Krull principal ideal theorem

In a Noetherian ring, every prime ideal minimal over a proper principal ideal has height at most one.

If an ideal of a Noetherian ring is generated by $n$ elements, every prime ideal minimal over it has height at most $n$. The proof inducts on $n$, using the Krull principal ideal theorem in a quotient and prime avoidance to choose a prime chain compatible with the induction.

If $R$ is Noetherian and ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊{\mathfrak{p}}_2$ are prime ideals, there are infinitely many primes strictly between ${\mathfrak{p}}_0$ and ${\mathfrak{p}}_2$. After quotienting by ${\mathfrak{p}}_0$ and localizing at ${\mathfrak{p}}_2$, this reduces to a local Noetherian domain of dimension at least two. If it had only finitely many height-one primes, prime avoidance would give a nonzero element of the maximal ideal outside all of them, contradicting the Krull principal ideal theorem.

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 $n+1$.