[proofplan] The direction $(2) \Rightarrow (1)$ is immediate: the minimal polynomial provides an integrality relation with coefficients in $\sqrt{\mathfrak{a}}$, and the [Reducing to the Radical Ideal](/theorems/2943) corollary converts $\sqrt{\mathfrak{a}}$-integrality to $\mathfrak{a}$-integrality. For $(1) \Rightarrow (2)$, we embed into an algebraic closure and show that every conjugate of $b$ (i.e., every root of the minimal polynomial) is also $\mathfrak{a}$-integral. Since the coefficients of the minimal polynomial are elementary symmetric functions of these conjugates, and the $\mathfrak{a}$-integral elements form a set closed under sums and products, the coefficients are $\mathfrak{a}$-integral. Because they lie in $\operatorname{Frac}(A)$ and $A$ is integrally closed, they belong to $\sqrt{\mathfrak{a}}$. [/proofplan] [step:Prove $(2) \Rightarrow (1)$: the minimal polynomial gives $\sqrt{\mathfrak{a}}$-integrality] Suppose the minimal polynomial of $b$ over $\operatorname{Frac}(A)$ is $f = T^n + (a_1/1)T^{n-1} + \cdots + (a_n/1)$ with $a_1, \ldots, a_n \in \sqrt{\mathfrak{a}}$. Since $f(b) = 0$ in $\operatorname{Frac}(B) \supset B$, in $B$ we have \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_n = 0. \end{align*} This is a monic polynomial with coefficients in $\sqrt{\mathfrak{a}}$ vanishing at $b$, so $b$ is $\sqrt{\mathfrak{a}}$-integral. By [Reducing to the Radical Ideal](/theorems/2943), $b$ is $\mathfrak{a}$-integral. [/step] [step:Establish algebraicity and identify the minimal polynomial] Suppose $b$ is $\mathfrak{a}$-integral: there exist $n \geq 1$ and $a_1, \ldots, a_n \in \mathfrak{a}$ with \begin{align*} b^n + a_1 b^{n-1} + \cdots + a_n = 0. \end{align*} Define $h := T^n + (a_1/1)T^{n-1} + \cdots + (a_n/1) \in \operatorname{Frac}(A)[T]$. Since $h(b) = 0$ and $h$ is monic of positive degree, $b$ is algebraic over $\operatorname{Frac}(A)$. Let $f \in \operatorname{Frac}(A)[T]$ denote the minimal polynomial of $b$ over $\operatorname{Frac}(A)$, so $f$ is monic, irreducible over $\operatorname{Frac}(A)$, and $f \mid h$ in $\operatorname{Frac}(A)[T]$. [/step] [step:Show every conjugate of $b$ is $\mathfrak{a}$-integral by transporting the integrality relation] Fix an algebraic closure $\Omega$ of $\operatorname{Frac}(B)$, and factor $f$ over $\Omega$ as \begin{align*} f = \prod_{i=1}^{\ell} (T - \alpha_i), \quad \alpha_1, \ldots, \alpha_\ell \in \Omega, \end{align*} where $\ell = \deg f$ and we may take $\alpha_1 = b$. Since $f$ is irreducible over $\operatorname{Frac}(A)$ and both $b = \alpha_1$ and each $\alpha_i$ are roots of $f$, for each $i$ there is a $\operatorname{Frac}(A)$-algebra isomorphism \begin{align*} \varphi_i: \operatorname{Frac}(A)[b] \xrightarrow{\,\sim\,} \operatorname{Frac}(A)[\alpha_i] \end{align*} sending $b \mapsto \alpha_i$ (and fixing $\operatorname{Frac}(A)$ pointwise). Since $h(b) = 0$ in $B$ and $h$ has coefficients in $\operatorname{Frac}(A)$, applying $\varphi_i$ gives $h(\alpha_i) = \varphi_i(h(b)) = 0$. Therefore \begin{align*} \alpha_i^n + a_1 \alpha_i^{n-1} + \cdots + a_n = 0 \end{align*} with $a_1, \ldots, a_n \in \mathfrak{a}$. This shows each $\alpha_i$ is $\mathfrak{a}$-integral. [guided] Why does this work? The minimal polynomial $f$ is irreducible over $\operatorname{Frac}(A)$, so any two roots of $f$ in $\Omega$ are conjugate over $\operatorname{Frac}(A)$. Concretely, for each root $\alpha_i$ of $f$, the map $b \mapsto \alpha_i$ extends to a $\operatorname{Frac}(A)$-algebra isomorphism $\operatorname{Frac}(A)[b] \cong \operatorname{Frac}(A)[\alpha_i]$, since both quotients $\operatorname{Frac}(A)[T]/(f)$ are isomorphic to these field extensions. The polynomial $h$ has coefficients in $\operatorname{Frac}(A)$, so $\varphi_i$ fixes all coefficients of $h$. Since $\varphi_i(b) = \alpha_i$ and $\varphi_i$ is a ring homomorphism, $\varphi_i(h(b)) = h(\alpha_i)$. Because $h(b) = 0$, we get $h(\alpha_i) = 0$. This means each $\alpha_i$ satisfies the same monic polynomial with coefficients $a_j \in \mathfrak{a}$, so each $\alpha_i$ is $\mathfrak{a}$-integral. [/guided] [/step] [step:Conclude that the coefficients of $f$ lie in $\sqrt{\mathfrak{a}}$] The coefficients of $f = \prod_{i=1}^\ell (T - \alpha_i)$ are, up to sign, the elementary symmetric polynomials $e_1(\alpha_1, \ldots, \alpha_\ell), \ldots, e_\ell(\alpha_1, \ldots, \alpha_\ell)$. Each $e_j$ is a sum of products of elements from $\{\alpha_1, \ldots, \alpha_\ell\}$. By [Integral Closure of an Ideal as Radical](/theorems/2877), applied to $A \subset \Omega$ and the ideal $\mathfrak{a}$, the set of $\mathfrak{a}$-integral elements in $\Omega$ equals $\sqrt{\mathfrak{a}\overline{A}^\Omega}$, where $\overline{A}^\Omega$ is the integral closure of $A$ in $\Omega$. In particular, this set is an ideal of $\overline{A}^\Omega$, hence closed under addition and multiplication. Since each $\alpha_i$ is $\mathfrak{a}$-integral (by the previous step), every sum of products of the $\alpha_i$ is $\mathfrak{a}$-integral. Therefore each coefficient of $f$ is $\mathfrak{a}$-integral. The coefficients of $f$ lie in $\operatorname{Frac}(A)$ (since $f \in \operatorname{Frac}(A)[T]$). Since $A$ is integrally closed, $\overline{A} = A$ in $\operatorname{Frac}(A)$, and [Integral Closure of an Ideal as Radical](/theorems/2877) applied to $A \subset \operatorname{Frac}(A)$ gives: the $\mathfrak{a}$-integral elements of $\operatorname{Frac}(A)$ equal $\sqrt{\mathfrak{a} \cdot A} = \sqrt{\mathfrak{a}}$. Therefore each coefficient of $f$ lies in $\sqrt{\mathfrak{a}}$. Writing $f = T^\ell + c_1 T^{\ell-1} + \cdots + c_\ell$ with $c_j \in \sqrt{\mathfrak{a}} \subset A$, the minimal polynomial has the required form. [guided] This is the step where both the structure of the minimal polynomial and the hypothesis that $A$ is integrally closed are consumed. The coefficients of $f$ are elementary symmetric functions of the roots $\alpha_1, \ldots, \alpha_\ell$. For example, the coefficient of $T^{\ell-1}$ is $-(\alpha_1 + \cdots + \alpha_\ell)$ and the constant term is $(-1)^\ell \alpha_1 \cdots \alpha_\ell$. We need these to be $\mathfrak{a}$-integral. By [Integral Closure of an Ideal as Radical](/theorems/2877), the $\mathfrak{a}$-integral elements in any ring extension of $A$ form the radical $\sqrt{\mathfrak{a}\overline{A}'}$ (where $\overline{A}'$ is the integral closure of $A$ in that extension). Crucially, this is an ideal, hence closed under sums and products. Since each $\alpha_i$ is $\mathfrak{a}$-integral, every polynomial expression in the $\alpha_i$ is also $\mathfrak{a}$-integral. In particular, each elementary symmetric polynomial $e_j(\alpha_1, \ldots, \alpha_\ell)$ is $\mathfrak{a}$-integral. Now the key point: the coefficients of $f$ lie in $\operatorname{Frac}(A)$, not just in $\Omega$. What does $\mathfrak{a}$-integrality mean for elements of $\operatorname{Frac}(A)$? Since $A$ is integrally closed, the integral closure of $A$ in $\operatorname{Frac}(A)$ is $A$ itself. So [Integral Closure of an Ideal as Radical](/theorems/2877) applied to $A \subset \operatorname{Frac}(A)$ gives: the $\mathfrak{a}$-integral elements of $\operatorname{Frac}(A)$ are $\sqrt{\mathfrak{a} \cdot A} = \sqrt{\mathfrak{a}}$. Since $\sqrt{\mathfrak{a}} \subset A$ (the radical of an ideal of $A$ is contained in $A$), the coefficients of $f$ lie in $\sqrt{\mathfrak{a}} \subset A$, as required. [/guided] [/step]