Northcott - Proposition 3 - Inductive Systems Maximal Elements are Prime Ideals

In summary, In the book Lessons on Rings and Modules and Multiplicities by D.G. Northcott, in Chapter 2, Proposition 3 is presented along with its proof. In the proof, there is a statement that all the maximal elements of $\Omega$ are prime ideals. The author goes on to use this statement to prove the proposition. However, the reader, Peter, does not understand why it follows that the ideal $C$ meets the set $S$. After further explanation and clarification, it is understood that since $P$ is maximal in the set of ideals that do not intersect $S$ and $C$ properly contains $P$, $C$ cannot be in $\Omega$ and must intersect $S$.
  • #1
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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 the proof Proposition 3 in Chapter 2 concerning the demonstration that all the maximal elements of \(\displaystyle \Omega\) are prime ideals.Proposition 3 and its proof read as follows:

https://www.physicsforums.com/attachments/3714
https://www.physicsforums.com/attachments/3715

In Northcott's proof above, we read the following:

" ... ... Let \(\displaystyle C\) consist of all elements that can be expressed in the form \(\displaystyle r \alpha + \pi\) where \(\displaystyle r \in R\) and \(\displaystyle \pi \in P\). It is a simple matter to check that \(\displaystyle C\) is an ideal containing \(\displaystyle P\). Indeed, since \(\displaystyle \alpha = 1 \alpha + 0\), \(\displaystyle \alpha \in C\) and therefore C strictly contains P. However P is maximal in \(\displaystyle \Omega\). It therefore follows that \(\displaystyle C\) meets \(\displaystyle S\). ... ... "

I do not fully understand why, in the above argument, it follows that \(\displaystyle C\) meets \(\displaystyle S\).

Can someone show formally and rigorously whhy, exactly, it follows that \(\displaystyle C\) meets \(\displaystyle S\).

Help will be appreciated ... ...

Peter
 
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  • #2
Hi Peter,

$P$ is maximal in the set of ideals that do not intersect $S$, and $C$ properly contains $P$, so $P$ can't be in this set, but it's still an ideal, so it must intersect $S$
 
  • #3
Fallen Angel said:
Hi Peter,

$P$ is maximal in the set of ideals that do not intersect $S$, and $C$ properly contains $P$, so $P$ can't be in this set, but it's still an ideal, so it must intersect $S$

Thanks for the help, Fallen Angel, but i need further help on this matter ...You write:

"... $P$ is maximal in the set of ideals that do not intersect $S$ ... "

Yes, understand this ...
You then write:

" ... $C$ properly contains $P$ ..."

Yes, follow this as well ...
BUT ... then you write:

" ... so $P$ can't be in this set ... "I do not really follow ... which set do you mean when you say "this set" ...Can you please clarify ...

... ... ... ... thinking further ... ...
... ... I suspect that you mean \(\displaystyle \Omega\) by "this set" ... but then \(\displaystyle P\) is a maximal element in this set ... so then doesn't it follow that \(\displaystyle P\) is in this set (is that correct?)

I am thinking it may be that a maximal element of \(\displaystyle \Omega\) need not be in \(\displaystyle \Omega\) ... ... I have been assuming that if \(\displaystyle P\) is a maximal element of \(\displaystyle \Omega\) then \(\displaystyle P \in \Omega\) ...

Can you clarify this matter ...

Peter
***EDIT***

It appears that my thoughts - see above:

" ... ... I am thinking it may be that a maximal element of \(\displaystyle \Omega\) need not be in \(\displaystyle \Omega\) ... ...

are not correct ...

On page 71, Northcott writes the following:" ... ... An element \(\displaystyle \mu\) of \(\displaystyle \Omega\) is called a maximal element if from \(\displaystyle x \in \Omega\) and \(\displaystyle \mu \le x\) follows \(\displaystyle x = \mu\). ... ... "So, from this definition we have that a maximal element of \(\displaystyle \Omega\) belongs to the set \(\displaystyle \Omega\).
 
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  • #4
Hi Peter,

You're right, there was a typo

Fallen Angel said:
Hi Peter,

$P$ is maximal in the set of ideals that do not intersect $S$, and $C$ properly contains $P$, so $\color{red}C\color{black}$ can't be in this set ($\color{red}\Omega \color{black}$ ), but it's still an ideal, so it must intersect $S$
 
  • #5
Fallen Angel said:
Hi Peter,

You're right, there was a typo

Thanks Fallen Angel, I think I get this now ... thanks to you!

Since \(\displaystyle C\) properly contains \(\displaystyle P\), then \(\displaystyle C\) cannot be in \(\displaystyle \Omega\) because \(\displaystyle P\) is maximal in \(\displaystyle \Omega\), and so no element in \(\displaystyle \Omega\) can contain \(\displaystyle P\) (by definition of maximality).

... ... and then, since \(\displaystyle C\) is an ideal that is not in \(\displaystyle \Omega\), then it must intersect \(\displaystyle S\) (since by definition \(\displaystyle \Omega\) is the set of all ideals which do not meet \(\displaystyle S\)).By the way ... thanks for all your help on this issue/problem ... ...

Peter
 

Related to Northcott - Proposition 3 - Inductive Systems Maximal Elements are Prime Ideals

1. What is the Northcott - Proposition 3 about?

The Northcott - Proposition 3 states that in an inductive system of ideals, the maximal elements are always prime ideals. This means that in a sequence of ideals where each element contains the previous one, the largest ideals are always prime.

2. Why is Northcott - Proposition 3 important in mathematics?

This proposition is important because it helps us better understand the behavior of prime ideals in inductive systems. It also has applications in algebraic geometry and commutative algebra.

3. What is an inductive system of ideals?

An inductive system of ideals is a sequence of ideals in a ring where each ideal contains the previous one. This means that the sequence is increasing and has no upper bound.

4. How does Northcott - Proposition 3 relate to other mathematical concepts?

This proposition is closely related to the concept of Noetherian rings, which are rings where every ideal can be generated by a finite number of elements. In Noetherian rings, every increasing sequence of ideals eventually stabilizes, and this is where Northcott - Proposition 3 applies.

5. Is Northcott - Proposition 3 true for all rings?

No, Northcott - Proposition 3 is only true for Noetherian rings. In non-Noetherian rings, there can be inductive systems of ideals where the maximal elements are not prime. However, this proposition does hold for many important classes of rings, such as polynomial rings and Dedekind domains.

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