[proofplan] We prove both inclusions. For the forward inclusion, we show that any $\mathfrak{a}$-integral element $b$ lies in $\overline{A}$ and satisfies $b^n \in \mathfrak{a}\overline{A}$, hence $b \in \sqrt{\mathfrak{a}\overline{A}}$. For the reverse inclusion, we take $b \in \sqrt{\mathfrak{a}\overline{A}}$ and use the Cayley--Hamilton theorem for modules (applied to the finitely generated $A$-module $A[x_1, \ldots, x_m]$ and the multiplication-by-$b^n$ endomorphism) to produce a monic polynomial for $b$ with coefficients in $\mathfrak{a}$. [/proofplan]

[step:Show that every $\mathfrak{a}$-integral element lies in $\sqrt{\mathfrak{a}\overline{A}}$]Suppose $b \in B$ is $\mathfrak{a}$-integral, so there exist $n \geq 1$ and $a_1, \ldots, a_n \in \mathfrak{a}$ such that \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_n = 0. \end{align*} This equation is a monic polynomial of degree $n$ over $A$ satisfied by $b$, so $b$ is integral over $A$, hence $b \in \overline{A}$. Since $b \in \overline{A}$ and each $a_i \in A \subset \overline{A}$, all the elements $b^0, b^1, \ldots, b^{n-1}$ lie in $\overline{A}$. Rearranging the integrality relation gives \begin{align*} b^n = -(a_1 b^{n-1} + a_2 b^{n-2} + \cdots + a_n). \end{align*} Each summand $a_i b^{n-i}$ is a product of an element $a_i \in \mathfrak{a}$ and an element $b^{n-i} \in \overline{A}$, so $a_i b^{n-i} \in \mathfrak{a}\overline{A}$. Therefore $b^n \in \mathfrak{a}\overline{A}$, which gives $b \in \sqrt{\mathfrak{a}\overline{A}}$.[/step]

[guided]We want to show that if $b$ satisfies a monic polynomial with coefficients in $\mathfrak{a}$, then $b$ belongs to the radical of the extended ideal $\mathfrak{a}\overline{A}$. Suppose $b^n + a_1 b^{n-1} + \cdots + a_n = 0$ with $a_i \in \mathfrak{a}$. First, this relation is itself a monic polynomial of degree $n$ over $A$ vanishing at $b$, so $b$ is integral over $A$, meaning $b \in \overline{A}$. This is important: it ensures that all the intermediate powers $b^0, b^1, \ldots, b^{n-1}$ are elements of $\overline{A}$ (since $\overline{A}$ is a ring). Now rearrange the relation: \begin{align*} b^n = -(a_1 b^{n-1} + a_2 b^{n-2} + \cdots + a_n). \end{align*} The right-hand side is a sum of terms of the form $a_i b^{n-i}$. Each $a_i$ lies in $\mathfrak{a}$ (by hypothesis of $\mathfrak{a}$-integrality) and each $b^{n-i}$ lies in $\overline{A}$ (as established above), so each $a_i b^{n-i} \in \mathfrak{a}\overline{A}$. Therefore $b^n \in \mathfrak{a}\overline{A}$, which by definition of the radical means $b \in \sqrt{\mathfrak{a}\overline{A}}$.[/guided]

[step:Show that every element of $\sqrt{\mathfrak{a}\overline{A}}$ is $\mathfrak{a}$-integral]Suppose $b \in \sqrt{\mathfrak{a}\overline{A}}$. Then $b^n \in \mathfrak{a}\overline{A}$ for some $n \geq 1$, so we may write \begin{align*} b^n = \sum_{i=1}^{m} a_i x_i \end{align*} with $a_i \in \mathfrak{a}$ and $x_i \in \overline{A}$. Since each $x_i$ is integral over $A$, by the [Tower Property of Integral Elements](/theorems/???) (the finiteness criterion: a ring generated over $A$ by finitely many integral elements is a finitely generated $A$-module), the $A$-subalgebra $M := A[x_1, \ldots, x_m] \subset \overline{A}$ is a finitely generated $A$-module.[/step]

[guided]We want to show that $b$ satisfies a monic polynomial with coefficients in $\mathfrak{a}$. The key idea is to use the Cayley--Hamilton theorem for modules: given a finitely generated $A$-module and an endomorphism whose image lands inside $\mathfrak{a}$ times the module, the endomorphism satisfies a monic polynomial with coefficients in $\mathfrak{a}$. Since $b^n = \sum_{i=1}^m a_i x_i$ with $a_i \in \mathfrak{a}$ and $x_i \in \overline{A}$, we need a finitely generated $A$-module containing these elements. Each $x_i$ is integral over $A$, so by the finiteness criterion (a ring generated over $A$ by finitely many integral elements is a finitely generated $A$-module), the $A$-subalgebra $M := A[x_1, \ldots, x_m]$ is finitely generated as an $A$-module.[/guided]

[step:Apply the Cayley--Hamilton theorem for modules to the multiplication-by-$b^n$ map]Define the $A$-linear map \begin{align*} f: M &\to M \\ z &\mapsto b^n z. \end{align*} We verify that $f(M) \subset \mathfrak{a}M$. For any $z \in M$, we have $f(z) = b^n z = \left(\sum_{i=1}^m a_i x_i\right) z = \sum_{i=1}^m a_i (x_i z)$. Since $M$ is a subring of $B$ containing each $x_i$, we have $x_i z \in M$ for all $z \in M$, so $f(z) \in \mathfrak{a}M$. Therefore $f(M) \subset \mathfrak{a}M$. By the [Cayley--Hamilton Theorem for Modules](/theorems/???), applied to the finitely generated $A$-module $M$ and the $A$-linear endomorphism $f$ with $f(M) \subset \mathfrak{a}M$, there exist $\ell \geq 1$ and $\alpha_1, \ldots, \alpha_\ell \in \mathfrak{a}$ such that \begin{align*} f^\ell + \alpha_1 f^{\ell - 1} + \cdots + \alpha_\ell = 0 \end{align*} in $\operatorname{End}_A(M)$. Evaluating both sides at $1_A \in M$ gives \begin{align*} (b^n)^\ell + \alpha_1 (b^n)^{\ell-1} + \cdots + \alpha_\ell = 0 \end{align*} in $B$. This is a monic polynomial of degree $n\ell$ in $b$ (since $(b^n)^j = b^{nj}$) with coefficients $\alpha_1, \ldots, \alpha_\ell \in \mathfrak{a}$. Therefore $b$ is $\mathfrak{a}$-integral.[/step]

[guided]The idea is to exploit the fact that the multiplication-by-$b^n$ endomorphism maps $M$ into $\mathfrak{a}M$, and then apply a determinant trick. Define $f: M \to M$ by $f(z) = b^n z$. We check $f(M) \subset \mathfrak{a}M$: for any $z \in M$, we have $f(z) = b^n z = \left(\sum_{i=1}^m a_i x_i\right)z$. Since $M = A[x_1, \ldots, x_m]$ is a subring of $B$, each product $x_i z$ lies in $M$, and each $a_i \in \mathfrak{a}$, so $f(z) = \sum_i a_i(x_i z) \in \mathfrak{a}M$. The [Cayley--Hamilton Theorem for Modules](/theorems/???) states: if $M$ is a finitely generated $A$-module, $\mathfrak{a}$ is an ideal of $A$, and $f: M \to M$ is an $A$-linear map with $f(M) \subset \mathfrak{a}M$, then $f$ satisfies a monic polynomial of degree $\ell = $ the number of generators of $M$, with coefficients in $\mathfrak{a}$. The hypotheses are verified: $M$ is a finitely generated $A$-module (established in the previous step), $\mathfrak{a}$ is an ideal of $A$, and $f(M) \subset \mathfrak{a}M$ (just verified). The conclusion gives $f^\ell + \alpha_1 f^{\ell-1} + \cdots + \alpha_\ell = 0$ with $\alpha_i \in \mathfrak{a}$. Evaluating at $1_A \in M$ (which lies in $M$ since $1_A \in A \subset A[x_1, \ldots, x_m] = M$) yields $(b^n)^\ell + \alpha_1(b^n)^{\ell-1} + \cdots + \alpha_\ell = 0$ in $B$. Writing $b^{n\ell} + \alpha_1 b^{n(\ell-1)} + \cdots + \alpha_\ell = 0$, this is a monic polynomial in $b$ of degree $n\ell$ with coefficients in $\mathfrak{a}$ (the intermediate coefficients are zero, which are also in $\mathfrak{a}$). Hence $b$ is $\mathfrak{a}$-integral.[/guided]