PlanetMath: finitely generated modules over a principal ideal domain

Proof. For $n=1$ this is clear, since $M$ is an ideal of $R$ and is generated by some element $a \in R$ . Now suppose that the statement is true for all submodules of $R^m, 1 \le m \le n-1$ .

For a submodule $M$ of $R^n$ we define $f\colon M\to R$ by $(k_1, \ldots, k_n) \mapsto k_1$ . The image of $f$ is an ideal $\mathfrak{I}$ in $R$ . If $\mathfrak{I}=\{0\}$ , then $M \subseteq \ker(f)=(0) \times R^{n-1}$ . Otherwise, $\mathfrak{I}=(g), g \ne 0$ . In the first case, elements of $\ker(f)$ can be bijectively mapped to $R^{n-1}$ by the function $\ker(f) \to R^{n-1}$ given by $(0, k_1, \ldots, k_{n-1}) \mapsto (k_1,\ldots, k_{n-1})$ ; so the image of $M$ under this mapping is a submodule of $R^{n-1}$ , which by the induction hypothesis is finitely generated and free.

Now let $x \in M$ such that $f(x) =gh$ and $y \in M$ with $f(y)=g$ . Then $f(x-hy)=f(x) -f(hy)=0$ , which is equivalent to $x-hy \in \ker(f) \cap R^n:=N$ which is isomorphic to a submodule of $R^{n-1}$ . This shows that $Rx+N=M$ .

Let $\{g_1, \ldots, g_s\}$ be a basis of $N$ . By assumption, $s \le n-1$ . We'll show that $\{x, g_1, \ldots, g_s\}$ is linearly independent. So let $rx +\sum_{i=1}^s r_ig_i=0$ . The first component of the $g_i$ are 0, so the first component of $rx$ must also be 0. Since $f(x)$ is a multiple of $g \ne 0$ and $0=r\cdot f(x)$ , then $r=0$ . Since $\{g_1, \ldots, g_s\}$ are linearly independent, $\{x, g_1, \ldots, g_s\}$ is a generating set of $M$ with $s+1\le n$ elements. $\qedsymbol$

Now let $M$ be a finitely generated torsion-free $R$ -module. Choose a maximal linearly independent subset $S$ of $M$ , and let $F$ be the submodule of $M$ generated by $S$ . Let $\{m_1,\dots,m_n\}$ be a set of generators of $M$ . For each $i=1,\dots,n$ there is a non-zero $r_i\in R$ such that $r_i\cdot m_i\in F$ . Put $r=\prod_{i=1}^n r_i$ . Then $r$ is non-zero, and we have $r\cdot m_i\in F$ for each $i=1,\dots,n$ . As $M$ is torsion-free, the multiplication by $r$ is injective, so $M \cong r \cdot M \subseteq F$ . So $M$ is isomorphic to a submodule of a free module, and is therefore free.

Proof of (II): Let $\pi\colon M \to M/T$ be defined by $a \mapsto a + T$ . Then $\pi$ is surjective, so $m_1, \ldots, m_t \in M$ can be chosen such that $\pi(m_i)=n_i$ , where the $n_i$ 's are a basis of $M/T$ . If $0_M=\sum_{i=1}^t a_im_i$ , then $0_n=\sum_{i=1}^t a_in_i$ . Since $n_1,\ldots,n_t$ are linearly independent in $N$ it follows $0=a_1=\ldots =a_t$ . So the submodule spanned by $m_1,\ldots,m_t$ of $M$ is free.

Now let $m$ be some element of $M$ and $\pi(m)=\sum_{i=1}^t a_in_i$ . This is equivalent to $m-\left(\sum_{i=1}^t a_in_i\right) \in \ker(\pi)=T$ . Hence, any $m\in M$ is a sum of the form $f+t$ , for some $f \in F$ and $t \in T$ . Since $F$ is torsion-free, $F \cap T=\{0\}$ , and it follows that $M=F \oplus T$ .