A set of prime factors of an integer in $\mathcal{O}_k$ I've got a basic question from Thm 2 (ch.7, $\S2$) of Lang's Algebraic Number Theory.
Let $k$ be a number field and $A$ its adele group. Let $S_{\infty}$ be the set of Archimedean absolute values of $k$. Theorem says that $k+ A_{S_{\infty}}=A$ where $A_{S_{\infty}}$ is the subgroup of $S_{\infty}$-adeles.  
During the proof, Lang states this: 
Given $x\in A$, let $m$ be a rational integer such that $mx$ has integral components for all non-Archimedean $v$. Let $S$ be the set of primes $\mathfrak p$ of $\mathcal O_k$ such that $\mathfrak p \mid m$. 
Question: what does $\mathfrak p \mid m$ mean here? $m$ is not necessarily prime, right?
He then uses this to show that $x-\alpha/m\in A_{S_{\infty}}$ where $\alpha\in \mathcal{O}_k$ satisfies $$mx\equiv\alpha \;\;(\text{mod }\mathfrak{p}^{\nu})$$ for all $\mathfrak p \in S$ and for some fixed large $\nu$. 
Many thanks for your help. 

This means the prime ideal $\frak p$ contains the ideal generated by $m$.
