Lemma 8

Suppose that the rank $n$ of $M$ is $\geqslant 2$; then there exists an element $x \neq 0$ of $M$ such that $M/Ax$ is torsion-free.

Let $y \neq 0$ be an element of $M$. By Lemma 7, the set $Y$ of $p \in P$ such that $y(p) = 0$ is finite. If $Y = \varnothing$, it follows from Lemma 7, applied to the sequence $(x_i)$ consisting of the single element $y$, that $M/Ay$ is torsion-free. Suppose therefore that $Y \# @$ and write $S = \bigcap_{p \in Y} (A - p)$; we know (no. 4, Lemma 2) that $S^{-1}A$ is a semi-local principal ideal domain whose maximal ideals are the $pS^{-1}A$, where $p \in Y$, the corresponding local rings being the $A_p$. Therefore

whence for all $p \in Y$. By Chapter II, § 1, no. 2, Proposition 6, there exists an element $z/s \in S^{-1}M$ ($z \in M, s \in S$) whose canonical images in the $M(p)$ for $p \in Y$ are all $\neq 0$. By definition of $S$ therefore $z(p) \neq 0$ for all $p \in Y$. We may further assume that $y$ and $z$ are linearly independent over $K$. For in the opposite case consider an element $t \in M$ which is linearly independent of $y$ (such exists since $n \geqslant 2$); on the other hand take an element $a \neq 0$ belonging to $\bigcap_{p \in Y} p$ (which is not reduced to 0 since $A$ is an integral domain) and write $z' = z + at$: clearly $y$ and $z'$ are linearly independent over $K$ and $z'(p) = z(p) \neq 0$ for all $p \in Y$.

Supposing therefore that $y$ and $z$ are linearly independent over $K$, let $Z$ be the set of $p \in P - Y$ such that $y(p)$ and $z(p)$ are linearly independent over $k(p)$; it follows from Lemma 7 that this set is finite. For all $p \in Z$, we may therefore write $z(p) = \lambda(p)y(p)$ where $\lambda(p) \in k(p)$. Now, $\operatorname{Card}(A/p) > 2$ for all $p \in P$; it therefore follows from Lemma 6 that there exists $b \in A$ such that, for all $p \in Z$, the canonical image of $b$ in $A/p$ is distinct from $\lambda(p)$. We now show that the element $x = z - by$ solves the problem; it suffices (by virtue of Lemma 7 applied with $m = 1$) to verify that $x(p) \neq 0$ for all $p \in P$. Now:

— if $p \in Y$, then $x(p) \neq 0$ by construction; — if $p \in Z$, then $x(p) = \mu_p y(p)$ where $\mu_p \neq 0$ by virtue of the choice of $b$ and hence $x(p) \neq 0$ since $y(p) \# 0$; — if $p \in P - (Y \cup Z)$, $y(p)$ and $z(p)$ are linearly independent and hence $x(p) \neq 0$.

Having established these lemmas, we pass to the proof of Theorem 6. We argue by induction on $n$, the case $n \leq 1$ being trivial since $M$ itself is then isomorphic to an ideal of $A$. Suppose therefore that $n \geq 2$; by virtue of Lemma 8 there exists a free submodule $L_0$ of $M$ of rank 1 such that $M/L_0$ is torsion-free; $M/L_0$ is therefore of rank $n - 1$. By the induction hypothesis there exists a free submodule $L$, of $M/L_0$ such that $(M/L_0)/L_1$ is isomorphic to an ideal of $A$. Let $L$ be the inverse image of $L_1$ in $M$; $L/L_0$ is isomorphic to $L_1$ and, since $L_1$ is free, $L$ is isomorphic to $L_0 \oplus L_1$ (Algebra, Chapter II, § 1, no. 11, Proposition 21) and hence free; as $M/L$ is isomorphic to $(M/L_0)/L_1$, the theorem has been shown.