Take the tour ×  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.  share|improve this
question add comment  1 Answer  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 $R^2$.  share|improve this
answer 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 add comment  Your
Answer