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Let $R$ be a commutative DVR, and let $M$ be the free $R$-module of finite rank $k\ge 2$. Let $N$ be a submodule of $M$ isomorphic to $R$.

Is it true that $N$ is a direct summand of $M$?

Thanks in advance.

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In most examples, R has principal ideals I whose corresponding quotients are torsion modules. it follows that I is not a summand of R, and that $I\oplus 0$ is not a summand of $R^2$.

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I'm sorry, but why $I\oplus 0$ cannot be a summand of $R^2$? $R$ cannot have torsion? – Rod Sep 11 '11 at 22:09
@Rod Right, $R$ can't have torsion because as a ring it is a domain. – Dylan Moreland Sep 11 '11 at 23:55
@Dylan Yes, I missed that. Thank you all. – Rod Sep 12 '11 at 10:07

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