MHB 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 $$\Omega$$ 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 $$\Omega$$ 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 $$\Omega$$ 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 $$\Sigma $$ is a non-empty totally ordered subset of $$\Omega$$ ... ... that is, $$\Sigma $$ is a collection of ideals that is totally ordered by inclusion ... ... hence the union, B, of the ideals in $$\Sigma $$ is also an ideal ... ... and since the ideals are totally ordered by inclusion, we have that $$B \in \Sigma $$ and since $$\Sigma $$ is a subset of $$\Omega$$, we have that $$B \in \Omega$$.

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

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