Lemma 4

Let W be a vector space of finite rank over L and let R be a lattice of W with respect to B. Then, for all $p \in P(A)$, $(R_{[A]})_p = \bigcap_{\mathfrak{p}/p} R_{\mathfrak{p}}$.

If $S = A - p$, the prime ideals of the ring $S^{-1}B$ are generated by the prime ideals of B not meeting S, in other words the ideals $\mathfrak{P}_i$ ($1 \leq i \leq m$) lying above p and the ideal (0); this shows that $S^{-1}B$ is a semi-local ring whose maximal ideals are the m, $= \mathfrak{P}_i(S^{-1}B)$ for $1 \leq i \leq m$; moreover the local ring $(S^{-1}B)_{\mathfrak{P}_i}$ is isomorphic to $B_{\mathfrak{p}_i}$ (Chapter 11, § 2, no. 5, Proposition 11) and hence is a discrete valuation ring. The ring $S^{-1}B$ is therefore a Dedekind domain (§ 2, no. 2, Theorem 1 (f)) and, as it is semi-local, it is a principal ideal domain (§ 2, no. 2, Proposition 1). This being so, $(R_{[A]})_p$ is equal to $S^{-1}R$ considered as an A,-module; by the above, $S^{-1}R$ is a free lattice of W with respect to $S^{-1}B$ and Theorem 2 of no. 2 may therefore be applied to it, giving $S^{-1}R = \bigcap_i (S^{-1}R)_{m_i}$: but $(S^{-1}R)_{m_i} = R_{\mathfrak{p}_i}$, which proves the lemma.

Returning to the proof of Proposition 19, by Lemma 4,

and the conclusion follows from no. 2, Theorem 2.