Proposition 1

Let $K$ be a field and $A$ a valuation ring of $K$. Then: (a) *Every ring $B$ such that $A \subset B \subset K$ is a valuation ring of $K$; (b) *The maximal ideal $m(B)$ of such a ring is contained in $A$ and it is a prime ideal of $A$; (c) *The mapping $\mathfrak{p} \mapsto A$, is a decreasing bijection of the set of prime ideals of $A$ onto the set of rings $B$ such that $A \subset B \subset K$; its inverse bijection is the mapping $B \mapsto m(B)$.

If $B$ is a ring such that $A \subset B \subset K$ and $x \in K - B$, then $x \in K - A$, whence $x^{-1} \in m(A) \subset B$, which proves both that $B$ is a valuation ring of $K$ and that $m(B) \subset m(A)$; as $m(B) = m(B) \cap A$ is a prime ideal of $A$, we have shown (a) and (b). Moreover, $A_{m(B)} \subset B$; conversely, if $x \in B - A$, then $x^{-1} \in A$ and $x^{-1} \notin m(B)$ and hence $x \in A_{m(B)}$; thus $A_{m(B)} = B$. Finally let $\mathfrak{p}$ be a prime ideal of $A$; we write $B = A_{\mathfrak{p}}$; then $m(B) \cap A = \mathfrak{p}$. (Chapter II, § 2, no. 5, Proposition 11) and $m(B) \subset A$ by (b); hence $m(B) = \mathfrak{p}$, which shows that the mappings $\mathfrak{p} \mapsto A$, and $B \mapsto m(B)$ of the statement are inverse bijections.