Finitely Generated Modules and Artinian Rings

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SUMMARY

The discussion centers on the relationship between finitely generated modules and Artinian rings, specifically proving that if $S$ is an Artinian ring and $R_S$ is a finitely generated $S$-module, then $R$ must also be Artinian. A lemma is referenced stating that if $R$ is Artinian, any finitely generated $R$-module $V$ is also Artinian. The participants confirm the validity of the proof found in "A Course in Ring Theory" by Passman, which utilizes several other lemmas to establish the conclusion.

PREREQUISITES
  • Understanding of Artinian rings and modules
  • Familiarity with finitely generated modules
  • Knowledge of ring theory concepts and terminology
  • Ability to follow mathematical proofs and lemmas
NEXT STEPS
  • Study the properties of Artinian rings and modules in detail
  • Learn about finitely generated modules and their implications in ring theory
  • Review the proof of the lemma regarding finitely generated modules in Artinian rings
  • Explore additional resources on ring theory, particularly "A Course in Ring Theory" by Passman
USEFUL FOR

Mathematicians, students of abstract algebra, and anyone studying ring theory who seeks to understand the implications of finitely generated modules in relation to Artinian rings.

Sudharaka
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Hi everyone, :)

Here's another question that I am struggling to complete. If you have any hints or suggestions for this one, I would be so grateful. :)

Question:

Let $S\subseteq R$ be rings and assume that $R_S$ is a finitely generated $S$-module. If $S$ is Artinian prove that $R$ is also Artinian.
 
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Sudharaka said:
Hi everyone, :)

Here's another question that I am struggling to complete. If you have any hints or suggestions for this one, I would be so grateful. :)

Question:

Let $S\subseteq R$ be rings and assume that $R_S$ is a finitely generated $S$-module. If $S$ is Artinian prove that $R$ is also Artinian.

I came up with an answer and it would be nice if someone can confirm it, or show mistakes in it. :)

There is a lemma that says, "Let $V$ be a finitely generated $R$-module. If $R$ is Artinian, then so is $V$." By this lemma, we know that $R_S$ ($R$ as a $S$-module) is Artinian. So if we take any decreasing chain of $R$-submodules of $R_R$;

\[R_R \supseteq W_1\supseteq \cdots \supseteq W_n \supseteq \cdots\]

Each $R$-submodule is also a $S$-submodule. Since $R_S$ is Artinian the above chain should stabilize at some point and therefore $R_R$ is also Artinian.
 
I don't see a problem with this, but...

Can you prove the lemma?
 
Deveno said:
I don't see a problem with this, but...

Can you prove the lemma?

Thank you so much for confirming. :) Yes I can prove it since I went through the proof which is in the textbook I am referring for Ring Theory (A Course in Ring Theory by Passman). The proof in lengthy and uses several other lemmas which are mentioned previously in the book, so I am not going to write down it here.

Thanks again for all your help. I sometimes wonder how some people (like you) grasp hard concepts in Ring Theory very easily whereas I have to go through the textbooks, internet, forums etc, for hours to figure them out. :)
 
Don't get discouraged...I found rings very hard, too.
 
Sudharaka said:
Thank you so much for confirming. :) Yes I can prove it since I went through the proof which is in the textbook I am referring for Ring Theory (A Course in Ring Theory by Passman). The proof in lengthy and uses several other lemmas which are mentioned previously in the book, so I am not going to write down it here.

Thanks again for all your help. I sometimes wonder how some people (like you) grasp hard concepts in Ring Theory very easily whereas I have to go through the textbooks, internet, forums etc, for hours to figure them out. :)

Thank you for this post Sudharaka.

It is heartening to find that other members find ring theory a challenge ...

... very rewarding when you achieve understanding of such a wonderful theory though ...
Peter
 
Peter said:
Thank you for this post Sudharaka.

It is heartening to find that other members find ring theory a challenge ...

... very rewarding when you achieve understanding of such a wonderful theory though ...
Peter

Hi Peter, :)

You are welcome, surely you'll see me posting a lot of ring theory questions.
 

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