= Solution
Put $K=\operatorname{Frac}R$, let
$$
f(T)=T^d+a_{d-1}T^{d-1}+\cdots+a_0
$$
be the <minimal polynomial of an algebraic element> $y$ over $K$, and let $B$ be the integral closure of $R$ in a finite normal extension containing all roots of $f$. Since $y$ is integral over $R$ and $R$ is <integrally closed domain>, every $a_i$ belongs to $R$.
Write $y=\sum_jp_jz_j$ with $p_j\in\mathfrak p$ and $z_j\in A$. Every $K$-embedding into the normal extension fixes the $p_j$ and sends each $z_j$ to an element integral over $R$. Thus every conjugate of $y$ lies in the extended ideal $\mathfrak pB$. Each nonleading coefficient of $f$ is, up to sign, an <elementary symmetric polynomial> in those conjugates, so it lies in $\mathfrak pB\cap R$.
For an <integral extension>, extension followed by contraction preserves a prime ideal:
$$
\mathfrak pB\cap R=\mathfrak p.
$$
Indeed, the determinant trick gives $r^m\in\mathfrak p$ for $r\in\mathfrak pB\cap R$, and primality then gives $r\in\mathfrak p$. Hence $a_i\in\mathfrak p$ for every $i<d$.
Solved by gpt-5.6-sol high.
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