[proofplan] All three parts exploit the exactness of localization. Part (1) is a direct element-level computation showing both sides consist of the same fractions. Part (2) uses the exactness of $S^{-1}(\cdot)$ applied to the pullback diagram of $N$ and $P$ inside $M$; alternatively, one inclusion is immediate and the reverse requires clearing denominators with an element of $S$. Part (3) applies the exact functor $S^{-1}(\cdot)$ to the short exact sequence $0 \to N \to M \to M/N \to 0$. [/proofplan] [step:Prove $S^{-1}(N + P) = S^{-1}N + S^{-1}P$] For the inclusion $S^{-1}(N + P) \subset S^{-1}N + S^{-1}P$: let $\frac{x}{s} \in S^{-1}(N + P)$ with $x \in N + P$ and $s \in S$. Write $x = n + p$ with $n \in N$ and $p \in P$. Then \begin{align*} \frac{x}{s} = \frac{n + p}{s} = \frac{n}{s} + \frac{p}{s} \in S^{-1}N + S^{-1}P. \end{align*} For the reverse inclusion $S^{-1}N + S^{-1}P \subset S^{-1}(N + P)$: let $\frac{n}{s_1} + \frac{p}{s_2} \in S^{-1}N + S^{-1}P$ with $n \in N$, $p \in P$, $s_1, s_2 \in S$. Then \begin{align*} \frac{n}{s_1} + \frac{p}{s_2} = \frac{s_2 n + s_1 p}{s_1 s_2}. \end{align*} Since $N$ and $P$ are $R$-submodules of $M$, we have $s_2 n \in N$ and $s_1 p \in P$, so $s_2 n + s_1 p \in N + P$. Since $s_1 s_2 \in S$ (as $S$ is multiplicative), $\frac{s_2 n + s_1 p}{s_1 s_2} \in S^{-1}(N + P)$. [/step] [step:Prove $S^{-1}(N \cap P) = S^{-1}N \cap S^{-1}P$] The inclusion $S^{-1}(N \cap P) \subset S^{-1}N \cap S^{-1}P$ is immediate: if $\frac{x}{s} \in S^{-1}(N \cap P)$ with $x \in N \cap P$, then $x \in N$ gives $\frac{x}{s} \in S^{-1}N$, and $x \in P$ gives $\frac{x}{s} \in S^{-1}P$. For the reverse inclusion, let $\frac{x}{t} \in S^{-1}N \cap S^{-1}P$. Then $\frac{x}{t} \in S^{-1}N$ and $\frac{x}{t} \in S^{-1}P$, so we can write \begin{align*} \frac{x}{t} = \frac{n}{s_1} \quad \text{and} \quad \frac{x}{t} = \frac{p}{s_2} \end{align*} for some $n \in N$, $p \in P$, $s_1, s_2 \in S$. The first equality gives $u_1(s_1 x - tn) = 0$ for some $u_1 \in S$, i.e., $u_1 s_1 x = u_1 t n \in N$. The second gives $u_2(s_2 x - tp) = 0$ for some $u_2 \in S$, i.e., $u_2 s_2 x = u_2 t p \in P$. Define $w := u_1 s_1 u_2 s_2 x \in M$. Then $w = u_2 s_2 (u_1 s_1 x) = u_2 s_2 (u_1 tn) \in N$ (since $u_1 t n \in N$ and $N$ is an $R$-submodule). Similarly, $w = u_1 s_1 (u_2 s_2 x) = u_1 s_1(u_2 tp) \in P$. Hence $w \in N \cap P$. Since $u_1 s_1 u_2 s_2 \in S$ (as $S$ is multiplicative), we have \begin{align*} \frac{x}{t} = \frac{u_1 s_1 u_2 s_2 x}{u_1 s_1 u_2 s_2 t} = \frac{w}{u_1 s_1 u_2 s_2 t} \in S^{-1}(N \cap P). \end{align*} [guided] The forward inclusion $S^{-1}(N \cap P) \subset S^{-1}N \cap S^{-1}P$ is immediate from $N \cap P \subset N$ and $N \cap P \subset P$. The reverse inclusion is more subtle. The difficulty is that an element of $S^{-1}N \cap S^{-1}P$ is represented as a fraction $\frac{x}{t}$ of $S^{-1}M$ that happens to also lie in both $S^{-1}N$ and $S^{-1}P$. This does not immediately mean the numerator $x$ lies in $N \cap P$ --- only that $x$ is "equivalent" to elements of $N$ and $P$ after clearing denominators. Concretely, $\frac{x}{t} = \frac{n}{s_1}$ means $u_1(s_1 x - tn) = 0$ for some $u_1 \in S$, giving $u_1 s_1 x = u_1 tn \in N$. Similarly, $\frac{x}{t} = \frac{p}{s_2}$ means $u_2 s_2 x = u_2 tp \in P$. The element $x$ itself may not lie in $N \cap P$, but the rescaled element $w := u_1 s_1 u_2 s_2 x$ does: $w = u_2 s_2 \cdot (u_1 s_1 x)$, and $u_1 s_1 x \in N$ (shown above), so $w \in N$ since $N$ is an $R$-submodule. By the same argument with the factors swapped, $w \in P$. Hence $w \in N \cap P$. Since $S$ is multiplicative, $u_1 s_1 u_2 s_2 t \in S$, and \begin{align*} \frac{x}{t} = \frac{u_1 s_1 u_2 s_2 x}{u_1 s_1 u_2 s_2 t} = \frac{w}{u_1 s_1 u_2 s_2 t} \in S^{-1}(N \cap P). \end{align*} The key technique is multiplying through by enough elements of $S$ to land in $N \cap P$ at the level of numerators. This is a recurring pattern in localization arguments. [/guided] [/step] [step:Prove $S^{-1}M / S^{-1}N \cong S^{-1}(M/N)$ via exactness of localization] Consider the short exact sequence of $R$-modules \begin{align*} 0 \to N \xrightarrow{\;\iota\;} M \xrightarrow{\;\pi\;} M/N \to 0, \end{align*} where $\iota: N \hookrightarrow M$ is the inclusion and $\pi: M \to M/N$ is the quotient map. Since the localization functor $S^{-1}(\cdot)$ is exact (by the [Localization Is Exact](/theorems/2847) theorem applied to both short exact sequences $0 \to N \to M$ and $M \to M/N \to 0$), the localized sequence \begin{align*} 0 \to S^{-1}N \xrightarrow{\;S^{-1}\iota\;} S^{-1}M \xrightarrow{\;S^{-1}\pi\;} S^{-1}(M/N) \to 0 \end{align*} is exact. The map $S^{-1}\iota$ is injective and its image is $S^{-1}N$ viewed as a submodule of $S^{-1}M$. The map $S^{-1}\pi$ is surjective with $\ker(S^{-1}\pi) = \operatorname{im}(S^{-1}\iota) = S^{-1}N$. By the first isomorphism theorem for $S^{-1}R$-modules, $S^{-1}\pi$ induces an isomorphism \begin{align*} S^{-1}M / S^{-1}N &\xrightarrow{\;\sim\;} S^{-1}(M/N), \\ \frac{m}{s} + S^{-1}N &\longmapsto S^{-1}\pi\!\left(\frac{m}{s}\right) = \frac{\pi(m)}{s} = \frac{m + N}{s}. \end{align*} [guided] Part (3) is the cleanest application of exactness. We start with the short exact sequence $0 \to N \xrightarrow{\iota} M \xrightarrow{\pi} M/N \to 0$ of $R$-modules, where $\iota$ is the inclusion and $\pi$ is the canonical projection. By the [Localization Is Exact](/theorems/2847) theorem, the functor $S^{-1}(\cdot)$ preserves exactness. Applying it to the entire short exact sequence: \begin{align*} 0 \to S^{-1}N \xrightarrow{S^{-1}\iota} S^{-1}M \xrightarrow{S^{-1}\pi} S^{-1}(M/N) \to 0 \end{align*} is exact. (Exactness at $S^{-1}N$ says $S^{-1}\iota$ is injective; exactness at $S^{-1}(M/N)$ says $S^{-1}\pi$ is surjective; and exactness at $S^{-1}M$ says $\ker(S^{-1}\pi) = \operatorname{im}(S^{-1}\iota)$.) The image of $S^{-1}\iota$ is exactly $S^{-1}N$ sitting inside $S^{-1}M$ (the localization of the inclusion). So $\ker(S^{-1}\pi) = S^{-1}N$, and the first isomorphism theorem yields \begin{align*} S^{-1}M / S^{-1}N \xrightarrow{\;\sim\;} S^{-1}(M/N). \end{align*} Tracing the map explicitly: a coset $\frac{m}{s} + S^{-1}N$ maps to $S^{-1}\pi(\frac{m}{s}) = \frac{\pi(m)}{s} = \frac{m + N}{s}$, which is the formula stated in the theorem. Why is this natural? The isomorphism is induced by the universal property of quotients applied to the $S^{-1}R$-linear surjection $S^{-1}\pi: S^{-1}M \to S^{-1}(M/N)$. It commutes with any $R$-module map out of $M$ that kills $N$, making it a natural transformation between the functors $S^{-1}((\cdot)/N)$ and $S^{-1}(\cdot)/S^{-1}N$. [/guided] [/step]