If \(F_1, F_2\) are free modules of a ring with the invariant dimension property, then \[\mathrm{rank}(F_1 \oplus F_2) = \mathrm{rank} F_1 + \mathrm{rank} F_2.\]
Solution
Let \(X\) be a basis of \(F_1\) and \(Y\) a basis of \(F_2\). Define \[ B=\{(x,0):x\in X\}\cup\{(0,y):y\in Y\} \subseteq F_1\oplus F_2. \]
1. The set \(B\) spans \(F_1\oplus F_2\).
Let \((u,v)\in F_1\oplus F_2\). Since \(X\) and \(Y\) are bases, there are finite expressions \[ u=\sum_{i=1}^m r_i x_i, \qquad v=\sum_{j=1}^n s_j y_j. \] Hence \[ (u,v) =\sum_{i=1}^m r_i(x_i,0) +\sum_{j=1}^n s_j(0,y_j), \] so \((u,v)\) lies in the span of \(B\).
2. The set \(B\) is linearly independent.
Suppose \[ \sum_{i=1}^m r_i(x_i,0) +\sum_{j=1}^n s_j(0,y_j) =(0,0). \] Comparing coordinates gives \[ \sum_{i=1}^m r_i x_i=0 \qquad\text{and}\qquad \sum_{j=1}^n s_j y_j=0. \] Linear independence of \(X\) and \(Y\) forces every \(r_i\) and \(s_j\) to be zero. Thus \(B\) is linearly independent.
3. Therefore \(B\) is a basis of \(F_1\oplus F_2\) and\[ \operatorname{rank}(F_1\oplus F_2) =\operatorname{rank}F_1+\operatorname{rank}F_2. \]
By 1 and 2, \(B\) is a basis. The two subsets defining \(B\) are disjoint and are in bijection with \(X\) and \(Y\), respectively, so \[ |B|=|X|+|Y|. \] Because the ring has the invariant dimension property, the rank of a free module is the cardinality of any basis. Hence \[ \operatorname{rank}(F_1\oplus F_2) =|B| =|X|+|Y| =\operatorname{rank}F_1+\operatorname{rank}F_2. \]