Suppose $R$ is an integral domain and $M$, $N$ are torsion-free $R$-modules. If $m$ and $n$ are nonzero elements of $M$ and $N$ respectively then does it follow that $m \otimes n \neq 0$ ?
If either $M$ or $N$ is free then it is indeed true, but can it be generalized to the case where neither module is free but both modules are torsion-free?
If you assume that torsionfree modules are flat, then this is true. If a torsionfree module is non flat, you should be able to get counterexamples. See https://mathoverflow.net/questions/51095/flat-module-and-torsion-free-module for reference.
So assume $R$ is a Prüfer domain, that is, every torsionfree module is flat.
Consider the inclusion map $j\colon nR\to N$, which is injective. Since $M$ is torsion-free, it is flat and therefore $$ M\otimes j\colon M\otimes nR\to M\otimes N $$ is injective. Similarly, $i\colon mR\to M$ is injective, so also the $mR\otimes nR\to M\otimes nR$ is injective. Therefore the composition $mR\otimes nR\to M\otimes N$ is injective. On the other hand, there are isomorphisms $R\to mR$ and $R\to nR$ defined by $r\mapsto mr$ and $r\mapsto nr$ respectively, because $M$ and $N$ are torsion-free, so $R\otimes R\to mR\otimes nR$ is an isomorphism. The element $1\otimes 1$ is definitely nonzero in $R\otimes R$.
(All tensor products are over the ring $R$.)