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.
Best Answer
Here is an answer using Tor:
Consider the exact sequence $$ 0\rightarrow\mathbb{Z}\rightarrow F\rightarrow K\rightarrow 0,$$
where the map from $\mathbb{Z}$ to $F$ sends the generator $1$ to $a$, and $K$ is the cokernel. Tensoring with $H$ we get the long exact sequence $$ \cdots Tor(F,H)\rightarrow Tor(K,H)\rightarrow H\rightarrow F\otimes H\rightarrow H\otimes K\rightarrow 0. $$
Now $Tor(F,H)=0$ since $F$ is torsion-free. So the kernel of $H\rightarrow F\otimes H$, given by sending $h$ to $a\otimes h$, is $Tor(K,H)$. This is not always non-zero, and the problem as stated is incorrect. Consider $H=\mathbb{Z}/2\mathbb{Z}$, $F=\mathbb{Z}$, $a=2$.