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I'm reading this article by A. Facchini and Z. Nazemian, in wich they discuss modules with chain conditions up to isomorphism. A couple of the main concepts are the following:

Definition We say that an $R$-module $M$ is isoartinian if for every descending chain $M\geq M_{1}\geq M_{2}\geq\cdots$ of submodules of $M$, there exists an index $n\geq 1$ such that $M_{n}$ is isomorphic to $M_{i}$, for every $i\geq n$

Definition We say that an $R$-module $M$ is isosimple if it is non zero and every non zero submodule of $M$, is isomorphic to $M$.

In sake of my question it will be necessary to know the following facts:

Fact 1 Every submodule $N$ of an isoartinian $R$-module $M$ is itself isoartinian.

Fact 2 Every isoartinian $R$-module $M$ contains an isosimple submodule $L$.

I'm trying to write down (with full details) the proof of Theorem 2.9, which establishes the following:

Theorem Any isoartinian $R$-module $M$ contains an essential submodule that is a direct sum of isosimple modules.

Proof Let $S$ denote the set of all families of independent isosimple submodules of $M$. By Zorn's Lemma, $S$ has a maximal member $W=\{V_{\lambda}\mid\lambda\in\Lambda\}$. Since every non-zero submodule of $M$ contains an isosimple module, from the maximality of $W$, it follows that $V=\oplus_{\lambda\in\Lambda}V_{\lambda}$ is essential on $M$.

Question Can anybody help me with an idea of what do they mean with the word independent in the preceding proof? It is quite evident that the independence is key to the aplication of Zorn's Lemma, which is where I'm stuck. Everything else in the proof is easy, but the autors don't mention what they mean by independent isosimple submodules.

Attempt I have tried using non-isomorphic isosimple submodules but it failed.

Any help will be tremendously appreciated.

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  • $\begingroup$ Assume that we have the following definition: a set of submodules is independent if the submodule they generate is naturally isomorphic to their direct sum. Then it is consistent with the proof. So it is likely be the right definition $\endgroup$
    – Luc Guyot
    Jul 11, 2019 at 18:22

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