 Solution

A Euclidean domain is a principal ideal domain. The structure theorem for finitely generated modules over a principal ideal domain therefore gives, for every finitely generated module $M$ over $R$,where the $d_i$ are nonzero nonunits and $d_1\mid d_2\mid \cdots \mid d_s$. The factors $d_i$ are unique up to multiplication by units; $r$ is uniquely determined.

The rank of a module is this integer $r$. Equivalently, with $K$ the field of fractions of $R$,The tensor product of modules kills each displayed torsion summand after extending scalars to $K$, while sending $R^r$ to $K^r$, explaining the equality.

If $M$ is a torsion-free module, no summand $R/(d_i)$ can occur: its nonzero element $1+(d_i)$ is killed by the nonzero scalar $d_i$. Hence $s=0$ andThis also covers $M=0$, with $r=0$.