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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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    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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    @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