Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 101 / 3

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 ${\mathfrak{p}}_0$.

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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 $i<d$.

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As an $A^{\prime }$-algebra, $A{\otimes }_R{A}^{\prime }$ is generated by the elements $a\otimes 1$. If $a$ obeys a monic relationover $R$, then $a\otimes 1$ obeys the same monic relation after applying the structure map $R\to A^{\prime }$. Thus every generator is an integral element. The subalgebra generated by finitely many integral elements is finite as a module, and therefore integral; each tensor involves only finitely many generators. Hence $A{\otimes }_R{A}^{\prime }$ is integral over $A^{\prime }$.

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If the coefficients of $p$ are integral over $R$, they generate a finite $R$-algebra $B$. Then $B[t_1,\dots ,t_n]$ is a finite $R[t_1,\dots ,t_n]$-module, so every one of its elements, including $p$, is integral.

Conversely, use the fact that the integral closure of a graded ring is graded. Give $A[t_1,\dots ,t_n]$ its ${\mathbb{N}}^n$-grading and regard $R[t_1,\dots ,t_n]$ as a graded subring. If $p$ is integral, each homogeneous component $a_{\alpha }{t}^{\alpha }$ is integral. Applying the evaluation homomorphism $t_1=\cdots =t_n=1$ shows that every coefficient $a_{\alpha }$ is integral over $R$.

Solved by gpt-5.6-sol high.