A module is flat if and only if the tensor product of 1 with any injective homomorphism from a finitely generated module is injective

Let R be a ring and let A be a unital right R-module. Prove that A is flat if and only if for all injective left R-module homomorphisms \psi : L \rightarrow M where L is finitely generated, the map 1 \otimes \psi : A \otimes_R L \rightarrow A \otimes_R M is injective.


Certainly if A is flat, then every such map 1 \otimes \psi is injective.

Suppose now that A is a right R-module such that for all injective left R-module homomorphisms \psi : L \rightarrow M with L finitely generated, 1 \otimes \psi is injective. Let \psi : L \rightarrow M be an arbitrary injective module homomorphism. Suppose (1 \otimes \psi)(\sum a_i \otimes \ell_i) = 0. Now there exists a finitely generated submodule L^\prime \subseteq L containing all of the \ell_i. (For example, the submodule generated by the \ell_i.) The restriction \psi^\prime of \psi to L^\prime is injective by our hypothesis, and certainly (1 \otimes \psi)(\sum a_i \otimes \ell_i) = (1 \otimes \psi^\prime)(\sum a_i \otimes \ell_i). Thus \sum a_i \otimes \ell_i = 0. In particular, \mathsf{ker}\ 1 \otimes \psi = 0, so that 1 \otimes \psi is injective.

Thus A is flat.

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