[proofplan] We construct the map $\varphi \colon S^{-1}R \otimes_R M \to S^{-1}M$ from the $R$-bilinear pairing $(\tfrac{r}{s}, m) \mapsto \tfrac{rm}{s}$ and verify it is $S^{-1}R$-linear. Surjectivity is immediate since every fraction $\tfrac{m}{s}$ is the image of $\tfrac{1}{s} \otimes m$. For injectivity, we first show that every element of $S^{-1}R \otimes_R M$ can be written as a pure tensor of the form $\tfrac{1}{s} \otimes m$ by clearing denominators, then show that $\varphi(\tfrac{1}{s} \otimes m) = 0$ implies $\tfrac{1}{s} \otimes m = 0$ by unwinding the equivalence relation in $S^{-1}M$. [/proofplan] [step:Construct $\varphi$ from the universal property of the tensor product] Define the map \begin{align*} \Phi \colon S^{-1}R \times M &\to S^{-1}M, \quad \left(\frac{r}{s},\; m\right) \mapsto \frac{rm}{s}. \end{align*} We verify $\Phi$ is $R$-bilinear. For the left variable, let $a \in R$: \begin{align*} \Phi\!\left(\frac{r_1}{s_1} + \frac{r_2}{s_2},\; m\right) = \Phi\!\left(\frac{s_2 r_1 + s_1 r_2}{s_1 s_2},\; m\right) = \frac{(s_2 r_1 + s_1 r_2)m}{s_1 s_2} = \frac{r_1 m}{s_1} + \frac{r_2 m}{s_2} = \Phi\!\left(\frac{r_1}{s_1}, m\right) + \Phi\!\left(\frac{r_2}{s_2}, m\right). \end{align*} For the right variable: $\Phi(\tfrac{r}{s}, m_1 + m_2) = \tfrac{r(m_1 + m_2)}{s} = \tfrac{rm_1}{s} + \tfrac{rm_2}{s}$. The $R$-balancing condition holds: $\Phi(\tfrac{r}{s} \cdot a, m) = \Phi(\tfrac{ra}{s}, m) = \tfrac{ram}{s} = \tfrac{r(am)}{s} = \Phi(\tfrac{r}{s}, am)$. By the universal property of the tensor product $S^{-1}R \otimes_R M$, there exists a unique $R$-linear (and hence abelian group) homomorphism \begin{align*} \varphi \colon S^{-1}R \otimes_R M \to S^{-1}M, \quad \frac{r}{s} \otimes m \mapsto \frac{rm}{s}. \end{align*} Moreover, $\varphi$ is $S^{-1}R$-linear: for $\tfrac{a}{t} \in S^{-1}R$, \begin{align*} \varphi\!\left(\frac{a}{t} \cdot \left(\frac{r}{s} \otimes m\right)\right) = \varphi\!\left(\frac{ar}{ts} \otimes m\right) = \frac{arm}{ts} = \frac{a}{t} \cdot \frac{rm}{s} = \frac{a}{t} \cdot \varphi\!\left(\frac{r}{s} \otimes m\right). \end{align*} [/step] [step:Prove surjectivity] Every element of $S^{-1}M$ has the form $\tfrac{m}{s}$ for some $m \in M$ and $s \in S$. We compute: \begin{align*} \varphi\!\left(\frac{1}{s} \otimes m\right) = \frac{1 \cdot m}{s} = \frac{m}{s}. \end{align*} Therefore every element of $S^{-1}M$ lies in the image of $\varphi$, and $\varphi$ is surjective. [/step] [step:Reduce every tensor to a pure tensor $\tfrac{1}{s} \otimes m$ by clearing denominators] Let $t = \sum_{i=1}^{\ell} \tfrac{r_i}{s_i} \otimes m_i$ be an arbitrary element of $S^{-1}R \otimes_R M$. Set $s = s_1 s_2 \cdots s_\ell \in S$ (a product of elements of the multiplicative set $S$, hence in $S$). For each $i$, define $t_i = \prod_{j \neq i} s_j$, so that $s = s_i t_i$. Using the $R$-balancing relation $\tfrac{r_i}{s_i} \otimes m_i = \tfrac{t_i r_i}{t_i s_i} \otimes m_i = \tfrac{t_i r_i}{s} \otimes m_i = \tfrac{1}{s} \otimes t_i r_i m_i$ (where the last equality uses $\tfrac{t_i r_i}{s} = \tfrac{1}{s} \cdot t_i r_i$ and the $R$-balancing $\tfrac{a}{s} \otimes m_i = \tfrac{1}{s} \otimes a m_i$ for $a = t_i r_i$): \begin{align*} t = \sum_{i=1}^{\ell} \frac{1}{s} \otimes t_i r_i m_i = \frac{1}{s} \otimes \sum_{i=1}^{\ell} t_i r_i m_i. \end{align*} Therefore every element of $S^{-1}R \otimes_R M$ is a pure tensor of the form $\tfrac{1}{s} \otimes m$. [guided] This reduction is the key to the injectivity argument. In a general tensor product $A \otimes_R M$, not every element is a pure tensor. However, the special structure of $S^{-1}R$ — where every finite collection of denominators can be brought to a common denominator — means that every element of $S^{-1}R \otimes_R M$ is in fact pure. Given $t = \sum_{i=1}^{\ell} \tfrac{r_i}{s_i} \otimes m_i$, the idea is to "clear denominators" by writing all fractions over a common denominator $s = s_1 \cdots s_\ell$. For each index $i$, define $t_i = \prod_{j \neq i} s_j$, the product of all denominators except $s_i$. Then $s = s_i t_i$ and \begin{align*} \frac{r_i}{s_i} = \frac{t_i r_i}{t_i s_i} = \frac{t_i r_i}{s}. \end{align*} Now we use the $R$-balancing property of the tensor product: for any $a \in R$, $\tfrac{a}{s} \otimes m = \tfrac{1}{s} \cdot a \otimes m = \tfrac{1}{s} \otimes am$ (since $\tfrac{a}{s} = \tfrac{1}{s} \cdot \tfrac{a}{1}$ and the balancing relation gives $\tfrac{a}{1} \otimes m = 1 \otimes am$, then scaling by $\tfrac{1}{s}$). More precisely, $\tfrac{t_i r_i}{s} \otimes m_i = \tfrac{1}{s} \otimes t_i r_i m_i$ because $\tfrac{t_i r_i}{s} = \tfrac{1}{s} \cdot (t_i r_i)$ and the $R$-balancing gives $(t_i r_i) \cdot \tfrac{1}{s} \otimes m_i = \tfrac{1}{s} \otimes (t_i r_i) m_i$. Therefore: \begin{align*} t = \sum_{i=1}^{\ell} \frac{1}{s} \otimes t_i r_i m_i = \frac{1}{s} \otimes \left(\sum_{i=1}^{\ell} t_i r_i m_i\right), \end{align*} using the linearity of the tensor product in the second variable. [/guided] [/step] [step:Prove injectivity by showing $\varphi(\tfrac{1}{s} \otimes m) = 0$ implies $\tfrac{1}{s} \otimes m = 0$] By the previous step, it suffices to show: if $\tfrac{1}{s} \otimes m \in \ker \varphi$, then $\tfrac{1}{s} \otimes m = 0$ in $S^{-1}R \otimes_R M$. Suppose $\varphi(\tfrac{1}{s} \otimes m) = 0$. Then $\tfrac{m}{s} = \tfrac{0}{1}$ in $S^{-1}M$. By the equivalence relation defining $S^{-1}M$, there exists $u \in S$ such that $u \cdot (1 \cdot m - s \cdot 0) = um = 0$ in $M$. We now compute in $S^{-1}R \otimes_R M$: \begin{align*} \frac{1}{s} \otimes m = \frac{u}{us} \otimes m = \frac{1}{us} \otimes um = \frac{1}{us} \otimes 0 = 0. \end{align*} Here the first equality uses $\tfrac{1}{s} = \tfrac{u}{us}$ in $S^{-1}R$ (which holds because $1 \cdot (us \cdot 1 - s \cdot u) = 0$), the second uses the $R$-balancing relation $\tfrac{u}{us} \otimes m = \tfrac{1}{us} \otimes um$, and the third uses $um = 0$. [guided] This is where the equivalence relation in $S^{-1}M$ connects back to the tensor product. The condition $\tfrac{m}{s} = \tfrac{0}{1}$ in $S^{-1}M$ does not mean $m = 0$; it means there exists $u \in S$ with $um = 0$. The question is: does this imply the tensor $\tfrac{1}{s} \otimes m$ vanishes? The answer is yes, and the mechanism is the $R$-balancing relation combined with the ability to multiply numerator and denominator of a fraction by the same element. Starting from $\tfrac{1}{s} \otimes m$, we multiply the fraction $\tfrac{1}{s}$ by $\tfrac{u}{u} = \tfrac{1}{1}$ (which does not change its value in $S^{-1}R$) to get \begin{align*} \frac{1}{s} \otimes m = \frac{u}{us} \otimes m. \end{align*} Now the $R$-balancing relation lets us move $u$ from the first factor to the second: \begin{align*} \frac{u}{us} \otimes m = \frac{1}{us} \otimes um. \end{align*} Since $um = 0$ in $M$, we get $\tfrac{1}{us} \otimes 0 = 0$. This shows $\ker \varphi = 0$, completing the proof that $\varphi$ is injective. [/guided] [/step] [step:Conclude that $\varphi$ is a natural isomorphism of $S^{-1}R$-modules] Since $\varphi$ is an $S^{-1}R$-linear map that is both surjective and injective, it is an isomorphism of $S^{-1}R$-modules: \begin{align*} S^{-1}R \otimes_R M \xrightarrow{\;\sim\;} S^{-1}M, \qquad \frac{r}{s} \otimes m \mapsto \frac{rm}{s}. \end{align*} Naturality in $M$: for any $R$-module homomorphism $\alpha \colon M \to N$, the diagram \begin{align*} S^{-1}R \otimes_R M &\xrightarrow{\;\varphi_M\;} S^{-1}M \\ \operatorname{id} \otimes \alpha \downarrow \quad &\qquad\quad \downarrow S^{-1}\alpha \\ S^{-1}R \otimes_R N &\xrightarrow{\;\varphi_N\;} S^{-1}N \end{align*} commutes: both compositions send $\tfrac{r}{s} \otimes m$ to $\tfrac{r \alpha(m)}{s}$. Therefore $\varphi$ is a natural isomorphism of functors from $R$-modules to $S^{-1}R$-modules. [/step]