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Let $R$ be a commutative ring with $1$.

An $R$-module $K$ has the 'S' property if $K/T \simeq K$ (i.e. isomorphic) implies that the submodule $T$ is trivial.

By Fitting's lemma any Noetherian module has the 'S' property. There exist non-Noetherian modules with this property. For example the infinite product of $Z_{2}\times Z_{3}\times Z_{5}\times...$ running over all of the primes has the 'S' property, but is not Noetherian.

I am curious if there is a characterization of these kinds of modules out there.

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    $\begingroup$ What do you mean $K/T=K$? Do you mean $K/T$ is isomorphic to $K$? $\endgroup$
    – Keenan Kidwell
    Commented Mar 23, 2010 at 20:31
  • $\begingroup$ Yes, sorry isomorphism. $\endgroup$
    – Johannes Wachs
    Commented Mar 23, 2010 at 20:33

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These are called Hopfian modules. I didn't see any particularly exciting general characterization, but there are several special case characterizations (that show up easily in a google or mathscinet search). There are also several papers devoted to giving "interesting" examples.

An exercise in Lam's Lectures on Modules and Rings asks one to prove that every finitely generated module over a commutative ring is Hopfian (so if the ring is not-noetherian, this is a generalization).

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