The cases $n=0$ and $n=d$ can indeed be finite: a <Noetherian ring> has finitely many minimal primes, and a semilocal ring may have finitely many maximal ideals. We prove that every intermediate height occurs infinitely often.

Because $d$ is a finite integer equal to the supremum of prime-chain lengths, there is a chain
$$
\mathfrak p_0\subsetneq\mathfrak p_1\subsetneq\cdots
\subsetneq\mathfrak p_d.
$$
Each $\mathfrak p_i$ has height exactly $i$: its displayed lower chain gives height at least $i$, while any longer lower chain could be extended by the remaining displayed primes and would contradict $\operatorname{ht}\mathfrak p_d=d$.

Suppose $0<n<d$. The three primes
$$
\mathfrak p_{n-1}\subsetneq\mathfrak p_n
\subsetneq\mathfrak p_{n+1}
$$
fall under <prime ideals between a three-prime chain>, so infinitely many primes $\mathfrak q$ satisfy
$$
\mathfrak p_{n-1}\subsetneq\mathfrak q
\subsetneq\mathfrak p_{n+1}.
$$
Every such $\mathfrak q$ has height exactly $n$: the lower inclusion gives height at least $n$, and height at least $n+1$ would, after appending $\mathfrak p_{n+1}$, contradict its height $n+1$. Therefore there are infinitely many height-$n$ primes. The assumed finiteness forces
$$
\boxed{n\in\{0,d\}}.
$$
This is <infinitude of intermediate-height prime ideals>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 101 / 5 / iii / Solution

If a Noetherian ring has finite <Krull dimension> $d$, then it has infinitely many prime ideals of each height $n$ with $0<n<d$. In a prime chain of maximal length, apply <prime ideals between a three-prime chain> to the terms of heights $n-1$, $n$, and $n+1$.

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