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?

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.
