Lemma 7

Let $x_i$ ($1 \leq i \leq m$) be linearly independent elements of $M$ (over $A$ or $K$, which amounts to the same) and let $L$ be the sub-A-module of $M$ generated by the $x_i$. Then, for almost all $p \in P$, the $x_i(p) \in M(p)$ are linearly independent over $k(p)$; for them to be linearly independent over $k(p)$ for all $p \in P$, it is necessary and sufficient that $M/L$ be torsion-free.

Let $x_1, \ldots, x_n$ be elements of $M$ which, with $x_1, \ldots, x_n$, form a basis of $V$ and let $N$ be the free sub-A-module of $M$ generated by the $x_i$ ($1 \leq i \leq n$). It follows from no. 3, Theorem 3 that $N = M$, for almost all $p \in P$; as $x_1(p), \ldots, x_n(p)$ form a basis of $N(p)$ over $k(p)$, this establishes the first assertion.

If $M/L$ is torsion-free, so is $(M/L)_p = M_p/L_p$ for all $p \in P$ (no. 1, Example 6) and, as $A_p$ is a principal ideal domain, $M_p/L_p$ is free. Therefore $M$, is the direct sum of $L$, and a free $A$-module $E$ of rank $n - m$; hence $M(p)$ is the direct sum of $L(p)$ and the vector $k(p)$-space $E/pE$ of rank $n - m$; therefore $L(p)$ is of rank $m$ and, as it is generated by the $x_i(p)$ ($1 \leq i \leq m$), the latter are linearly independent.

Conversely, suppose that the $x_i(p)$ ($1 \leq i \leq m$) are linearly independent over $k(p)$ for all $p \in P$. Then $L$, is a direct factor of $M$, for all $p$ (Chapter II, § 3, no. 2, Corollary 1 to Proposition 5) and therefore $M_p/L_p = (M/L)_p$ is torsion-free for all $p \in P$. We conclude that $P \cap \operatorname{Ass}(M/L) = \varnothing$ by Chapter IV, § 1, no. 2, Corollary to Proposition 5. But as L is reflexive, it follows from no. 2, Proposition 7 (i) that the only prime ideal which can belong to $\operatorname{Ass}(M/L)$ is the ideal (0); hence $M/L$ is torsion-free.