Why Are Maximal Elements in Omega Prime Ideals?

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In summary: Thanks so so much for the clarification ... still reflecting on what you have written ... ...Thanks again for your help ...
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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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"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.

ThePerfectHacker said:
"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

Peter said:
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.

ThePerfectHacker said:
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

1. What is the difference between a prime ideal and a maximal ideal?

A prime ideal is a subset of a ring that is closed under addition and multiplication, and also has the property that if the product of two elements in the ideal is in the ideal, then at least one of the elements must be in the ideal. A maximal ideal, on the other hand, is an ideal that is not a subset of any other proper ideal. In other words, it is the largest possible ideal in a given ring.

2. How can prime ideals and maximal ideals be used in ring theory?

Prime ideals and maximal ideals play an important role in the study of ring theory. They help to classify rings into different types, and also provide a way to study the structure and properties of rings. For example, prime ideals are used to define prime rings, which have important applications in algebraic geometry and number theory.

3. Are prime ideals and maximal ideals always distinct?

No, in some cases, a prime ideal and a maximal ideal can be the same. For example, in a commutative ring with unity, every maximal ideal is also a prime ideal. In other cases, such as in the ring of integers, there can be distinct prime and maximal ideals.

4. Can a prime ideal or maximal ideal be generated by a single element?

Yes, both prime ideals and maximal ideals can be generated by a single element. A prime ideal can be generated by any of its elements, while a maximal ideal can be generated by any element that is not in the ideal itself.

5. How are prime ideals and maximal ideals related to the concept of irreducibility?

In a commutative ring, a prime ideal is equivalent to an irreducible ideal, and a maximal ideal is equivalent to a maximal irreducible ideal. This means that a prime ideal cannot be factored into smaller ideals, and a maximal ideal cannot be properly contained in any other irreducible ideal. This relationship is important in understanding the structure of rings and their factorization properties.

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