Why Are Maximal Elements in Omega Prime Ideals?

[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 Proposition 3 in Chapter 2.

Proposition 3 and its proof read as follows:

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In the last sentence of the statement of Proposition 3, we read the following:

" ... ... Then \(\displaystyle \Omega\), together with the relation \(\displaystyle \le \) is a non-empty inductive system and all its maximal elements are prime ideals."My question/problem is as follows:

\(\displaystyle \Omega\) is a set of ideals and so its maximal elements would be maximal ideals ... ... BUT ... in the Corollary to Proposition 1 in this chapter, Northcott has already shown that "Every maximal ideal is a prime ideal" ... SO ... it appears that this part of Proposition 3 is unnecessary/redundant ...

But this makes me feel I am missing something or misunderstanding something ... ...

Can someone please clarify this situation ... that is why is Northcott apparently proving that all maximal ideals are prime ideals twice over ... ?

Hope someone can help ... ...
In order for MHB members to fully understand the above, I am providing Proposition 1 and its Corollary, as follows:

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[ThePerfectHacker]

"Maximal element" with respect to the ordering $\leq$, is different from saying "maximal ideal". The confusion is that the word 'maximal', is being used in two different contexts.

 

[Math Amateur]

"Maximal element" with respect to the ordering $\leq$, is different from saying "maximal ideal". The confusion is that the word 'maximal', is being used in two different contexts.

Thanks for your help ThePerfectHacker ... but i think I need a bit more help in order to fully understand your statement ...

Since \(\displaystyle \Omega\) is a set of ideals, surely then the maximal elements in \(\displaystyle \Omega\) are maximal ideals ...

Can you help further?

Thanks again for your help,

Peter

 

[ThePerfectHacker]

Thanks for your help ThePerfectHacker ... but i think I need a bit more help in order to fully understand your statement ...

Since \(\displaystyle \Omega\) is a set of ideals, surely then the maximal elements in \(\displaystyle \Omega\) are maximal ideals ...

The theorem says that if $\Omega$ is a set of ideals where $\leq$ simply means containment then all the maximal elements are prime ideals.

What is a "maximal element"? It is an ideal $I \in \Omega$ with the property that $I$ is not contained in any other ideal (other than itself) from $\Omega$. We say that $I$ is a "maximal element with respect the ordering $\leq$" i.e. it is not contained in anything else.

This is not the same as saying $I$ is a "maximal ideal", because a maximal ideal is an ideal not contained in any ideal. The property of $I$ is that it is not contained in any ideal of $\Omega$. It is a different concept.

 

[Math Amateur]

The theorem says that if $\Omega$ is a set of ideals where $\leq$ simply means containment then all the maximal elements are prime ideals.

What is a "maximal element"? It is an ideal $I \in \Omega$ with the property that $I$ is not contained in any other ideal (other than itself) from $\Omega$. We say that $I$ is a "maximal element with respect the ordering $\leq$" i.e. it is not contained in anything else.

This is not the same as saying $I$ is a "maximal ideal", because a maximal ideal is an ideal not contained in any ideal. The property of $I$ is that it is not contained in any ideal of $\Omega$. It is a different concept.

Thanks so so much for the clarification ... still reflecting on what you have written ... ...

Thanks again for your help ...

Peter