MHB Why Are Maximal Elements in Omega Prime Ideals?

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SUMMARY

The discussion centers on the interpretation of Proposition 3 from D.G. Northcott's "Lessons on Rings and Modules and Multiplicities," specifically regarding the relationship between maximal elements in the set of ideals $$\Omega$$ and prime ideals. It clarifies that "maximal element" refers to an ideal that is not contained in any other ideal from $$\Omega$$, while "maximal ideal" is a broader term not limited to the context of $$\Omega$$. The theorem asserts that all maximal elements in $$\Omega$$ are indeed prime ideals, which is not redundant but a necessary distinction in the context of the ordering $$\leq$$ defined by containment.

PREREQUISITES
  • Understanding of ring theory and ideal theory
  • Familiarity with the concepts of prime ideals and maximal ideals
  • Knowledge of the ordering relation in set theory
  • Basic comprehension of inductive systems in mathematics
NEXT STEPS
  • Study the definitions and properties of prime ideals in ring theory
  • Explore the differences between maximal elements and maximal ideals in depth
  • Review the proof of Proposition 1 and its corollary in Northcott's text
  • Investigate the implications of inductive systems in algebraic structures
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Mathematicians, particularly those specializing in abstract algebra, students studying ring theory, and anyone seeking to deepen their understanding of ideals in algebraic structures.

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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:

View attachment 3680
View attachment 3681

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

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

$$\Omega$$ 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:

View attachment 3682
View attachment 3683
 
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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 $$\Omega$$ is a set of ideals, surely then the maximal elements in $$\Omega$$ 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 $$\Omega$$ is a set of ideals, surely then the maximal elements in $$\Omega$$ 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
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

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