Proposition 5

Let $A$ be a principal ideal domain, and $f$ a linear map of rank $r$ from a free $A$-module $L$ of rank $m$ into a free $A$-module $L'$ of rank $n$. Then there exist bases $(e_i)$ ($1 \leq i \leq m$) of $L$ and $(e'_j)$ ($1 \leq j \leq n$) of $L'$ such that $f(e_i) = \alpha_i e'_i$ for $1 \leq i \leq r$ and $f(e_i) = 0$ for $i > r$, where the $\alpha_i$ are nonzero elements of $A$, each of which divides the next; the ideals $A \alpha_i$ are the invariant factors of $f(L)$ in $L'$, and are thus uniquely determined.

Let $L_0 = f^{-1}(0)$ be the kernel of $f$; the quotient $L/L_0$ is isomorphic to the module $f(L)$, which is a submodule of $L'$ and so free (VII, p. 15, Cor. 2); thus $L_0$ has a complement $L_1$ in $L$ (II, p. 218, Prop. 21), and the restriction of $f$ to $L_1$ is an isomorphism from $L_1$ onto $f(L) = M'$. If the ideals $A \alpha_i$ ($1 \leq i \leq r$) are the invariant factors of $M'$ in $L'$, then Th. 1 of VII, p. 18 shows that there exists a basis $(e'_j)$ ($1 \leq j \leq n$) of $L'$ such that $(\alpha_i e'_i)$ ($1 \leq i \leq r$) is a basis of $M'$. Since the restriction of $f$ to $L_1$ is an isomorphism from $L_1$ onto $M'$, there exists a basis $(e_i)$ ($1 \leq i \leq r$) of $L_1$ such that $f(e_i) = \alpha_i e'_i$. This basis extends to a basis $(e_s)$ ($1 \leq k \leq m$) of $L$ by taking $(e_s)$ ($r+1 \leq s \leq m$) to be a basis of the kernel $L_0$.