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Factor principal ideals in a field with class number 2

  1. Apr 14, 2012 #1
    1. The problem statement, all variables and given/known data

    Let K be a number field. Let a be an irreducible element of K. Prove that if <a>(the ideal generated by a) is not a prime ideal, then it is a product of two prime ideals.


    2. Relevant equations



    3. The attempt at a solution

    I'm actually quite clueless...:frown: I have been thinking maybe it is possible for me to obtain some information about the extension K:Q from the condition that the class number is 2, like the degree of the extension, etc..?
    Any help is appreciated!
     
  2. jcsd
  3. Apr 14, 2012 #2

    morphism

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    The class number of K and the degree of the extension K/Q have nothing to do with each other. (There are quadratic extensions of Q with arbitrarily large class number.)

    In your problem I assume by an "irreducible element of K" you really mean an irreducible element in the ring of integers of K. Anyway, just start with what you're given: (a) is not prime, so it factors into a product of prime ideals. Some of these will be principal, some not. Try to use the fact that the class group is Z/2Z to simplify your factorization.
     
  4. Apr 14, 2012 #3
    Thanks for your reply!
    Yes. This is what I meant.
    Could they really be principal? Would that imply that a has a proper factor, which is a contradiction?
     
  5. Apr 14, 2012 #4

    morphism

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    Good! Can you use this observation to finish off the solution?
     
  6. Apr 14, 2012 #5
    Figured it out! Finally can take a break after 8 hours in the library. :approve:

    Thanks so much. :biggrin:
     
  7. Apr 14, 2012 #6

    morphism

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    Well done!
     
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