I'm in trouble with the proof of the following
Lemma If $A$ is a torsion-free abelian group, $\Bbb{P}$ is the set of all primes and $\Bbb{Z}_{(p)}$ denotes the set of all rational numbers whose denominator is prime to $p\in\Bbb{P}$, then $A=\bigcap_{p\in\Bbb{P}}[\Bbb{Z}_{(p)}\otimes A]$.
Proof If $x\in\bigcap_{p\in\Bbb{P}}[\Bbb{Z}_{(p)}\otimes A]$, for some $p\in\Bbb{P}$, we have $x=\sum_{i=1}^nq_ia_i$, with $q_i\in\Bbb{Z}_{(p)},a_i\in A$. There is an integer $s$, prime to $p$, s.t. $sx\in A$. Let $s$ have prime factors $p_1,\dots,p_m$ and
select an integer $s_j$, prime to $p_j$, s.t. $s_jx\in A$ for $j=1,\dots m$.
As the integers $s,s_1,\dots,s_m$ are relatively prime, it is an easy matter to establish the inclusion $x\in A$.
My question is (possibly) a number-theoretic one: how can I choose (and so prove the existence of) the $s_j$ for $j=1,\dots,m$?
Abelian torsion free means $A$ injects into $\Bbb{Q} \otimes_{\Bbb{Z}} A$ and its image is $1\otimes_{\Bbb{Z}} A $.
For $b \in \Bbb{Q} \otimes_{\Bbb{Z}} A$ then $$b = \sum_{j=1}^J q_j \otimes_{\Bbb{Z}} a_j, \qquad q_j \in \Bbb{Q}, a_j \in A$$ With $N$ the lcm of the denominators of the $q_j$ then $$Nb = \sum_{j=1}^J N q_j \otimes_{\Bbb{Z}} a_j=1\otimes_{\Bbb{Z}}\sum_{j=1}^J N q_j a_j \in 1\otimes_{\Bbb{Z}} A $$
Let $I$ be the set of integers such that $nb \in 1\otimes_{\Bbb{Z}} A$, since $\Bbb{Z}$ is a PID $I = M \Bbb{Z}$ for some non-zero $M$.
If $M \ne \pm 1$ then $b \not \in 1\otimes_{\Bbb{Z}} A \cong A$, there is some $p | M$ so that $$b\not \in \bigcup_{l \ge 1, p \nmid l} l^{-1} \otimes_{\Bbb{Z}} A= \Bbb{Z}_{(p)}\otimes_{\Bbb{Z}} A$$ and hence $$\bigcap_p \Bbb{Z}_{(p)}\otimes_{\Bbb{Z}} A=1\otimes_{\Bbb{Z}} A \cong A$$