Exercise 8, § 4

Let $A$ be a local integral domain of dimension 1 (Exercise 7) and $K$ its field of fractions. Show that for $A$ to be the intersection of valuation rings (of $K$) of height 1, it is necessary and sufficient that $A$ be completely integrally closed. Prove successively that: (a) A subring of $K$ containing $A$ and the inverse of an element $\neq 0$ of the maximal ideal $m$ of $A$ is equal to $K$ (use the fact that every ideal of $A$ distinct from 0 and $A$ is $m$-primary). (b) If $B \supseteq A$ is a subring of $K$ distinct from $K$, there exists a valuation ring $V$ of height 1 such that $B \subset V$ (use (a)). (c) Let $z \in K - A$; show that $zm \subset m$ is impossible arguing as in Exercise 7 (b). Deduce that there exists a valuation ring $V$ of height 1 such that $A \subset V$ and $z \notin V$. (If $zm \subset A$, use (a); otherwise, there is an $x \in m$ such that

observe that $z \notin A[z^{-1}]$ and use (b) to show the existence of $V$.)