Theorem 1

If $M$ is a lattice of $V$, then $M^* = \bigcap_{p \in P} M_p^*$.

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.