= Krull principal ideal theorem
{c}
{wiki}
= Principal ideal theorem
{synonym}
In a <Noetherian ring>, every <prime ideal> minimal over a proper principal ideal has height at most one.
For a proof, suppose instead that $\mathfrak p_0\subsetneq\mathfrak q\subsetneq\mathfrak p$ 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, $0\subsetneq\mathfrak q\subsetneq\mathfrak p$ with $\mathfrak p$ maximal, and $\mathfrak p$ is the only prime containing $(a)$. For the symbolic powers
$$
\mathfrak q^{(n)}=\mathfrak q^nR_{\mathfrak q}\cap R,
$$
each $\mathfrak q^{(n)}$ 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 $a\notin\mathfrak q$ gives
$$
\mathfrak q^{(n)}=\mathfrak q^{(n+1)}+a\mathfrak q^{(n)}.
$$
The <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 $\mathfrak qR_{\mathfrak q}$ stabilize at a nonzero ideal, contradicting the <Krull intersection theorem> for the Noetherian local domain $R_{\mathfrak q}$.
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