I am trying to show that if $F,H$ are abelian groups with $F$ free abelian, and if $a \in F$ and $h \in H$ are non-zero, then $a \otimes h \ne 0$ in $F \otimes H$.
This is specifically in a section describing the derived functor Tor. Of course, that doesn't mean the solution has to involve that, but there is probably a way. I know that $F$ free abelian means that $F$ is torsion free and hence $\mbox{Tor}(F,A)=0$.
I was trying to use a formulation of $\mbox{Tor}$ in terms of exact sequences. If:
$0 \to R \stackrel{i}{\hookrightarrow} F \to A \to 0$ is an exact sequence then $\mbox{Tor}(A,B) =\mbox{ker}(i \otimes 1_b)$
Seemed to me if I picked the right sequence I could get that $\mbox{Tor}=0$ implies that the kernel is trivial, which would give the result, but I can't get this to work
Edit It appears that this is false from the answers below. Here is a link to the question.