Let $R$ be a commutative ring with unity, let $P$ be a projective and finitely generated $R$-module, and let $(M_\alpha)_{\alpha}$ be a (possibly infinite) family of $R$-modules. I want to prove that the canonical map
$$(\prod_\alpha M_\alpha) \otimes_R P \to \prod_\alpha M_\alpha \otimes P,$$
$$(m_\alpha)_\alpha \otimes_R p \mapsto (m_\alpha \otimes p)_\alpha,$$
is an isomorphism.
Attempt: The claim holds if $P = \bigoplus_{i=1}^k R$ is a free and finitely generated $R$-module, since then
$$(\prod_\alpha M_\alpha) \otimes_R P = (\prod_\alpha M_\alpha) \otimes_R (\bigoplus_{i=1}^k R) \cong \bigoplus_{i=1}^k (\prod_\alpha M_\alpha) \otimes_R R \cong $$
$$\prod_{i=1}^k \prod_\alpha M_\alpha \cong \prod_\alpha \bigoplus_{i=1}^k M_\alpha \otimes_R R \cong \prod_\alpha M_\alpha \otimes_R P.$$
Further, the above map is clearly natural in $P$. According to the literature, this observation should imply that the claim holds for projective finitely generated $P$, but I do not see how. Could anybody help me please? (Any definition of a projective module can be used here.)
Best Answer
Let $P\oplus Q=R^k$. Then you have a split exact sequence $0\to P\to R^k\to Q\to0$ which produces the commutative diagram with split exact rows $$\require{AMScd} \begin{CD} 0 @>>> \Bigl(\prod_\alpha M_\alpha\Bigr) \otimes_R P @>>> \Bigl(\prod_\alpha M_\alpha\Bigr) \otimes_R R^k @>>> \Bigl(\prod_\alpha M_\alpha\Bigr) \otimes_R Q @>>> 0 \\ @. @VVV @VVV @VVV \\ 0 @>>> \prod_\alpha (M_\alpha \otimes P) @>>> \prod_\alpha (M_\alpha \otimes R^k) @>>> \prod_\alpha (M_\alpha \otimes Q) @>>> 0 \end{CD} $$ where the middle vertical row is an isomorphism. A simple diagram chasing shows that that the left and right arrows are, respectively, injective and surjective.
Exchange the roles of $P$ and $Q$.