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

Question

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

Answer 1

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 $R$.

Answer 2

$R$ 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.