Height-one localizations of normal Noetherian domains are DVRs

By [L1] and [L3], the ring $R_{\mathfrak{p}}$ is a Noetherian local domain. By [F1], its Krull dimension is $\dim (R_{\mathfrak{p}})=\mathrm{ht}(\mathfrak{p})=1$.

The localisation remains integrally closed. Let $x$ lie in the fraction field of $R_{\mathfrak{p}}$ and be integral over $R_{\mathfrak{p}}$. Write a monic equation $x^n+\frac{a_{n-1}}{s_{n-1}}{x}^{n-1}+\cdots +\frac{a_0}{s_0}=0$ with each $s_i\mathrm{\notin }\mathfrak{p}$. Put $s=s_0\cdots s_{n-1}$. Then $sx$ is integral over $R$, so [F3] gives $sx\in R$. Since $s\mathrm{\notin }\mathfrak{p}$, we conclude $x=(sx)\mathrm{/}s\in R_{\mathfrak{p}}$. Thus $R_{\mathfrak{p}}$ is integrally closed.

The maximal ideal of $R_{\mathfrak{p}}$ is nonzero. Choose $0\mathrm{\ne }a\in \mathfrak{p}$. If $a\mathrm{/}1=0$ in $R_{\mathfrak{p}}$, then some $s\mathrm{\notin }\mathfrak{p}$ satisfies $sa=0$, impossible in the domain $R$. Hence $a\mathrm{/}1\in \mathfrak{p}{R}_{\mathfrak{p}}$ is nonzero, so $R_{\mathfrak{p}}$ is not a field.

Steps 1.1, 1.2, and 1.3 verify condition 3 of [L4] for the ring $R_{\mathfrak{p}}$. Therefore $R_{\mathfrak{p}}$ is a discrete valuation ring.