Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 5 / b

The Krull dimension $\dim R$ is the supremum of lengths $n$ of strict chainsof prime ideals. The transcendence degree ${\mathrm{trdeg}}_{k}L$ is the cardinality of a transcendence basis of $L/k$.

By Noether normalization lemma, there are algebraically independent $y_1,\dots ,y_d\in A$ such that $A$ is finite, hence integral, over $k[y_1,\dots ,y_d]$. Their fraction field has transcendence degree $d$, and $L$ is algebraic over it, so ${\mathrm{trdeg}}_{k}L=d$. Going up and incomparability show that an integral extension preserves Krull dimension, while a polynomial ring in $d$ variables over a field has dimension $d$. Hence

A chain of length $s$ in $R/\mathfrak{p}$ lifts to a chainin $R$. Since $R$ is a domain and $\mathfrak{p}\ne 0$, prepending $(0)⊊\mathfrak{p}$ gives a chain of length $s+1$. Therefore $s+1\le r$ and

Going up lifts every prime chain in $A$ to one in $B$, so $\dim B\ge \dim A$. Conversely, contracting a strict chain of primes of $B$ gives a chain in $A$, and the incomparability theorem for integral extensions ensures that no strict inclusion contracts to equality. Thus $\dim A\ge \dim B$, and