Direct Sums and Rank (12.1.3)

Dummit and Foote Abstract Algebra, section 12.1, exercise 3:

Let $R$ be an integral domain and let $A$ and $B$ be $R$-modules of rank $m$ and $n$ respectively. Show $A ⊕ B$ is an $R$-module of rank $m+n$.

Proof: Letting $a_1,...,a_m$ and $b_1,...,b_n$ be the fulfilling linearly independent elements, we see $(a_1,0),...,(0,b_n)$ are linearly independent. Observing the homomorphism $(x,y) ↦ (x \text{ mod }R^m, y \text{ mod }R^n)$ admitted by these former elements we see the kernel is the submodule generated by the latter. By 12.1.2, we will show the quotient is torsion and thus $A ⊕ B$ is of rank $m+n$. Take $\overline{(a,b)}$. Also by 12.1.2 we have the respective images of $A$ and $B$ under the homomorphism are those of $A$ and $B$ under the homomorphisms provided by their own linearly independent elements, and as such there are nonzero $r_1,r_2$ such that $r_1\overline{a}=r_2\overline{b}=0$, so that $r_1r_2$ is nonzero and $r_1r_2\overline{(a,b)}=0$.$~\square$