Lemma 1

Let $p$ and $p'$ be two prime ideals of $A$ such that (0) is the only prime ideal of $A$ contained in $p \cap p'$. Then, for every sub-A-module $E$ of $V$, $(E_p)_{p'} = K.E.$

Let $S$ be the multiplicative subset $(A - p)(A - p')$ of $A$; by Chapter II, $§ 2$, no. 3, Proposition 7, $(E_p)_{p'} = S^{-1}E$. Further, $A \subset S^{-1}A \subset K$; the prime ideals of $S^{-1}A$ correspond to the prime ideals $q$ of $A$ such that $q \cap S = \varnothing$ (Chapter II, $§ 2$, no. 5, Proposition 11) and by hypothesis (0) is the only prime ideal of $A$ not meeting $S$; hence $S^{-1}A = K$ and $S^{-1}E = K.E.$

We now return to the proof of (ii). If $p \in P$ is distinct from the $p_i$ ($1 \leq i \leq h$), Lemma 1 applied to $N(p_i)$ gives $((N(p_i))_{p_i})_p = (K.N(p_i))_p = V$, since the $p_i$ and $p$ are of height 1. Then

(Chapter II, $§ 2$, no. 4). On the other hand, if $p$ is equal to $p_i$ ($1 \leq i \leq h$), then $(N(p_i))_{p_j} = V$ for $i \neq j$ by the argument as above and $(N(p_i))_{p_i} = N(p_i)$, whence

We have therefore proved that $Q_p = N(p)$ for all $p \in P$. Then

is reflexive and satisfies the relations $N_p = Q_p = N(p)$ for all $p \in P$; the uniqueness property follows immediately from Theorem 2 of no. 2.