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Let $R$ be a ring with unity and $M$ any unitary right $R$-module. We say that $M$ is finitely cogenerated if whenever $\lbrace A_\lambda \rbrace_{\lambda\in \Lambda}$ is a collection of submodules of $M$ with $\bigcap_{\lambda\in \Lambda} A_\lambda=0$ there exists a finite subset $\Lambda'\subset \Lambda$ such that $\bigcap_{\lambda\in \Lambda'} A_\lambda=0$. It's proven that a module is finitely cogenerated iff the intersection of any chain of nonzero submodules is nonzero.

For an essential submodule $B\subseteq M$, we write $B \subseteq^{\mathrm{ess}}M$.

Assume that whenever $K \subseteq^{\mathrm{ess}} N \subseteq M$ then $N/K$ is finitely cogenerated. How to prove that this implies that $M/A$ is finitely cogenerated for any submodule $A$ with $\mathrm{soc}(M) \subseteq A \subseteq M$, where $\mathrm{soc}(M)$ is the socle of $M$ which is the sum of all simple submodules (or, the intersection of all essential submodules).

I need any help. Thanks in advance.

Lemmon
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Hussein Eid
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  • You make $soc(A)\subseteq A\subseteq M$ sound like a restriction, but it is true for all submodules $A$... was this intended? I wonder if you meant $soc(M)$ or something else. – rschwieb Jun 06 '23 at 18:07
  • I've voted to close for improvement because this is a poorly titled PSQ question. Feel free to ping me if improvements are made and I can cast alternative votes then. – rschwieb Jun 06 '23 at 18:09
  • Oh! I edited the question. It was intended to be $soc(M)$ not $soc(A)$. Can you please now remove your vote for closing the question?!. @rschwieb – Hussein Eid Jun 06 '23 at 20:23
  • @HusseinEid. Thanks for fixing the statement! I would like to change the vote, but the post has not yet grown beyond a Problem Statement Question. What are the dead ends you've encountered in your attempts? Much like others i have a lot of desire to help you solve problems but I have little desire to do them entirely from scratch for you... – rschwieb Jun 06 '23 at 20:25
  • Unfortunately, I don't even know how to begin. I thought of taking a complement of $soc(M)$, say $K$, in $A$ so that $soc(M)\oplus K \subseteq^{ess} A$. – Hussein Eid Jun 06 '23 at 20:27
  • @HusseinEid What about supposing $M/A$ is not f.cog, then looking at $A\subseteq E(A)$, the injective hull? $E(A)$ will be a summand in $M$, and $E(A)/A$ would be f.cog, and both could be useful. – rschwieb Jun 07 '23 at 14:58
  • @HusseinEid Can you also please improve the title of your question and transfer your comment work to the body of the question? – rschwieb Jun 07 '23 at 14:59
  • May you suggest please an appropriate title of the question?! @rschwieb – Hussein Eid Jun 07 '23 at 15:38
  • Use your question. Like "Show that if $K\subseteq_eN\subseteq M$ implies $N/K$ is f.cog for all $N,K$, then $M/A$ is f.cog for any $A\supseteq soc(M)$." You even have 24 characters leftover: you might be able to spell out "cogenerated" – rschwieb Jun 07 '23 at 18:14
  • Why $E(A)$ is a summand of $M$?!. Is it even a submodule of $M$?!. It is known that $E(A) \subseteq E(M)$ and hence $E(A)$ is a summand of $E(M)$. @rschwieb – Hussein Eid Jun 08 '23 at 02:36
  • @HusseinEid yes, you're right: might not even be a submodule. A slip on my part. – rschwieb Jun 08 '23 at 02:46
  • You might get more views if, say, you lifted a finger to use the title I typed out for you. – rschwieb Jun 08 '23 at 14:09

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