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Integral bases and Discriminants

  1. Feb 22, 2012 #1
    dz86e.jpg

    Here is my solution, I think I have part i) done OK, but I'm not sure about how to proceed with part ii).

    I suppose I need to show that both determinants of the base change matrices Cij and Dij are = ±1?


    Thanks
     
    Last edited: Feb 22, 2012
  2. jcsd
  3. Feb 22, 2012 #2

    morphism

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    You pretty much have it... Just combine both your equations into
    [tex] \Delta(\alpha_1,\ldots,\alpha_n) = (\det D_{ij})^2 (\det C_{ij})^2 \Delta(\alpha_1,\ldots,\alpha_n). [/tex]
    What can you conclude from this?
     
  4. Feb 22, 2012 #3
    that [itex](\det D_{ij})^2 (\det C_{ij})^2 = 1 [/itex]?

    so the determinants are either ±1, so we can conclude the statement?
     
  5. Feb 22, 2012 #4

    morphism

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    Yes, because one discriminant is (det)^2 times the other!
     
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