Proposition 2

*Let K be a field, B a valuation ring of K and h, the place & K associated with B (with values in $\kappa(B)$). Then the mapping $A \mapsto h_B(A)$ defines a bijection of the set $\mathscr{A}$ of valuation rings of K contained in B onto the set $\mathscr{A}'$ of valuation rings of $\kappa(B)$.

If $A \in \mathscr{A}$, then $h_B(A) \in \mathscr{A}'$: for if $x' = h_B(x)$ (where $x \in B$) is an element of $\kappa(B) - h_B(A)$, then $x \notin A$, hence $x^{-1} \in A$ and $h_B(x)^{-1} \in h_B(A)$. On the other hand, for $A \in \mathscr{A}$, $A \supset m(B)$ (Proposition 1 (b)) and hence the mapping,

$A \mapsto h_B(A)$ is injective. Finally, let $A' \in \mathscr{A}'$ and $A = h_B(A') \subset B$; we shall show, which will complete the proof, that $A \in \mathscr{A}$; if $x \in K - A$, then either $x \notin B$, or $x \in B$; if $x \notin B$, then $x^{-1} \in m(B) \subset A$; if $x \in B$, then $h_B(x) \in \kappa(B)$ and $h_B(x) \notin A$, hence $h_B(x^{-1}) \in A$ and we conclude again that $x^{-1} \in A$; hence $A \in \mathscr{A}$.