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Primary Ideals, prime ideals and maximal ideals.

  1. Nov 8, 2013 #1
    I have a problem in understanding the proof to Dummit and Foote Section 15.2, Proposition 19 regarding primary ideals. I hope someone can help.

    My problem is with Proposition 19 part 4 - but note that part 4 relies on part 2 - see attachment.

    The relevant sections of Proposition 19 read as follows: (see attachment)

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

    Proposition 19.

    Let R be a commutative ring with 1

    ... ...

    (2) The ideal Q is primary if and only if every zero divisor in R/Q is nilpotent.

    ... ...

    (4) If Q is an ideal whose radical is a maximal ideal. then Q is a primary ideal

    ... ... etc

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

    The proof of (4) above proceeds as follows:

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

    Proof. (see attachment)

    ... ...

    To prove (4) we pass to the quotient ring R/Q: by (2) it suffices to show that every zero divisor in this quotient ring is nilpotent.

    We are reduced to the situation where Q = (0) and M = rad Q = rad (0), which is the nilradical, is a maximal ideal.

    Since the nilradical is contained in every prime ideal (Proposition 12), it follows that M is the unique prime ideal, so also the unique maximal ideal.

    ... ... etc (see attachment)

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

    I have two problems with the proof above.

    (1) I do not completely follow the statement:

    "We are reduced to the situation where Q = (0) and M = rad Q = rad (0), which is the nilradical, is a maximal ideal."

    I know this is something to do with zero divisors in R/Q, but why/how do we end up discussing Q = (0) exactly? Can someone please clarify? ( I suspect it is simple but I cannot see the connection :-( )


    (2) I am puzzled by the statement "it follows that M is the unique prime ideal, so also the unique maximal ideal"

    This statement appears to indicate that in R we have that prime ideals are maximal ideals.

    I thought that we could only assume that maximal ideals were prime ideals (D&F Corollary 14 page 256) but not the converse? Can someone please clarify.

    Would appreciate some help.

    Peter



    [Note: D&F Corollary 14, page 256 reads as follows:

    Corollary 14. Assume R is commutative. Every maximal ideal is a prime ideal.]
     

    Attached Files:

  2. jcsd
  3. Nov 9, 2013 #2
    You need to prove that every zero divisor is nilpotent in ##R/Q##. This is equivalent to saying that ##\{0\}## is primary, since ##(R/Q)/\{0\} \cong R/Q##.
    Now, you know that the radical of ##Q## is maximal in ##R##. You can say that this implies that the radical of ##\{0\}## is maximal in ##R/Q##. So you know the radical of ##\{0\}## is maximal, and you need to prove ##\{0\}## is primary. This is exactly the theorem with ##Q=\{0\}##.

    Assume that there are other maximal ideals than ##M##. Then these other maximal ideals are also prime ideals. Thus there are other prime ideals than ##M##. But ##M## was the unique prime ideal.
     
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