Show that $R \otimes_R M\cong M$ is isomorphic to $M$, for every left $R$-module $M$, $R$ a ring

Question

My professor says it is trivial, but I cannot still see the triviality :(

Let $R$ be a ring. Show that $R \otimes_R M\cong M$ for a left $R$-module $M$.

Any ideas how to address this?

Answer 1

Since $R$ is an $R-R$ bimodule $R\otimes_R M$ is a left $R$ module. The assignment $(r,m) \rightarrow rm $ defines a middle linear map $R\times M\rightarrow M$. So there is a group homomorphism $\alpha: R\otimes_R M \rightarrow M$ such that$\alpha(r\otimes m)=rm$. Verify that $\alpha$ is in fact a homomorphism of left R-modules. Then verify that the map $\beta: M\rightarrow R\otimes_R M$ given by $m \rightarrow 1_{R} \otimes m$ is an $R$-module homomorphism such that $\alpha\beta=1_{M}$ and $\beta\alpha=1_{R\otimes_R M}$.Hence $\alpha:R\otimes_R M \cong M$. The isomorphism $A\otimes_R R \cong A$ is constructed similarly. (This proof is from Hungerford, Graduated Texts in Mathematics, 212 pp.)