Let $A$ be an abelian group (i.e., a $\mathbb Z$-module), and consider the tensor product $\mathbb Q\otimes_{\mathbb Z} A$. I want to show that the element $\frac{1}{d}\otimes n=0$ iff there's a positive integer $r$ such that $rn=0$.
I have shown that this tensor product is isomorphic to $U^{-1}A$, where $U$ is the set of nonzero integers, which allows me to extend the $\mathbb Z$-module $A$ to a $\mathbb Q$-module. The isomorphism is given by $\frac{a}{b}\otimes n\mapsto b^{-1}(an)$, and the inverse by $b^{-1}n\mapsto \frac{1}{b}\otimes n$.
Thus, if $\frac{1}{d}\otimes n=0$, then it follows that $d^{-1}n=0$. But this means that $n=0$, right? I'm a bit confused as to why I need to find an $r$...perhaps my confusion stems from not exactly understanding the equivalence relation on $U^{-1}A$? I think the relation is $u_1^{-1}n_1=u_2^{-1}n_2$ iff $u_2n_1=u_1n_2$. But this means that, since $0=1^{-1}\cdot 0$, $d^{-1}n=0$ iff $n=1$...do I have the wrong equivalence relation?