Let \(F_1,F_2\) be free modules over a ring for which free-module rank is well-defined. Prove \[ \operatorname{rank}(F_1\oplus F_2) = \operatorname{rank}(F_1)+\operatorname{rank}(F_2). \]
Solution
Let \(\mathcal B_1\) and \(\mathcal B_2\) be bases of \(F_1\) and \(F_2\). Define \[ \mathcal B = \{(v,0):v\in\mathcal B_1\} \cup \{(0,w):w\in\mathcal B_2\}. \]
Every \((x,y)\in F_1\oplus F_2\) can be written uniquely as \[ (x,y) = \sum_i r_i(v_i,0)+\sum_j s_j(0,w_j), \] using the unique basis expansions of \(x\) and \(y\). Thus \(\mathcal B\) spans and is linearly independent, hence is a basis.
Therefore \[ \operatorname{rank}(F_1\oplus F_2) =|\mathcal B| =|\mathcal B_1|+|\mathcal B_2| =\operatorname{rank}(F_1)+\operatorname{rank}(F_2). \] For finite ranks this is ordinary integer addition; in general it is cardinal addition.