MHB Northcott - Proposition 3 - Inductive Systems Maximal Elements are 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 the proof Proposition 3 in Chapter 2 concerning the demonstration that all the maximal elements of $$\Omega$$ 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 $$C$$ consist of all elements that can be expressed in the form $$r \alpha + \pi$$ where $$r \in R$$ and $$\pi \in P$$. It is a simple matter to check that $$C$$ is an ideal containing $$P$$. Indeed, since $$\alpha = 1 \alpha + 0$$, $$\alpha \in C$$ and therefore C strictly contains P. However P is maximal in $$\Omega$$. It therefore follows that $$C$$ meets $$S$$. ... ... "

I do not fully understand why, in the above argument, it follows that $$C$$ meets $$S$$.

Can someone show formally and rigorously whhy, exactly, it follows that $$C$$ meets $$S$$.

Help will be appreciated ... ...

Peter

 

[Fallen Angel]

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 $S$

 

[Math Amateur]

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 $$\Omega$$ by "this set" ... but then $$P$$ is a maximal element in this set ... so then doesn't it follow that $$P$$ is in this set (is that correct?)

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

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 $$\Omega$$ need not be in $$\Omega$$ ... ...

are not correct ...

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

 

[Fallen Angel]

Hi Peter,

You're right, there was a typo

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 $S$

 

[Math Amateur]

Hi Peter,

You're right, there was a typo

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

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

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

Peter