# Finite rank free modules over PIDs

I have two finite rank free modules $M,N$ over a principal ideal domain (the ring of formal complex power series to be exact) and a homomorphism between these two modules $\phi:M\rightarrow N$. Under what conditions does the kernel $\ker \phi$ have a complement $C$ in $M$ such that I can write $M=C\oplus \ker\phi$.?

I guess this question might be considered very elementary by many, so I'd also be happy to just be given a reference to a good text book. Searching Google did not turn up anything useful.

Thanks in advance for any replies.

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This will always be the case. Since $N$ is free, $M/\ker\phi$ is torsion-free, hence free, so there is a section $M/\ker\phi\rightarrow M$; you can take $C$ to be the image of this section. By the way, in the future you should use math.stackexchange.com – Kevin Ventullo May 22 '13 at 2:00
Adding to Kevin's comment, this would still be true over a Dedekind ring, since $M/ker\phi$ is a submodule of a free module and hence projective. – Steven Landsburg May 22 '13 at 2:44