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Calculate the discriminant of a basis [Number Theory]

  1. Feb 21, 2012 #1
    Question:

    df91ts.png

    The needed proposition and two examples:

    1zb9px4.png


    This is as far as I have got:

    zn0m8h.jpg

    I need to reduce this (I think) so I can represent is as a matrix! Any idea on how to do this?

    Thanks
     
  2. jcsd
  3. Feb 21, 2012 #2

    morphism

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    How about rewriting your last equation as
    [tex](\zeta - 1)Dp(\zeta) = p\zeta^{p-1}[/tex] and then taking norms of both sides?
     
  4. Feb 21, 2012 #3
    I'm unsure on how to calculate [itex]N(p\zeta^{p-1})[/itex]

    [itex]N(\zeta - 1) = 5[/itex] though.
     
  5. Feb 21, 2012 #4

    morphism

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    [itex]N(\zeta - 1)[/itex] isn't 5 - not unless p=5.

    As for [itex]N(p \zeta^{p-1})[/itex], this is just [itex]N(p)N(\zeta)^{p-1}[/itex], and [itex]N(p)[/itex] and [itex]N(\zeta)[/itex] are really easy to compute.
     
  6. Feb 21, 2012 #5
    Providing [itex]N(\zeta - 1)=p[/itex] (is this easy to prove?) then I also get [itex]N(p)=p^{2}[/itex] and [itex]N(\zeta)=1[/itex]

    So my overall answer should be p?
     
  7. Feb 21, 2012 #6

    morphism

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    [itex]N(p)[/itex] should be [itex]p^{p-1}[/itex].

    And as for computing [itex]N(\zeta-1)[/itex], you can either do it the determinant way (and do a bunch of row operations), or you could use the differentiation trick you used above except with [itex]t=1[/itex] isntead of [itex]t=\zeta[/itex].
     
  8. Feb 21, 2012 #7
    Hmm when t=1 I get the denominator as 0, am I not using it right?
     
  9. Feb 21, 2012 #8

    morphism

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    Notice that
    [tex]t^p - 1 = (t-1)(t-\zeta)\cdots(t-\zeta^{p-1})[/tex] so the t-1 in the denominator gets canceled off.

    Edit:
    Actually, we don't want to be looking at (t^p-1)/(t-1), rather just at t^p-1. Differentiate both sides of the equation above and plug in t=1.
     
  10. Feb 21, 2012 #9
    Just to be clear you want me to differentiate both sides of [itex]t^p - 1 = (t-1)(t-\zeta)\cdots(t-\zeta^{p-1})[/itex]?

    Unsure on how to do the RHS without getting messy!
     
  11. Feb 21, 2012 #10

    morphism

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    Yes, that's what I want you to differentiate. It won't be too messy (especially if you plug in t=1 after you differentiate). :)
     
  12. Feb 21, 2012 #11
    Hmm I'll admit I'm not sure how to go about doing the RHS!

    Do I need to? Why can't I use just the differential of the LHS?
     
  13. Feb 21, 2012 #12

    morphism

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    If you differentiate the RHS and plug in t=1, everything will immediately vanish except for the first term, which becomes
    [tex](1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1}).[/tex]
    This really just follows from the product rule of differentiation.
     
  14. Feb 21, 2012 #13
    Ah I see, sorry I've never used the product rule for more than 2 terms before, I get it now!

    so [itex]p = (1-\zeta)(1-\zeta^2)\cdots(1-\zeta^{p-1})[/itex]

    i suppose we want to rearrange this?
     
  15. Feb 21, 2012 #14

    morphism

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    No real rearrangement is necessary. Just notice that the RHS is also [itex](\zeta - 1)(\zeta^2 - 1) \cdots (\zeta^{p-1}-1)[/itex] (because p is odd), which is just the product of all the conjugates of [itex]\zeta-1[/itex], a.k.a. _____
     
  16. Feb 21, 2012 #15
    Norm!

    So the norm is p? Havn't we got a few too many conjugates though?
     
  17. Feb 21, 2012 #16

    morphism

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    Yes, the norm is p, and no we have the right amount of conjugates! This is because the conjugates of [itex]\zeta[/itex] are [itex]\zeta[/itex] itself and [itex]\zeta^2, \ldots, \zeta^{p-1}[/itex]. Try to actually write down a rigorous proof, using minimal polynomials. It should be easy.
     
  18. Feb 21, 2012 #17
    Ok thanks :)

    Is N(p) = p^p-1 since we can write the matrix as a matrix with only p along the diagonal?
     
  19. Feb 22, 2012 #18

    morphism

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    Yup!
     
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