Let \(R\) be a nonzero commutative ring and \(M\) a free \(R\)-module. The rank of \(M\) is the cardinality of a basis of \(M\).
Let \(R\) be a nonzero commutative ring and \(M\) a free \(R\)-module. Any two bases of \(M\) have the same cardinality.
Since \(R\neq0\), \(R\) has a maximal ideal \(\mfm\), and \(k\coloneqq R/\mfm\) is a field. If \((e_i)_{i\in I}\) is a basis of \(M\), then \(M\cong R^{(I)}\), so \(M/\mfm M\cong (R/\mfm)^{(I)}\) and the images of the \(e_i\) form a basis of the \(k\)-vector space \(M/\mfm M\). Hence \(\abs{I}=\dim_k M/\mfm M\), which does not depend on the basis.
If \(R\) is an integral domain, the rank of \(M\) is the supremum of the cardinalities of \(R\)-linearly independent families in \(M\). With \(K\) the field of fractions of \(R\) and \((e_i)_{i\in I}\) a basis, \(M\cong R^{(I)}\subseteq K^{(I)}\); a family that is \(R\)-linearly independent in \(R^{(I)}\) is \(K\)-linearly independent in \(K^{(I)}\), after clearing denominators in a \(K\)-linear relation, so it has at most \(\abs{I}\) elements, and the basis itself is an independent family of cardinality \(\abs{I}\).