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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.
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 $R^{m},1\leq m\leq n1$.
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 $M\subseteq\ker(f)=(0)\times R^{{n1}}$. Otherwise, $\mathfrak{I}=(g),g\neq 0$. In the first case, elements of $\ker(f)$ can be bijectively mapped to $R^{{n1}}$ by the function $\ker(f)\to R^{{n1}}$ given by $(0,k_{1},\ldots,k_{{n1}})\mapsto(k_{1},\ldots,k_{{n1}})$; 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 $xhy\in\ker(f)\cap R^{n}:=N$ which is isomorphic to a submodule of $R^{{n1}}$. This shows that $Rx+N=M$.
Let $\{g_{1},\ldots,g_{s}\}$ be a basis of $N$. By assumption, $s\leq n1$. 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 finitely generated modules over a principal ideal domain 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 torsionfree, that is, $\operatorname{tor}(M/\operatorname{tor}(M))=\{0\}$. In particular, if $M$ is torsionfree, 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 nonzero torsion elements (which is obvious if $M=\operatorname{tor}(M)$).
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 $\{m_{1},\dots,m_{n}\}$ be a set of generators of $M$. For each $i=1,\dots,n$ 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 $i=1,\dots,n$. 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 $\pi\colon M\to M/T$ 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 torsionfree, $F\cap T=\{0\}$, and it follows that $M=F\oplus T$.
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Addendum
The statement could be made more precise in the following manner:
If M is a finitely generated module over a PID R, then
M \cong R^n \oplus R/(q_1) \oplus ... \oplus R/(q_r)
where q_1  q_1  ... q_r, the ideals (q_i) are uniquely determined.
Here, R^n is the free module F which is the torsion free part of M. Therefore, n is uniquely determined too. The other factors represent a direct sum decomposition of M_{tor} (the torsion submodule of M).