finitely generated modules over a principal ideal domain
Let $R$ be a principal ideal domain^{} and let $M$ be a finitely generated^{} $R$ module.
Lemma.
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 ${R}^{m},1\le m\le n1$.
For a submodule $M$ of ${R}^{n}$ we define $f:M\to R$ by $({k}_{1},\mathrm{\dots},{k}_{n})\mapsto {k}_{1}$. The image of $f$ is an ideal $\Im $ in $R$. If $\Im =\{0\}$, then $M\subseteq \mathrm{ker}(f)=(0)\times {R}^{n1}$. Otherwise, $\Im =(g),g\ne 0$. In the first case, elements of $\mathrm{ker}(f)$ can be bijectively mapped to ${R}^{n1}$ by the function $\mathrm{ker}(f)\to {R}^{n1}$ given by $(0,{k}_{1},\mathrm{\dots},{k}_{n1})\mapsto ({k}_{1},\mathrm{\dots},{k}_{n1})$; so the image of $M$ under this mapping is a submodule of ${R}^{n1}$, 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 $f(xhy)=f(x)f(hy)=0$, which is equivalent^{} to $xhy\in \mathrm{ker}(f)\cap {R}^{n}:=N$ which is isomorphic^{} to a submodule of ${R}^{n1}$. This shows that $Rx+N=M$.
Let $\{{g}_{1},\mathrm{\dots},{g}_{s}\}$ be a basis of $N$. By assumption^{}, $s\le n1$. We’ll show that $\{x,{g}_{1},\mathrm{\dots},{g}_{s}\}$ is linearly independent^{}. So let $rx+{\sum}_{i=1}^{s}{r}_{i}{g}_{i}=0$. 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 $\{{g}_{1},\mathrm{\dots},{g}_{s}\}$ are linearly independent, $\{x,{g}_{1},\mathrm{\dots},{g}_{s}\}$ is a generating set of $M$ with $s+1\le n$ elements. ∎
Corollary.
If $M$ is a finitely generated $R$module over a PID generated by $s$ elements and $N$ is a submodule of $M$, then $N$ can be generated by $s$ or fewer elements.
Proof.
Let $\{{g}_{1},\mathrm{\dots},{g}_{s}\}$ be a generating set of $M$ and $f:{R}^{s}\to M$, $({r}_{1},\mathrm{\dots},{r}_{s})\mapsto {\sum}_{i=1}^{s}{r}_{i}{g}_{i}$. Then the inverse image^{} ${N}^{{}^{\prime}}$ of $N$ is a submodule of ${R}^{s}$, and according to lemma Lemma. can be generated by $s$ or fewer elements. Let ${n}_{1},\mathrm{\dots},{n}_{t}$ be a generating set of ${n}^{{}^{\prime}}$; then $t\le s$, and since $f$ is surjective^{}, $f({n}_{1}),\mathrm{\dots},f({n}_{t})$ is a generating set of $N$. ∎
Theorem.
Let $M$ be a finitely generated module over a principal ideal domain $R$.
 (I)

Note that $M/\mathrm{tor}(M)$ is torsionfree, that is, $\mathrm{tor}(M/\mathrm{tor}(M))=\{0\}$. In particular, if $M$ is torsionfree, then $M$ is free.
 (II)

Let $\mathrm{tor}(M)$ be a proper submodule of $M$. Then there exists a finitely generated free submodule $F$ of $M$ such that $M=F\oplus \mathrm{tor}(M)$.
Proof of (I): Let $T=\mathrm{tor}(M)$. For $m\in M$, $\overline{m}$ denotes the coset modulo $T$ generated by $m$. Let $m$ be a torsion element of $M/T$, so there exists $\alpha \in R\setminus \{0\}$ such that $\alpha \cdot \overline{m}=0$, which means $\alpha \cdot \overline{m}\subseteq T$. But then $\alpha \cdot m$ is a member of $T$, and this implies that $M/T$ has no nonzero torsion elements (which is obvious if $M=\mathrm{tor}(M)$).
Now let $M$ be a finitely generated torsionfree $R$module. Choose a maximal linearly independent subset $S$ of $M$, and let $F$ be the submodule of $M$ generated by $S$. Let $\{{m}_{1},\mathrm{\dots},{m}_{n}\}$ be a set of generators^{} of $M$. For each $i=1,\mathrm{\dots},n$ there is a nonzero ${r}_{i}\in R$ such that ${r}_{i}\cdot {m}_{i}\in F$. Put $r={\prod}_{i=1}^{n}{r}_{i}$. Then $r$ is nonzero, and we have $r\cdot {m}_{i}\in F$ for each $i=1,\mathrm{\dots},n$. As $M$ is torsionfree, 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 $\pi :M\to M/T$ be defined by $a\mapsto a+T$. Then $\pi $ is surjective, so ${m}_{1},\mathrm{\dots},{m}_{t}\in M$ can be chosen such that $\pi ({m}_{i})={n}_{i}$, where the ${n}_{i}$’s are a basis of $M/T$. If ${0}_{M}={\sum}_{i=1}^{t}{a}_{i}{m}_{i}$, then ${0}_{n}={\sum}_{i=1}^{t}{a}_{i}{n}_{i}$. Since ${n}_{1},\mathrm{\dots},{n}_{t}$ are linearly independent in $N$ it follows $0={a}_{1}=\mathrm{\dots}={a}_{t}$. So the submodule spanned by ${m}_{1},\mathrm{\dots},{m}_{t}$ of $M$ is free.
Now let $m$ be some element of $M$ and $\pi (m)={\sum}_{i=1}^{t}{a}_{i}{n}_{i}$. This is equivalent to $m\left({\sum}_{i=1}^{t}{a}_{i}{n}_{i}\right)\in \mathrm{ker}(\pi )=T$. 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 torsionfree, $F\cap T=\{0\}$, and it follows that $M=F\oplus T$.
Title  finitely generated modules over a principal ideal domain 

Canonical name  FinitelyGeneratedModulesOverAPrincipalIdealDomain 
Date of creation  20130322 13:55:22 
Last modified on  20130322 13:55:22 
Owner  yark (2760) 
Last modified by  yark (2760) 
Numerical id  21 
Author  yark (2760) 
Entry type  Topic 
Classification  msc 13E15 