Lemma 10.107.10. 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 jth 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
Comments (1)
Comment #9756 by Félix on