Noetherian Rings and Modules: Theorem 2.2 - Cohn - Section 2.2 Chain Conditions

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Discussion Overview

The discussion revolves around Theorem 2.2 from P.M. Cohn's "Introduction to Ring Theory," specifically focusing on the implications of the theorem regarding chain conditions in Noetherian and Artinian rings and modules. Participants are examining the relationship between parts (b) and (c) of the theorem and seeking clarification on the existence of maximal submodules within a collection of submodules.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Peter expresses an understanding that if part (b) of Theorem 2.2 holds, then any non-empty collection of submodules in \(M\) may have multiple maximal members, suggesting that different starting points in the collection can lead to different maximal submodules.
  • Some participants challenge Peter's reasoning, suggesting that it may be more beneficial to follow the proof in the text to clarify the situation regarding maximal members.
  • Peter later indicates that he has resolved his issue after revisiting the text, acknowledging the assistance he received from other participants in understanding the material.

Areas of Agreement / Disagreement

There is disagreement regarding Peter's interpretation of the implications of the theorem. Some participants do not follow his reasoning and suggest alternative approaches, indicating that the discussion remains unresolved on this specific point.

Contextual Notes

Participants reference the proof of the theorem and its implications, but there is no consensus on the interpretation of the existence of maximal submodules in the context of the theorem.

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I am reading P.M. Cohn's book: Introduction to Ring Theory (Springer Undergraduate Mathematics Series) ... ...

I am currently focused on Section 2.2: Chain Conditions ... which deals with Artinian and Noetherian rings and modules ... ...

I need help with understanding a feature of the Theorem and its proof ...

Theorem 2.2 and its proof read as follows:https://www.physicsforums.com/attachments/4900
View attachment 4901Part (c) of the above theorem effectively states that if Part (b) holds then any non-empty collection of submodules in $$M$$ has a maximal member ... ...

Now my understanding of the proof of $$\text{ (b) } \Longrightarrow \text{ (c) }$$ (which I wish someone to confirm) implies that any non-empty collection of submodules of $$M$$ may actually have several or, indeed, many maximal members ... that is members that are maximal submodules of $$M$$ ... since, following the proof of $$\text{ (b) } \Longrightarrow \text{ (c) }$$, we may start with different members of the collection $$\mathscr{C}$$ and build different strictly ascending chains which may end up having different maximal submodules ... ...

Is my analysis correct ... ... ?

I would appreciate it if someone would confirm my analysis is correct ... and/or ... point out any errors or shortcomings ...

Hope someone can help ... ...

Peter
=================================================

in order for MHB readers to appreciate the definitions and context to Theorem 2.2 in Cohn, I am providing Cohn's brief introduction to Section 2.2 Chain Conditions ... which reads as follows:View attachment 4902
 
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I am sorry, Peter, but I cannot follow this reasoning. Maybe it is best to follow the proof in the text. Or you can try this: suppose you have a non-empty collection $\mathscr{C}$ of submodules of $M$ that does not have a maximal member, what happens then?
 
steenis said:
I am sorry, Peter, but I cannot follow this reasoning. Maybe it is best to follow the proof in the text. Or you can try this: suppose you have a non-empty collection $\mathscr{C}$ of submodules of $M$ that does not have a maximal member, what happens then?
Thanks for reminding me of this issue, Steenis ...

I just revisited the Cohn text ... and have resolved the issue ...

By the way ... your help in the past has been critical and crucial to my understanding of ring and module theory ...

So thank you ...!

Peter
 
You are welcome, Peter, it helps me too.
 

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