PlanetMath: finitely generated modules over a principal ideal domain

Proof. For

this is clear, since

is an ideal of

and is generated by some element $a \in R$ . Now suppose that the statement is true for all submodules of $% latex2html id marker 553 $ R^m, 1 \le m \le n-1$$ .

For a submodule of we define $% latex2html id marker 559 $ f\colon M\to R$$ by $(k_1, \ldots, k_n) \mapsto k_1$ . Since is surjective, the image of is an ideal $% latex2html id marker 567 $ \mathfrak{I}$$ in . If $% latex2html id marker 571 $ \mathfrak{I}=\{0\}$$ , then $M \subseteq \ker(f)=(0) \times R^{n-1}$ . Otherwise, $% latex2html id marker 575 $ \mathfrak{I}=(g), g \ne 0$$ . 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 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 and $y \in M$ with . Then , 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 .

Let $\{g_1, \ldots, g_s\}$ be a basis of . By assumption, $% latex2html id marker 609 $ s \le n-1$$ . We'll show that $\{x, g_1, \ldots, g_s\}$ is linearly independent. So let $rx +\sum_{i=1}^s r_ig_i=0$ . The first component of the are 0, so the first component of must also be 0. Since is a multiple of $g \ne 0$ and $0=r\cdot f(x)$ , then . Since $\{g_1, \ldots, g_s\}$ are linearly independent, $\{x, g_1, \ldots, g_s\}$ is a generating set of with $% latex2html id marker 633 $ s+1\le n$$ elements. $\qedsymbol$

Now let

be a finitely generated torsion-free

-module with generating set $\{g_1, \ldots, g_n\}$ . The homomorphism $% latex2html id marker 746 $ f\colon R^n \to N$$ defined by $(r_1, \ldots, r_n) \mapsto \sum_{i=1}^n r_ig_i$ is injective since

is torsion-free. Let $\{e_1, \ldots, e_n\}$ be the standard basis of

. Then the elements $f(e_1),\ldots, f(e_n)$ are linearly independent in

. Now let $N=M/\operatorname{tor}(M)$ where $\operatorname{tor}(M)=\{0\}$ . Then the cosets can be identified with the elements of

, and the statement follows.

Proof of (II): Let $% latex2html id marker 766 $ \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

's are a basis of

. 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

it follows $0=a_1=\ldots =a_t$ . So the submodule spanned by $m_1,\ldots,m_t$ of

is free.

Now let

be some element of

and $\pi(m)=\sum_{i=1}^t a_in_i$ . This is equivalent to $% latex2html id marker 800 $ 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

, for some $f \in F$ and $t \in T$ . Since

is torsion-free, $F \cap T=\{0\}$ , and it follows that $M=F \oplus T$ .