Corollary

Suppose that $A$ is a discrete valuation ring. Let $\hat{A}$ be its completion and let $\hat{K}$ be the field of fractions of $\hat{A}$ (Chapter VI, § 5, no. 3). The mapping $\phi$, which maps every lattice $M$ of $V$ to the lattice $\hat{A}M$ of $\hat{V} = V \otimes_K \hat{K}$ with respect to $\hat{A}$, is bijective and its inverse maps every lattice $M'$ of $\hat{V}$ with respect to $\hat{A}$ to its intersection $M' \cap V$ ($V$ being canonically identified with a vector sub-K-space of $\hat{V}$).

If $L$ is a free lattice of $V$, the lattices $aL$ (for $a \in A, a \neq 0$) form a fundamental system of neighbourhoods of 0 for a topology $\mathscr{T}$ on $V$ (compatible with its $A$-module structure), which (when a basis of $L$ over $A$ is taken) is identified with the product topology on $K^n$); by virtue of Proposition 2, a fundamental system of neighbourhoods of 0 for $\mathscr{T}$ also consists of all the lattices of $V$ with respect to $A$; clearly $\hat{V}$ is the completion of $V$ with respect to $\mathscr{T}$. Moreover, if $m$ is the maximal ideal of $A$, the topology $\mathscr{T}$ induces on every lattice $M$ of $V$ with respect to $A$ the $m$-adic topology since $M$ is a finitely generated $A$-module (Chapter III, § 3, no. 2, Theorem 2) and $\hat{A}M$ is the completion of $M$ with respect to this topology (Chapter III, § 2, no. 12, Proposition 16); moreover, as $M$ is open (and therefore closed) in $V$, $\hat{A}M \cap V = M$, which proves again the fact that $\dagger$ is injective (which follows directly from Proposition 4, (ii), since $A$ is a faithfully flat $A$-module). Finally, if $M'$ is a lattice of $\hat{V}$ with respect to $\hat{A}$, $M = M' \cap V$ is a lattice of $V$ with respect to $A$, for every element of $\hat{A}$ is the product of an element of $A$ and an invertible element of $\hat{A}$ and hence it follows from Proposition 2 that there exist $a, b$ in $A - \{0\}$ such that $a \hat{A}L \subset M' \subset b \hat{A}L$, whence $aL \subset M' \cap V \subset bL$. Moreover $M'$ is open in $V$ and, as $V$ is dense in $\hat{V}$, $M'$ is the completion of $M' \cap V = M$; this proves that $\phi$ is surjective, whence the corollary.

Example (6) Let $S$ be a multiplicative subset of $A$ not containing 0; we apply Proposition 4 to $B = S^{-1}A$; then $L = K$, $BM = S^{-1}M$; hence $S^{-1}M$ is a lattice of $V$ with respect to $S^{-1}A$. Moreover: