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The Stacks project

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

What is K in this context?


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