The symbolic-power step inside the principal ideal theorem

Since $\mathfrak{m}$ is both maximal and minimal among primes containing $(x)$, it is the only such prime. Hence the quotient $\bar{R}=R\mathrm{/}(x)$ is Noetherian by [L2] and has the single prime $\bar{\mathfrak{m}}$. Its nilradical is therefore $\bar{\mathfrak{m}}$, so [L3] gives ${\bar{\mathfrak{m}}}^N=0$ for some $N$. Each layer ${\bar{\mathfrak{m}}}^j\mathrm{/}{\bar{\mathfrak{m}}}^{j+1}$ is a finite-dimensional vector space over $R\mathrm{/}\mathfrak{m}$: it is finitely generated by [L2] and is annihilated by $\bar{\mathfrak{m}}$. A descending chain of ideals in $\bar{R}$ induces descending chains in these finitely many finite-dimensional layers, so all layers, and therefore the original chain, stabilize. Thus $\bar{R}$ is Artinian.

Let $\mathfrak{p}⊊\mathfrak{m}$ be prime. Then $x\mathrm{\notin }\mathfrak{p}$. For $r\ge 1$, define the symbolic power ${\mathfrak{p}}^{(r)}:={\mathfrak{p}}^r{R}_{\mathfrak{p}}\cap R\mathrm{.}$ These form a descending chain. By step 1.1, the ideals $({\mathfrak{p}}^{(r)}+(x))\mathrm{/}(x)$ in $\bar{R}$ stabilize, so for some $r$, ${\mathfrak{p}}^{(r)}+(x)={\mathfrak{p}}^{(r+1)}+(x)\mathrm{.}$ If $a\in {\mathfrak{p}}^{(r)}$, write $a=b+xc$ with $b\in {\mathfrak{p}}^{(r+1)}$. In $R_{\mathfrak{p}}$ the element $x$ is a unit, while $a-b\in {\mathfrak{p}}^r{R}_{\mathfrak{p}}$; hence $c\in {\mathfrak{p}}^r{R}_{\mathfrak{p}}\cap R={\mathfrak{p}}^{(r)}$. Therefore ${\mathfrak{p}}^{(r)}={\mathfrak{p}}^{(r+1)}+x{\mathfrak{p}}^{(r)}\mathrm{.}$

The quotient module $M={\mathfrak{p}}^{(r)}\mathrm{/}{\mathfrak{p}}^{(r+1)}$ is finitely generated by [L2], and step 2.1 says $M=xM$. Apply [L5] with $I=(x)$. It gives $(1-a)M=0$ for some $a\in (x)\subseteq \mathfrak{m}$. Fact [L6] makes $1-a$ a unit, so $M=0$ and ${\mathfrak{p}}^{(r)}={\mathfrak{p}}^{(r+1)}$. Localizing this equality at $\mathfrak{p}$ gives $(\mathfrak{p}{R}_{\mathfrak{p}}{)}^r=(\mathfrak{p}{R}_{\mathfrak{p}}{)}^{r+1}\mathrm{.}$

Apply [L5] in the local ring $R_{\mathfrak{p}}$ to the finite module $(\mathfrak{p}{R}_{\mathfrak{p}}{)}^r$ and the ideal $\mathfrak{p}{R}_{\mathfrak{p}}$. Step 3.1 says that ideal times the module is the module, so [L6] gives $(\mathfrak{p}{R}_{\mathfrak{p}}{)}^r=0$. Because $R_{\mathfrak{p}}$ is a domain, this forces $\mathfrak{p}{R}_{\mathfrak{p}}=0$, and contraction through [L4] gives $\mathfrak{p}=(0)$. Thus every prime strictly below $\mathfrak{m}$ is zero, and [L1] yields $\mathrm{ht}(\mathfrak{m})\le 1$.