Theorem 3

Let $A$ be a subring of a field $K$. The integral closure $A'$ of $A$ in $K$ is the intersection of the valuation rings of $K$ which contain $A$; if $A$ is local, $A'$ is the intersection of the valuation rings of $K$ which dominate $A$.

Let $x$ be an element of $A'$ and $V$ a valuation ring of $K$ containing $A$; as $x$ is integral over $V$, there exists a prime ideal $p'$ of $V[x]$ such that $p' \cap V = m(V)$ (Chapter V, § 2, no. 1, Theorem 1); clearly then the local ring $(V[x])_{p'}$ dominates $V$ and hence is equal to it; whence $x \in V$. Conversely let $y$ be an element of $K$ which is not integral over $A$; then there exists a maximal ideal 2.2 of $A[y^{-1}]$ which contains $y^{-1}$ (no. 2, Lemma 1); there exists also a valuation ring $V$ of $K$ which dominates $(A[y^{-1}])_m$ (no. 2, Corollary to Theorem 2); as $y^{-1} \in m(V)$, $y \notin V$. Further $2.2 \cap A$ is a maximal ideal of $A$ (no. 2, Lemma 1); hence, if $A$ is local, $2.2 \cap A = m(A)$ and $V$ dominates $A$.