Why is a submodule of a free module over a PID is free?

Question

Rotman - Advanced modern algebra p.650 Theorem 9.8

Let $R$ be a PID and $M$ be a free $R$-module and $N$ be an $R$-submodule of $M$.

Let $\beta$ be an $R$-basis for $M$ and well-order it.

Now, define $M_a'=\langle x\in\beta:x<a \rangle$ and $M_a=\langle x\in \beta: x\leq a\rangle$.

Then, define $N_a=M_a\cap N$ and $N_a'=M_a'\cap N$ for all $a\in\beta$.

Then, $N_a/N_a'=N_a/(N_a\cap M_a')\cong (N_a+M_a')/M_a' \subset M_a/M_a'\cong R$.

Hence $N_a/N_a'$ can be embedded into $R$ and since $R$ is a PID, $N_a/N_a'\cong \langle p\rangle $ for some $p\in R$.

But how the above argument asserts that, if $p$ is nonzero, there exists $h\in N_a$ such that $N_a=N_a'\oplus \langle h \rangle$ and $\langle h \rangle\cong R$?

I can only see that there exists $h$ such that $N_a=N_a'\oplus \langle h \rangle$, but I'm not sure why is $\langle h\rangle\cong R$..

Answer 1

Since the submodule is torsion free, the mapping from $R$ to $\langle h \rangle$ by $r \mapsto rh$ is an isomorphism.

Answer 2

This is because, as $Rp$ is a free $R$-module, the short exact sequence: $$0\to N'_a\to N_a\to Rp\to 0$$ splits.