I am studying the Cohn's book "Free ideal Rings and Localization in General Rings", and there is something he takes for granted and I cannot find its reason in any part of the book. The question is:
A ring $R$ is said a (left) $n$-fir iff every left ideal generated by at most $n$ elements is free of unique rank (thinking of the ideal as a left $R$-module). Then, if $m\leq n$, Cohn used the fact that $R^m$ has unique rank (as a left $R$-module). I cannot figure why.
If $R$ has an ideal $I$ that is free of rank $k>1$ then it has a free ideal of rank $n$ for all $n\geq0$, since $I\cong R^k$ has a subideal isomorphic to $I\oplus R^{k-1}\cong R^{2k-1}$, and so on.
So in this case, if $R$ is also an $n$-fir, it has ideals isomorphic to $R,R^2,\dots,R^n$ that all have unique rank.
If $R$ does not have an ideal that is free of rank $2$, but is a domain (as any $n$-fir is), then any two non-zero elements $a,b\in R$ must satisfy $Ra\cap Rb\neq\{0\}$, and so $R$ is a left Ore domain, and these are known to satisfy the invariant basis number property (IBN) for other reasons.