TITLE: Showing $R(m\otimes n)$ is free if $Rm$ and $Rn$ are free
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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.