Show that tensoring by a free module of finite rank preserves arbitrary products
I know that there are some more general statements but I would like to have an explicit proof for this particular case. I tried to use the fact that free modules are flat but without success. In particular I can't see how a possible proof can fail if the rank is not finite.
Note that $$ \Bigl(\prod_{\lambda}M_\lambda\Bigr)\otimes_R R^n \cong \Bigl(\Bigl(\prod_{\lambda}M_\lambda\Bigr)\otimes_R R\Bigr)^{\!n} \cong \Bigl(\prod_{\lambda}M_\lambda\Bigr)^{\!n} \color{red}{\cong} \Bigl(\prod_{\lambda}M_\lambda^{n}\Bigr) \cong \Bigl(\prod_{\lambda}M_\lambda\otimes_R R^{n}\Bigr) $$
It's a general result that tensoring with finitely presented modules commutes with products.
For non finitely generated free modules, the red isomorphism does not hold.
About the red isomorphism (finite case), the map is $$ \bigl((m^{(k)}_\lambda)_{\lambda}\bigr)_{1\le k\le n} \mapsto \bigl((m^{(k)}_\lambda)_{1\le k\le n}\bigr)_{\lambda} $$ It's a special case of $$ \prod_{\mu}\Bigl(\prod_{\lambda}M_{\lambda,\mu}\Bigr) \cong \prod_{\lambda}\Bigl(\prod_{\mu}M_{\lambda,\mu}\Bigr) $$