I'm reading this article : https://drexel28.wordpress.com/2012/01/25/tensor-product-of-free-modules/
I don't get a theorem in the above page:
Theorem: Let $M$&$N$ be free left $R$-modules. Then $M\otimes_R N$ is a free left $R$-module.
If $R$ is not commutative, which right $R$-module structure does $M$ have to form the tensor product $M\otimes_R N$?
Does this article assumes that $R$ is commutative and $M$ is equipped with the natural $(R,R)$-bimodule structure?
Being left $R$-free module, $M\simeq R^{(I)}$ for some index set $I$, thus there is a natural ($R$,$R$)-bimodule structure on $M$ (Choosing different isomorphism $M\stackrel{\sim}{\longrightarrow} R^{(I)}$, we may get different right $R$-module structure if $R$ is not commutative, but they all are ($R$,$R$)-isomorphic. Thus $M\otimes_R N$ can be equipped with a left $R$-module structure (not very natural though) which makes it a free left $R$-module.