Krull principal ideal theorem

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

For a proof, suppose instead that ${\mathfrak{p}}_0⊊\mathfrak{q}⊊\mathfrak{p}$ and that $\mathfrak{p}$ is minimal over $(a)$. Quotient by ${\mathfrak{p}}_0$ and localize at $\mathfrak{p}$; one may assume that $R$ is a Noetherian local domain, $0⊊\mathfrak{q}⊊\mathfrak{p}$ with $\mathfrak{p}$ maximal, and $\mathfrak{p}$ is the only prime containing $(a)$. For the symbolic powerseach ${\mathfrak{q}}^{(n)}$ is $\mathfrak{q}$-primary. The quotient $R/(a)$ is a zero-dimensional Noetherian ring and hence Artinian, so the descending chain $({\mathfrak{q}}^{(n)}+(a))/(a)$ stabilizes. For large $n$, writing an element of ${\mathfrak{q}}^{(n)}$ modulo ${\mathfrak{q}}^{(n+1)}$ and using $a\notin \mathfrak{q}$ givesThe Nakayama lemma applied to ${\mathfrak{q}}^{(n)}/{\mathfrak{q}}^{(n+1)}$ gives ${\mathfrak{q}}^{(n)}={\mathfrak{q}}^{(n+1)}$. Localizing at $\mathfrak{q}$ would therefore make the powers of the nonzero maximal ideal $\mathfrak{q}{R}_{\mathfrak{q}}$ stabilize at a nonzero ideal, contradicting the Krull intersection theorem for the Noetherian local domain $R_{\mathfrak{q}}$.

If a principal prime ideal $(x)$ in a Noetherian local ring has positive height, then the ring is a domain. Localizing at $(x)$ shows that every annihilator of a power of $x$ lies in $(x)$; stabilization of the annihilator chain then proves that $x$ is a nonzerodivisor and that zero is prime.

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 $I$ is an $\mathfrak{m}$-primary ideal generated by $n>0$ elements in a Noetherian local ring and $L=\mathrm{length}(R/I)$, then $I^j/I^{j+1}$ is a quotient of $\left(\genfrac{}{}{0ex}{}{j+n-1}{n-1}\right)$ copies of $R/I$. Summing their lengths gives the displayed bound. Since $I\subseteq \mathfrak{m}$, it also bounds $\mathrm{length}(R/{\mathfrak{m}}^t)$. Together with the prime-chain lower bound for local length, localization at a minimal prime proves the Krull height theorem.

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 $n+1$.