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Let R be a commutative pid, and let M be the free R-module of finite rank k.

Given a non-zero proper submodule N of M, does there always exist a projection P such that ker(P)=N? If so, how can we construct such a projection?

EDIT: By projection, I mean an idempotent endomorphism.

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    Please clarify a little bit! Sometimes the quotient map $\pi:M\rightarrow M/N$ is called a projection. Sometimes it is required that a projection should be idempotent, i.e. $P:M\rightarrow M$, and $P^2=P$. Which is the case here? I suspect the latter, but in that case there are counterexamples...2011-09-11
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    Sure, sorry. I'll edit the question.2011-09-11

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Hint: If there exists an element $m\in M$ such that $m\notin N$, but $rm\in N$ for some non-zero $r\in R$, what can you say about $P(m)$?

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    Since $R$ is a domain, $P(m)$ would be $0$ by linearity. I don't see how could we have such a pair $(m,r)$. I want also to tell you that I'm actually interested when $R$ is dvr.2011-09-11
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    $N=\pi M$ for $\pi =$ the generator of the maximal ideal of a DVR $R$?2011-09-11
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    Oohh, I see... but what if $N$ is small, say $N$ isomorphic to $R$?2011-09-11
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    $M=R$, $N=\pi R\simeq R$.2011-09-11
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    If $k=2$, do you think that it is true that $N$ is a direct summand of $M$, at least?2011-09-11
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    Well, I think I should ask a new question. Thank you for your help.2011-09-11
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    You can, of course, imbed $\pi R\subset R \subset R^k$ in several ways. The ultimate criterion is that a projection exists, iff and only if $N$ is a direct summand of $M$. In the case of a PID (and finitely generated $M$) this is equivalent to $M/N$ not having any torsion. Over a PID torsion free + f.g. implies free, hence projective, hence a direct summand. I'm off air (past my bed time).2011-09-11
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    @Rod: Sorry. I missed your last comment. Yes, N has to be a summand. For an idempotent endomorphism of $M$ you always have $$M=\ker P \oplus \mathrm{Im} P.$$2011-09-11