Northcott - Proposition 3 - Non-empty Inductive System

[Math Amateur]

I am reading D.G. Northcott's book: Lessons on Rings and Modules and Multiplicities.

I am currently studying Chapter 2: Prime Ideals and Primary Submodules.

I need help with an aspect of the proof Proposition 3 in Chapter 2 concerning showing that \(\displaystyle \Omega\) is a non-empty inductive system.

Proposition 3 and its proof read as follows:

View attachment 3684
View attachment 3685

Since my question relates to \(\displaystyle \Omega\) as a non-empty inductive system I am providing Northcott's definition of an inductive system, together with Zorn's Lemma for good measure ... ...

https://www.physicsforums.com/attachments/3686

I am puzzled by the role of S in the proof of \(\displaystyle \Omega\) as a non-empty inductive system because the proof seems to me to be independent of the existence and nature of S.My argument (without referring to S) is as follows:We have that \(\displaystyle \Sigma \) is a non-empty totally ordered subset of \(\displaystyle \Omega\) ... ... that is, \(\displaystyle \Sigma \) is a collection of ideals that is totally ordered by inclusion ... ... hence the union, B, of the ideals in \(\displaystyle \Sigma \) is also an ideal ... ... and since the ideals are totally ordered by inclusion, we have that \(\displaystyle B \in \Sigma \) and since \(\displaystyle \Sigma \) is a subset of \(\displaystyle \Omega\), we have that \(\displaystyle B \in \Omega\).

So if the above is correct, then the proof appears to follow without considering S ... ... Can someone please critique my analysis ...

Obviously I am missing something ... indeed, I suspect that the weak link is the assertion that because the ideals are totally ordered by inclusion, we have that \(\displaystyle B \in \Sigma \) ... ... but i cannot really see the error in this assertion ...

Hope someone can help ...

Peter

 

[Fallen Angel]

Hi Peter,

$B$ doesn't need to be on $\Sigma$, but $B$ is the union of some ideals non intersecting $S$, so any element $b\in B$ is also an element of some $I\in \Sigma$, so $b\notin S$ (because $I\cap S=\emptyset $).

Hence $B\in \Omega$

 

[Math Amateur]

Hi Peter,

$B$ doesn't need to be on $\Sigma$, but $B$ is the union of some ideals non intersecting $S$, so any element $b\in B$ is also an element of some $I\in \Sigma$, so $b\notin S$ (because $I\cap S=\emptyset $).

Hence $B\in \Omega$

Thanks for the help Fallen Angel ... ...

Still reflecting on this matter ...

Thanks again,

Peter

 

[Math Amateur]

Hi Peter,

$B$ doesn't need to be on $\Sigma$, but $B$ is the union of some ideals non intersecting $S$, so any element $b\in B$ is also an element of some $I\in \Sigma$, so $b\notin S$ (because $I\cap S=\emptyset $).

Hence $B\in \Omega$

Hi Fallen Angel,

Thanks again for your help ...I have been thinking over what you have said ... ...

We have that \(\displaystyle \Sigma\) is a non-empty, totally ordered subset of \(\displaystyle \Omega\) ... ... so B would be equal to the largest set in \(\displaystyle \Sigma\) since the total order means the sets \(\displaystyle X_i\) in \(\displaystyle \Sigma\) are such that \(\displaystyle X_1 \subseteq X_2 \subseteq X_3 \subseteq X_4 \ ... \ ... \subseteq X_n\) ... so essentially \(\displaystyle B = X_n\) and hence \(\displaystyle B \in \Sigma\) ... ...

Can you critique my analysis ... pointing out any errors or misunderstandings ...

Peter

 

[Fallen Angel]

Hi Peter,

You are assuming that $\Sigma$ is finite.

If $\Sigma$ is infinite $B$ can be out of it. For example, consider the open sets family $\{U_{n}\}_{n\in \Bbb{N}}$ where $U_{n}=(\frac{1}{n},1)$.

Then $(0,1)=\displaystyle\cup_{n\in \Bbb{N}}U_{n} \neq U_{j}, \ \forall j\in \Bbb{N}$

 

[Math Amateur]

Hi Peter,

You are assuming that $\Sigma$ is finite.

If $\Sigma$ is infinite $B$ can be out of it. For example, consider the open sets family $\{U_{n}\}_{n\in \Bbb{N}}$ where $U_{n}=(\frac{1}{n},1)$.

Then $(0,1)=\displaystyle\cup_{n\in \Bbb{N}}U_{n} \neq U_{j}, \ \forall j\in \Bbb{N}$

Hi Fallen Angel ... yes, you are right ... had not thought through the infinite case ... ...

Thanks for all your help regarding this issue ... ...

Peter