb{Z}[\zeta]$ and that there is a unit $\epsilon \in\mathbb{Z}[\zeta]$ such that $$ p \ = \ \epsilon (1-\zeta)^{p-1} $$
My own thoughts
Every $x \in (1-\zeta)$ can be written as $x'(1-\zeta)$ for some $x' \in \mathbb{Z}[\zeta]$. This allows me to describe its elements (in a ugly way), since $x' = a_0 +a_1 \zeta + \cdots + a_{p-2}\zeta^{p-2}$ for some $a_0, \cdots , a_n \in \mathbb{Z} $. Rewriting gives us: $$ x \quad = \quad (1-\zeta)(a_0+a_1\zeta + \cdots + a_{p-2}\zeta^{p-2}) $$ Which is $$ (a_0 + \cdots + a_{p-2}\zeta^{p-2}) \ - \ (a_0\zeta+ \cdots + a_{p-3}\zeta^{p-2}) \ + \ a_{p-2}(1+\zeta + \cdots + \zeta^{p-2}) $$ So $$ x \quad = \quad (a_0 + a_{p-2}) \ + \ (a_0-a_1+a_{p-2})\zeta \ + \ (a_1-a_2+a_{p-2})\zeta^2 \ + \ \cdots \ + \ (a_{p-3}-a_{p-2}+a_{p-2}) \zeta^{p-2} $$ $x$ has a striking property: the sum of its coefficients is $a_n + (p-2)a_{p-2}$. I don't think that it entirely describes the ideal though. I have to show that $a,b \notin (1-\zeta) \Rightarrow ab \notin (1- \zeta)$ but I don't see how I could do that.
Now I tell you what I tried to find an $\epsilon$ as described above. I expanded: $$ (1-\zeta)^{p-1} \quad = \quad \sum_{0 \leq k \leq p-1}\frac{(p-1)!}{k!(p-1-k)!}(-\zeta)^k \quad = \quad \sum_{0 \leq k \leq p-1}\left( \frac{(p-1)!}{k!(p-1-k)!}-1 \right) (-\zeta)^k $$ The last step is true by the fact that we can subtract $1+\zeta+ \zeta^2+\cdots + \zeta^{p-1}$ since this sum equals zero. In that way $1$ and $\zeta^{p-2}$ vanish, but I still don't see what kind of $\epsilon$ I could choose.
Here is what I tried to show that the statement given in the answer is true. I looked at the product: $\prod_{1 \leq k \leq p-1} (1- \zeta^k)$ If we take an automorphic map $\sigma$ that fixes $\mathbb{Q}$, it has to be described by $\sigma \ : \ \zeta \mapsto \zeta^n$ for some $n \in \mathbb{N}$ such that $ggd(n,p) = 1$.By symmetry the given product has to be fixed, so it has to belong to $\mathbb{Q}$. Still busy...
I try to leave the details:
Look at the quotient $\mathbb Z[\zeta]/(1-\zeta)$. As an abelian group it is generated by the classes of $1$ and the powers of $\zeta$. Since those classes coincide in the quotient, it is a cyclic group, generated by $1$. Since we have $p = \prod_{k=1}^{p-1}(1-\zeta^k)$ (left to the reader), we see $p \cdot 1 = 0$ in $\mathbb Z[\zeta]/(1-\zeta)$, in particular we deduce $\mathbb Z[\zeta]/(1-\zeta) \cong \mathbb Z/p\mathbb Z$, hence the first assertion.
For the second assertion you should note that $(1-\zeta^k)$ ($1 \leq k \leq p-1$) and $(1-\zeta)$ are the same elements up to a unit. Then have another look at the above product.