When are two simple tensors $m' \otimes n'$ and $m \otimes n$ equal? (tensor product over modules)

Question

Suppose that $M$ is a right R-module and $N$ is a left $R$-module. We can construct $M \underset{R}\otimes N$ and give it an Abelian group structure by considering the free R-module $K$ generated by the relations $\{(x+x',y)-(x,y)-(x',y),(x,y+y')-(x,y)-(x,y'),(xr,y)-(x,ry)\}$ and then quotienting $M \times N$ by $K$. If $R$ is commutative this Abelian group turns into an $R$-module itself.

Now, the question is, what does $m' \otimes n' = m \otimes n$ mean in this situation? So, if I were given two simple tensors, how could I know if these two were equal or not?

I've been thinking about this for a while, but it's still pretty vague for me. A simpler question is when $m \otimes n=0$ in $M \underset{R}\otimes N$?

Answer 1

There are two trivial answers and one more profound answer:

1) $m \otimes n = m' \otimes n'$ means that $(m,n) - (m',n')$ lies in the mentioned submodule of bilinear relations

2) $m \otimes n = m' \otimes n'$ means that $\beta(m,n)=\beta(m',n')$ for all $R$-bilinear maps $\beta : M \times N \to T$, where $T$ is any abelian group.

3) We have the following theorem:

Choose generating sets $E$ of $M$ and $F$ of $N$. Every element of $M \otimes_R N$ can be written as $\sum_{e \in E} e \otimes n(e)$, where $n : E \to N$ is a function with finite support. It vanishes if and only if there is a map $\lambda : E \times F \to R$ ("matrix") with finite support such that

$$\forall e \in E ~: ~ n(e) = \sum_{f \in F} \lambda(e,f) \cdot f$$ $$\forall f \in F~ : ~ 0 = \sum_{e \in E} e \cdot \lambda(e,f) $$

Reference: Pierre Mazet, Caracterisation des epimorphismes par relations et generateurs

In any case, you see that it is very hard to decide if $m \otimes n = m' \otimes n'$ holds or not. But this doesn't cause any problems when working with the tensor product, since its universal property is the only thing what matters and with which you can prove properties of the tensor product.

Here is some further evidence that it is hard to determine when two pure tensors are equal: If $R$ is a commutative ring, let's call an $R$-module $M$ symtrivial if $m \otimes n = n \otimes m$ holds for all $m,n \in M$. If $M$ is finitely generated, then Nakayama's Lemma implies that $M$ is symtrivial if and only if $M$ is cyclic locally, i.e. there is an open covering $\mathrm{Spec}(R)=\bigcup_i D(f_i)$ such that each $R_{f_i}$-module $M_{f_i}$ is cyclic. This absolutely fails when $M$ is not finitely generated. So how to classify arbitrary symtrivial $R$-modules? I hope to see an answer some day. Meanwhile, Will Sawin has classified symtrivial modules in case $R$ is a Dedekind domain at mathoverflow. From this classification you see that (even for $R=\mathbb{Z}$) there are lots of symtrivial modules.