A minimal prime over a principal nonzerodivisor has height one

By [L1], $\mathrm{ht}(\mathfrak{p})\le 1$.

By [L4], the nilradical is a finite intersection of minimal primes. Since $\mathfrak{p}$ contains that intersection, it contains one of those primes; choose $\mathfrak{q}\subseteq \mathfrak{p}$. The localization $R_{\mathfrak{q}}$ is a nonzero Noetherian local ring with only one prime, $\mathfrak{q}{R}_{\mathfrak{q}}$. Its nilradical is therefore that maximal ideal and is nilpotent by [L4]. If $x\in \mathfrak{q}$, the image $x\mathrm{/}1$ would be nilpotent in $R_{\mathfrak{q}}$, contradicting that multiplication by the nonzerodivisor $x$ remains injective after localization. Thus $x\mathrm{\notin }\mathfrak{q}$ and $\mathfrak{q}\mathrm{\ne }\mathfrak{p}$.

Apply [L2] to this $\mathfrak{q}$. In the resulting Noetherian local domain $A=(R\mathrm{/}\mathfrak{q}{)}_{\mathfrak{p}\mathrm{/}\mathfrak{q}}$, the image of $x$ is nonzero and lies in the maximal ideal. If that maximal ideal had height $0$, then [L3] would make $A$ a field, forcing $x\mathrm{/}1$ to be a unit, contradiction. Hence the maximal ideal of $A$ has height at least $1$, so $\mathfrak{p}$ has height at least $1$.

Steps 1.1 and 2.1 give $\mathrm{ht}(\mathfrak{p})=1$.