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Primigenial ring ideals question

  1. Oct 5, 2007 #1

    learningphysics

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    I'm going through this proof in Allan Clark's Elements of Abstract Algebra to prove that a primigenial ring is a Dedekind Domain.

    A primigenial ring is one in which every proper ideal can be written as a product of proper prime ideals.

    There's a step in the proof that I'm not able to understand...

    We're given p an invertible proper prime ideal in a primigenial ring R. a is an element in R - p.

    He proves that p + (a) = p^2 + (a)

    then...
    [tex]p = p \cap (p^2 + (a))[/tex]... no problem here...

    then the next step:

    [tex]p \cap (p^2 + (a)) \subset p^2 + (a)p[/tex]. I'm not sure how he does this step... I'd appreciate any help or hints. Thanks a bunch!
     
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  3. Oct 6, 2007 #2

    Hurkyl

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    Well, you can always try the direct approach: what is the form of a general element of p^2 + (a)? And which of those are in [itex]p \cap (p^2 + (a))[/itex]?
     
  4. Oct 6, 2007 #3

    learningphysics

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    Ah... thanks Hurkyl. I think I have it now, but I'd appreciate if you just check my reasoning...

    a general element of p^2 + (a) will be of the form p1p2 + ra, where r is any element of the ring...

    so if p1p2 + ra is an element of the ideal p... and since p1p2 is an element of the ideal p, then ra must be an element of the ideal p (since the ideal is an additive group)...

    ra belongs to p. so r belongs p:(a) which is an ideal...

    (a)*[p:(a)] is a subset of the ideal p.

    since p is prime and (a) is not a subset of p, p:(a) must belong to p. so r is an element of p and so:

    p1p2 + ra is an element of p^2 + (a)p...
     
  5. Oct 6, 2007 #4

    Hurkyl

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    Not quite: recall that the product IJ doesn't consist only of numbers of the form ij, but also of all sums of such numbers.

    But that's no biggie here; it's good enough to simply write an element as q + ra, where q is in p^2.


    It's somewhat quicker to just invoke the definition of p being prime. I'm always hazy on the properties of the colon ideal, but this part of your argument sounds plausible.
     
  6. Oct 6, 2007 #5

    learningphysics

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    Thanks again Hurkyl! I'm learning a lot from this thread...

    From your post I take it that we can go directly from:

    ra belongs to p, and since p is prime r must belong to p? how does one justify this step... I'm guessing it's something very trivial that I'm missing...

    The definition of prime ideal I've learnt is that if a product of two ideals belongs to a prime ideal... then at least one of them belongs to the prime ideal... that's why I brought up the colon ideal...

    perhaps there is another equivalent definition of prime ideal?
     
  7. Oct 6, 2007 #6

    Hurkyl

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    The statement I usually see as the definition involves elements -- but it's easily translated into the corresponding principal ideals. Since [itex](r) (a) \subseteq p[/itex]...
     
  8. Oct 6, 2007 #7

    learningphysics

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    cool. Thanks Hurkyl!
     
  9. Oct 6, 2007 #8

    learningphysics

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    :redface: I see my mistake now... yes, the book gives the definition of prime ideal with elements... ie ab is an element of a prime ideal, then a is an element or b is an element of the ideal...

    there is a theorem afterwards that says if we have two ideals a and b such that ab is a subset of a prime ideal p, then a is a subset of p or b is a subset of p...

    I was mixing up this theorem with the actual definition of prime ideal...
     
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