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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.

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    $\begingroup$ Title answer: always, see here. $\endgroup$ Mar 7, 2019 at 14:11
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    $\begingroup$ @DietrichBurde This is not true, $\mathbb{Q}$ is not isomorphic to $\mathbb{Z} \oplus \mathbb{Q}/\mathbb{Z}$ as abelian groups. $\endgroup$
    – Junkyards
    Mar 7, 2019 at 14:18
  • $\begingroup$ You said always which I did not find clear in this context, as you pointed out the title specifically. Here this works because of this specific context, since you can find the map $\rho$ which is given as an assumption in your link. $\endgroup$
    – Junkyards
    Mar 7, 2019 at 14:25
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    $\begingroup$ @Junkyards Thank you for your help. My comment should be formulated more precisely, you are right. $M$ is a free $R$-module of rank $n$. $\endgroup$ Mar 7, 2019 at 14:34

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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 $\{e_{k+1},...,e_n\}$).

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