Question in proof from James Milne's Algebraic Number Theory I'm having difficulty understanding a step in a proof from J.S. Milne's Algebraic Number Theory (link). Here $\zeta$ is a $p$th root of unity and $\mathfrak p = (1-\zeta^i)$ for any $1\leq i\leq p-1$ is the unique prime ideal dividing $(p)$, and it's been assumed that $p$ does not divide $x, y,$ or $z$.

LEMMA 6.9 The elements $x+\zeta^iy$ of $\mathbb Z[\zeta]$ are relatively prime in pairs.
PROOF. We have to show that there does not exist a prime ideal $\mathfrak q$ dividing $x+\zeta^iy$ and $x+\zeta^jy$ for $i\neq j$. Suppose there does. Then $\mathfrak q\mid ((\zeta^j-\zeta^i)y) = \mathfrak py$ and $\mathfrak q\mid ((\zeta^i-\zeta^j)x) = \mathfrak px$. By assumption, $x$ and $y$ are relatively prime, and therefore $\mathfrak q = \mathfrak p$. 

Thus $x+y \equiv x+\zeta^iy\equiv 0 \mod \mathfrak p$.

Hence $x + y\in \mathfrak p\cap\mathbb Z=(p)$. But $z^p=x^p+y^p\equiv x+y=0\mod p$
  which contradicts our hypotheses. 

My problem is in the middle line which I have isolated. Why is there an $i$ such that $x+\zeta^iy\in\mathfrak p$?

By the assumption (to be contradicted) there is some prime $\frak{q}$ that divides $x+ \zeta^i y$ and $x+\zeta^j y$. In particular $\frak{q}$ divides $(x+ \zeta^i y)$, which is to say $x+\zeta^i y \in \frak{q}$. The first part of the proof shows $\frak{q}$ has to be $\frak{p}$. 
