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Math Help - Primary Ideals, prime ideals and maximal ideals.

  1. #1
    Super Member Bernhard's Avatar
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    Primary Ideals, prime ideals and maximal ideals.

    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.]
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  2. #2
    Member Haven's Avatar
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    Re: Primary Ideals, prime ideals and maximal ideals.

    Hello once again

    In the first statement, you have probably seen statements like:
    P is a prime ideal of R if and only if R/P is an integral domain.
    This is because the condition [TEX]ab\in P[TEX] implies a\in P or b\in P
    becomes ab=0 implies a=0 or b=0 in R/P, as we are modding out by elements of P.

    Similarly the primary condition becomes the condition that all zero divisors are nilpotent and it is a similar proof to the statement above. By passing to the quotient ring R/P, and using the ideal correspondence theorem, we see P corresponds to the zero ideal in R/P. So it suffices to prove the statement for the zero ideal in R/P.

    Now the radical of the zero ideal, actually consists of all nilpotent elements, by definition.


    For the second point, we appeal to the statement that requires Zorn's Lemma: Any proper ideal is contained in some maximal ideal. So P must be contained in some maximal ideal Q. All maximal ideals are prime, so Q is prime and Q contains P. As P is the only prime it follows that P=Q and so P is maximal.
    Thanks from Bernhard
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