Let $R$ be a ring and $M$ be a left $R$-module. Assume $M$ is free and admits a finite basis. Does it follow that all linearly independent subsets of $M$ can be extended to bases? Under which circumstances can this be done?
My attempt:
Let $L$ be a linearly independent subset of $M$. Let $B$ be a basis of $M$. Either $L$ spans $M$ or it doesn't. If $L$ spans, we're done. If $L$ doesn't, then we can choose $b$ to be an element of $B$ not spanned by $L$. We seek to show that $L \cup \{b\}$ is linearly independent. If it weren't, then we would have that $\sum_i \lambda_i l_i = \lambda b$ for $\lambda \neq 0$ (where the $l_i$ all belong to $l$). But I can't divide by $\lambda$, so now I'm stuck.
Well, $\{2\}$ is a linearly independent set in $R=\mathbb Z$ but you can't extend it to a basis of $\mathbb Z$ because then it would have too many elements...
It seems like a necessary and sufficient condition for a free submodule $B$ of a free module $A$ to have a free complement $C$ is that $A/B$ needs to be a free module.