Theorem (Krull's Principal Ideal Theorem): Let $R$ be a Noetherian ring, $x \in R$ a non-unit, and $\mathfrak{p}$ a prime ideal minimal over $(x)$. Then $\operatorname{ht}(\mathfrak{p}) \leq 1$.

Step 1: Reduction to the local case.

Replacing $R$ by $R_\mathfrak{p}$, we may assume $(R, \mathfrak{p})$ is a Noetherian local ring with $\mathfrak{p}$ the unique minimal prime over $(x)$. We need to show $\operatorname{ht}(\mathfrak{p}) \leq 1$, i.e., there is no chain $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \mathfrak{p}$ of prime ideals.

Step 2: Suppose for contradiction that $\operatorname{ht}(\mathfrak{p}) \geq 2$.

Then there exists a chain $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \mathfrak{p}$. We will derive a contradiction.

Since $\mathfrak{p}$ is minimal over $(x)$, the ring $R/(x)$ has a unique minimal prime, namely $\mathfrak{p}/(x)$, so $\dim R/(x) = 0$. In particular, $\mathfrak{p}/(x)$ is both the maximal and the unique minimal prime of $R/(x)$.

Step 3: Symbolic powers.

Consider the $n$-th symbolic power of $\mathfrak{p}_1$: $\mathfrak{p}_1^{(n)} = \mathfrak{p}_1^n R_{\mathfrak{p}_1} \cap R = \{r \in R : sr \in \mathfrak{p}_1^n \text{ for some } s \notin \mathfrak{p}_1\}$

Define ideals $\mathfrak{a}_n = (\mathfrak{p}_1^{(n)}, x) / (x)$ in the ring $\bar{R} = R/(x)$. Since $\bar{R}$ is a zero-dimensional Noetherian local ring (Artinian), the descending chain $\mathfrak{a}_1 \supseteq \mathfrak{a}_2 \supseteq \cdots$ stabilizes: there exists $N$ such that $\mathfrak{a}_n = \mathfrak{a}_N$ for all $n \geq N$.

Step 4: Unraveling the stabilization.

The equality $\mathfrak{a}_n = \mathfrak{a}_N$ for $n \geq N$ means: for every $r \in \mathfrak{p}_1^{(N)}$, we can write $r = a + xb$ for some $a \in \mathfrak{p}_1^{(n)}$ and $b \in R$. Thus $\mathfrak{p}_1^{(N)} \subseteq \mathfrak{p}_1^{(n)} + (x) \cdot R \quad \text{for all } n \geq N$

Step 5: Working in $R_{\mathfrak{p}_1}$.

Localize at $\mathfrak{p}_1$. In $R_{\mathfrak{p}_1}$, the ideal $(x)$ is not contained in $\mathfrak{p}_1 R_{\mathfrak{p}_1}$ (otherwise $x \in \mathfrak{p}_1 \subsetneq \mathfrak{p}$, contradicting minimality of $\mathfrak{p}$ over $(x)$; actually we need to be more careful — since $\mathfrak{p}$ is minimal over $(x)$ and $\mathfrak{p}_1 \subsetneq \mathfrak{p}$, the element $x$ is not in $\mathfrak{p}_1$). Wait — we need $x \notin \mathfrak{p}_1$.

Indeed, since $\mathfrak{p}$ is minimal over $(x)$ and $\mathfrak{p}_1 \subsetneq \mathfrak{p}$, if $x \in \mathfrak{p}_1$ then $\mathfrak{p}_1 \supseteq (x)$, contradicting the minimality of $\mathfrak{p}$ over $(x)$. So $x \notin \mathfrak{p}_1$.

In $R_{\mathfrak{p}_1}$, the element $x$ becomes a unit. Therefore: $\mathfrak{p}_1^{(N)} R_{\mathfrak{p}_1} \subseteq \mathfrak{p}_1^{(n)} R_{\mathfrak{p}_1} + (x) R_{\mathfrak{p}_1} = \mathfrak{p}_1^n R_{\mathfrak{p}_1} + R_{\mathfrak{p}_1} = R_{\mathfrak{p}_1}$

This gives $\mathfrak{p}_1^N R_{\mathfrak{p}_1} \subseteq \mathfrak{p}_1^n R_{\mathfrak{p}_1}$ for all $n$, hence $\mathfrak{p}_1^N R_{\mathfrak{p}_1} \subseteq \bigcap_n \mathfrak{p}_1^n R_{\mathfrak{p}_1}$.

Step 6: Applying the Krull Intersection Theorem.

By the Krull intersection theorem (valid in Noetherian local rings), $\bigcap_n \mathfrak{p}_1^n R_{\mathfrak{p}_1} = 0$. Therefore $\mathfrak{p}_1^N R_{\mathfrak{p}_1} = 0$, which means $\mathfrak{p}_1 R_{\mathfrak{p}_1}$ is nilpotent. Since $R_{\mathfrak{p}_1}$ is a Noetherian local ring with nilpotent maximal ideal, it has dimension $0$.

But $\mathfrak{p}_0 R_{\mathfrak{p}_1}$ is a prime ideal strictly contained in $\mathfrak{p}_1 R_{\mathfrak{p}_1}$, giving $\dim R_{\mathfrak{p}_1} \geq 1$ — a contradiction.

Therefore $\operatorname{ht}(\mathfrak{p}) \leq 1$. $\square$