MHB Why Is the Containment xR ⊆ x1R Proper in Lemma 4.3.10?

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The discussion centers on understanding the proof of Lemma 4.3.10 from Paul E. Bland's "Rings and Their Modules," specifically the proper containment of xR in x1R. The goal is to construct an increasing chain of ideals, demonstrating that each inclusion is proper, which is crucial for showing that the chain must terminate in a Noetherian ring. The termination at x_nR implies that x_n is primitive, as any non-primitive element would allow for an extension of the chain, leading to a contradiction. Participants confirm that if x_n is not primitive, it can be expressed as a product involving a non-unit, thus contradicting the assumption of termination. This reinforces the importance of the proper containment in the overall proof structure.
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I am reading Paul E. Bland's book, "Rings and Their Modules".

I am focused on Section 4.3: Modules Over Principal Ideal Domains ... and I need some help to fully understand the proof of Lemma 4.3.10 ... ...

Lemma 4.3.10 and its proof read as follows:https://www.physicsforums.com/attachments/8283
View attachment 8284I do not follow the strategy of the proof ... for example Bland writes:

" ... ... Thus, $$xR \subseteq x_1 R$$ and we claim that this containment is proper ... ... "But ... why is Bland proving that this containment is proper ... what is the point ... how is this furthering the proof ...?Peter
 
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He wants to construct an increasing chain
$$xR \subsetneq x_1R \subsetneq x_2R \subsetneq \cdots $$

in $F$, in which each containment is proper.
But he clains that $F$ is noetherian (by prop.4.2.11) and therefore this chain must terminate.
 
steenis said:
He wants to construct an increasing chain
$$xR \subsetneq x_1R \subsetneq x_2R \subsetneq \cdots $$

in $F$, in which each containment is proper.
But he clains that $F$ is noetherian (by prop.4.2.11) and therefore this chain must terminate.
Thanks for the help, Steenis ...

Appreciate your assistance ...

Just a further point ...

At the end of Bland's proof we read ...

" ... ... If the chain terminates at $$x_n R$$ then $$x_n$$ is primitive ... ... " Why/how does the chain terminating at $$x_n R$$ imply that $$x_n$$ is primitive ... ...?Peter***EDIT***

Thinking about it a bit more I suspect that if $$x_n$$ is not primitive then we can extend the chain by showing $$x_n R \subsetneq x_{n+1} R $$ ... but ... contradiction! ... the chain terminates at $$x_n$$ ...

Is that correct ... ?

Peter
 
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Yes correct. If $x_n$ is not primitive then $x_n = x_{n+1} b$ with $x_{n+1} \in F$ and $b \in R$ is not a unit. And then $x_n R \subsetneq x_{n+1} R$. The same argument is used a few times in the proof.
 
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