# finitely generated modules over a principal ideal domain

generated by \PMlinkescapephrasegenerating set

###### 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.

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 $rx$ 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 FinitelyGeneratedModulesOverAPrincipalIdealDomain 2013-03-22 13:55:22 2013-03-22 13:55:22 yark (2760) yark (2760) 21 yark (2760) Topic msc 13E15