Proposition 6

Let K be a field, $v$ a valuation on K, A its ring, L an algebraic extension of K and $A'$ the integral closure of A in L. Let 23 be the set of valuation rings on L which extend $v$ and $\mathfrak{M}'$ the set of maximal ideals of $A'$. Then the mapping $V \mapsto m(V) \cap A'$ is a bijection of 23 onto $\mathfrak{M}'$ and $m' \mapsto A'_{m'}$ is the inverse bijection.

Every maximal ideal $m'$ of $A'$ is such that $m' \cap A$ is the maximal ideal of A (Chapter V, § 2, no. 1, Proposition 1) and $A'_{m'}$ is dominated by a valuation ring V of L (which is therefore the ring of a valuation on L extending $v$) ($§ 1$, no. 2, Corollary to Theorem 2). The field L is the union of a directed family of sub-extensions K, of L which are finite over K and it will suffice, in order to see that $V = A'_{m'}$, to prove that $V \cap K = A'_{m'} \cap K$, for all $\alpha$. Now, if we write $A'_\alpha = A' \cap K_\alpha$, $A'_\alpha$ is the integral closure of A in K, and hence is the intersection of the rings of the valuations on K, which extend $v$ and these rings $V_\alpha$ are finite in number and are the local rings $(A'_\alpha)_{i|m_\alpha}$ of $A'_\alpha$ ($1 \leq i \leq n$), where the $m'_{i\alpha}$ are the distinct maximal ideals of $A'_\alpha$ (no. 3, Remark); but $m' \cap A'_\alpha$ is one of the $m'_{i\alpha}$ and $V \cap K_\alpha$ is therefore equal to the corresponding local ring $(A'_\alpha)_{m'_{i\alpha}} \subset A'_{m'}$, which completes the proof that $V = A'_{m'}$. Conversely, if $V \in \mathfrak{V}$, then $A' \subset V$ (§ 3, no. 3, Proposition 6) and, if $m' = m(V) \cap A'$, then $m' \cap A = m$, hence $m'$ is a maximal ideal of $A'$ (Chapter V, § 2, no. 1, Proposition 1) and the above argument shows that $V = A'_{m'}$.