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The question is:

Assume that $R$ is a (commutative) ring. Under what conditions on $R$-modules $M,N$ does the tensor product $M\otimes_RN$ consist of elementary tensors only? That is, every element of $M\otimes_RN$ is of the form $m\otimes n$ for some $m\in M, n \in N$.

This question is not really motivated by anything, apart from the fact that it seems somewhat more comfortable to work with elementary tensors than consider sums of those. This is for example the case of the localization of a module over commutative ring (i.e. $M\otimes_{R}S^{-1}R$; here every element is of the form $m\otimes\frac{1}{s}$ for suitable $s\in S$), where this fact can be useful.

In the example above, $S^{-1}R$ is always flat as an $R$-module. So, does it suffice for example to assume that $N$ is a flat $R$ module? Or, more generally, that $\mathrm{Tor}_i^{R}(M,N)$ are trivial?

Thanks in advance for any help.

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    $\begingroup$ No, it doesn't suffice to assume flatness or vanishing Tor; consider the case that $R$ is a field and $M, N$ have dimension greater than $1$. You should expect that this almost never happens. In any case, there's no reason to be afraid of non-elementary tensors; almost anything you'd ever want to do to a tensor is linear or multilinear anyway. $\endgroup$ May 19, 2014 at 18:51
  • $\begingroup$ @Qiaochu Yuan: I see, this is certainly true, thanks. Are there at least some other classes of modules (apart from the localiztions of the ring) with the considered property? (Or: what is so special about the localizations? Recently I was told that a localization of a ring is basically a direct limit over some set of copies of $R$. Is this the property that is somewhat relevant?) $\endgroup$ May 21, 2014 at 21:24
  • $\begingroup$ The special property of the localization that makes this work is that every pair of elements is an $R$-multiple of a common element; this is like having rank $1$ (and maybe is equivalent to it?). $\endgroup$ May 22, 2014 at 0:24

1 Answer 1

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If we speak about unital rings and the definition of a module implies $1\cdot m = m$, then there is an answer for an important case where $R$ is a field. In this case, modules are vector spaces.

$M\otimes_F N$ consists of elementary tensors if and only if at least one of $M$, $N$ has dimension not greater than one.

But it might be obvious for the original poster. What about a general commutative ring, indeed?

$M\otimes_R N$ consists of elementary tensors only if at least one of $M$, $N$ has rank not greater than one.

I follow the definition of the rank of a module as “the maximal cardinality of linearly independent subset”. One can easily check that if ranks of both $M$ and $N$ are 2 or greater, then there are tensors in $M\otimes_R N$ that can’t be expressed as elementary tensors.

Now we found a necessary condition, but it is not sufficient. There are even rank $0$ ${\mathbb Z}$-modules whose tensor product doesn’t consist of elementary tensors only, such as $$ ({\mathbb Z}_2\oplus{\mathbb Z}_2)\otimes_{\mathbb Z} ({\mathbb Z}_2\oplus{\mathbb Z}_2).$$

On the other hand, rank = 1 does not imply that the module is one-dimensional in the same sense as a vector space. There are many important cases of rank 1 modules that are not simplistic ones (such as the ring over itself, or a quotient ring over the ring). For example, rational numbers ${\mathbb Q}$ form a rank 1 ${\mathbb Z}$-module that is not finitely generated, and not one-dimensional in a reasonable sense at all. For instance, in the sense invented by me, Incnis Mrsi.

There are cases where $M\otimes_R N$ consists of elementary tensors, but neither $M$ nor $N$ is one-dimensional in a reasonable sense.

The simplest example is $$ {\mathbb Q}\otimes_{\mathbb Z} {\mathbb Q} = {\mathbb Q}. $$

Though, some “one-dimensional in a very broad sense” definition (but stronger than rank ≤ 1), that captures all such cases, might exist.

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