Krull's principal ideal theorem

Let $A$ be a commutative ring . Recall that the height $\operatorname{ht}(\mathfrak p)$ of a prime ideal $\mathfrak p\subset A$ measures the maximal length of a strictly increasing chain of prime ideals ending at $\mathfrak p$, and that $A$ is Noetherian if it satisfies the ascending chain condition on ideals.

A prime ideal $\mathfrak p$ is said to be minimal over an ideal $I\subseteq A$ if $I\subseteq \mathfrak p$ and there is no prime $\mathfrak q$ with $I\subseteq \mathfrak q\subsetneq \mathfrak p$.

Theorem (Krull’s principal ideal theorem).
Let $A$ be a Noetherian ring and let $f\in A$. If $\mathfrak p$ is a prime ideal minimal over the principal ideal $(f)$, then

Equivalently, in the prime spectrum $\operatorname{Spec}(A)$, every irreducible component of the closed set $V(f)$ has codimension at most $1$ (codimension computed via height ).

A common generalization (often called Krull’s height theorem) says: if an ideal $I$ can be generated by $r$ elements, then any prime ideal minimal over $I$ has height at most $r$.

Examples

1. A reducible hypersurface in a polynomial ring.
   Let $A=k[x,y]$ for a field $k$, and take $f=xy$. The primes minimal over $(xy)$ are $(x)$ and $(y)$, since $(xy)=(x)\cap (y)$ in this case. Each of these primes has height $1$ (e.g. $(0)\subsetneq (x)$ is a prime chain of length $1$), in agreement with the theorem.

2. An arithmetic example.
   In $A=\mathbb Z$, take $f=12$. The primes minimal over $(12)$ are $(2)$ and $(3)$, since every prime ideal containing $(12)$ must contain one of $(2)$ or $(3)$. Each has height $1$ because $\mathbb Z$ has a prime chain $(0)\subsetneq (p)$ and no longer chain ending at $(p)$.

3. A principal prime ideal.
   In $A=k[x,y,z]$, the ideal $(x)$ is prime, hence it is minimal over itself. Since $(0)\subsetneq (x)$ is a chain of primes, $\operatorname{ht}(x)\ge 1$, and Krull’s principal ideal theorem forces $\operatorname{ht}(x)=1$.