I need to prove the following:
Let $R$ be an arbitrary ring (according to Lang, Bourbaki etc., so, with $1$). Let $M$ be a free $R$-module, and let $A$ be one of it's bases. If $|\mathbb{N}| \leq |A|$, then all bases of $M$ have the same cardinality.
Any ideas on how to prove it? Obviously, every element of $A$ is a unique finite sum of elements of $B$, and vice versa.
If a free module has an infinite basis, it is not finitely generated; so any two bases must be infinite.
Suppose $B$ and $C$ are bases. For each $b\in B$, there is a finite subset $C(b)$ of $C$ such that $b$ is a linear combination of the elements in $C(b)$. Thus we have a map $b\mapsto C(b)$; then $|B|\ge|\{C(b):b\in B\}|$. Since each $C(b)$ is finite, we have $$ \Bigl|\bigcup_{b\in B}C(b)\Bigr|\le \aleph_0|B|=|B| $$ Now, prove that $$ \bigcup_{b\in B}C(b)=C $$ By symmetry, $|B|\le|C|$.