finitely generated modules over a principal ideal domain

generated bygenerating set

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 $M$ isfree and finitely generated by $s\mathrm{\le }n$ 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 truefor all submodules of $R^m,1\le m\le n-1$.

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 $M\subseteq \ker (f)=(0)\times R^{n-1}$.Otherwise, $\Im =(g),g\ne 0$.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$(0,k_1,\mathrm{\dots },k_{n-1})\mapsto (k_1,\mathrm{\dots },k_{n-1})$;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 $\{g_1,\mathrm{\dots },g_s\}$ be a basis of $N$.By assumption, $s\le n-1$.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 $R$-moduleover a PID generated by $s$ elementsand $N$ 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 torsion-free, that is, $\mathrm{tor}(M/\mathrm{tor}(M))=\{0\}$.In particular, if $M$ is torsion-free, 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 non-zero torsion elements(which is obvious if $M=\mathrm{tor}(M)$).

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 $\{m_1,\mathrm{\dots },m_n\}$ be a set of generators of $M$.For each $i=1,\mathrm{\dots },n$ 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 $i=1,\mathrm{\dots },n$.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 $\pi :M\to M/T$ 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 $m-\left({\sum }_{i=1}^t{a}_i{n}_i\right)\in \ker (\pi )=T$.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 $M=F\oplus T$.