# Testing whether an element of a tensor product of modules is zero

Let $A$ be a commutative ring and let $F$, $E$ be $A$-modules. Then $E$ is the direct limit of its finitely generated submodules. Serge Lang, in his Algebra, p. 604, remarks that a technique to test whether an element of $F \otimes E$ is zero, is to examine its image in $F \otimes N$, as $N$ varies over the finitely generated submodules of $E$. Could somebody please explain why this is a valid technique? What is the image of an element of $F \otimes E$ in $F \otimes N$?

## 1 Answer

There is the following (quite unknown ...) criterion when an element of the tensor product $$M \otimes_A N$$ of two $$A$$-modules $$M,N$$ vanishes:

Choose generating sets $$E$$ of $$M$$ and $$F$$ of $$N$$. Every element of $$M \otimes_A N$$ 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 $$\lambda : E \times F \to A$$ with finite support such that

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

You should think $$\lambda$$ as a finite matrix which gives a reason why $$\sum_{e \in E} e \otimes n(e)$$ vanishes, namely why it is optained from $$0 \otimes 0$$ by applying bilinear relations.

There is a nice proof for this using the general properties of the tensor product, appearing in Pierre Mazet, Caracterisation des epimorphismes par relations et generateurs. There it is also used to give a full characterization of epimorphisms in the category of commutative rings (which, however, is useless in practice; in my opinion only the corollary about the cardinality is interesting).

As for the question about Lang's comment: It is already explained very well, so I don't really know what I should add. Since $$M$$ is the directed colimit of its finitely generated submodules $$M_i$$, it follows that $$M \otimes_A N$$ is the directed colimit of the $$M_i \otimes_A N$$ (which are, however, not submodules in general!). This means that every element of $$M \otimes_A N$$ lies in the image of $$M_i \otimes_A N \to M \otimes_A N$$ for some $$i$$, and that two elements of $$M_i \otimes_A N$$ and $$M_j \otimes_A N$$ become equal in $$M \otimes_A N$$ iff they become equal in $$M_k \otimes_A N$$ for some $$k \geq i,j$$. In particular, an element of $$M \otimes N$$, which vanishes, already vanishes in some $$M_i \otimes N$$ for $$i$$ large enough. This is often used for establishing general results in homological algebra which reduce something concerning tensor products for the finitely generated case. For example, $$M$$ is flat if and only if $$\mathrm{Tor}_{>0}(M,N)=0$$ for all finitely generated $$N$$. And $$M$$ is flat as soon as every finitely generated submodule of $$M$$ is flat. For example, this implies that torsionfree $$\mathbb{Z}$$-modules are flat (try to prove this directly). But note that Lang does not give any criterion when an element in the tensor product precisely vanishes.

• How is this not well-known? This is Lemma 6.4 in Eisenbud. Am I missing something? – Alex Youcis Mar 9 '13 at 21:08
• Oh, thank you. Meanwhile I've also found it in Bourbaki's Algèbre. So let's just say that this criterion is not used very often. Ok? – Martin Brandenburg Dec 23 '13 at 20:09
• @AlexYoucis You are right. However Eisenbud's proof is too messy: he only consider generators for the module $N$, and he uses additional, unnecessary subscripts. – Matemáticos Chibchas Oct 1 '15 at 0:58