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.

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