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

  • #1
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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:
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Proposition 19.

... ...

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

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

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The proof of (5) above reads as follows:


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Proof.

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

Then [itex] Q \subseteq M [/itex] so [itex] rad \ Q \subseteq rad \ M = M [/itex].

... ... 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 [itex] M \subseteq rad \ M [/itex]

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

How do we know that [itex] rad \ M \ne R [/itex]?

Would appreciate some help.

Peter
 

Attachments

  • Page 682 of Dummit and Foote.pdf
    75.8 KB · Views: 138

Answers and Replies

  • #2
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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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  • #3
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Thanks R136a1!

But just thinking over this ...

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

[itex] rad \ M = \{a \in R \ | \ a^k \in M [/itex] for some [itex] k \ge 1 [/itex]

So then [itex] 1 \in rad \ M \Longrightarrow 1^k \in M [/itex] for some [itex] k \ge 1 [/itex]

[itex] \Longrightarrow \ 1 \in M [/itex]

[itex] \Longrightarrow \ M = R [/itex]

Thus would mean [itex] M^n = R [/itex] so [itex] Q = R [/itex] 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
 
  • #4
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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 [itex] M \ne R [/itex] and the only ideals containing M are M and R."
 
  • #5
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Yes, it is a contradiction because ##M=R## is not possible by definition.
 

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