I read the proof in Lang's Algebra (page 880) that submodules of free $R$-modules for $R$ a PID are themselves free. I can't figure out where the proof requires commutativity of $R$. Is it possible to remove this assumption? In division rings, there are no zero divisors, and the only ideals are $0$ and $R$, so in particular every ideal is principal. Over division rings every module is free, so the result that submodules of free modules are free holds in this case.
Is it true in general that rings $R$ without zero divisors where every left ideal is principal have the property that submodules of free left $R$-modules are free? This part of the question is less important, but if commutativity is required, then where in Lang's proof does he assume it?
I haven't looked at the proof in Lang, but the result is true even over noncommutative rings. More generally, if $R$ is left hereditary (i.e. all left ideals are projective), then every submodule of a free left $R$-module is isomorphic to a direct sum of ideals in $R$. The result in question follows from this.