Question on quotient ring Let $\zeta := \zeta_p = e^{2\pi i/p}\in \mathbb{C}$ and set $R = \mathbb{Z}[\zeta] = \left\{\sum_{i=0}^{p-1} a_i\zeta^i\mid a_i \in \mathbb{Z}\right\}$. Let $\mathfrak{p} = (1-\zeta)$ be an ideal of $R$. Now consider the quotient ring $\overline{R} := R/\mathfrak{p}$. Then (since $1-\zeta \in \mathfrak{p})$, $\overline{1} = \overline{\zeta}$, so also $\overline{\zeta^i} = \overline{1}$. Why does this imply that $\overline{R} = \overline{\mathbb{Z}} = (\mathbb{Z}+\mathfrak{p})/\mathfrak{p}$?
 A: Basically, the substitution of $1$ for $\zeta$ leaves only the integers behind.
$\mathbb{Z}[\langle \zeta\rangle]/(1-\zeta) = \left\{\overline{\sum_{i=0}^{p-1} a_i\zeta^i}\mid a_i \in \mathbb{Z}\right\}=\left\{\overline{\sum_{i=0}^{p-1} a_i}\mid a_i \in \mathbb{Z}\right\}=\{\overline{z}\mid z\in \Bbb Z\}$.
You are basically looking at the augmentation map of the group ring for a cyclic group onto its coefficient ring.

Addendum: Apparently in my haste I and fellow commenters overlooked that this is not merely the group ring of the cyclic group of order $p$! Direct your thanks, votes and accept checks to Jyrki Lahtonen's solution, for he was the first to catch this.
To make sure this answer does not stand incorrectly: $\zeta$ additionaly satisfies the relation given by its minimal polynomial, bringing $p\in (1-\zeta)$. Now $x$ does not have any such relation in $\Bbb Z[x]$, so $\Bbb Z[x]/(x^n-1)$ turns out differently! Since $p\in (1-\zeta)$, the quotient boils down to $\Bbb Z/(p\Bbb Z)$
A: The first thing (also explained by other answerers) is to notice that $\zeta^i\equiv1\pmod{\mathfrak{p}}$ for all integers $i$. May be I misunderstood, but I think the question asks for more?
The other point is that the minimal polynomial of $\zeta_p$ is 
$$
m(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1.
$$
As $m(\zeta_p)=0$, we also have that
$$
0=m(\zeta_p)=\zeta_p^{p-1}+\zeta_p^{p-2}+\cdots+\zeta_p+1\equiv1+1+\cdots+1=p\pmod{\mathfrak{p}}.
$$
Or, in other words, the prime $p\in\mathfrak{p}$.
Taking this into account gives finally
$$
\overline{R}\cong\Bbb{Z}/p\Bbb{Z}.
$$
A: $R \cong \mathbb{Z}[x]/(x^{p-1}+\dotsc+x+1)$ with $\zeta \mapsto x$. Hence, $$R/(1-\zeta) \cong \mathbb{Z}[x]/(x-1,x^{p-1}+\dotsc+x+1)=\mathbb{Z}/(1^{p-1}+\dotsc+1+1)=\mathbb{Z}/(p).$$
