The Stacks project

Proof. If $M$ is a submodule of $R^{\oplus n}$, then $M$ is torsion free. For the converse, assume $M$ is torsion free. Let $K$ be the fraction field of $R$. Then $M \otimes _ R K$ is a finite dimensional $K$-vector space. Choose a basis $e_1, \ldots , e_ r$ for this vector space. Let $x_1, \ldots , x_ n$ be generators of $M$. Write $x_ i = \sum (a_{ij}/b_{ij}) e_ j$ for some $a_{ij}, b_{ij} \in R$ with $b_{ij} \not= 0$. Set $b = \prod _{i, j} b_{ij}$. Since $M$ is torsion free the map $M \to M \otimes _ R K$ is injective and the image is contained in $R^{\oplus r} = R e_1/b \oplus \ldots \oplus Re_ r/b$. $\square$