Suppose that $M=R^n$ and $R$ a PID, where $n\geq 1,$ and suppose $N$ is a submodule of $M$. A complement of $N$ in $M$ is a submodule $P$ of $M$ so that $M=N\oplus P$ (internal). If $A\in M_{n\times n}(R)$, the nullspace of $A$ is the submodule $\{x\in M:Ax=0\}$. Prove that
$N$ has a complement in $M \iff N$ is the nullspace of some $A\in M_{n\times n}(R)$.
My Thoughts:
So I've been able to show that if $N$ has a complement in $M$, then $N$ is the nullspace of some $A\in M_{n\times n}(R)$. But I'm having a lot of trouble with the converse.
I ended up trying to show that $M\simeq N\oplus M/N$ (External), but either I'm doing it wrong or I'm on the wrong track. Any help with this problem is greatly appreciated.
HINT
Use the theorem of elementary divisors: if $N$ is a submodule of the free module $R^n$ ($R$ a PID) there exists a basis $\{e_1,...,e_n\}$ of $R^n$, an integer $k\leq n$ and elements $r_1\mid\cdots\mid r_k$ of $R$ (here $\mid$ denotes divisibility) such that $$ \{r_1e_1,...,r_ke_k\} $$ is a basis of $N$.
The modules that have a complement are those such that $r_1=\cdots=r_k=1$ (the complement has basis $\{e_{k+1},...,e_n\}$).