Why does $N(\mathfrak p)$ belong to $\mathfrak p$ ?
$N(\mathfrak p)$ is the norm of the prime ideal $\mathfrak p\in\mathcal O$ defined as $N(\mathfrak p)=|\mathcal O/\mathfrak p|=$(say $p$)
Now since every prime ideal is maximal in the ring of integers, we have $\mathcal O/\mathfrak p$ is a field, hence of prime characteristic $p\in\mathbb Z$, is then $p\mathcal O$ prime, then I could say: If $p\mathcal O$ and $\mathfrak p$ share any elements then by maximality $p\mathcal O\subset \mathfrak p$ and thus $p\in\mathfrak p$, if not again by maximality $p\mathcal O+ \mathfrak p=\mathcal O$... Can I use a cardinality argument here ?