Every $R$-linear map of $S^{-1} R$ Modules is automatically a $R_S$-linear map

Question

I am currently learning a little bit about localization in rings. I am stuck with this simple looking question:

"Every $R$-linear map of $S^{-1} R$ Modules is automatically a $S^{-1} R$-linear map"

Here, $R$ is a commutative and unitary ring and $S\subseteq R$ is a multiplicative monoid of $R$. But what can I try now? Applying the basic definition:

Let $\phi : M \rightarrow N$, with $M,N$ being $S^{-1} R$ Modules, $\phi$ is $R$-linear, so: $$ \phi(\lambda m + n) = \lambda \phi(m) + \phi(n) \hspace{3mm}\forall m \in M, n \in N, \lambda \in R $$

But what can I do now? I have to show, that its also equal for $\lambda \in R_S$, but how can I do that? Can someone give me a tip?

Answer 1

Let $\phi$ as in your problem; for convenience we denote $(1\big/s)m$ by $m\big/s$ for any $s\in S$, $m\in M$. Because $\phi$ is $R$-linear, it suffices to show that $\phi(m\big/s)=\phi(m)\big/s$ for any $m\in M, s\in S$. (Why? Explanation below, but think it through yourself first.)

Suppose the above equality holds. Since $\phi$ is $R$-linear it is in particular additive, so to show $S^{-1}R$-linearity we need only show that $\phi(\lambda m)=\lambda\phi(m)$ for any $m\in M,\lambda\in S^{-1}R$. But every element of $S^{-1}R$ is of the form $r\big/s$ for some $r\in R,s\in S$, and we have $$\phi((r\big/s)m)=\phi(rm\big/s)=\phi(rm)\big/s=r\phi(m)\big/s=(r\big/s)\phi(m)$$ where the second equality holds by the equation above and the third equality holds by $R$-linearity.

Now, by $R$-linearity we have that $s\phi(m\big/s)=\phi(sm\big/s)=\phi(m)$, and so multiplying both sides by $1\big/s$ gives the desired equality.