I'm trying to generalize this but I'm somehow stuck.
Let $k$ be a number field, let $S$ be a finite set of places of $k$ containing all the archimedean places, and let $\mathcal{O}_S$ be the ring of $S$-integers of $k$. I would like to prove that if $x \in \mathcal{O}_S$ and $p$ is a prime number that divides $N_k(x)$ and is not an $S$-unit, then there is a prime ideal $P$ of $\mathcal{O}_S$ which divides both $x\mathcal{O}_k$ and $p\mathcal{O}_k$.
My idea is the following: $x\mathcal{O}_k = P_1 \cdots P_n / Q_1 \cdots Q_m$, where $P_i, Q_j$ are prime ideals of $\mathcal{O}_S$ and $Q_1, \ldots, Q_m \in S$ (as finite places). Then $N_k(x) = N(P_1) \cdots N(P_n) / N(Q_1) \cdots N(Q_m)$ (but I'm not sure if this is true, since $x\mathcal{O}_k$ is not an integral ideal), hence $p$ divides some $N(P_i)$, from which it follows that $P_i$ divides both $p\mathcal{O}_k$ and $x\mathcal{O}_k$ (but who guarantees that it is a prime of $\mathcal{O}_S$?
Thank for your help.