## 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.