Over an integral domain, the rank of a direct sum is the sum of the ranks

Let R be an integral domain and let A and B be (left, unital) R-modules. Prove that \mathsf{rank}(A \oplus B) = \mathsf{rank}(A) + \mathsf{rank}(B), where the rank of a module is the largest possible cardinality of a linearly independent subset.

Suppose A has rank n and B has rank m. By the previous exercise, there exist free submodules A_1 \subseteq A and B_1 \subseteq B having free ranks n and m, respectively, such that the quotients A/A_1 and B/B_1 are torsion. Note that A_1 \oplus B_1 \subseteq A \oplus B is free. By this previous exercise, we have (A \oplus B)/(A_1 \oplus B_1) \cong_R (A/A_1) \oplus (B/B_1). Note that since R is an integral domain, finite direct sums of torsion modules are torsion. Thus (A \oplus B)/(A_1 \oplus B_1) is torsion. Since A_1 \oplus B_1 is free and has free rank n+m, by this previous exercise, A \oplus B has rank n+m.

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