Simple tensors and projective modules For a (unital) ring $R$, let $M$ and $N$ be a projective right, and left, $R$-module respectively.
Does it hold that $0 \neq m \otimes n \in M \otimes_R N$ for all non-zero $m \in M$, $n\in N$?
If so, then why?
 A: It does not work even for projective, or free, modules, and the example is already given in comments to the linked more general question by John Palmieri. Let me recap it:
Take $R$ to be a non-reduced commutative ring, e.g. $R=k[X]/(X^2)$, and let $x \in R$ be a nonzero element with $x^2=0.$ Then $R\otimes_R R \simeq R$ by the "multiplication map" $a\otimes b \mapsto ab$, and so the element $x \otimes x$ corresponds under this isomorphism to $x^2=0$.

In fact, the case $R\otimes R$ is in some sense the essential case to consider: Suppose that $M\oplus M' \simeq R^{\oplus I}, N\oplus N'\simeq R^{\oplus J}$. Then there is an injective map
$$M \otimes_R N \hookrightarrow_{\oplus} R^{\oplus I}\otimes_R R^{\oplus J} \simeq (R\otimes_R R)^{\oplus (I\times J)} \simeq R^{\oplus (I\times J)}.$$
So the original elementary tensor $m\otimes n$ induces bunch of elements $(m_{i}n_{j})_{i \in I, j\in J}$ of $R$, and $m\otimes n$ is nonzero iff at least one of the products $m_{i}n_{j}$ is nonzero.
Given the fact that at least one of the $m_i$'s and  at least one of the $n_j$'s
are nonzero, it follows that $m\otimes n \neq 0$ provided that $R$ has no non-zero divisors.
