# proof of finitely generated torsion-free modules over Prüfer domains

Let $M$ be a finitely generated    torsion-free module over a Prüfer domain $R$ with field of fractions  $k$. We show that $M$ is isomorphic   to a direct sum    (http://planetmath.org/DirectSum) of finitely generated ideals in $R$.

We shall write $k\otimes M$ for the vector space  over $k$ generated by $M$. This is just the localization  (http://planetmath.org/LocalizationOfAModule) of $M$ at $R\setminus\{0\}$ and, as $M$ is torsion-free, the natural map $M\rightarrow k\otimes M$ is one-to-one and we can regard $M$ as a subset of $k\otimes M$.

As $M$ is finitely generated, the vector space $k\otimes M$ will finite dimensional (http://planetmath.org/Dimension2), and we use induction  on its dimension  $n$. Supposing that $n>0$, choose any basis $e_{1},\ldots,e_{n}$ and define the linear map $f\colon k\otimes M\rightarrow k$ by projection   (http://planetmath.org/Projection) onto the first component   ,

 $f(x_{1}e_{1}+\cdots+x_{n}e_{n})=x_{1}.$

Restricting to $M$, this gives a nonzero map $M\rightarrow k$. Furthermore, as $M$ is finitely generated, $f(M)$ will be a finitely generated fractional ideal  in $k$. Choosing any nonzero $c\in R$ such that $\mathfrak{a}\equiv cf(M)\subseteq R$,

 $g\colon M\rightarrow\mathfrak{a},\ g(u)=cf(u)$

defines a homorphism from $M$ onto the nonzero and finitely generated ideal $\mathfrak{a}$. As $R$ is Prüfer and invertible ideals are projective, $g$ has a right-inverse $h\colon\mathfrak{a}\rightarrow M$. Then $h$ has the left-inverse $g$ and is one-to-one, so defines an isomorphism     between $\mathfrak{a}$ and its image (http://planetmath.org/ImageOfALinearTransformation). We decompose $M$ as the direct sum of the kernel of $g$ and the image of $h$,

 $M=\operatorname{ker}(g)\oplus\operatorname{Im}(h)\cong\operatorname{ker}(g)% \oplus\mathfrak{a}.$

Projection from the finitely generated module $M$ onto $\operatorname{ker}(g)$ shows that it is finitely generated and,

 $\operatorname{dim}(k\otimes\operatorname{ker}(g))=\operatorname{dim}(k\otimes M% )-\operatorname{dim}(k\otimes\mathfrak{a})=n-1.$

So, the result follows from applying the induction hypothesis to $\operatorname{ker}(g)$.

Title proof of finitely generated torsion-free modules over Prüfer domains ProofOfFinitelyGeneratedTorsionfreeModulesOverPruferDomains 2013-03-22 18:36:14 2013-03-22 18:36:14 gel (22282) gel (22282) 4 gel (22282) Proof msc 13F05 msc 13C10