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.