[proofplan]
By the [Integral Closure of an Ideal as Radical](/theorems/2877) theorem, $b$ is $\mathfrak{a}$-integral iff $b \in \sqrt{\mathfrak{a}\overline{A}}$, and $b$ is $\sqrt{\mathfrak{a}}$-integral iff $b \in \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$, where $\overline{A}$ is the integral closure of $A$ in $B$. It therefore suffices to prove $\sqrt{\mathfrak{a}\overline{A}} = \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$. One inclusion follows from $\mathfrak{a} \subset \sqrt{\mathfrak{a}}$; the reverse uses the general fact that extending a radical ideal through a ring homomorphism lands inside the radical of the extended ideal.
[/proofplan]
[step:Reformulate both conditions using the Integral Closure of an Ideal as Radical theorem]
Let $\overline{A}$ denote the integral closure of $A$ in $B$. By [Integral Closure of an Ideal as Radical](/theorems/2877), the integral closure of $\mathfrak{a}$ in $B$ equals $\sqrt{\mathfrak{a}\overline{A}}$, and the integral closure of $\sqrt{\mathfrak{a}}$ in $B$ equals $\sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$. Therefore:
\begin{align*}
b \text{ is } \mathfrak{a}\text{-integral} &\iff b \in \sqrt{\mathfrak{a}\overline{A}}, \\
b \text{ is } \sqrt{\mathfrak{a}}\text{-integral} &\iff b \in \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}.
\end{align*}
The claim reduces to showing $\sqrt{\mathfrak{a}\overline{A}} = \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$.
[/step]
[step:Prove $\sqrt{\mathfrak{a}\overline{A}} \subset \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$ using the inclusion $\mathfrak{a} \subset \sqrt{\mathfrak{a}}$]
Since $\mathfrak{a} \subset \sqrt{\mathfrak{a}}$, extending to $\overline{A}$ gives $\mathfrak{a}\overline{A} \subset \sqrt{\mathfrak{a}}\,\overline{A}$. Taking radicals preserves the inclusion (if $I \subset J$ then $\sqrt{I} \subset \sqrt{J}$), so
\begin{align*}
\sqrt{\mathfrak{a}\overline{A}} \subset \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}.
\end{align*}
[/step]
[step:Establish the general fact that extending a radical lands inside the radical of the extension]
[claim:Extension of radical ideals]
Let $\varphi: R \to R'$ be a ring homomorphism and $I$ an ideal of $R$. Then $\varphi(\sqrt{I}) \cdot R' \subset \sqrt{I \cdot R'}$, where $I \cdot R'$ denotes the ideal of $R'$ generated by $\varphi(I)$.
[/claim]
[proof]
A general element of $\varphi(\sqrt{I}) \cdot R'$ is a finite sum $\sum_{j=1}^k r'_j \varphi(x_j)$ with $r'_j \in R'$ and $x_j \in \sqrt{I}$. For each $j$, there exists $n_j \geq 1$ with $x_j^{n_j} \in I$. Set $N := n_1 + \cdots + n_k$. We show $\left(\sum_{j=1}^k r'_j \varphi(x_j)\right)^N \in I \cdot R'$.
Expand $\left(\sum_{j=1}^k r'_j \varphi(x_j)\right)^N$ by the multinomial theorem. Each monomial in the expansion has the form
\begin{align*}
c \cdot (r'_1)^{m_1} \cdots (r'_k)^{m_k} \cdot \varphi(x_1)^{m_1} \cdots \varphi(x_k)^{m_k}
\end{align*}
where $m_1 + \cdots + m_k = N$ and $c$ is a multinomial coefficient. Since $m_1 + \cdots + m_k = N = n_1 + \cdots + n_k$, at least one index $j$ satisfies $m_j \geq n_j$ (otherwise $m_j \leq n_j - 1$ for all $j$, giving $\sum m_j \leq \sum(n_j - 1) = N - k < N$, a contradiction). For that index, $\varphi(x_j)^{m_j} = \varphi(x_j^{m_j})$ and $x_j^{m_j} \in I$ (since $m_j \geq n_j$ and $x_j^{n_j} \in I$). Therefore each monomial in the expansion lies in $I \cdot R'$, so $\left(\sum_j r'_j \varphi(x_j)\right)^N \in I \cdot R'$.
This shows $\sum_j r'_j \varphi(x_j) \in \sqrt{I \cdot R'}$.
[/proof]
[/step]
[step:Prove $\sqrt{\sqrt{\mathfrak{a}}\,\overline{A}} \subset \sqrt{\mathfrak{a}\overline{A}}$ using the extension-of-radical fact]
Apply the claim to the inclusion homomorphism $\varphi: A \hookrightarrow \overline{A}$ and the ideal $I = \mathfrak{a}$. The claim gives
\begin{align*}
\varphi(\sqrt{\mathfrak{a}}) \cdot \overline{A} = \sqrt{\mathfrak{a}}\,\overline{A} \subset \sqrt{\mathfrak{a}\overline{A}}.
\end{align*}
Taking radicals of both sides and using $\sqrt{\sqrt{J}} = \sqrt{J}$ for any ideal $J$:
\begin{align*}
\sqrt{\sqrt{\mathfrak{a}}\,\overline{A}} \subset \sqrt{\sqrt{\mathfrak{a}\overline{A}}} = \sqrt{\mathfrak{a}\overline{A}}.
\end{align*}
[guided]
This is the non-trivial direction. The issue is that $\sqrt{\mathfrak{a}}$ is larger than $\mathfrak{a}$, so extending $\sqrt{\mathfrak{a}}$ to $\overline{A}$ could a priori produce a larger ideal than $\sqrt{\mathfrak{a}\overline{A}}$. But the claim shows that extending a radical lands inside the radical of the extension: $\sqrt{\mathfrak{a}}\,\overline{A} \subset \sqrt{\mathfrak{a}\overline{A}}$.
Why does this hold? An element $x \in \sqrt{\mathfrak{a}}$ satisfies $x^n \in \mathfrak{a}$ for some $n$. When we multiply $x$ by elements of $\overline{A}$ and raise to a high enough power, the multinomial expansion forces at least one factor to have a sufficiently high power of some $x_j$ to land in $\mathfrak{a}$, hence in $\mathfrak{a}\overline{A}$. The pigeonhole argument in the claim makes this precise.
Once we have $\sqrt{\mathfrak{a}}\,\overline{A} \subset \sqrt{\mathfrak{a}\overline{A}}$, taking radicals gives $\sqrt{\sqrt{\mathfrak{a}}\,\overline{A}} \subset \sqrt{\sqrt{\mathfrak{a}\overline{A}}} = \sqrt{\mathfrak{a}\overline{A}}$, using the idempotence of the radical operation ($\sqrt{\sqrt{J}} = \sqrt{J}$ for any ideal $J$, since $x^m \in \sqrt{J}$ implies $x^{mn} \in J$).
[/guided]
[/step]
[step:Combine both inclusions to conclude]
From the second and fourth steps, we have
\begin{align*}
\sqrt{\mathfrak{a}\overline{A}} \subset \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}} \subset \sqrt{\mathfrak{a}\overline{A}},
\end{align*}
so $\sqrt{\mathfrak{a}\overline{A}} = \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$. Therefore $b \in \sqrt{\mathfrak{a}\overline{A}}$ if and only if $b \in \sqrt{\sqrt{\mathfrak{a}}\,\overline{A}}$, which means $b$ is $\mathfrak{a}$-integral if and only if $b$ is $\sqrt{\mathfrak{a}}$-integral.
[/step]
