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Show that an integral domain $R$ is principal if and only if every submodule of a cyclic $R$-module is also cyclic.

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Hints. If $M$ is a cyclic $R$-module, then $M\simeq R/I$, $I$ ideal of $R$. A submodule of $M$ corresponds to $J/I$, with $J$ an ideal of $R$ containing $I$. Use now that $R$ is a principal ideal domain.

For the converse consider the cyclic $R$-module $R$.

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  • $\begingroup$ is it true that if T is a PID, then R/I is a PID, such that every ideal of R/I is an ideal? If not, I don't see how this proof works? I don't quite understand this proof. Can you explain more? Thanks $\endgroup$ – user10024395 Mar 17 '15 at 3:10
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$R$ is a cyclic $R$-module, and so if every submodule, i.e ideal, is cyclic $R$ is a PID.

Vice-versa, if $R$ is a PID and $M$ a cyclic $R$-module, then $M = \langle m \rangle$ and so $$M \cong R/\operatorname{Ann}(m) $$ so every submodule of $M$ corresponds to a principal ideal $I \subseteq R$ with $\operatorname{Ann}(m) \subseteq I $ and so is cyclic.

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  • $\begingroup$ can you explain more about why why submodule of M corresponds to a principal ideal $I \subseteq R$ with $Ann(m)\subseteq I$ and so is cyclic? Thank you $\endgroup$ – user10024395 Mar 24 '15 at 14:50

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