What does it mean for the $(S^{-1}R)$-module $S^{-1}M$ to be compatible with the $R$-module structure?

Question

From Algebra, Chapter $0$ by Aluffi:

I think $S^{-1}M$ will be an $S^{-1}R$-module naturally since $M$ is an $R$-module: we define the action of $a/s_0 \in S^{-1}R$ on $m/s \in S^{-1}M$ by $$\frac{a}{s_0} \cdot \frac{m}{s} := \frac{a\cdot m}{s_0s}.$$

What does it mean for the $(S^{-1}R)$-module $S^{-1}M$ to be compatible with the $R$-module structure?

Answer 1

Say you have two rings $R, T$ and a morphism $i:R\to T$ and an $R$-module $M$, a $T$-module $N$, and an abelian group morphism $f:M\to N$.

Then $f$ is "compatible" with all this data if when we give $N$ the induced $R$-module structure ($r\cdot n := i(r)n$) $f$ is an $R$-module morphism.

Here, $i:R\to S^{-1}R$ is the canonical map and $f:M\to S^{-1}M$ is also the canonical map.

Concretely, it boils down to $(r/1)(m/1) = (rm)/1$ for all $r\in R, m\in M$