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

  1. Nov 9, 2013 #1
    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 [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

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


    The proof of (5) above reads as follows:


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

    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
     

    Attached Files:

  2. jcsd
  3. Nov 9, 2013 #2
    Try to use contradiction. Assume that ##\textrm{rad}(M) = R##. Then ##1\in \text{rm}(M)##. Now use the definition of the radical.
     
  4. Nov 10, 2013 #3
    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
     
  5. Nov 10, 2013 #4
    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."
     
  6. Nov 10, 2013 #5
    Yes, it is a contradiction because ##M=R## is not possible by definition.
     
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