## Direct sums of free modules are free

Let $R$ be a ring with 1 and let $\{F_i\}_I$ be a family of free unital left $R$-modules. Prove that $\bigoplus_I F_i$ is free.

Let $A_i = \{a_{i,j}\}$ be a free basis for each $F_i$. We can assume that the $A_i$ are disjoint. (For example, isomorphically replace each $F_i$ by $F \times \{i\}$.) Let $A = \bigcup_I A_i$. Let $F(A)$ denote the free $R$-module on $A$.

We have the natural inclusion $A \rightarrow F(A)$. We also have a natural inclusion $\iota : A \rightarrow \bigoplus_I F_i$, where $a_{i,j} \mapsto (b_t)$, where $b_t = a_{i,j}$ if $t = i$ and $0$ otherwise. By the universal property of free modules, there is a unique $R$-module homomorphism $\Psi : F(A) \rightarrow \bigoplus_I F_i$ such that $\Psi(a_{i,j}) = \iota(a_{i,j})$. We claim that $\Psi$ is an $R$-module isomorphism.

Suppose $x \in \mathsf{ker}\ \Psi$. Say $x = \sum r_{i,j}a_{i,j}$. Now $0 = \Psi(x)_i = \sum_j r_{i,j}a_{i,j} \in F_i$. Since $F_i$ is free on $A_i$, we have $r_{i,j} = 0$ for all $j$, for all $i$. Thus $x = 0$, so that $\mathsf{ker}\ \Psi = 0$, and thus $\Psi$ is injective.

Now suppose $(\sum_j r_{i,j} a_{i,j}) \in \bigoplus_I F_i$. Since only finitely many of the $r_{i,j}$ are nonzero, $\sum_{i,j} r_{i,j} a_{i,j} \in F(A)$. Certainly $\Psi(\sum_{i,j} r_{i,j}a_{i,j}) = (\sum_j r_{i,j} a_{i,j})$, so that $\Psi$ is surjective.

Thus $\bigoplus_I F_i \cong_R F(A)$, and thus $\bigoplus_I F_i$ is free as an $R$-module.