If $Rm$ is free and $N$ is free is $ m \otimes n \neq 0$?

Question

This post is a follow up to the counterexample presented in the following questions If $Rm$ is free, how do you show $m \otimes n \neq 0$?. The hope now is that we can eliminate the pathological behavior presented in the counterexamples by imposing the condition that one of the modules we are working with be free. In particular I hope the following question formulated below is sufficient to avoid the nice counter arguments cited above.

Let $R$ be a commutative ring with identity. Let $M, N$ be $R$-modules with $m \in M$ and $n\in N$ with $n \neq 0$.

If $Rm$ is free and $N$ is free is $ m \otimes n \neq 0$?

Answer 1

Let $\{n_i:i\in I\}$ be a basis for $N$, and let $\{\phi_i:i\in I\}\subseteq\hom(N,R)$ be the dual basis.

Since $n\neq0$, there exists $i_0\in I$ such that $\phi_{i_0}(n)\neq0$. Now consider the composition $f:M\otimes N\to M$ of the map $\mathrm{id}_M\otimes\phi_{i_0}:M\otimes N\to M\otimes R$ with the canonical isomorphism $m\otimes r\in M\otimes R\mapsto mr\in R$.

Then $f(m\otimes n)=m\phi_{i_0}(n)\neq0$, so $m\otimes n\neq0$.

Of course, the same works if $N$ is only projective and $\{(n_i,\phi_i)\in N\times\hom(N,R):i\in I\}$ is a 'dual basis' on $N$.

More generally, the statememt works if $N$ is simply flat. Indeed, that $m$ be free in your sense means that the map $\mu:r\in R\mapsto rm\in M$ is injective so if $N$ is flat, then so is $\mu\otimes \mathrm{id}_N:R\otimes N\to M\otimes N$ is also injective. The element $1\otimes n$ of $R\otimes N$ is not zero: it follows that $m\otimes n=\mu(1\otimes n)$ is also not zero.