Let $(e_i)_{i \in I}$ be a basis for $L$, and let $p_i$ be the coordinate functions corresponding to this basis. Choose a well-ordering (Set Theory, III, p. 153) of $I$ and let $L_\alpha$ denote the submodule generated by the $e_i$, for $\alpha \leq i$; put $M_\alpha = M \cap L_\alpha$. The coordinate function $p_i$ maps $M_\alpha$ onto an ideal $a_\alpha$ of $A$; since $a_\alpha$ is a projective $A$-module, there exists (II, p. 231, Prop. 4) a submodule $N_\alpha$ of $M$, such that the map $x \mapsto p_i(x)$ from $N_\alpha$ into $a_\alpha$ is bijective. Let $M'_\alpha$ be the submodule of $L$ generated by the $N_\alpha$, for $\alpha \leq i$; we will show that $M'_\alpha = M_\alpha$ for all $\alpha$, which will imply that $M$ is generated by the family $(N_\alpha)_{\alpha < i}$. In fact, suppose $M'_\lambda = M_\lambda$ for all $\lambda < i$; then $p_\lambda(x) \in a_\lambda$, for all $x \in M_\lambda$; thus there exists $y \in N_\lambda$ such that $x - y$ is a linear combination of finitely many elements $e_i$, with $\alpha < \lambda$; in other words $x - y$ is an element of $M_\alpha$, for some $\lambda < i$; the inductive hypothesis shows that $x - y \in M'_\lambda \subset M'_\alpha$, that is that $x \in M'_\alpha$, and so $M'_\alpha = M_\alpha$. It remains to show that the sum of the $N_\alpha$ is direct; now suppose there exists a linear relation $\sum a_\alpha = 0$, with $a_\alpha \in N_\alpha$, where the $a_\alpha$ (all but finitely many of which are zero) are not all zero. Let $p$ be the greatest index $\alpha$ such that $a_\alpha \neq 0$; since $p_\mu(a_\lambda) = 0$ for $\lambda < \mu$, we have $p_\mu(a_\mu) = p_\mu\left( \sum a_\alpha \right) = 0$, so $a_\mu = 0$, contradicting the choice of $p$.

Indeed every projective $A$-module is a submodule of a free $A$-module (II, p. 231, Prop. 4), and Th. 1 applies.

PROPOSITION I. — If L is a free module of finite rank n over a principal ideal domain A, then every submodule M of L is a free module of rank $\leq n$. Indeed M is a free module by Cor. 2 to Th. 1, and it has rank $\leq n$ by the previous remark, or by the following lemma: