Proposition 3

*Let $L$ be a free module over a principal ideal domain $A$ and let $M$ be a nonzero submodule of $L$. Let $x$ be an element of $M$ such that $c_L(x)$ is maximal among the contents of elements of $M$, let $c$ be a content of $x$ in $L$, and let $f$ be a linear form on $L$ such that $f(x) = c$.

a) $L$ is the direct sum of $A(x/c)$ and the kernel $K$ of $f$. b) $M$ is the direct sum of $Ax$ and $K \cap M$. c) $g(M) \subset Ac$ for every linear form $g$ on $L$.

Put $y = x/c$ ; clearly $Ay \cap K = \{0\}$, since $f(y) = 1$. Furthermore, for each $u \in L$ we have

with $f(u)y \in Ay$ and $u - f(u)y \in K$ ; this proves a). Now note that for $u \in M$ we have $f(u) \in Ac$ : indeed, let $u \in M$, and let $d$ be a gcd of $f(u)$ and $c$ ; then there exist $\lambda, \mu \in A$ such that $d = Xf(u) + \mu c = f(Xu + \mu x)$ ; hence the content of the element $\lambda u + \mu x$ of $M$ divides $d$ ; by the maximality of $c$, this implies that $d$ is an associate of $c$, so $f(u) \in Ac$. Hence for all $u$ in $M$, we can write which shows b). Finally, let $g$ be a linear form on $L$ ; by a) there exists a scalar $\alpha \in A$ and a linear form $h$ on $K$ such that $g(u) = \alpha f(u) + h(u - f(u)\,y)$ ; thus by

b) we have $g(M) \subset Ac + h(K \cap M)$. To prove c), it is therefore sufficient to prove that $h(K \cap M) \subset Ac$ for every linear form $h$ on $K$, or equivalently for every linear form $h$ on $L$ such that $h(x) = 0$; now, if $u \in K \cap M$ and $d$ is a gcd of $h(u)$ and $c$, then there exist $A, \mu \in A$ with $d = \lambda h(u) + \mu c$; then $(f + h)(Au + \mu x) = d$, which implies as above that $h(u) \in Ac$, and c) follows.