direct product commutes with tensor product?

Question

Let $(A_i)_{i\in I}$ be a family of right $R$-modules and $M$ be a left $R$-module, where $I$ is an index set. The natural homomorphism $$\varphi:(\prod_{i\in I}A_i)\otimes_RM\to \prod_{i\in I}(A_i\otimes_RM)$$ given by $(a_i)\otimes m\mapsto(a_i\otimes m)$ is not always a bijection. It is easy to obtain $\varphi$ is a surjection provided $M$ is finitely generated. If we assume that $M$ is finitely presented, can we prove $\varphi$ is a bijection?

Answer 1

$M$ being finitely presented means that there is an exact sequence $$R^m \to R^n \to M \to 0.$$ This gives a commutative diagram $$\require{AMScd}\begin{CD} \bigl(\prod_i A_i\bigr) \otimes_R R^m @>>>\bigl(\prod_i A_i\bigr) \otimes_R R^n @>>> \bigl(\prod_i A_i\bigr) \otimes_R M @>>> 0\\ @VVV @VVV @VVV \\ \prod_i (A_i \otimes_R R^m) @>>> \prod_i (A_i \otimes_R R^n) @>>> \prod_i (A_i \otimes_R M) @>>> 0 \end{CD}$$ with exact rows since tensoring is right exact. The first and second vertical arrows are easily seen to be isomorphisms. Then by the five lemma, the right vertical arrow is an isomorphism, too.