[proofplan] We show the natural map $\varphi: M \to S^{-1}(M_R)$ defined by $m \mapsto \tfrac{m}{1}$ is an isomorphism of $S^{-1}R$-modules. The key observation is that since $M$ is already an $S^{-1}R$-module, every element $s \in S$ acts invertibly on $M$, which forces the localization map to be bijective. Surjectivity follows because any fraction $\tfrac{m}{s}$ equals the image of $\tfrac{1}{s} \cdot m$. Injectivity follows because $\tfrac{m}{1} = 0$ in $S^{-1}(M_R)$ implies $um = 0$ for some $u \in S$, and the invertibility of $u$ in the $S^{-1}R$-module structure forces $m = 0$. [/proofplan] [step:Define the map $\varphi$ and verify $S^{-1}R$-linearity] Define the map \begin{align*} \varphi: M &\to S^{-1}(M_R) \\ m &\mapsto \frac{m}{1}. \end{align*} This is the canonical localization map applied to the restriction $M_R$. We verify that $\varphi$ is $S^{-1}R$-linear. Let $\tfrac{r}{s} \in S^{-1}R$ and $m \in M$. The $S^{-1}R$-module structure on $M$ gives an element $\tfrac{r}{s} \cdot m \in M$. We compute: \begin{align*} \varphi\!\left(\frac{r}{s} \cdot m\right) = \frac{\frac{r}{s} \cdot m}{1}. \end{align*} On the other hand, the $S^{-1}R$-action on $S^{-1}(M_R)$ gives \begin{align*} \frac{r}{s} \cdot \varphi(m) = \frac{r}{s} \cdot \frac{m}{1} = \frac{rm}{s}, \end{align*} where the last equality uses the $S^{-1}R$-module structure on $S^{-1}(M_R)$. To see that $\frac{\frac{r}{s} \cdot m}{1} = \frac{rm}{s}$, observe that the $R$-module action on $M_R$ satisfies $r \cdot m = \frac{r}{1} \cdot m$ (restricting the $S^{-1}R$-action). In $S^{-1}(M_R)$, we have $\frac{rm}{s} = \frac{r \cdot m}{s}$. Also $\frac{\frac{r}{s} \cdot m}{1} = \frac{s \cdot (\frac{r}{s} \cdot m)}{s} = \frac{(\frac{rs}{s}) \cdot m}{s} = \frac{r \cdot m}{s}$, where the second equality uses the definition of fractions in $S^{-1}(M_R)$ and the third uses $\frac{rs}{s} = \frac{r}{1}$ in $S^{-1}R$. Therefore $\varphi$ is $S^{-1}R$-linear. [guided] We need to check that $\varphi$ respects the $S^{-1}R$-module structure. The module $M$ is an $S^{-1}R$-module by hypothesis, and $S^{-1}(M_R)$ inherits an $S^{-1}R$-module structure as a localization. The $R$-module $M_R$ is obtained from $M$ by restricting scalars along the ring map $\iota: R \to S^{-1}R$, $r \mapsto \frac{r}{1}$. So for $r \in R$ and $m \in M$, the $R$-action on $M_R$ is $r \cdot m = \frac{r}{1} \cdot m$ (the $S^{-1}R$-action applied to the image of $r$). Let $\frac{r}{s} \in S^{-1}R$ and $m \in M$. We must show $\varphi(\frac{r}{s} \cdot m) = \frac{r}{s} \cdot \varphi(m)$. The left side is $\frac{\frac{r}{s} \cdot m}{1}$. The right side is $\frac{r}{s} \cdot \frac{m}{1} = \frac{r \cdot m}{s} = \frac{rm}{s}$, using the $S^{-1}R$-module structure on $S^{-1}(M_R)$. To equate these, note that in $S^{-1}(M_R)$: \begin{align*} \frac{\frac{r}{s} \cdot m}{1} = \frac{s \cdot (\frac{r}{s} \cdot m)}{s} = \frac{r \cdot m}{s} = \frac{rm}{s}, \end{align*} where the first equality uses the equivalence $\frac{x}{1} = \frac{sx}{s}$ in any localization, and the second uses $s \cdot (\frac{r}{s} \cdot m) = (s \cdot \frac{r}{s}) \cdot m = r \cdot m$ in $M$ (since $\frac{s}{1} \cdot \frac{r}{s} = \frac{r}{1}$ in $S^{-1}R$, and the $R$-action on $M_R$ is $r \cdot m = \frac{r}{1} \cdot m$). So $\varphi$ is $S^{-1}R$-linear. [/guided] [/step] [step:Prove surjectivity using the invertibility of elements of $S$ in $M$] Let $\frac{m}{s} \in S^{-1}(M_R)$ with $m \in M$ and $s \in S$. Since $M$ is an $S^{-1}R$-module, the element $\frac{1}{s} \in S^{-1}R$ acts on $m$ to give $\frac{1}{s} \cdot m \in M$. We claim $\varphi(\frac{1}{s} \cdot m) = \frac{m}{s}$. Indeed, $\varphi(\frac{1}{s} \cdot m) = \frac{\frac{1}{s} \cdot m}{1}$. In $S^{-1}(M_R)$, we compute: \begin{align*} \frac{\frac{1}{s} \cdot m}{1} = \frac{s \cdot (\frac{1}{s} \cdot m)}{s} = \frac{m}{s}, \end{align*} where the first equality uses $\frac{x}{1} = \frac{sx}{s}$ in the localization and the second uses $s \cdot (\frac{1}{s} \cdot m) = (\frac{s}{1} \cdot \frac{1}{s}) \cdot m = \frac{1}{1} \cdot m = m$ in the $S^{-1}R$-module $M$, then restricting to the $R$-action. Since every element of $S^{-1}(M_R)$ is of the form $\frac{m}{s}$, the map $\varphi$ is surjective. [guided] Why is $\varphi$ surjective? A general element of $S^{-1}(M_R)$ is a fraction $\frac{m}{s}$ with $m \in M$ and $s \in S$. We need to find an element of $M$ that maps to $\frac{m}{s}$. The natural candidate is $\frac{1}{s} \cdot m$, where $\frac{1}{s} \in S^{-1}R$ acts on $m$ via the $S^{-1}R$-module structure of $M$. This element is well-defined precisely because $M$ is an $S^{-1}R$-module, so $\frac{1}{s}$ is an available scalar. We compute $\varphi(\frac{1}{s} \cdot m) = \frac{\frac{1}{s} \cdot m}{1}$. In $S^{-1}(M_R)$, two fractions $\frac{x}{s_1}$ and $\frac{y}{s_2}$ are equal if and only if there exists $u \in S$ with $u(s_2 x - s_1 y) = 0$ in $M_R$. We check $\frac{\frac{1}{s} \cdot m}{1} = \frac{m}{s}$: this requires $u(s \cdot (\frac{1}{s} \cdot m) - 1 \cdot m) = 0$ for some $u \in S$. Taking $u = 1$, we need $s \cdot (\frac{1}{s} \cdot m) - m = 0$. In the $S^{-1}R$-module $M$, $s \cdot (\frac{1}{s} \cdot m) = \frac{s}{1} \cdot \frac{1}{s} \cdot m = 1 \cdot m = m$, so $s \cdot (\frac{1}{s} \cdot m) - m = 0$ in $M_R$. Therefore $\frac{\frac{1}{s} \cdot m}{1} = \frac{m}{s}$, confirming surjectivity. [/guided] [/step] [step:Prove injectivity using the invertibility of elements of $S$ in $M$] Suppose $\varphi(m) = 0$, i.e., $\frac{m}{1} = \frac{0}{1}$ in $S^{-1}(M_R)$. By the definition of equality in the localization, there exists $u \in S$ such that $u \cdot m = 0$ in $M_R$ (equivalently, $u(1 \cdot m - 1 \cdot 0) = um = 0$). Since $M$ is an $S^{-1}R$-module and $u \in S \subset R$, the element $u$ acts as $\frac{u}{1} \in S^{-1}R$ on $M$. The element $\frac{u}{1}$ is a unit in $S^{-1}R$ with inverse $\frac{1}{u}$. Therefore $u$ acts invertibly on $M$: \begin{align*} m = \frac{1}{u} \cdot (u \cdot m) = \frac{1}{u} \cdot 0 = 0. \end{align*} Hence $\ker \varphi = \{0\}$, and $\varphi$ is injective. [guided] If $\varphi(m) = \frac{m}{1} = 0$ in $S^{-1}(M_R)$, the definition of the localization gives some $u \in S$ with $um = 0$ in $M_R$. This is where the proof could end if $M$ were just an $R$-module --- we would only know that $m$ is "killed by something in $S$." But $M$ is an $S^{-1}R$-module, which is much stronger. Since $u \in S$, the element $\frac{u}{1}$ is a unit in $S^{-1}R$ (with inverse $\frac{1}{u}$). The $S^{-1}R$-module structure of $M$ means we can "divide by $u$": applying $\frac{1}{u}$ to both sides of $um = 0$: \begin{align*} m = \frac{1}{u} \cdot (u \cdot m) = \frac{1}{u} \cdot 0 = 0. \end{align*} This is the essential difference between localizing an $R$-module (where the localization map need not be injective) and localizing a module that already has an $S^{-1}R$-structure (where every element of $S$ acts invertibly, so the torsion that the localization was meant to kill is already absent). [/guided] [/step] [step:Verify the inverse map and conclude] The map $\psi: S^{-1}(M_R) \to M$ defined by $\psi(\frac{m}{s}) = \frac{1}{s} \cdot m$ is well-defined: if $\frac{m}{s} = \frac{m'}{s'}$ in $S^{-1}(M_R)$, then $u(s'm - sm') = 0$ for some $u \in S$. In $M$, applying $\frac{1}{uss'} \in S^{-1}R$: \begin{align*} \frac{1}{s} \cdot m - \frac{1}{s'} \cdot m' = \frac{1}{uss'} \cdot u(s'm - sm') = \frac{1}{uss'} \cdot 0 = 0. \end{align*} By the surjectivity and injectivity arguments above, $\psi \circ \varphi = \operatorname{id}_M$ and $\varphi \circ \psi = \operatorname{id}_{S^{-1}(M_R)}$. Therefore $\varphi$ is an isomorphism of $S^{-1}R$-modules with inverse $\psi$. [/step]