= Rank of a module over an integral domain
{title2=$\operatorname{rank}_R M=\dim_{\operatorname{Frac}(R)}(\operatorname{Frac}(R)\otimes_R M)$}
= Rank of a module
{synonym}
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 $M\cong R^r\oplus\bigoplus_iR/(d_i)$, 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.
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