Showing $R(m\otimes n)$ is free if $Rm$ and $Rn$ are free Let $R$ be a commutative ring with identity. 
The following is a statement I came across about the submodule $Rt$ generated by a decomposable tensor $t=m\otimes n$ being free, given that $Rm$ and $Rn$ are free.  I am not sure if the converse is true but I would be interested in seeing a counterexample.
Let $M, N$ be $R$-modules, and let $m$ be in $M$ and $n$ be in $N$.  Suppose also that $Rm$ and $Rn$ are free.  

Is $R(m \otimes n)$ free?

 A: I think this works. Let $A=k[x,y,z]/(x^2,xy,xz)$, let $M$ be the quotient of the free $A$-module generated by $e_1$ and $e_2$ subject to the relations $$xe_1=ye_2 \qquad ze_2=0$$ and let $N$ be the  quotient of the free $A$-module generated by $f_1$ and $f_2$ subject to the relation $$yf_1=zf_2.$$  Then $e_1$ is free in $M$ and $f_1$ is free in $N$, yet $$x\cdot e_1\otimes f_1 = xe_1\otimes f_1 = ye_2\otimes f_1 = e_2\otimes yf_1 = e_2\otimes zf_2 = ze_2\otimes f_2 = 0.$$
Let me check using Macaulay2:
First, construct our base ring
i1 : R = QQ[x,y,z]/(x*z,x*y,x*x);

Next, $M$ as a quotient of the free module $F=R^2$
i2 : F = R^2;

i3 : M = F / (x*F_0 - y*F_1, z*F_1);

and then $N$, also as a quotient of $F$,
i4 : N = F / (y*F_0 - z*F_1);

The element $e_1$, the image of the first generator of $F$ in $N$ is free:
i5 : kernel map(M, R^1, {{1}, {0}})

o5 = image 0

Likewise, $f_1$,the image of the first generator of $F$ in $N$ is free:
                             1
o5 : R-module, submodule of R

i6 : kernel map(N, R^1, {{1}, {0}})

o6 = image 0
                             1
o6 : R-module, submodule of R

Finally, $e_1\otimes f_1$ is not free in $M\otimes N$:
i7 : kernel map(M**N, R^1, {{1}, {0}, {0}, {0}})

o7 = image | x |

                             1
o7 : R-module, submodule of R

This not only shows that $x$ kills $e_1\otimes f_1$ but that in fact it generates its (one-dimensional) annihilator.
N.B. I constructed this by first deciding the relations which define the modules, and then iteratively computing kernels using and adding relations to the ring until I got $e_1$ and $f_1$ to be free.
A: This is a minor complement to Mariano's answer. The goal is to make the computation as easy and visual as possible. 
Let $K$ be a field, let $X,Y$ be indeterminates, and let $x,y$ be the canonical images of $X,Y$ in 
$$A:=\frac{K[X,Y]}{(X^2,XY,Y^2)}\quad.
$$ 
Using the diagram 
$$
e_1\stackrel{y}{\to}e_2\stackrel{x}{\leftarrow}e_3\stackrel{y}{\to}e_4 
$$ 
define the $A$-module $E$ as follows: 


*

*$\{e_1,e_2,e_3,e_4\}$ is a $K$-basis of $E$; 

*the first arrow means $ye_1=e_2$; 

*the absence of an $x$-arrow emanating from $e_1$ means $xe_1=0$; 
and so on. 
This is indeed an $A$-module because the arrows are uncomposable, and $Ae_3$ is free because two arrows emanate from $e_3$. 
Let $F$ be the $A$-module attached in a similar way to the diagram 
$$ 
f_2\stackrel{x}{\leftarrow}f_3\stackrel{y}{\to}f_4\stackrel{x}{\leftarrow}f_5. 
$$ 
In particular $Af_3$ is free. 
Now compute 
$$
x(e_3\otimes f_3)=xe_3\otimes f_3=ye_1\otimes f_3=e_1\otimes yf_3=e_1\otimes xf_5=xe_1\otimes f_5=0. 
$$ 
So $A(e_3\otimes f_3)$ is not free, although $Ae_3$ and $Af_3$ are. 
