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

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 $m \otimes n \neq 0$?

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It suffices that N is flat. If you want a submodule of a tensor product to behave reasonably, then in general you need to tensor product with a flat module. – Jack Schmidt Oct 3 '11 at 3:51

## 1 Answer

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.

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@MaraianoSuarez-Alvarez Thanks for the answer. Is the map $f(m \otimes n) \neq 0$ imply $m \otimes n \neq 0$ because otherwise one of the maps in the composition would be zero? – user7980 Oct 3 '11 at 6:29

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@user7980: if $f:X\to Y$ is a group homomorphism and $x\in X$ is such that $f(x)$ is not zero, then $x$ itself is not zero... – Mariano Suárez-Alvarez♦ Oct 3 '11 at 6:30
@MaraianoSuarez-Alvarez if $m = 0$ dosent $m \phi_{i_0} (n) = 0$ ? – user7980 Oct 3 '11 at 6:33
Is it possible to rule this case out becasuse $\{ 0 \}$ has no basis? – user7980 Oct 3 '11 at 6:38
@user7980: $\{0\}$ has a basis: the empty set... In any case, I don't understand what your two last comments come from. – Mariano Suárez-Alvarez♦ Oct 3 '11 at 6:45
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