[proofplan] For (1a), we reduce an integral equation modulo $\mathfrak{b}$ to produce an integral equation in the quotient. For (1b), we divide an integral equation by a power of $s$ to produce an integral equation in the localisation. For (2), one inclusion follows from (1b); for the reverse, we clear denominators in a monic equation over $S^{-1}A$ to produce an element of $\overline{A}$ that represents the given fraction. [/proofplan] [step:Prove (1a): reduce an integral equation modulo $\mathfrak{b}$] The composite ring homomorphism $A \hookrightarrow B \twoheadrightarrow B/\mathfrak{b}$ has kernel $\{a \in A : a \in \mathfrak{b}\} = \mathfrak{b} \cap A = \mathfrak{b}^c$. By the first isomorphism theorem for rings, this composite factors through an injective ring homomorphism $A/\mathfrak{b}^c \hookrightarrow B/\mathfrak{b}$, so we may view $A/\mathfrak{b}^c$ as a subring of $B/\mathfrak{b}$. Let $\bar{b} = b + \mathfrak{b} \in B/\mathfrak{b}$. Since $B$ is integral over $A$, there exist $a_1, \ldots, a_n \in A$ such that \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_n = 0 \quad \text{in } B. \end{align*} Applying the surjection $\pi: B \to B/\mathfrak{b}$ to both sides: \begin{align*} \bar{b}^n + \bar{a}_1 \bar{b}^{n-1} + \cdots + \bar{a}_n = \bar{0} \quad \text{in } B/\mathfrak{b}, \end{align*} where $\bar{a}_i = a_i + \mathfrak{b}^c \in A/\mathfrak{b}^c$. This is a monic polynomial of degree $n$ over $A/\mathfrak{b}^c$ satisfied by $\bar{b}$. Since $\bar{b} \in B/\mathfrak{b}$ was arbitrary, $B/\mathfrak{b}$ is integral over $A/\mathfrak{b}^c$. [/step] [step:Prove (1b): divide an integral equation by $s^n$ to obtain integrality in the localisation] Let $b/s \in S^{-1}B$ with $b \in B$ and $s \in S$. Since $B$ is integral over $A$, there exist $a_1, \ldots, a_n \in A$ such that \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_n = 0 \quad \text{in } B. \end{align*} Applying the canonical localisation map $\iota: B \to S^{-1}B$, $\iota(x) = x/1$, to both sides gives \begin{align*} \frac{b^n}{1} + \frac{a_1 b^{n-1}}{1} + \cdots + \frac{a_n}{1} = \frac{0}{1} \quad \text{in } S^{-1}B. \end{align*} Dividing by $s^n/1 = (s/1)^n$ (which is a unit in $S^{-1}B$ since $s \in S$): \begin{align*} \left(\frac{b}{s}\right)^n + \frac{a_1}{s}\left(\frac{b}{s}\right)^{n-1} + \cdots + \frac{a_n}{s^n} = \frac{0}{1}. \end{align*} Each coefficient $a_k / s^k$ lies in $S^{-1}A$ (since $a_k \in A$ and $s^k \in S$, as $S$ is multiplicatively closed). This is a monic equation of degree $n$ for $b/s$ over $S^{-1}A$. Since $b/s \in S^{-1}B$ was arbitrary, $S^{-1}B$ is integral over $S^{-1}A$. [guided] We want to show every element of $S^{-1}B$ is integral over $S^{-1}A$. The idea is straightforward: take the integral equation for $b$ over $A$ and "localise" it. Let $b/s \in S^{-1}B$. Since $b \in B$ is integral over $A$, we have $b^n + a_1 b^{n-1} + \cdots + a_n = 0$ with $a_i \in A$. This equation holds in $B$, hence also in $S^{-1}B$ after applying the localisation map. We need an equation for $b/s$, not for $b/1$. Dividing the entire equation by $(s/1)^n$, and using that $s \in S$ implies $s/1$ is a unit in $S^{-1}B$: \begin{align*} \left(\frac{b}{s}\right)^n + \frac{a_1}{s}\left(\frac{b}{s}\right)^{n-1} + \frac{a_2}{s^2}\left(\frac{b}{s}\right)^{n-2} + \cdots + \frac{a_n}{s^n} = 0. \end{align*} The coefficient of $\left(\frac{b}{s}\right)^{n-k}$ is $\frac{a_k}{s^k}$. Since $a_k \in A$ and $s^k \in S$ (because $S$ is a multiplicative set containing $s$), each coefficient $a_k/s^k$ lies in $S^{-1}A$. The leading coefficient is $1$, so this is a monic equation over $S^{-1}A$. [/guided] [/step] [step:Prove (2), forward inclusion: $S^{-1}\overline{A}$ is integral over $S^{-1}A$] By hypothesis, $\overline{A}$ is the integral closure of $A$ in $B$, so $B$ restricted to $\overline{A}$ gives $\overline{A}$ integral over $A$. In particular, $A \subset \overline{A} \subset B$ and $\overline{A}$ is integral over $A$. Applying part (1b) to the integral extension $A \subset \overline{A}$ with multiplicative set $S \subset A$, we obtain that $S^{-1}\overline{A}$ is integral over $S^{-1}A$. Since $S^{-1}\overline{A} \subset S^{-1}B$ (the localisation functor preserves inclusions), every element of $S^{-1}\overline{A}$ is integral over $S^{-1}A$ and lies in $S^{-1}B$. Hence $S^{-1}\overline{A}$ is contained in the integral closure of $S^{-1}A$ in $S^{-1}B$. [/step] [step:Prove (2), reverse inclusion: every element of $S^{-1}B$ integral over $S^{-1}A$ lies in $S^{-1}\overline{A}$] Let $b/s \in S^{-1}B$ (with $b \in B$, $s \in S$) be integral over $S^{-1}A$. Then there exist $a_1/s_1, \ldots, a_n/s_n \in S^{-1}A$ (with $a_i \in A$, $s_i \in S$) such that \begin{align*} \left(\frac{b}{s}\right)^n + \frac{a_1}{s_1}\left(\frac{b}{s}\right)^{n-1} + \cdots + \frac{a_n}{s_n} = 0 \quad \text{in } S^{-1}B. \end{align*} Set $t := s_1 s_2 \cdots s_n \in S$ (since $S$ is multiplicatively closed). Multiplying both sides by $(st)^n$: \begin{align*} (tb)^n + a_1 \frac{t}{s_1} s (tb)^{n-1} + a_2 \frac{t^2}{s_1 s_2} s^2 (tb)^{n-2} + \cdots + a_n \frac{t^n}{s_1 \cdots s_n} s^n = 0. \end{align*} For each $1 \leq k \leq n$, the factor $\frac{t^k}{s_1 \cdots s_k} = s_{k+1} \cdots s_n \cdot \frac{t^{k-1}}{s_1 \cdots s_{k-1}} \cdot \frac{1}{1}$ is a product of elements in $S$, and $a_k s^k \in A$ (since $a_k \in A$ and $s \in A$). More precisely, define $c_k := a_k s^k \cdot \frac{t^k}{s_1 \cdots s_k}$. Since $t = s_1 \cdots s_n$, we have $\frac{t^k}{s_1 \cdots s_k}$ is a product of elements of $A$ (it is a monomial in $s_1, \ldots, s_n$), so $c_k \in A$. Rewriting, the equation becomes a monic polynomial equation for $tb \in B$ with coefficients in $A$: \begin{align*} (tb)^n + c_1 (tb)^{n-1} + \cdots + c_n = 0, \quad c_k \in A. \end{align*} Therefore $tb$ is integral over $A$, so $tb \in \overline{A}$. It follows that \begin{align*} \frac{b}{s} = \frac{tb}{ts} \in S^{-1}\overline{A}, \end{align*} since $tb \in \overline{A}$ and $ts \in S$. [guided] We must show that any element of $S^{-1}B$ that is integral over $S^{-1}A$ already lies in $S^{-1}\overline{A}$. The strategy is to clear denominators from the monic equation to land back in $B$ with a monic equation over $A$. Let $b/s \in S^{-1}B$ satisfy the monic equation \begin{align*} \left(\frac{b}{s}\right)^n + \frac{a_1}{s_1}\left(\frac{b}{s}\right)^{n-1} + \cdots + \frac{a_n}{s_n} = 0 \end{align*} with $a_i \in A$, $s_i \in S$. The denominators $s_1, \ldots, s_n$ are the obstacle: we need a single element of $S$ that clears them all. Set $t := s_1 \cdots s_n \in S$. Multiply both sides by $(st)^n$. The leading term becomes $(st)^n \cdot (b/s)^n = (tb)^n$. For the $k$-th coefficient term, we get \begin{align*} (st)^n \cdot \frac{a_k}{s_k} \cdot \left(\frac{b}{s}\right)^{n-k} = a_k \cdot \frac{t^k}{s_k} \cdot s^k \cdot \frac{t^{n-k} b^{n-k}}{t^{n-k}} = a_k \cdot \frac{t^k}{s_k} \cdot s^k \cdot (tb)^{n-k} \cdot t^{-n+k} \cdot t^{n-k}. \end{align*} Let us be more careful. Rewrite $(b/s)^{n-k} = b^{n-k}/s^{n-k}$. Multiplying the $k$-th term by $(st)^n$: \begin{align*} \frac{a_k}{s_k} \cdot \frac{b^{n-k}}{s^{n-k}} \cdot s^n t^n = a_k \cdot \frac{t^n}{s_k} \cdot s^k \cdot b^{n-k}. \end{align*} We want to express this as $c_k \cdot (tb)^{n-k}$ for some $c_k \in A$. Since $(tb)^{n-k} = t^{n-k} b^{n-k}$, we need $c_k = a_k \cdot s^k \cdot t^n / (s_k \cdot t^{n-k}) = a_k \cdot s^k \cdot t^k / s_k$. Now $t^k / s_k = (s_1 \cdots s_n)^k / s_k$; in fact, since $t = s_1 \cdots s_n$ and all $s_i \in S \subset A$, the expression $t^k / s_k$ is an element of $A$ (it is a product of elements of $A$). So $c_k = a_k s^k \cdot (t^k / s_k) \in A$. The resulting equation $(tb)^n + c_1(tb)^{n-1} + \cdots + c_n = 0$ is monic with coefficients in $A$, so $tb \in \overline{A}$. Therefore $b/s = tb/(ts) \in S^{-1}\overline{A}$, since $tb \in \overline{A}$ and $ts \in S$. [/guided] [/step] [step:Combine the inclusions to complete part (2)] From the forward inclusion, $S^{-1}\overline{A}$ is contained in the integral closure of $S^{-1}A$ in $S^{-1}B$. From the reverse inclusion, every element of $S^{-1}B$ integral over $S^{-1}A$ lies in $S^{-1}\overline{A}$. Together, $S^{-1}\overline{A}$ equals the integral closure of $S^{-1}A$ in $S^{-1}B$. [/step]