Is Every Isomorphism on Tensor Product of Modules an R-module Isomorphism?

Question

We have an R-module M. We also have S, a multiplicatively closed subset of R. I want to prove that there exists a unique R-module isomorphism f: S$^{-1}$R$⊗$M $\to$S$^{-1}$M, defined by f(${\frac rs}$⊗$_R$m) = ${rm\over s}$. I used the Universal Property of Tensor Products to show f is a unique group homomorphism. I also proved f is both injective and surjective.

My question is, does my proof show that f is an R-module isomorphism or a group isomorphism? If my proof shows the later, how can I go ahead to prove f is an R-module isomorphism?

Thanks.

Answer 1

Let $R$ be a ring and $S\subset R$ a multiplicatively closed set; let $M$ be a $R$-module. Denote by $u:M\to S^{-1}R\otimes_R M$ the $R$-linear map defined by $m\mapsto \frac 11\otimes m$.

$S^{-1}R\otimes_R M$ has the following universal property: for any $R$-linear map $f:M\to N$, where the codomain $N$ is a $S^{-1}R$-module, there is a unique $S^{-1}R$-linear map $g:S^{-1}R\otimes_R M\to N$ such that $g\circ u=f$.

Sketch of proof. If $g$ exists, it necessarily maps $\frac 11\otimes m$ to $f(m)$, so in order to be $S^{-1}R$-linear $g$ maps $\frac rs\otimes m=\frac rs\cdot (\frac 11\otimes m)$ to $\frac rs\cdot f(m)$: this shows the uniqueness. To show that such $g$ exists, see that the set map $S^{-1}R\times M\to N$, defined by $(\frac rs, m)\mapsto \frac rs\cdot f(m)$, is $S^{-1}R$-linear in the first coordinate and $R$-linear in the second one.

Take $N:=S^{-1}M$ and let $f:M\to S^{-1}M$ be the $R$-linear map defined by $m\mapsto \frac m1$. Then the arguments above state that $S^{-1}R\otimes_RM\to S^{-1}M:\frac rs\otimes m\mapsto \frac {r\cdot m}s$ is the unique $S^{-1}R$-linear isomorphism between $S^{-1}R\otimes_RM$ and $S^{-1}M$.

In fact, it is also the unique $R$-linear isomorphism, since any $R$-linear map between $S^{-1}R$-modules is automatically $S^{-1}R$ linear.