 Rank of a module over an integral domain

For a module $M$ over an integral domain $R$ with field of fractions $K$, its rank is the dimension of $K{\otimes }_{R}M$. The tensor product of modules here extends scalars to a field, turning the question into a dimension calculation. For a free module $R^r$, the result is $r$.

For a finitely generated module over a principal ideal domain, the structure theorem for finitely generated modules over a principal ideal domain gives $M\cong R^r\oplus {⨁}_{i}R/(d_i)$, with $d_i$ nonzero nonunits. Each torsion element of a module vanishes after extension of scalars, because its annihilating nonzero scalar becomes invertible in $K$. Thus $K{\otimes }_{R}M\cong K^r$, proving that the rank is the number of free summands. The rank does not record the torsion submodule: for example, $R\oplus R/(d)$ and $R$ have the same rank.