Modules over Principal Idea Domains .... and Primitive Elements .... Bland, Lemma 4.3.10 .... ....

[Math Amateur]

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, \(\displaystyle xR \subseteq x_1 R\) 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

 

[steenis]

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.

 

[Math Amateur]

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 \(\displaystyle x_n R\) then \(\displaystyle x_n\) is primitive ... ... " Why/how does the chain terminating at \(\displaystyle x_n R\) imply that \(\displaystyle x_n\) is primitive ... ...?Peter***EDIT***

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

Is that correct ... ?

Peter

 

[steenis]

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.