1

Suppose that $M=R^n$, where $n\geq 1$ and $R$ is a PID and suppose $N\leq M$. A \emph{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 \emph{nullspace} of $A$ is the submodule $\{x\in M\mid Ax=0\}$. Show that $$N\text{ has a complement in }M\Leftrightarrow N\text{ is the null space of some }A\in M_{n\times n}(R).$$ Do all submodules of $M$ have a complement in $M$?

From the RHS to the LHS, I know that $N$ is a free module. And I try to expand the basis of $N$ to the basis of $M$ but I don't know whether I can do this. From the LFS to the RHS, I think should I give out an $A$ whose solution space is generated by the basis of $N$?

Can anyone help? Thanks!

Boga
  • 11

1 Answers1

1

No. A direct sum decomposition of a ring $R=I\oplus J$ corresponds to a pair of orthogonal idempotents, which do not exist except $0$ and $1$ in integral domains or local rings, for instance.

Furthermore, a submodule of a free module is not necessarily free, except in PIDs. It isn't even a projective module in general.

Bernard
  • 175,478
  • I see, $2\mathbb{Z}$ is a counter example. Can you help me with the proof of the existence of complement is equivalent to its being a null space of some A? – Boga Oct 12 '15 at 23:52
  • I can't: it is false. The situation is very different from the situation in vector spaces, due to the lack of invertible elements in rings. A direct summand of a free module is a projective module, but is not a free module itself in general. – Bernard Oct 13 '15 at 08:27