[proofplan] We prove the four parts sequentially. Part (1) shows that every ideal $\mathfrak{b}$ of $S^{-1}R$ satisfies $\mathfrak{b}^{ce} = \mathfrak{b}$, using the fact that any element $\frac{r}{s} \in \mathfrak{b}$ can be recovered from its numerator $r \in \mathfrak{b}^c$ by dividing by $s$. Part (2) characterizes contracted ideals via the formula $\mathfrak{a}^{ec} = \bigcup_{s \in S} (\mathfrak{a} : s)$, where $(\mathfrak{a} : s) = \{r \in R : rs \in \mathfrak{a}\}$: the condition $\mathfrak{a}^{ec} = \mathfrak{a}$ holds precisely when $(\mathfrak{a} : s) = \mathfrak{a}$ for all $s \in S$, which is the no-zero-divisor condition. Part (3) is a direct computation: if $\mathfrak{a}$ meets $S$, then $\frac{1}{1}$ lies in the extended ideal; conversely, $\frac{1}{1} \in \mathfrak{a}^e$ forces an element of $\mathfrak{a} \cap S$ to exist. Part (4) combines the previous parts with the observation that contraction preserves primality and that $\mathfrak{p}^e$ is prime when $\mathfrak{p} \cap S = \varnothing$, using the isomorphism $(S^{-1}R)/\mathfrak{p}^e \cong S^{-1}(R/\mathfrak{p})$. [/proofplan]

[step:Prove that every ideal of $S^{-1}R$ is extended]Let $\mathfrak{b}$ be an ideal of $S^{-1}R$. The contraction is $\mathfrak{b}^c = \iota^{-1}(\mathfrak{b}) = \{r \in R : \frac{r}{1} \in \mathfrak{b}\}$, and the extension of an ideal $\mathfrak{a} \trianglelefteq R$ is $\mathfrak{a}^e = S^{-1}R \cdot \iota(\mathfrak{a}) = \{\frac{a}{s} : a \in \mathfrak{a}, s \in S\}$. We claim $\mathfrak{b}^{ce} = \mathfrak{b}$. The inclusion $\mathfrak{b}^{ce} \subset \mathfrak{b}$ holds in general: if $\frac{a}{s} \in \mathfrak{b}^{ce}$ with $a \in \mathfrak{b}^c$ and $s \in S$, then $\frac{a}{1} \in \mathfrak{b}$, so $\frac{a}{s} = \frac{1}{s} \cdot \frac{a}{1} \in \mathfrak{b}$ (since $\mathfrak{b}$ is an ideal of $S^{-1}R$). For the reverse inclusion $\mathfrak{b} \subset \mathfrak{b}^{ce}$: let $\frac{r}{s} \in \mathfrak{b}$. Then $\frac{r}{1} = \frac{s}{1} \cdot \frac{r}{s} \in \mathfrak{b}$ (since $\frac{s}{1} \in S^{-1}R$ and $\mathfrak{b}$ is an ideal). Hence $r \in \mathfrak{b}^c$, and $\frac{r}{s} \in \mathfrak{b}^{ce}$. Therefore $\mathfrak{b} = \mathfrak{b}^{ce}$, so $\mathfrak{b}$ is the extension of $\mathfrak{b}^c$.[/step]

[guided]An ideal $\mathfrak{b}$ of $S^{-1}R$ is called "extended" if $\mathfrak{b} = \mathfrak{a}^e$ for some ideal $\mathfrak{a}$ of $R$, where $\mathfrak{a}^e = \{\frac{a}{s} : a \in \mathfrak{a},\, s \in S\}$ is the ideal of $S^{-1}R$ generated by the image of $\mathfrak{a}$. We show $\mathfrak{a} = \mathfrak{b}^c$ works, i.e., $\mathfrak{b}^{ce} = \mathfrak{b}$. For $\mathfrak{b}^{ce} \subset \mathfrak{b}$: any element of $\mathfrak{b}^{ce}$ has the form $\frac{a}{s}$ with $a \in \mathfrak{b}^c$ (meaning $\frac{a}{1} \in \mathfrak{b}$) and $s \in S$. Since $\mathfrak{b}$ is an ideal of $S^{-1}R$ and $\frac{1}{s} \in S^{-1}R$, we get $\frac{a}{s} = \frac{1}{s} \cdot \frac{a}{1} \in \mathfrak{b}$. For $\mathfrak{b} \subset \mathfrak{b}^{ce}$: take any $\frac{r}{s} \in \mathfrak{b}$. We need to show $r \in \mathfrak{b}^c$, i.e., $\frac{r}{1} \in \mathfrak{b}$. Since $\frac{r}{s} \in \mathfrak{b}$ and $\frac{s}{1} \in S^{-1}R$, the product $\frac{s}{1} \cdot \frac{r}{s} = \frac{sr}{s} = \frac{r}{1}$ lies in $\mathfrak{b}$. (The equality $\frac{sr}{s} = \frac{r}{1}$ holds because $1 \cdot (1 \cdot sr - s \cdot r) = 0$.) So $r \in \mathfrak{b}^c$, and then $\frac{r}{s} \in \mathfrak{b}^{ce}$. The argument here is essentially that every element of $S$ becomes a unit in $S^{-1}R$, so we can always "clear denominators" and "reintroduce denominators" freely. This is why every ideal of $S^{-1}R$ comes from an ideal of $R$ --- the localization ring has no "new" ideal structure beyond what $R$ already provides.[/guided]

[step:Characterize contracted ideals via the colon formula]For an ideal $\mathfrak{a} \trianglelefteq R$, define the colon ideal $(\mathfrak{a} : s) := \{r \in R : rs \in \mathfrak{a}\}$ for each $s \in S$. [claim:The contraction of the extension equals the union of colon ideals] $\mathfrak{a}^{ec} = \bigcup_{s \in S} (\mathfrak{a} : s)$. [/claim] [proof] An element $r \in R$ lies in $\mathfrak{a}^{ec}$ if and only if $\frac{r}{1} \in \mathfrak{a}^e$, i.e., $\frac{r}{1} = \frac{a}{s}$ for some $a \in \mathfrak{a}$ and $s \in S$. By the definition of equality in $S^{-1}R$, this means there exists $u \in S$ with $u(sr - a) = 0$, i.e., $usr = ua \in \mathfrak{a}$. Since $us \in S$ (as $S$ is multiplicative), this gives $r \in (\mathfrak{a} : us)$, so $r \in \bigcup_{s' \in S} (\mathfrak{a} : s')$. Conversely, if $r \in (\mathfrak{a} : s')$ for some $s' \in S$, then $rs' \in \mathfrak{a}$, so $\frac{r}{1} = \frac{rs'}{s'} \in \mathfrak{a}^e$, giving $r \in \mathfrak{a}^{ec}$. [/proof] Now $\mathfrak{a}$ is contracted (meaning $\mathfrak{a}^{ec} = \mathfrak{a}$) if and only if $(\mathfrak{a} : s) = \mathfrak{a}$ for every $s \in S$ (since $\mathfrak{a} \subset (\mathfrak{a} : s)$ always holds, the condition is $(\mathfrak{a} : s) \subset \mathfrak{a}$ for all $s \in S$). The condition $(\mathfrak{a} : s) \subset \mathfrak{a}$ for all $s \in S$ means: for every $s \in S$ and $r \in R$, if $rs \in \mathfrak{a}$ then $r \in \mathfrak{a}$. Passing to the quotient $R/\mathfrak{a}$: write $\bar{s}$ and $\bar{r}$ for the images of $s$ and $r$ in $R/\mathfrak{a}$. The condition "$rs \in \mathfrak{a} \implies r \in \mathfrak{a}$" becomes "$\bar{r}\bar{s} = 0 \implies \bar{r} = 0$." This says that $\bar{s}$ is not a zero divisor in $R/\mathfrak{a}$. The ideal $\mathfrak{a}$ is contracted if and only if this holds for every $s \in S$, i.e., the image of $S$ in $R/\mathfrak{a}$ contains no zero divisors.[/step]

[guided]The contraction-extension of $\mathfrak{a}$ is the set of elements of $R$ whose images in $S^{-1}R$ lie in $\mathfrak{a}^e$. Unwinding the definitions: $r \in \mathfrak{a}^{ec}$ iff $\frac{r}{1} = \frac{a}{s}$ for some $a \in \mathfrak{a}$, $s \in S$, iff there exists $u \in S$ with $usr \in \mathfrak{a}$. Setting $s' = us \in S$, this is $r \in (\mathfrak{a} : s')$ for some $s' \in S$, giving the formula $\mathfrak{a}^{ec} = \bigcup_{s \in S} (\mathfrak{a} : s)$. The ideal $\mathfrak{a}$ is always contained in $\mathfrak{a}^{ec}$ (since $(\mathfrak{a} : 1) = \mathfrak{a}$ and $1 \in S$). So $\mathfrak{a}$ is contracted iff $\mathfrak{a}^{ec} \subset \mathfrak{a}$, iff $(\mathfrak{a} : s) \subset \mathfrak{a}$ for all $s \in S$. Now $(\mathfrak{a} : s) \subset \mathfrak{a}$ says: whenever $rs \in \mathfrak{a}$, we have $r \in \mathfrak{a}$. In the quotient ring $R/\mathfrak{a}$, this becomes: $\bar{r}\bar{s} = 0 \implies \bar{r} = 0$, i.e., $\bar{s}$ is a non-zero-divisor in $R/\mathfrak{a}$. (Note: if $\bar{s} = 0$, then $s \in \mathfrak{a}$, and the condition $(\mathfrak{a} : s) \subset \mathfrak{a}$ would require every $r \in R$ to lie in $\mathfrak{a}$ (since $rs \in \mathfrak{a}$ for all $r$), forcing $\mathfrak{a} = R$, which is the degenerate case.) The condition for all $s \in S$ says the image of $S$ in $R/\mathfrak{a}$ consists entirely of non-zero-divisors.[/guided]

[step:Prove $\mathfrak{a}^e = S^{-1}R$ if and only if $\mathfrak{a} \cap S \neq \varnothing$] Suppose $x \in \mathfrak{a} \cap S$. Then $\frac{1}{1} = \frac{x}{x} \in \mathfrak{a}^e$ (since $x \in \mathfrak{a}$ and $x \in S$), so $\mathfrak{a}^e = S^{-1}R$. Conversely, suppose $\mathfrak{a}^e = S^{-1}R$. Then $\frac{1}{1} \in \mathfrak{a}^e$, so $\frac{1}{1} = \frac{a}{s}$ for some $a \in \mathfrak{a}$ and $s \in S$. By the definition of equality in $S^{-1}R$, there exists $u \in S$ with $u(s \cdot 1 - 1 \cdot a) = 0$, i.e., $us = ua$. Since $a \in \mathfrak{a}$, we have $ua \in \mathfrak{a}$, so $us \in \mathfrak{a}$. Since $u, s \in S$ and $S$ is multiplicative, $us \in S$. Therefore $us \in \mathfrak{a} \cap S$. [/step]

[step:Establish the bijection with $\operatorname{Spec}(S^{-1}R)$ via contraction and extension]We construct the bijection in two parts: define the maps, then prove they are mutually inverse. **Contraction preserves primality.** Let $\mathfrak{q} \in \operatorname{Spec}(S^{-1}R)$. The contraction $\mathfrak{q}^c = \iota^{-1}(\mathfrak{q})$ is a prime ideal of $R$ (since the preimage of a prime ideal under a ring homomorphism is prime). By part (3), $\mathfrak{q} \neq S^{-1}R$ implies $\mathfrak{q}^c \cap S = \varnothing$ (if some $s \in \mathfrak{q}^c \cap S$, then $\frac{s}{1} \in \mathfrak{q}$ and $\frac{1}{s} \in S^{-1}R$, so $\frac{1}{1} = \frac{1}{s} \cdot \frac{s}{1} \in \mathfrak{q}$, contradicting $\mathfrak{q} \neq S^{-1}R$). So contraction defines a map $\operatorname{Spec}(S^{-1}R) \to \{\mathfrak{p} \in \operatorname{Spec}(R) : \mathfrak{p} \cap S = \varnothing\}$. **Extension preserves primality for primes disjoint from $S$.** Let $\mathfrak{p} \in \operatorname{Spec}(R)$ with $\mathfrak{p} \cap S = \varnothing$. By part (3), $\mathfrak{p}^e \neq S^{-1}R$. We show $\mathfrak{p}^e$ is prime by exhibiting $S^{-1}R / \mathfrak{p}^e$ as an integral domain. [claim:The quotient $S^{-1}R / \mathfrak{p}^e$ is isomorphic to $\bar{S}^{-1}(R/\mathfrak{p})$] Let $\bar{S}$ denote the image of $S$ under the projection $R \to R/\mathfrak{p}$. There is an isomorphism of rings $S^{-1}R / \mathfrak{p}^e \cong \bar{S}^{-1}(R/\mathfrak{p})$. [/claim] [proof] Consider the composite ring homomorphism \begin{align*} \varphi: R \xrightarrow{\;\iota\;} S^{-1}R \xrightarrow{\;\pi\;} S^{-1}R / \mathfrak{p}^e, \end{align*} where $\pi$ is the canonical projection. The kernel of $\varphi$ is $\iota^{-1}(\mathfrak{p}^e) = \mathfrak{p}^{ec}$. Since $\mathfrak{p} \cap S = \varnothing$ and $\mathfrak{p}$ is prime, the image of $S$ in $R/\mathfrak{p}$ consists of nonzero elements of the integral domain $R/\mathfrak{p}$, hence contains no zero divisors. By part (2), $\mathfrak{p}$ is contracted, so $\mathfrak{p}^{ec} = \mathfrak{p}$. By the first isomorphism theorem, $\varphi$ induces an injective ring homomorphism $\bar{\varphi}: R/\mathfrak{p} \hookrightarrow S^{-1}R / \mathfrak{p}^e$. For every $s \in S$, the element $\bar{\varphi}(\bar{s}) = \pi(\frac{s}{1})$ is a unit in $S^{-1}R / \mathfrak{p}^e$ (since $\frac{1}{s}$ maps to its inverse). By the universal property of localization, $\bar{\varphi}$ extends uniquely to a ring homomorphism $\bar{S}^{-1}(R/\mathfrak{p}) \to S^{-1}R / \mathfrak{p}^e$, sending $\frac{\bar{r}}{\bar{s}} \mapsto \pi(\frac{r}{s})$. This map is surjective: every element of $S^{-1}R / \mathfrak{p}^e$ has the form $\pi(\frac{r}{s})$ (since elements of $S^{-1}R$ are fractions $\frac{r}{s}$), which is the image of $\frac{\bar{r}}{\bar{s}}$. This map is injective: if $\pi(\frac{r}{s}) = 0$, then $\frac{r}{s} \in \mathfrak{p}^e$, so $\frac{r}{s} = \frac{a}{t}$ for some $a \in \mathfrak{p}$, $t \in S$. There exists $v \in S$ with $v(tr - sa) = 0$, i.e., $vtr = vsa \in \mathfrak{p}$. Since $v, t \notin \mathfrak{p}$ (as $\mathfrak{p} \cap S = \varnothing$) and $\mathfrak{p}$ is prime, $r \in \mathfrak{p}$, giving $\bar{r} = 0$ in $R/\mathfrak{p}$, hence $\frac{\bar{r}}{\bar{s}} = 0$ in $\bar{S}^{-1}(R/\mathfrak{p})$. [/proof] Since $R/\mathfrak{p}$ is an integral domain and $\bar{S}$ consists of nonzero elements of $R/\mathfrak{p}$, the localization $\bar{S}^{-1}(R/\mathfrak{p})$ is a subring of $\operatorname{Frac}(R/\mathfrak{p})$, hence an integral domain. By the isomorphism above, $S^{-1}R / \mathfrak{p}^e$ is an integral domain, so $\mathfrak{p}^e$ is a prime ideal of $S^{-1}R$.[/step]

[guided]We need to show $\mathfrak{p}^e$ is prime. The most transparent way is to identify the quotient $S^{-1}R / \mathfrak{p}^e$ and check it is an integral domain. The key idea: localizing and then quotienting should be the same as quotienting and then localizing. More precisely, $S^{-1}R / \mathfrak{p}^e \cong \bar{S}^{-1}(R/\mathfrak{p})$, where $\bar{S}$ is the image of $S$ in $R/\mathfrak{p}$. Why is $\bar{S}^{-1}(R/\mathfrak{p})$ an integral domain? Because $R/\mathfrak{p}$ is an integral domain (since $\mathfrak{p}$ is prime), and $\bar{S}$ consists of nonzero elements of $R/\mathfrak{p}$ (since $\mathfrak{p} \cap S = \varnothing$ means no element of $S$ maps to zero). Localizing an integral domain at a multiplicative set of nonzero elements embeds it into its fraction field, so the result is still an integral domain. What if $\mathfrak{p} \cap S \neq \varnothing$? Then some $s \in S$ maps to $\bar{s} = 0$ in $R/\mathfrak{p}$, and $\bar{S}$ contains $0$. Localizing at a multiplicative set containing $0$ gives the zero ring, which corresponds to $\mathfrak{p}^e = S^{-1}R$ (the extended ideal is the whole ring). This is consistent with part (3).[/guided]

[step:Verify the maps are mutually inverse]We check that the two maps $\mathfrak{p} \mapsto \mathfrak{p}^e$ and $\mathfrak{q} \mapsto \mathfrak{q}^c$ are inverse to each other. **$(\mathfrak{q}^c)^e = \mathfrak{q}$ for $\mathfrak{q} \in \operatorname{Spec}(S^{-1}R)$:** By part (1), every ideal of $S^{-1}R$ satisfies $\mathfrak{b}^{ce} = \mathfrak{b}$. In particular, $(\mathfrak{q}^c)^e = \mathfrak{q}$. **$(\mathfrak{p}^e)^c = \mathfrak{p}$ for $\mathfrak{p} \in \operatorname{Spec}(R)$ with $\mathfrak{p} \cap S = \varnothing$:** Since $\mathfrak{p}$ is prime and $\mathfrak{p} \cap S = \varnothing$, the image of $S$ in $R/\mathfrak{p}$ consists of nonzero elements of the integral domain $R/\mathfrak{p}$, hence contains no zero divisors. By part (2), $\mathfrak{p}$ is contracted, i.e., $\mathfrak{p}^{ec} = \mathfrak{p}$. Therefore the maps $\mathfrak{p} \mapsto \mathfrak{p}^e$ and $\mathfrak{q} \mapsto \mathfrak{q}^c$ define mutually inverse bijections between $\{\mathfrak{p} \in \operatorname{Spec}(R) : \mathfrak{p} \cap S = \varnothing\}$ and $\operatorname{Spec}(S^{-1}R)$.[/step]

[guided]The two directions of the inverse check use different parts of the theorem. For the first direction, $(\mathfrak{q}^c)^e = \mathfrak{q}$: this is immediate from part (1), which says every ideal of $S^{-1}R$ is extended. The identity $\mathfrak{b}^{ce} = \mathfrak{b}$ applied to $\mathfrak{b} = \mathfrak{q}$ gives $(\mathfrak{q}^c)^e = \mathfrak{q}$. For the second direction, $(\mathfrak{p}^e)^c = \mathfrak{p}$: this uses part (2). We need $\mathfrak{p}$ to be contracted, i.e., $\mathfrak{p}^{ec} = \mathfrak{p}$. By part (2), this requires the image of $S$ in $R/\mathfrak{p}$ to contain no zero divisors. Since $\mathfrak{p}$ is prime, $R/\mathfrak{p}$ is an integral domain, and $\mathfrak{p} \cap S = \varnothing$ means every element of $S$ maps to a nonzero element of $R/\mathfrak{p}$. A nonzero element of an integral domain is never a zero divisor, so the condition is satisfied. This completes the proof: the two maps are mutually inverse bijections, establishing the order-preserving correspondence between prime ideals of $R$ disjoint from $S$ and prime ideals of $S^{-1}R$.[/guided]