I asked this question on [math.stackexchange](https://math.stackexchange.com/q/2348425/275190), but without much success.


Assume that $R$ is a ring (commutative, with unit) and that $M$, $N$ are two $R$-modules. Let $(e\_i)\_{i\in I}$ and $(f\_j)\_{j\in J}$ be families of elements in $M$ and $N$, respectively. In $M\otimes\_RN$, consider the family $$(e\_i\otimes f\_j)\_{(i,j)\in I\times J}\;.$$


It is not difficult to prove that if $(e\_i)$ and $(f\_j)$ are generating families, then $(e\_i\otimes f\_j)$ generates $M\otimes\_RN$. And if both are bases, then so is $(e\_i\otimes f\_j)$.


However, it does not follow immediately that if $(e\_i)$ and $(f\_j)$ are linearly independent families (not necessarily generating), then so is $(e\_i\otimes f\_j)$. My question is therefore, does anybody know a counterexample? Or is it true and there is a proof?


Note: If $R$ was a field, then one can extend $(e\_i)$ and $(f\_j)$ to bases and see that $(e\_i\otimes f\_j)$ becomes part of a basis. If $R$ is an integral domain mit quotient field $\Bbbk=Q(R)$, then $(e\_i)$ and $(f\_j)$ remain linear independent in $M\otimes\_R\Bbbk$, $N\otimes\_R\Bbbk$, and we are in the situation above. So $R$ has to be "weird" enough to start with.



linear independent families in a tensor product
[This answer](https://mathoverflow.net/a/232343/22989) to a related question gives a way of constructing counterexamples.


For a similar but more concrete example, let $k$ be a field and $R=k[x,y]/(x^2,xy,y^2)$, so $R$ is a $3$-dimensional algebra over $k$, spanned by $1$, $x$ and $y$.


Let $M$ be the $5$-dimensional module with basis $\{a,b,c,d,e\}$, where $x$ acts by
$$ax=0, bx=d, cx=e, dx=0, ex=0$$
and $y$ acts by
$$ay=d, by=e, cy=0, dy=0, ey=0.$$


Then $\operatorname{Ann}\_R(b)=\{0\}$, so $\{b\}\subset M$ is linearly independent.


However, in $M\otimes\_RM$,
$$(b\otimes b)x=(bx)\otimes b=d\otimes b=(ay)\otimes b=a\otimes (by)=a\otimes e=a\otimes(cx)=(ax)\otimes c=0\otimes c=0,$$
so $\{b\otimes b\}\subset M\otimes\_RM$ is not linearly independent.


