TITLE: finite abelian groups tensor product.
QUESTION [5 upvotes]: Is the following question obvious ?
Let $G$ be an abelian group, such that for any finite abelian group $A$, we have 
$G\otimes_{\mathbf{Z}}A=0$, does it mean that $G$ is a $\mathbf{Q}$-vector space ?

REPLY [8 votes]: You have $G\otimes(\Bbb Z/n\Bbb Z)=0$ for all 
 $n\in \Bbb N$. That means $G/nG=0$, so $G=nG$, that is all elements of $G$ are divisible by $n$. Then $G$ is a divisible Abelian group. Conversely if $G$ is a divisible Abelian group, then
 $G\otimes(\Bbb Z/n\Bbb Z)=0$ and so $G\otimes A=0$ for all finitely generated
Abelian groups.
But not all divisible Abelian groups are $\Bbb Q$-modules: they may have torsion.
As an example, let $G=\Bbb Q/\Bbb Z$.