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Let $M$ be an $R$-module ($R$ commutative ring with unity) such that for every non-zero submodule $N$ of $M$ , $M/N$ is not isomorphic to $M$; then what can we say about $M$? Can we say $M$ is finitely generated? If in general $M$ is not finitely generated, then does some condition on $R$ implies $M$ is finitely generated? I am only able to prove that if $M$ is free then $M$ is finitely generated. Please help. thanks in advance

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Here's an example that's not finitely generated.

Let $R=\mathbb{Z}$ and $$M=\bigoplus_{p\text{ prime}}\mathbb{Z}/p\mathbb{Z}.$$

Then $M$ has elements of order $p$ for every prime $p$, but if $N$ is a non-zero submodule, then for some prime $p$, $M/N$ has no elements of order $p$, so $M\not\cong M/N$.

Or, again with $R=\mathbb{Z}$, take $M=\mathbb{Q}$. Every non-zero group endomorphism of $\mathbb{Q}$ is an isomorphism, so it's not isomorphic to a proper quotient of itself.

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    $\begingroup$ This can be generalized to any ring having infinitely many simple modules, because a decomposition into a direct sum of simple modules “complements direct summands”. $\endgroup$ – egreg Jul 21 '17 at 20:05
  • $\begingroup$ @Jeremy Rickard : Thanks for the example . Do you know of any conditions on $R$ which forces any such $M$ to be finitely generated ? $\endgroup$ – user Jul 22 '17 at 8:12
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    $\begingroup$ @users The only examples I've thought of are artinian rings of finite representation type (i.e., with only finitely many isomorphism classes of indecomposable modules). Then any module $M$ is a direct sum of copies of the indecomposables, and if the module is not finitely generated then some indecomposable $N$ must occur infinitely often as a direct summand, and so $M\cong M\oplus N$ and $M$ is isomorphic to a proper quotient of itself. $\endgroup$ – Jeremy Rickard Jul 22 '17 at 13:20
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    $\begingroup$ @users I've added an even more elementary example. $\endgroup$ – Jeremy Rickard Jul 22 '17 at 18:28
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    $\begingroup$ @JeremyRickard : Sorry , that should have been " The fraction field of a domain $R$ cannot be a finitely generated module over $R$ unless $R$ itself is a 'field' " ... $\endgroup$ – user Jul 23 '17 at 7:46

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