Vector space over $\mathbb{Q}$ or $\mathbb{Z}$? I am looking at the following:

Show that a torsion-free divisible group $G$ is a vector space over $\mathbb{Q}$.

I have no problem verifying the axioms of vector spaces after noting that the term divisible group $G$ implies that a solution to $ny=x$ exist where $n$ is an integer and is unique by the fact that the group $G$ is torsion-free.
What does vector spaces over $\mathbb{Q}$ mean? And why does the question say that it is a vector spaces over $\mathbb{Q}$ instead of $\mathbb{Z}$? 
 A: A vector space is defined over a field. If you want to define something over a ring, it's known as a module.
A: 
And why does the question say that it is a vector spaces over $\mathbb Q$ instead of $\mathbb Z$?

Notice that an Abelian group is a module over $\mathbb Z$ in a canonical way. The problem you quote says that this $\mathbb Z$-module structure extends canonically to a $\mathbb Q$-module or $\mathbb Q$-vector space structure provided that your group is also torsion-free and divisible .
A: Because it is talking about vector spaces over $\mathbb{Q}$ (the rational numbers). This is know as a $\mathbb{Q}$-vector space or more generally an $F$-vector space (where $F$ is an arbitrary field). See this for more details.
A: 
What does vector spaces over $\mathbb{Q}$ means?

Not sure, but are you asking what one actually means by a vector space $V$ over a field $F$? The field is the set of acceptable scalars for your vector space. So if the vector space is over $\mathbb{Q}$, you couldn't have $\pi$ as a scalar, for example, whereas you could have $3/4$.
