finitely generated modules over a principal ideal domain

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Let $R$ be a principal ideal domain and let $M$ be a finitely generated $R$ - module.

Lemma.

Let $M$ be a submodule of the $R$ -module $R^{n}$ . Then $M$ is free and finitely generated by $s\leq n$ elements.

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 $R^{m},1\leq m\leq n-1$ .

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, $\mathfrak{I}=(g),g\neq 0$ . 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, $s\leq n-1$ . We’ll show that $\{x,g_{1},\ldots,g_{s}\}$ 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 $r x$ must also be 0. Since $f(x)$ is a multiple of $g\neq 0$ 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 $s+1\leq n$ 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 $\{g_{1},\ldots,g_{s}\}$ be a generating set of $M$ and $f\colon R^{s}\to M$ , $(r_{1},\ldots,r_{s})\mapsto\sum_{i=1}^{s}r_{i}g_{i}$ . 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 $n_{1},\ldots,n_{t}$ be a generating set of $n^{{}^{\prime}}$ ; then $t\leq s$ , and since $f$ is surjective, $f(n_{1}),\ldots,f(n_{t})$ is a generating set of $N$ . ∎

Theorem.

Let $M$ be a finitely generated module over a principal ideal domain $R$ .

(I): Note that $M/\operatorname{tor}(M)$ is torsion-free, that is, $\operatorname{tor}(M/\operatorname{tor}(M))=\{0\}$ . In particular, if $M$ is torsion-free, then $M$ is free.
(II): Let $\operatorname{tor}(M)$ be a proper submodule of $M$ . Then there exists a finitely generated free submodule $F$ of $M$ such that $M=F\oplus\operatorname{tor}(M)$ .

Proof of (I): Let $T=\operatorname{tor}(M)$ . 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 non-zero torsion elements (which is obvious if $M=\operatorname{tor}(M)$ ).

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 $0_{M}=\sum_{i=1}^{t}a_{i}m_{i}$ , then $0_{n}=\sum_{i=1}^{t}a_{i}n_{i}$ . 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 $\pi(m)=\sum_{i=1}^{t}a_{i}n_{i}$ . This is equivalent to $m-\left(\sum_{i=1}^{t}a_{i}n_{i}\right)\in\ker(\pi)=T$ . 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 $M=F\oplus T$ .

Title	finitely generated modules over a principal ideal domain
Canonical name	FinitelyGeneratedModulesOverAPrincipalIdealDomain
Date of creation	2013-03-22 13:55:22
Last modified on	2013-03-22 13:55:22
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	21
Author	yark (2760)
Entry type	Topic
Classification	msc 13E15