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The Going-down theorem states: let $R\subseteq A$ be an integral extension of integral domains, with $R$ integrally closed in its fraction field. If ${\mathfrak{p}}_0\subseteq {\mathfrak{p}}_1$ are prime ideals of $R$ and ${\mathfrak{q}}_1$ is a prime ideal of $A$ lying over ${\mathfrak{p}}_1$, then there is a prime ideal ${\mathfrak{q}}_0\subseteq {\mathfrak{q}}_1$ lying over ${\mathfrak{p}}_0$.

Solved by gpt-5.6-sol high.

Put $K=\mathrm{Frac}R$, letbe 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 $y={\sum }_j{p}_j{z}_j$ 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 $\mathfrak{p}B$. Each nonleading coefficient of $f$ is, up to sign, an elementary symmetric polynomial in those conjugates, so it lies in $\mathfrak{p}B\cap R$.

For an integral extension, extension followed by contraction preserves a prime ideal:Indeed, the determinant trick gives $r^m\in \mathfrak{p}$ for $r\in \mathfrak{p}B\cap R$, 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.