Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 101 / 6 / d / Solution

We use three theorems for an integral extension $S\subseteq R$, none requiring Noetherian hypotheses. The Lying-over theorem says that each prime ideal $\mathfrak{p}$ of $S$ is $\mathfrak{q}\cap S$ for some prime ideal $\mathfrak{q}$ of $R$. The going-up theorem says that if $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ in $S$ and $\mathfrak{q}\cap S=\mathfrak{p}$, there is ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap S={\mathfrak{p}}^{\prime }$. The incomparability theorem for integral extensions says that comparable prime ideals of $R$ with the same contraction to $S$ are equal.

Take any strict chainin $R$. The contraction of an ideal operation gives ${\mathfrak{p}}_j={\mathfrak{q}}_j\cap S$, which are prime ideals of $S$, by the proof in Question 1(b), and remain strictly increasing by the incomparability theorem for integral extensions. Thus every chain length in $R$ occurs in $S$, giving $\dim R\le \dim S$.

Conversely, take any strict chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n$ in $S$. Use the Lying-over theorem to choose ${\mathfrak{q}}_0$ over ${\mathfrak{p}}_0$, then repeatedly use the going-up theorem to obtain ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ with the prescribed contractions. Each inclusion must be strict, since its contractions are distinct. Hence every chain length in $S$ occurs in $R$, giving $\dim S\le \dim R$.

Taking suprema proves integral extensions preserve Krull dimension:The argument works equally well when the dimensions are infinite, since it compares all finite chain lengths.