Theorem 2

If $M$ is a lattice of $V$, the following conditions are equivalent: (a) $M$ is reflexive. (b) $M = \bigcap_{p \in P} M_p$. (c) $\operatorname{Ass}(V/M) \subset P$.

The equivalence of (a) and (b) follows from the Corollary to Theorem 1. If (b) holds, $V/M$ is canonically identified with a sub-$A$-module of the product $\prod_{p \in P} (V/M_p)$; but in fact, it is contained in the direct sum $\bigoplus_{p \in P} (V/M_p)$: for if $L \subset M$ is a free lattice and $(e_i)_{1 \leq i \leq n}$ a basis of $L$, each of the coordinates $x_i$ of a point $x \in V$ with respect to $(e_i)$ belongs to $A$, except for a finite number of values of $p \in P$. The relation $V/M \subset \bigoplus_{p \in P} (V/M_p)$ then implies:

As $V/M_p$ is an $A_p$-module, an element of $A - p$ cannot annihilate an element $\neq 0$ of $V/M_p$, since the elements of $A - p$ are invertible in $A$; the elements of $\operatorname{Ass}(V/M_p)$ are therefore contained in $p$ and are $\neq 0$, since $V/M_p$ is a torsion $A$-module; as $p$ is of height 1, necessarily $\operatorname{Ass}(V/M_p) = \{ p \}$ if $V/M_p \neq (0)$ and $\operatorname{Ass}(V/M_p) = \varnothing$ if $V/M_p = (0)$; hence $\operatorname{Ass}(V/M) \subset P$.

Finally, if condition (c) holds, then

On the other hand, if $p \in P$, then it has been seen in the proof of the Corollary to Theorem 1 that $M_p^{**} = M_p$, whence $p \notin \operatorname{Ass}(M^{**}/M)$ (Chapter IV, $§ 1$, no. 3, Corollary 1 to Proposition 7). We conclude that Ass(M/M) = ∅, whence M = M (Chapter IV, § 1, no. 1, Corollary 1 to Proposition 2).