On Artinian rings

Question

Let $R$ be a ring, and for every $R$-module $M$, suppose that we have the following condition:

If $M$ is cogenerated by any finitely generated $R$-module $N$, then $M$ embeds in a finite direct sum of copies of $N$.

Does this assumption imply that $R$ is right Artinian?

Assuming that "$M$ is cogenerated by $N$" means that $M$ embeds in a direct product of copies of $N$, then if $M$ is the direct product of a large family of copies of $N$, then it's cogenerated by $N$ but is too large to embed in a finite direct sum of copies of $N$. — Jeremy Rickard, Commented Aug 2, 2017 at 9:13
Thanks for your comments. But this assumption can be happen. — Najmeh Dehghani, Commented Aug 2, 2017 at 12:19

Najmeh Dehghani · Accepted Answer · 2017-08-02 12:21:06Z

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I found the answer of the above question. The answer is this: Under this assumption R does not need to be right Artinian. Thank you.

answered Aug 2, 2017 at 12:21

Najmeh Dehghani

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