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This is a follow up to the post Showing $ m\otimes n$ is free given $m,n$ are free.

I am curious about conditions that would eliminate the possibility for a counterexample in the case that was outlined above.

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

If $Rm$ is free, how do you show $m \otimes n \neq 0$?

I was thinking of showing the contrapositive... Does the problem reduce to showing that if $m \otimes n = 0 $ then $Rm$ is not free?

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Let $R = {\mathbf Z}$, $M = {\mathbf Q}$, and $N = {\mathbf Z}/k{\mathbf Z}$ for any $k > 1$. Then any nonzero rational number $r$ is torsion-free in $M$ (please don't use the word "free" when you mean "torsion-free", since "free element" is not standard terminology) but for any $n$ in $N$ the tensor product $r \otimes n$ is $0$ since ${\mathbf Q} \otimes_{\mathbf Z} N = 0$. Perhaps you want to impose some condition on $N$ in your question (to make it self-contained). –  KCd Oct 2 '11 at 2:04
@KCd, he is considering rings possibly with zero divisors, so "torsion-free" is a bit strange, though. –  Mariano Suárez-Alvarez Oct 2 '11 at 2:37
My example is still a counterexample to the boxed question. user7980 should include some hypothesis on N which would eliminate my example. –  KCd Oct 2 '11 at 2:52
@user7980: The example given by KCd has $N$ finitely generated, so how "finitely generated" suffice? And you can take $M$ finitely generated (even free): take $R=\mathbb{Z}$, $M=\mathbb{Z}$, $m=k$ for $k\gt 0$, $N=\mathbb{Z}/k\mathbb{Z}$, and $n=1$. Then $m\otimes n = 0$. –  Arturo Magidin Oct 2 '11 at 4:34
@user7980: I have never seen "free" applied to an element or a set like you are doing; perhaps you could simply say "has trivial annihilator", which is more standard? –  Arturo Magidin Oct 2 '11 at 4:35

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