Proposition 6

(i) If M_1 and M are reflexive lattices of V, so is M, n M. (ii) If W is a vector subspace of V and M is a reflexive lattice of V, M n W is a reflexive lattice of W. (iii) Let V, W be two vector spaces of finite rank over K and M (resp. N) a lattice of V (resp. W). If N is reflexive, the lattice N: M of Hom_K(V, W) (no. 1, Proposition 3) is reflexive.

(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.