A question on the Chinese Remainder Theorem This is a question from Lang's ANT, Thm 2 (ch.7, $\S2$).
Let $k$ be a number field and $A$ its adele group.   
In the proof, Lang states 
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 $\mathfrak p \mid m$. We can find $\alpha\in \mathcal{O}_k$ s.t. $$mx\equiv\alpha \;\;(\text{mod }\mathfrak{p}^{n})$$ for all $\mathfrak p \in S$ and for some fixed large $n$, by the Chinese Remainder Theorem.
This congruence is confusing me-- for a start, I assume Lang means $mx_v$ for all non-arch $v$ instead of $mx$? 
Second, isn't $mx_v$ an element of $\mathcal{O}_{k_v}$, so not necessarily in $\mathcal{O}_k$ which is what we need in order to apply the CRT? 
Any help will be very much appreciated.

Yes, Lang means $mx_{\mathfrak p}$.
Secondly, you are right that $mx_{\mathfrak p}$ is an element of $\mathcal O_{k_{\mathfrak p}}$. However, since $\mathcal O_k$ is dense in $\mathcal O_{k_{\mathfrak p}}$, we can choose an element of $\mathcal O_k$ which is congruent to it modulo $\mathfrak p^n$, and apply the C.R.T. to these elements. As i707107 points out, there are finitely many primes being considered, so there is no problem after that.
