Primary Ideals, prime ideals and maximal ideals - D&F Section 15.2

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The discussion focuses on Proposition 19 Part (5) from Dummit and Foote's Algebra, specifically regarding maximal ideals and primary ideals. It establishes that if \( M \) is a maximal ideal and \( Q \) is an ideal such that \( M^n \subseteq Q \subseteq M \) for some \( n \ge 1 \), then \( Q \) is a primary ideal with \( \text{rad}(Q) = M \). The key point of confusion addressed is the proof that \( \text{rad}(M) = M \), which is confirmed by demonstrating that assuming \( \text{rad}(M) = R \) leads to a contradiction, as it violates the definition of a maximal ideal.

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I am studying Dummit and Foote Section 15.2. I am trying to understand the proof of Proposition 19 Part (5) on page 682 (see attachment)

Proposition 19 Part (5) reads as follows:
----------------------------------------------------------------------------------------------------------------------------

Proposition 19.

... ...

(5) Suppose M is a maximal ideal and Q is an ideal with M^n \subseteq Q \subseteq M for some n \ge 1.

Then Q is a primary idea, with rad Q = M

----------------------------------------------------------------------------------------------The proof of (5) above reads as follows:----------------------------------------------------------------------------------------------

Proof.

Suppose M^n \subseteq Q \subseteq M for some n \ge 1 where M is a maximal idea.

Then Q \subseteq M so rad \ Q \subseteq rad \ M = M.

... ... etc

----------------------------------------------------------------------------------------------

My problem is as follows:

Why can we be sure that rad M = M?

I know that M is maximal and so no ideal in R can contain M. We also know that M \subseteq rad \ M

Thus either rad M = M (the conclusion D&F use) or rad M = R?

How do we know that rad \ M \ne R?

Would appreciate some help.

Peter
 

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Try to use contradiction. Assume that ##\textrm{rad}(M) = R##. Then ##1\in \text{rm}(M)##. Now use the definition of the radical.
 
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Thanks R136a1!

But just thinking over this ...

Is the following thinking along the right track ...?

rad \ M = \{a \in R \ | \ a^k \in M for some k \ge 1

So then 1 \in rad \ M \Longrightarrow 1^k \in M for some k \ge 1

\Longrightarrow \ 1 \in M

\Longrightarrow \ M = R

Thus would mean M^n = R so Q = R also ...

This result is not a contradiction but it leads to the collapse of the conditions of the Proposition to triviality ...

Can you confirm my reasoning ... or indeed point out errors/inadequacies in my thinking

Peter
 
Just thinking further ... maybe in my reasoning in the last post I have indeed achieved a contradiction since my reasoning (if correct!) establishes that M = R .,,, where of course M is a maximal ideal by assumption ... but by D&F's definition of a maximal ideal, this is not possible ... so contradiction!

Can someone confirm that this is correct?

Note: Definition of maximal ideal, Dummit and Foote, page 253:

"An ideal M in an arbitrary ring R is called a maximal ideal if M \ne R and the only ideals containing M are M and R."
 
Yes, it is a contradiction because ##M=R## is not possible by definition.
 

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