The conditions are obviously necessary, for a free sub-A-module of V with the same rank as V generates V. Conversely, if KM = V, M contains a basis $(a_i)_{1 \leq i \leq n}$ of V over K and hence it contains the free sub-A-module L, generated by the $a_i$; on the other hand, if $M \subset M_1$, where M, is a sub-A-module of V generated by a finite number of elements b, and $(e_i)_{1 \leq i \leq n}$ is a basis of V over K, there exists an element s $\neq 0$ of A such that each of the b, is a linear combination of the $s^{-1}e_i$ with coefficients in A; if L, is the free sub-A-modules of V generated by the $s^{-1}e_i$, then $M \subset L_2$.

Remark (1). Recall that for every sub-A-module M of V the canonical mapping $M \otimes_A K \to V$ is injective and has image KM (Algebra, Chapter II, § 7, no. 10, Proposition 26); to say that KM = V means therefore that this mapping is bijective.

If $L_1, L_2$ are two free lattices of V such that $L, \subset M \subset L,$, the relations $xM \subset M_1 \subset yM$ imply $xL_1 \subset M, \subset yL_2$ and $xL_1$ and $yL_2$ are free lattices; conversely, if M, is a lattice and $(e_i)_{1 \leq i \leq n}$ a basis of L, over A, the relation KM, = V implies the existence of $x = a/s \in K^*$ (where a and s are non-zero elements of A) such that $xe_i \in M,$ for all $i$, whence $xM \subset xL_2 \subset M_1$ and a fortiori $aM \subset M,$; exchanging the roles of M and M, it can be similarly shown that there exists $b \neq 0$ in A such that $bM_1 \subset M$.

(ii) If W is a vector subspace of V and M is a lattice of V, $M \cap W$ is a lattice of W.

(iii) Let $V, V_1, \ldots, V_k$ be vector spaces of finite rank over $K$ and let

be a multilinear mapping whose image generates $V$. If $M_i$ is a lattice of $V_i$ for $1 \leq i \leq k$, the sub-A-module of $V$ generated by $f$ ($(M, x \ldots x M_k)$) is a lattice of $V$.

(iv) Let $V$ and $W$ be two vector spaces of finite rank over $K$, $M$ a lattice of $V$ and $N$ a lattice of $W$. The sub-A-module $N$: $M$ of $\mathrm{Hom}_K(V, W)$, consisting of the K-linear mappings $f$ such that $f(M) \subset N$, is a lattice of $\mathrm{Hom}_K(V, W)$.

(i) By virtue of Proposition 2, there exist non-zero $a$ and $b$ in $A$ such that $aM_1 \subset M_2 \subset b^{-1}M_1$; we conclude that $M_1 \cap M_2$ and $M_1 + M_2$ lie between $aM_1$ and $b^{-1}M_1$ and are therefore lattices by virtue of Proposition 2.

(ii) Let $S$ be a complement of $W$ in $V$, $L_w$ a free lattice of $W$ and $L_s$ a free lattice of $S$, so that $L = L_w \oplus L_s$ is a free lattice of $V$. Then there exist $x, y$ in $K^*$ such that $xL \subset M \subset yL$. We deduce that $xL_w \subset M \cap W \subset yL_w$, which shows that $M \cap W$ is a lattice of $W$ (Proposition 2).

(iii) As $KM_i = V$, clearly by linearity $f$ ($M_1 x \ldots x M_n$) generates the vector K-space $V$; on the other hand, for all $i$, there exists a finitely generated sub-A-module $N_i$ of $V$, such that $M_i \subset N_i$; the sub-A-module $N$ of $V$ generated by $f(N_1 x \ldots x N_n)$ is finitely generated and contains $M$ and hence $M$ is a lattice of $V$ (Proposition 1).

(iv) Let $P$ (resp. $Q$) be a free lattice of $V$ (resp. $W$) containing $M$ (resp. contained in $N$); obviously $N : M \supset Q : P$. Now it is immediate that $Q : P$ is isomorphic to $\mathrm{Hom}_A(P, Q)$, hence is a free A-module of rank $(\mathrm{rg}_A P)(\mathrm{rg}_A Q)$ (Algebra, Chapter II, § 1, no. 6, Corollary 1 to Proposition 6) and therefore a lattice of $\mathrm{Hom}_K(V, W)$. Similarly, if $P'$ (resp. $Q'$) is a free lattice of $V$ (resp. $W$) contained in $M$ (resp. containing $N$), then $Q' : P' \supset N : M$ and $Q' : P'$ is a lattice of $\mathrm{Hom}_K(V, W)$; whence the conclusion.

Remark (2) Proposition 3 (i) shows that the set $R(V)$ of lattices of $V$ is lattice-ordered with respect to inclusion; moreover, if $M$ is a fixed lattice of $V$, the $xM$, where $x$ runs through $K^*$, form a subset of $R(V)$ which is both coinitial and cofinal (Set Theory, Chapter III, § 1, no. 7).

(3) In the notation of Proposition 3 (iv), the canonical mapping

which maps every K-linear mapping $f \in N : M$ to the A-linear mapping from $M$ to $N$ which has the same graph as $f|_M$, is bijective, for every A-linear mapping $g : M \to N$ can be imbedded in a K-linear mapping

and it has been seen that $M \otimes_A K$ and $N \otimes_A K$ are respectively identified with $V$ and $W$.

In particular, if we take $W = K$, $N = A$, $\operatorname{Hom}_K(V, W)$ is just the dual vector K-space $V^*$ of $V$ and $A$: $M$ is identified with the dual A-module $M^*$ of $M$; we shall henceforth make this identification and we shall say that $M^*$ is the dual lattice of $M$: it is therefore the set of $x^* \in V^*$ such that $\langle x, x^* \rangle \in A$ for all $x \in M$.

(4) Let $S$ be a (not necessarily associative) K-algebra of finite rank with a unit element; then the bilinear mapping $(x, y) \mapsto xy$ of $S \times S$ to $S$ is (left and right) non-degenerate. If $M$ and $N$ are lattices of $S$ with respect to $A$, so are $M.N$ (Proposition 3 (iii)) and the set of $x \in S$ such that $xM \subset N$ (Corollary to Proposition 3). Note that there exists a sub-A-algebra of $S$ containing the unit element of $S$ which is a lattice of $S$; for consider a basis $(e_i)_{1 \leq i \leq n}$ of $S$ such that $e_1$ is the unit element of $S$ and let $e_i e_j = \sum_k c_{ijk} e_k$ be the multiplication table of $S$ ($1 \leq i \leq n, 1 \leq j \leq n$), so that $c_{1jk} = \delta_{jk}, c_{i1k} = \delta_{ik}$ (Kronecker symbols). Let $s \in A$ be a non-zero and such that $c'_{ijk} = s \cdot c_{ijk} \in A$ for all triplets of indices $(i, j, k)$; if we write $e'_i = s^{-1} e_i$ for $i \geq 2$, then

for $i \geq 2$ and $j \geq 2$; the lattice of $S$ with basis $e_1$ and the $e'_i$ ($2 \leq i \leq n$) is a sub-A-algebra of $S$ with unit element $e_1$.

(5) Let $V$ be a finite-dimensional vector space over $K$ and $f$ a non-degenerate bilinear form on $V$. If $M$ is a lattice of $V$, it follows from the Corollary to Proposition 3 that the set $M^*_f$ of $x \in V$ such that $f(x, y) \in A$ for ally $\in M$ is also a lattice of $V$; if $s_f: V \to V^*$ is the linear mapping left associated with $f$ (which is bijective), $s_f(M^*_f)$ is just the dual lattice $M^*$ of $M$.

(i) For every lattice $M$ of $V$ with respect to $A$, the image $BM$ of $M_{(B)} = M \otimes_A B$ in $V_{(L)} = V \otimes_K L$ is a lattice of $V_{(L)}$ with respect to $B$.

(ii) Suppose further that $B$ is a flat $A$-module. Then the canonical mapping $M_{(B)} \to BM$ is bijective. If further $B$ is faithfully flat, the mapping which maps every lattice $M$ of $V$ with respect to $A$ to the lattice $BM$ of $V_{(L)}$ with respect to $B$ is injective.

(i) As $KM = V$, clearly $L . (BM) = V_{(L)}$; on the other hand $M$ is contained in a finitely generated sub-$A$-module $M_1$ of $V$ and hence $BM$ is contained in $BM_1$ which is a finitely generated $B$-module; whence assertion (i) (Proposition 1).

(ii) $V_{(L)} = V \otimes_K L = V \otimes_A L$ (Chapter II, § 2, no. 7, Proposition 18) and, as $L$ is a flat $B$-module, it is also a flat $A$-module (Chapter I, § 2, no. 7, Corollary 3 to Proposition 8). As $B$ is a flat $A$-module, the canonical mapping $M \otimes_A B \to V \otimes_A B$ is injective; on the other hand, since $V$ is a free $K$-module and $K$ a flat $A$-module, $V$ is a flat $A$-module (Chapter I, § 2, no. 7, Corollary 3 to Proposition 8) and hence the canonical mapping $V \otimes_A B \to V \otimes_A L$ is injective, which establishes the first assertion. To see also that the relation $BM_1 = BM_2$ implies $M_1 = M_2$ for two lattices $M_1, M_2$ of $V$ with respect to $A$ when $B$ is a faithfully flat $A$-module, note first that $BM_1 \cap BM_2 = B(M_1 \cap M_2)$ (Chapter I, § 2, no. 6, Proposition 6); we may therefore confine our attention to the case where $M_1 \subset M$, and our assertion then follows from Chapter I, § 3, no. 1, Proposition 3 applied to the canonical injection $M_1 \to M_2$.

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:

(1)

in $\operatorname{Hom}_K(V, W)$.

Clearly the left hand side of (1) is contained in the right hand side. Conversely, let $f \in S^{-1}N : S^{-1}M$ and let $(x_i)_{1 \leq i \leq n}$ be a system of generators of $M$. There exists $s \in S$ such that $f(x_i) \in s^{-1}N$ for all $i$ and hence $sf \in N : M$, which proves the proposition.

Clearly $M^*$ is contained in each of the $M_p^*$. Conversely, suppose that $x^* \in \bigcap M_p^*$; if $x \in M$, then $\langle x, x^* \rangle \in \bigcap_{p \in P} A_p$ and, as $A = \bigcap_{p \in P} A_p$ (§ 1, no. 6, Theorem 4), $x^* \in M^*$.

Theorem 1 applied to $M^*$ shows that $M^{**} = \bigcap_{p \in P} M_p^{**}$. But as $A$, is a principal ideal domain ($§ 1$, no. 6, Theorem 4), $M$, is a finitely generated free $A$-module and hence $M_p^{**}$ is canonically identified with $M$, (Algebra, Chapter II, $§ 2$, no. 7, Proposition 14), whence the corollary.

For any lattice $M$ with respect to $A$, the canonical mapping $c_M : M \to M^{**}$ (Algebra, Chapter 11, $§ 2$, no. 7) identifies an element $x \in M$ with itself, for $x$ is the unique element $y$ of $V = V^{**}$ such that $\langle x, x^* \rangle = \langle y, x^* \rangle$ for all $x^* \in M^*$, since $M^*$ generates $V^*$. We shall say that $M$ is reflexive if $M^{**} = M$ (loc. cit.). As we have above $M^* = (M^*)^{**}$, it is seen that the dual of any lattice $M$ is always reflexive.

Remark (1) Let $M$ be a finitely generated $A$-module; it is immediate that the dual $M^*$ of $M$, identified with a sub-$A$-module of $\operatorname{Hom}_A(M, K)$ is a lattice of the vector $K$-space $\operatorname{Hom}_A(M, K)$; in particular, every finitely generated reflexive $A$-module is isomorphic to a lattice of a suitable vector $K$-space.

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).

The condition is obviously necessary and, if it is fulfilled, then

As M ⊂ M** = \bigcap_p M_p, certainly M ⊂ N.

Examples (1) Everyfree lattice is reflexive. (2) Take V = K. For a fractional ideal a of K to be a reflexive lattice, it is necessary and sufficient that it be a divisorial ideal by virtue of criterion (b) of Theorem 2 and § 1, no. 4, Propositions 5 and 7. (3) Let M be a lattice with respect to A; if S is a multiplicative subset of A not containing 0, Proposition 5 of no. 1 shows that S^{-1}(M*) = (S^{-1}M)*; if M is reflexive, S^{-1}M is therefore a reflexive lattice with respect to S^{-1}A.

(i) (M, n M_2)_p = (M_1)_p n (M_2)_p for all p ∈ P (Chapter II, § 2, no. 4, Theorem 1). If M_1 = \bigcap_{p \in P} (M_1)_p and M_2 = \bigcap_{p \in P} (M_2)_p, then

whence the conclusion by virtue of Theorem 2. (ii) Similarly (M n W)_p = M_p n W_p = M_p n W, whence

which proves (ii). (iii) As M is finitely generated, it follows from no. 1, Proposition 5 that (N: M)_p = N_p M_p; moreover, the relation N = \bigcap_{p \in P} N_p implies:

For if $f \in \bigcap_{p \in P} (N_p : M_p)$ and $x \in M$, then $f(x) \in \bigcap_{p \in P} N_p = N$, whence $f \in N : M$; this shows that $N : M$ is reflexive.

Remarks (2) If $M_1$ and $M_2$ are reflexive lattices of $V$, the lattice $M_1 + M_2$ is not necessarily reflexive (cf. § 1, Exercise 2). (3) If $M$ is a finitely generated $A$-module and $T$ its torsion submodule, the dual $M^*$ of $M$ is the same as the dual of $M/T$, since, for every linear form $f$ on $M$, the image $f(T)$ is a torsion submodule of $A$ and hence zero. As $M/T$ is isomorphic to a lattice of a vector space over $K$, it is seen that the dual of every finitely generated $A$-module is reflexive.

As $A$ is Noetherian, $M$ is also finitely generated; if we write $V = M_{(K)}$, $W = N_{(K)}$, $M$ (resp. $N$) is canonically identified with a lattice of $V$ (resp. $W$) (no. 1, Proposition 1). Consider the two exact sequences:

(i) We deduce (Chapter IV, § 1, no. 1, Proposition 3) that: If $M$ is reflexive, then $\operatorname{Ass}(V/M) \subset P$ (Theorem 2); on the other hand, clearly $\operatorname{Ass}(W/V)$ is, either empty, or reduced to $\{0\}$; whence (i). (ii) Similarly: The hypotheses therefore imply that $\operatorname{Ass}(V/M) \subset P \cup \{\{0\}\}$. But $V/M$ is a torsion $A$-module and hence $\{0\} \notin \operatorname{Ass}(V/M)$; Theorem 2 then shows that $M$ is reflexive.

€or x \in M, x^* \in M^*. As M is a quotient of a finitely generated free R-module L, M* is isomorphic to a sub-R-module of the dual L* and L* is free and finitely generated; since $\mathbf{R}$ is Noetherian, M* is therefore also a finitely generated R-module, whence an isomorphism $\omega_{M^*}: (M^{**})_{(S)} \to ((M^*)_{(S)})^*$ such that

for $x^* \in M$ and $x^{**} \in M^{**}$. On the other hand, there is an isomorphism $^t\omega_M: (M_{(S)})^{**} \to ((M^*)_{(S)})^*$, whence by composition a canonical isomorphism:

such that, in the above notation:

(1)

We consider now the canonical homomorphism $c_M: M \to M^{**}$ and show that the composite homomorphism :

(2)

is just the canonical homomorphism $c_{M_{(S)}}$. This follows immediately from (1) which gives the relations :

and from the fact that the elements $\omega_M(x^* \otimes 1)$ generate $(M_{(S)})^*$. This being so, the hypothesis that M is reflexive means that $c_M$ is bijective, hence so is $c_M \otimes 1$ and therefore $\psi = c_{M_{(S)}}$ is bijective, which shows the proposition.

(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: