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 $M\to k\otimes M$ 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 $e_1,\mathrm{\dots },e_n$ and define the linear map $f:k\otimes M\to k$ by projection (http://planetmath.org/Projection) onto the first component,

$$f(x_1{e}_1+\mathrm{\cdots }+x_n{e}_n)=x_1.$$

Restricting to $M$, this gives a nonzero map $M\to k$. 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 $𝔞\equiv cf(M)\subseteq R$,

$$g:M\to 𝔞,g(u)=cf(u)$$

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

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

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

$$\dim (k\otimes \ker (g))=\dim (k\otimes M)-\dim (k\otimes 𝔞)=n-1.$$

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