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If $M_1$ , $M_2$ are Noetherian modules such that $M_1$ is isomorphic to a submodule of $M_2$ and $M_2$ is isomorphic to a submodule of $M_1$ , then is $M_1 \cong M_2$ ?

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    $\begingroup$ Context? Thoughts? $\endgroup$ – rschwieb Nov 20 '16 at 17:01
  • $\begingroup$ @rschwieb : it is just a curiosity ... and I haven't really gotten anywhere , any progress with it ... $\endgroup$ – user228168 Nov 20 '16 at 17:29
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No. For instance, let $k$ be a field and $R=k[x,y]$, and consider the $R$-modules $M_1=R$ and $M_2=(x,y)\subset R$. Then $M_2$ is a submodule of $M_1$, and $M_1$ is isomorphic to the submodule of $M_2$ generated by $x$ (or any other nonzero element). But $M_1$ and $M_2$ are not isomorphic.

More generally, if $R$ is any Noetherian domain with a nonprincipal ideal $I\subset R$, $M_1=R$ and $M_2=I$ gives a counterexample similarly.

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