I'm studying torsion-free abelian groups and I know (see Fuchs, "Infinite Abelian Groups", vol. $2$, pp $154$) that, if $\mathbb{Z}_p$ is the set of $p$- adic integers and $\mathbb{Z}_{(p)}$ denotes the subset of $\mathbb{Q}$ whose elements have denominator prime to $p$, then the following hold: for any torsion-free abelian group $A$ and $p$ prime number $$A=\bigcap_p \, (\mathbb{Z}_{(p)}\otimes A)$$ and $$\mathbb{Z}_{(p)}\otimes A=(\mathbb{Q}\otimes A)\cap (\mathbb{Z}_p\otimes A)$$.
Now, I'm considering the following question: can I say more (eventually assuming that $A$ has finite rank, say $n$)?
I mean that, using the facts I have stated above, $$A=(\mathbb{Q}\otimes A) \cap\bigcap_p (\mathbb{Z}_p\otimes A)$$ $$\subseteq \mathbb{Q}^{rank(A)}\cap\bigcap_p (\mathbb{Z}_p\otimes A)$$.
Does the equality hold at least in the finite rank case or this is always a strict inclusion?
Furthermore, can I write $A=\mathbb{Q}^n\cap(\mathbb{Z}_p\otimes A)$ in the finite rank case?