The Stacks project

Proof. Since $M$ is finite, $M \otimes _ R K$ is a finite dimensional vector space over the fraction field $K$ of $R$. Thus the rank $r$ is finite.

Choose generators $x_1, \ldots , x_ n$ of $M$. Choose a basis $e_1, \ldots , e_ r$ for $M \otimes _ R K$. We may write $x_ i \otimes 1 = \sum (a_{ij}/b_{ij}) e_ j$ for some $a_{ij} , b_{ij} \in R$ with $b_{ij} \not= 0$. Setting $b = \prod b_{ij}$ and replacing $a_{ij}$ by $a_{ij}b/b_{ij}$ we may assume that $x_ i \otimes 1 = \sum (a_{ij}/b) e_ j$. Note that if we have a relation $\sum r_ i x_ i = 0$ in $M$ then $\sum r_ ia_{ij} = 0$ in $R$ for all $j = 1, \ldots , r$.

Proof of (2). The last comment above tells us the map $M \to \bigoplus Re_ j$ sending $x_ i$ to $\sum a_{ij} e_ j$. is well defined. The fact that $e_1, \ldots , e_ r$ forms a basis of $M \otimes _ R K$ tells us that this map becomes an isomorphism upon tensoring by $K$. Hence this is a map as in (2).

Proof of (3). The cokernel of the map $M \to R^{\oplus r}$ constructed in (2) is finite and torsion. Hence we can find a $g \in R$, $g \not= 0$ which annihilates this cokernel. In other words, such that the map $M_ g \to R_ g^{\oplus r}$ is surjective. Then we can choose a splitting $M_ g = N \oplus R_ g^{\oplus r}$. Then $N$ is a finite $R_ g$-module with $N \otimes _{R_ g} K = 0$. In other words, $N$ is torsion and finite over $R_ g$. Thus we can find a $g' \in R$ which annihilates $N$. Setting $f= gg'$ we see that (3) holds.

To prove (4) choose a basis $y_1, \ldots , y_ r$ of $M_ f$ where $f$ is as in (3). Then write $y_ j = m_ j/f^{t_ j}$ for some $m_ j \in M$ and $t_ j \geq 0$. The map $R^{\oplus r} \to M$ sending the $j$th basis vector to $m_ j$ will be as in (4). Details omitted. $\square$