## 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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