Let $M$ be a $\mathbb Z$-module and consider $M \otimes_\mathbb Z \mathbb Q$. Then $m\otimes q = 0$ for some $q \in \mathbb Q\setminus \{0\}$, if and only if $m$ is a torsion element.
The "if" direction is quite straightforward: if $m \otimes z = 0$, then: $$m\otimes q = m \otimes \frac{zq}{z}=zm\otimes \frac{q}{z} = 0$$
The other direction is where I'm having trouble: I have shown that if $ q = a/b$ Then: $$\begin{alignat}{} m \otimes q = 0 &\Rightarrow \quad m\otimes\frac{a}{b} = 0 \\&\Rightarrow \frac {abm}{b}\otimes\frac{1}{b}= 0 \\&\Rightarrow abm \otimes 1= 0 \end{alignat}$$ Does this necessarily imply that $m$ is a torsion element?
Thanks.
Note that $m\otimes q=0$ for nonzero $q\in\Bbb Q$ iff $m\otimes1=0$ in $\Bbb Q$.
We can construct a module isomorphic to $\Bbb Q$ by a localisation construction. Define an equivalence relation on $M\times\Bbb N$ as follows: $[m,t]\sim[m',t']$ iff $t'm-t'm$ is torsion. It's routine to prove it's an equivalence relation, and that $[m,t]+[m',t']=[mt'+m't,tt']$ defines an addition on the equivalence classes making them into a group $M'$. This is a $\Bbb Q$-module, and $\phi:m\mapsto[m,1]$ is a homomorphism from $M$ to $M'$. Then $\phi$ induces a map from $M\otimes \Bbb Q$ to $M'$ and if $m$ is non-torsion, then $\phi(m)\ne0$ and as this is an image of $m\otimes1$ then $m\otimes 1\ne0$.