This is problem 2.26(iii) in Rotman's Homological Algebra.
1. Problem 2.26(iii) (R a ring with IBN)
2. Partial Proof
Let $W \subset X$ be finite and define $$W' = \{b \in B : \text{supp}(b) \subset W \} $$
By construction, if $b \in W'$ then $b \in R^{(W)}$ so that $R^{(W')} \subset R^{(W)}$. Let $W_0 = \{ w \in W: \text{supp}(w) \notin W' \}$. The set $W' \cup W_0$ may not be linearly independent, but removing a finite number of elements (only from $W_0$) will achieve this--call this set $\hat W$.
Now we have $R^{(\hat W)} = R^{(W)}$ so that $|\hat W| = |W|$ (because $R$ is IBN). Since $W' \subset \hat W$, $W'$ is finite as desired.
(I need to show that $|B| = \text{Finn}(X) = |X|$).
We have shown that any two bases of a free module have the same cardinality when $R$ is IBN. Let $\phi : R^{(X)} \xrightarrow{\sim} R^{(Y)}$ be the assumed isomorphism. Then $\phi(X)$ is a basis of $R^{(Y)}$ so that
$$ |X| = |\phi(X)| = |Y| $$
3. Questions
I'm not sure if I properly extended the basis of $W'$ to $W$--I'm obviously trying to show that when $R$ has IBN, a free submodule of a module with finite basis must itself have finite basis.
Why does $|B| = |\text{Finn}(X)|$ I know that if $W = \{x_1, \ldots x_n\}$, each $x_i$ is a unique linear combination of elements in $B$ and also each $b$ is a unique linear combination of elements in $X$. But I don't know if there's a bijection $b \mapsto \text{supp}(b) =: W$.
Why does $|X| = |\text{Finn}(X)|$ when $|X|$ is infinite?
