Lemma 10.106.10. Let $R$ be a ring. Let $M$, $N$ be $R$-modules. Let $\{ x_ i\} _{i \in I}$ be a set of generators of $M$. Let $\{ y_ j\} _{j \in J}$ be a set of generators of $N$. Let $\{ m_ j\} _{j \in J}$ be a family of elements of $M$ with $m_ j = 0$ for all but finitely many $j$. Then

\[ \sum \nolimits _{j \in J} m_ j \otimes y_ j = 0 \text{ in } M \otimes _ R N \]

is equivalent to the following: There exist $a_{i, j} \in R$ with $a_{i, j} = 0$ for all but finitely many pairs $(i, j)$ such that

\begin{align*} m_ j & = \sum \nolimits _{i \in I} a_{i, j} x_ i \quad \text{for all } j \in J, \\ 0 & = \sum \nolimits _{j \in J} a_{i, j} y_ j \quad \text{for all } i \in I. \end{align*}

**Proof.**
The sufficiency is immediate. Suppose that $\sum _{j \in J} m_ j \otimes y_ j = 0$. Consider the short exact sequence

\[ 0 \to K \to \bigoplus \nolimits _{j \in J} R \to N \to 0 \]

where the $j$th basis vector of $\bigoplus \nolimits _{j \in J} R$ maps to $y_ j$. Tensor this with $M$ to get the exact sequence

\[ K \otimes _ R M \to \bigoplus \nolimits _{j \in J} M \to N \otimes _ R M \to 0. \]

The assumption implies that there exist elements $k_ i \in K$ such that $\sum k_ i \otimes x_ i$ maps to the element $(m_ j)_{j \in J}$ of the middle. Writing $k_ i = (a_{i, j})_{j \in J}$ and we obtain what we want.
$\square$

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