## Free modules over two sets of the same cardinality are isomorphic

Let $R$ be a ring with 1, and let $A$ and $B$ be sets of the same cardinality. Prove that the free modules $F(A)$ and $F(B)$ are isomorphic.

We begin with a lemma.

Lemma: Let $R$ be a ring with 1 and let $\theta : A \rightarrow B$ be a set map. Note that $\iota \circ \theta : A \rightarrow F(B)$, where $F(B)$ is the free $R$ module on $B$ and $\iota : B \rightarrow F(B)$ is the inclusion map. By the universal prperty of free modules, the induced map $\Theta : F(A) \rightarrow F(B)$ is an $R$-module homomorphism. If $\theta$ is injective, then so is $\Theta$, and if $\theta$ is surjective, then so is $\Theta$. Proof: Suppose $\theta$ is injective, and let $x \in \mathsf{ker}\ \Theta$. Now $\Theta(x) = \Theta(\sum r_i \cdot a_i) = \sum r_i \Theta(a_i)$ $= \sum r_i \theta(a_i) = 0$. Since $\theta$ is injective, the $\theta(a_i)$ are distinct. Since $F(B)$ is free on $B \supseteq \mathsf{im}\ \theta$, we have $r_i = 0$ for each $i$. Thus $x = 0$, and so $\mathsf{ker}\ \Theta = 0$. So $\Theta$ is injective. Suppose now that $\theta$ is surjective, and let $y = \sum r_i b_i \in F(B)$. Since $\theta$ is surjective, we have $b_i = \theta(a_i)$ for some $a_i \in A$ (not necessarily distinct). Then $y = \sum r_i \cdot \theta(a_i) = \Theta(\sum r_i a_i)$, so that $\Theta$ is surjective. $\square$

Since in our case $\theta$ is both injective and surjective, $\Theta$ is bijective, hence an $R$-module isomorphism.