Recall that a group is abelian iff it is a $\Bbb{Z}$-module. I know that if $A$ is a torsion-free abelian group, $\Bbb{Q}\otimes A$ is the divisible hull of $A$, up to isomorphism.
Now, suppose $A\subseteq \Bbb{Q}^n$ has rank $n\ge1$ and denote by $\Bbb{Q}\otimes A$ the divisible hull of $A$. Can I state the equality $\Bbb{Q}\otimes A=\Bbb{Q}^n$?
If $n=1$, then for any $r\in\Bbb{Q}, r\ne 0$, the $\Bbb{Z}$-submodule generated by $r$ equals $\Bbb{Q}$. In general, whenever $x_1,\dots,x_n\in A$ are $\Bbb{Z}$-linearly independent, then $A\supseteq \Bbb{Z}x_1\oplus\dots\oplus\Bbb{Z}x_n$, from which I get $\Bbb{Q}\otimes A\supseteq \Bbb{Q}^n$, the latter one being divisible of rank $n$.
Is it correct or are there some mistakes/imprecisions? The main point is that I may confuse equality with isomorphism.
Thank you in advance for your help.