Theorem 3

Let V be a vector space of finite rank over K and M a lattice of V with respect to A.

(i) Let N be a lattice of V with respect to A; then, for every prime ideal p of A, $N_p$ is a lattice of V with respect to A, and, for almost all $p \in P, N_p = M_p$.

(ii) Conversely, suppose given for all $p \in P$ a lattice $N(p)$ of V with respect to A, such that $N(p) = M_p$ for almost all $p \in P$. Then $N = \bigcap N(p)$ is a reflexive lattice of A with respect to A and it is the only reflexive lattice $N'$ of V with respect to A such that $N'_p = N(p)$ for all $p \in P$.

(i) The first assertion follows from no. 1, Proposition 4. Moreover, there exist $x, y$ in $\mathbf{K}^*$ such that $xN \subset M \subset yN$ (no. 1, Proposition 2); we know that, for almost all $p \in P, v_p(x) = v_p(y) = 0$ ($§ 1$, no. 6, Theorem 4), which shows that $x$ and $y$ are invertible in $A$, and hence $M_p = N_p$.

(ii) We may replace $M$ by $x^{-1}M$ where $x \neq 0$ in $A$ and assume that $N(p) \subset M$, for all $p \in P$. Let $p_1, \ldots, p_h$ be the elements of $P$ such that $N(p) = M$, for $p$ distinct from the $p_i$ ($1 \leq i \leq h$); we write:

As each of the $N(p_i)$ contain a free lattice with respect to $A_{p_i}$, it contains a fortiori a lattice of $V$ with respect to $A$, hence $Q$ contains a lattice of $V$ with respect to $A$ (no. 1, Proposition 3) and, as $Q$ is contained in $M$, $Q$ is a lattice with respect to $A$. To prove that $Q_p = N(p)$ for all $p \in P$, we shall use the following lemma: