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Let $A$ be a commutative ring, let $E,F$ be $A$-modules. Let $x\in E$ and $y\in F$. If $x\otimes y=0$ in $E\otimes F$, then there exist finitely generated submodules $M\subset E$ containing $x$ and $N\subset F$ containing $y$ such that $x\otimes y=0$ in $M\otimes N$.

Now, I already know the proof of this statement, see also here. $E$ and $F$ are the direct limits of their finitely generated submodules, so $E\otimes F$ is the direct limit of the modules of the form $M\otimes N$ with $M\subset E$ an $N\subset F$ finitely generated submodules.

Suppose we didn't know about direct limits. What would be the easiest proof of this statement? In other words: is the concept of direct limit essential here or a sledgehammer to crack a nut?

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The proof using directed ${\bf co}$limits is certainly the most easiest one. And when you define the tensor product via its universal property, you certainly have to use this universal property in order to say something about it. It easily implies that the tensor product commutes with colimits in each variable, so it is quite natural to use directed colimits here. And a direct proof using the universal property seems to be out of sight: If $x \otimes y = 0$, this means that $\beta(x,y)=0$ for all bilinear maps $\beta$ defined on $E \times F$. But why should this be true for bilinear maps defined on $M \times N$ for some finitely generated submodules $M,N$ of $E,F$? How to produce them from an "for all ..." assumption?

When you prefer the usual (otherwise useless) construction of the tensor product as the free module on the product modulo the submodule generated by bilinear relations, then the claim can be checked directly: If $x \otimes y=0$, this means that $(x,y)$ lies in the submodule, so it is a finite linear combination of bilinear relations. Only finitely elements of $E$ resp. $F$ are used here. So if $M$ and $N$ are the submodules generated by these elements, we also have that $(x,y)$ is a linear combination of bilinear relations in $M \times N$ and therefore $x \otimes y = 0$ in $M \otimes N$.

A similar proof works with an alternative construction of the tensor product.

The moral of this proof is: The reason for $x \otimes y=0$ has finite length, so involves only finitely many elements. For example, we have $1 \otimes 1 = 0$ in $\mathbb{Q} \otimes \mathbb{Z}/2$, the reason is that $(1,1) = \bigl((2 \cdot \frac{1}{2},1) - (\frac{1}{2},2 \cdot 1)\bigr) + \bigl((\frac{1}{2},0)-0 (\frac{1}{2},0)\bigr)$. Hence this already holds in $\langle \frac{1}{2} \rangle \otimes \mathbb{Z}/2$.

Once again: The proof using directed colimits is more conceptual.

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