A local domain has a dominating valuation overring

Let $S$ be the set of local subrings $B\subseteq K$ that dominate $A$, ordered by $B\le B^{\mathrm{\prime }}$ iff $A\subseteq B\subseteq B^{\mathrm{\prime }}$ and $B\cap {\mathfrak{m}}_{B^{\mathrm{\prime }}}={\mathfrak{m}}_B$. Then $S$ is nonempty since $A\in S$.

If $B\le B^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}\le B^{\mathrm{\prime }\mathrm{\prime }}$ then $B\le B^{\mathrm{\prime }\mathrm{\prime }}$: indeed ${\mathfrak{m}}_B=B\cap {\mathfrak{m}}_{B^{\mathrm{\prime }}}=B\cap (B^{\mathrm{\prime }}\cap {\mathfrak{m}}_{B^{\mathrm{\prime }\mathrm{\prime }}})=B\cap {\mathfrak{m}}_{B^{\mathrm{\prime }\mathrm{\prime }}}$. So $\le$ is a partial order on $S$.

The empty chain has upper bound $A$. Every nonempty chain in $S$ has an upper bound: if $\{B_i\}$ is a chain, let $B={\bigcup }_i{B}_i\subseteq K$, a subring of $K$. The union $\mathfrak{n}={\bigcup }_i{\mathfrak{m}}_{B_i}$ is an ideal of $B$ (any two elements lie in a common $B_i$ by comparability), it is proper since $1\mathrm{\notin }\mathfrak{n}$, and every $x\in B\setminus \mathfrak{n}$ lies in some $B_i\setminus {\mathfrak{m}}_{B_i}$, hence is a unit of $B_i$ and so of $B$. Therefore $B$ is local with maximal ideal $\mathfrak{n}$, it dominates each $B_i$, and it is an upper bound in $S$.

By [F5] the poset $S$ has a maximal element $V$.

We show $\mathrm{Frac}(V)=K$ by [F3]. Suppose first that some $t\in K$ is transcendental over $\mathrm{Frac}(V)$, so that $V[t]\subseteq K$ is a polynomial ring and $V[t]$ is a domain with $t\mathrm{\notin }\mathrm{Frac}(V)$. The ideal $\mathfrak{p}=(t)+{\mathfrak{m}}_{V}V[t]$ of $V[t]$ is prime because its quotient is $V[t]\mathrm{/}(t,{\mathfrak{m}}_V)\cong V\mathrm{/}{\mathfrak{m}}_V$, a field, and it is proper; the localization $V[t{]}_{\mathfrak{p}}$ is a local ring whose maximal ideal $\mathfrak{p}V[t{]}_{\mathfrak{p}}$ pulls back to $\mathfrak{p}\cap V={\mathfrak{m}}_V$, so it dominates $V$, while $t\in V[t{]}_{\mathfrak{p}}$ and $t\mathrm{\notin }V$ (as $t$ is transcendental over $\mathrm{Frac}(V)$) make it distinct from $V$. Then $V[t{]}_{\mathfrak{p}}$ dominates $V$ and is strictly larger in $S$, contradicting maximality of $V$.

Suppose next that some $t\in K$ is algebraic over $\mathrm{Frac}(V)$. Clearing denominators in a polynomial equation for $t$ over $\mathrm{Frac}(V)$ gives a nonzero $a\in V$ with $at$ integral over $V$ by [F6]. Then $A^{\mathrm{\prime }}=V[at]\subseteq K$ is integral over $V$ by [F9], and by [F4] (with $\ker (V\to A^{\mathrm{\prime }})=0$) there is a prime ${\mathfrak{m}}^{\mathrm{\prime }}$ of $A^{\mathrm{\prime }}$ with ${\mathfrak{m}}^{\mathrm{\prime }}\cap V={\mathfrak{m}}_V$; the localization $A_{{\mathfrak{m}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ is a local ring dominating $V$. Since $t=(at)\mathrm{/}a$ lies in $\mathrm{Frac}(A_{{\mathfrak{m}}^{\mathrm{\prime }}}^{\mathrm{\prime }})$, the ring $A_{{\mathfrak{m}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ is distinct from $V$ whenever $t\mathrm{\notin }\mathrm{Frac}(V)$, again contradicting maximality.

We show that $V$ is a valuation ring of $K$ by verifying [F1]. First, if $x\in K$ is integral over $V$, then $x\in V$: the ring $V[x]$ is integral over $V$ by [F9], so by [F4] there is a prime of $V[x]$ over ${\mathfrak{m}}_V$, whose localization dominates $V$ and hence equals $V$ by maximality; as $x\in V[x]$ lies in that localization, $x\in V$, and thus $V$ is integrally closed in $K$.

Step 4.2 applies to every $t\in K$ algebraic over $\mathrm{Frac}(V)$ and step 4.1 to every transcendental one; since each $t\in K\setminus \mathrm{Frac}(V)$ falls into one of the two cases and both contradict the maximality of $V$, no such $t$ exists, so $\mathrm{Frac}(V)=K$.

Now let $x\in K^{\times }$ and suppose $x\mathrm{\notin }V$; we show $x^{-1}\in V$. Let $A^{\mathrm{\prime }}=V[x]\subseteq K$, nonzero since $1\in V$, and suppose ${\mathfrak{m}}^{\mathrm{\prime }}$ is a prime of $A^{\mathrm{\prime }}$ lying over ${\mathfrak{m}}_V$, that is ${\mathfrak{m}}^{\mathrm{\prime }}\cap V={\mathfrak{m}}_V$; then $A_{{\mathfrak{m}}^{\mathrm{\prime }}}^{\mathrm{\prime }}$ dominates $V$, so $A_{{\mathfrak{m}}^{\mathrm{\prime }}}^{\mathrm{\prime }}=V$ by maximality of $V$, and since $x\in A^{\mathrm{\prime }}$ this would give $x\in V$, a contradiction. Hence no prime of $A^{\mathrm{\prime }}$ lies over ${\mathfrak{m}}_V$. A prime ${\mathfrak{m}}^{\mathrm{\prime }}$ of $A^{\mathrm{\prime }}$ contains ${\mathfrak{m}}_V{A}^{\mathrm{\prime }}$ exactly when it lies over ${\mathfrak{m}}_V$: one implication is immediate, and conversely ${\mathfrak{m}}_V{A}^{\mathrm{\prime }}\subseteq {\mathfrak{m}}^{\mathrm{\prime }}$ gives ${\mathfrak{m}}_V\subseteq {\mathfrak{m}}^{\mathrm{\prime }}\cap V$, hence ${\mathfrak{m}}^{\mathrm{\prime }}\cap V={\mathfrak{m}}_V$ because ${\mathfrak{m}}_V$ is maximal in $V$. So the set of primes of $A^{\mathrm{\prime }}$ containing ${\mathfrak{m}}_V{A}^{\mathrm{\prime }}$ is empty. If ${\mathfrak{m}}_V{A}^{\mathrm{\prime }}$ were a proper ideal of the nonzero ring $A^{\mathrm{\prime }}$, then [F7] would produce a maximal ideal of $A^{\mathrm{\prime }}$ containing it, which is prime by [F8] and would lie over ${\mathfrak{m}}_V$, a contradiction. Hence ${\mathfrak{m}}_V{A}^{\mathrm{\prime }}=A^{\mathrm{\prime }}$.

Thus $1={\sum }_{i=0}^d{t}_i{x}^i$ for some $t_i\in {\mathfrak{m}}_V$ and $d\ge 0$; note $t_0\in {\mathfrak{m}}_V$ so $1-t_0$ is a unit of $V$. Multiplying the relation by $x^{-d}$ gives $(1-t_0)x^{-d}=t_1{x}^{-(d-1)}+\cdots +t_d$, hence, after dividing by the unit $1-t_0$, a monic polynomial equation for $x^{-1}$ with coefficients in $V$. So $x^{-1}$ is integral over $V$ by [F6], and step 4.3 gives $x^{-1}\in V$.

Since $x\in K^{\times }$ with $x\mathrm{\notin }V$ was arbitrary, step 6.1 shows that for every $x\in K^{\times }$ at least one of $x$, $x^{-1}$ lies in $V$; this is the criterion [F1], so $V$ is a valuation ring of $K$ with $\mathrm{Frac}(V)=K$ by step 5.1, and it dominates $A$ because it dominates the intermediate rings down to $A$.