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

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