Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a submodule. Is $N$ always a direct summand of $F$?
I think the answer is yes, $N$ is also free of rank $m\leq n$ since we are working over a PID, and then $N\simeq R^m$, thus $F/N\simeq R^{n-m}$ which is a free $R$-module, so $N$ is a direct summand of $F$.
I'm just a little unsure since I can't find a reference for this online, and it seems like it should be a commonly documented fact if it is indeed true.
Best Answer
No. Let $R$ be a principal ideal domain that is not a field, and let $p\in R$ be a prime element. Then the ideal $I=pR$ cannot be a summand of $R$. Indeed, if there were an $R$-module isomorphism $R\cong I\oplus M$ for some $R$-module $M$, then we would have an injection $R/I\cong M\hookrightarrow R$, which is impossible as the source is a non-zero torsion $R$-module.
Your error is in concluding that the quotient of two free modules is free. This is not so, as the above example shows.