linear independent families in a tensor product

Question

I asked this question on math.stackexchange, 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.

Answer 1

This answer 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.

Answer 2

Where $M$ and $N$ are both free of finite type, I think on can proceed as follows. We start with a remark : if $M$ is a free $R$-module of finite type and $x \in M$ such that $\forall u \in M^*, \ u(x) = 0$, then $x=0$. Indeed, write $x = \sum_{j=1}^{p} \mu_i.x_i$ where $x_1, \ldots, x_p$ is a basis of $M$. Then $\mu_i = x_i^*(x) = 0$ for all $i$ and $x = 0$. (note that the linear form $x_i^*$ is well-defined as $\{x_1, \ldots, x_p\}$ is a basis of $M$.)

Now we prove that if both families $\{e_i,i \in I\}$ and $\{f_j, j \in J\}$ are linearly independent, then $\{e_i \otimes f_j, (i,j) \in I \times J \}$ is linearly independent in $M \otimes N$. Let $(\lambda_{i,j}) \in R^{|I| \times |J|}$ such that: $$\sum_{i \in I, j \in J} \lambda_{i,j}.e_i \otimes f_j = 0.$$ For any $u \in M^*$ and $v \in N^*$, the application: $$ u \otimes v : M \times N \longrightarrow R$$ defined by $\forall (m,n) \in M\times N, \ u \otimes v(m,n) = u(m)v(n)$ is bilinear. The universal property of the tensor product then ensures that: $$\sum_{i \in I, j \in J} \lambda_{i,j} u(e_i)v(f_j) = 0,$$ that is: $$ u \left(\sum_{i \in I} \left(\sum_{j \in J} \lambda_{i,j} v(f_j)\right).e_i \right) = 0.$$ As this is true for all $u \in M^*$, the preliminary remark implies that: $$\sum_{i \in I} \left(\sum_{j \in J} \lambda_{i,j} v(f_j)\right).e_i = 0$$ By linear independence of the $e_i$, we get: $$\forall i \in I, \ \sum_{j \in J} \lambda_{i,j} v(f_j) = 0,$$ That is: $$\forall i \in I, \ v\left(\sum_{j \in J} \lambda_{i,j}.f_j \right) = 0.$$ As it is true for all $v \in N^*$, the preliminary remark again implies that: $$\forall i \in I, \ \sum_{j \in J} \lambda_{i,j}.f_j = 0.$$ By linear independence of the $f_j$, we finally get $\forall i \in I, \ \forall j \in J, \lambda_{i,j} = 0$.

I have been looking for counter-examples when $R$ is not an integral domain and $M$ and $N$ are not free, but I haven't been able to find one. Perhaps someone will be luckier than me.