 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 $R_{\mathfrak{q}}$.