Q:

Divisible abelian $q$-group of finite rank

What does "finite rank" mean in the context of divisible abelian $q$-group? 
A divisible abelian $q$-group of finite rank is always a Prüfer $q$-group or it can be also a finite product of Prüfer $q$-group? 

Thanks for all future answers to my questions.  

A:

The Prüfer rank of an abelian group is the cardinality of a maximal linearly independent subset over $\mathbb{Z}$.
So if you got your Prüfer group $\mathbb{Z}(p^\infty)$, the maximum size of a linearly independent subset is going to be $1$.  Using the representation of a Prüfer $p$-group as $\mathbb{Z}(p^\infty)=\{e^{\frac{2\pi i m}{p^n}}:m,n\in \mathbb{Z}\}$, we take the set $\{e^{2 \pi i a p^{-n}},e^{2 \pi i b p^{-m}}\}$ and try to solve $$\left(e^{2 \pi i a p^{-n}}\right)^{x_1}\left(e^{2 \pi i b p^{-m}}\right)^{x_2}=e^{2 i \pi  \left(a p^{-n}x_1+b p^{-m} x_2\right)}=1$$
So we need to pick nontrivial $x_1$ and $x_2$ to satisfy $a p^{-n}x_1+b p^{-m} x_2=0$, which we can certainly do.   Thus $\mathbb{Z}(p^\infty)$ has rank $1$.  If you take a finite direct product of Prüfer groups, you'll still have finite rank because rank is additive; each component will just contribute $1$ to the total rank of the group.