Theorem 1

— Let $L$ be a free module over a principal ideal domain $A$, and let $M$ be a submodule of $L$ of finite rank $n$. Then there exist a basis $B$ of $L$, $n$ elements $e_i$ of $B$, and $n$ nonzero elements $\alpha_i$ of $A$ ($1 \leq i \leq n$) such that: a) the $\alpha_i e_i$ form a basis of $M$; b) $a_i$ divides $\alpha_{i-1}$, for $1 \leq i \leq n-1$.

Moreover the module $M'$ generated by the $(e_i)$ and the principal ideals $A\alpha_i$ are uniquely determined by the above conditions; the module $M'/M$ is the torsion submodule of $L/M$, and is isomorphic to the direct sum of the $A$-modules $A/A\alpha_i$; finally $L/M$ is the direct sum of $M'/M$ and a free module isomorphic to $L/M'$.

1) Existence of the $e_i$ and the $\alpha_i$.

If $M = \{0\}$ then the theorem is trivial. If $M \neq \{0\}$ then it follows from Prop. 3 that there exist an element $e_1$ of $L$, a nonzero element $a$, of $A$, and a submodule $L_1$ of $L$, such that $L$ is the direct sum of $Ae_1$ and $L_1$, such that $M$ is the direct sum of $A\alpha_1 e_1$ and the submodule $M_1 = M \cap L_1$ of $L$, and such that $g(M) \subset A\alpha_1$ for every linear form $g$ on $L$.

Now we can proceed by induction on the rank $n$ of $M$. Since $L_1$ is a free module (VII, p. 15, Cor. 2) and $M_1$ has rank $n-1$, there exist a basis $B_1$ of $L_1$, $(n-1)$ elements $e_2, \ldots, e_n$ of $B_1$, and nonzero elements $\alpha_2, \ldots, a$, of $A$ such that $(\alpha_2 e_2, \ldots, \alpha_n e_n)$ is a basis of $M_1$, and $\alpha_i$ divides $\alpha_{i+1}$ for $2 \leq i \leq n-1$. If $L'$ is the submodule of $L$, generated by the elements of $B_1$ distinct from $e_1, \ldots, e_n$, then $L$ is the direct sum of $L'$ and the module $M'$ generated by $e_1, \ldots, e_n$; now $(e_1, \ldots, e_n)$ is a basis for $M'$, and $(\alpha_1 e_1, \ldots, \alpha_n e_n)$ is a basis for $M$. It now remains only to show that $a_i$ divides $a_i$; but $A\alpha_2$ has the form $g(M_1)$, where $g$ is the linear form on $L$ defined by $g(e_2) = 1, g(e_i) = 0$ for $i \neq 2$, and $g(L') = (0)$; and we have seen above that $g(M_1) \subset A\alpha_1$.

2) Uniqueness properties.

Since the $a_i$ are nonzero, the module $M'$ is the set of $x \in L$ such that $\beta x \in M$ for some $\beta \neq 0$ in $A$; in other words $M'/M$ is the torsion submodule of $L/M$. This uniquely determines $M'$.

It is clear that $M'/M$ is isomorphic to the direct sum of the $n$ cyclic modules $A/A\alpha_i$ (II, p. 204, formula (26)). Let $r$ be the number of the ideals $A\alpha$, which are distinct from $A$: so that the first $n-r$ ideals $A\alpha_i$ are equal to $A$, and the last rare distinct from $A$. Then $M'/M$ is also isomorphic to the direct sum of the modules $A/A\alpha_n, \ldots, A/A\alpha_{n-r+1}$, where $A\alpha_n \subset A\alpha_{n-1} \subset \ldots \subset A\alpha_{n-r+1} \neq A$.

Thus the conditions of the Cor. to Prop. 2 (VII, p. 16) are satisfied: the ideals $Aa_i$ ($1 \leq i \leq n$) are thus uniquely determined by the module $M'/M$.

Since $L$ is the direct sum of $M'$ and $L'$, it follows that $L/M$ is the sum of $M'/M$ and $(L' + M)/M$, and this sum is direct since $(L' + M) \cap M' = M$; on the other hand $(L' + M)/M$ is isomorphic to $L'/(M \cap L')$ (I, p. 41, Th. 4, c)), that is to $L'$, which shows that $(L' + M)/M$ is a free module isomorphic to $L/M'$.