A chain of length $s$ in $R/\mathfrak p$ lifts to a chain
$$
\mathfrak p=\mathfrak p_0\subsetneq\cdots\subsetneq\mathfrak p_s
$$
in $R$. Since $R$ is a domain and $\mathfrak p\ne0$, prepending $(0)\subsetneq\mathfrak p$ gives a chain of length $s+1$. Therefore $s+1\leq r$ and
$$
\boxed{\dim(R/\mathfrak p)\leq r-1<r.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-101/5/b/ii/solution