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Primes in ring of Gauss integers - help

  1. Jan 16, 2005 #1
    Primes in ring of Gauss integers - help!!

    I'm having a very difficult time solving this question, please help!
    So i'm dealing with the ring [tex]R=\field{Z}[\zeta][/tex] where
    [tex]\zeta=\frac{1}{2}(-1+\sqrt{-3})[/tex]
    is a cube root of 1.
    Then the question is:
    Show the polynomial [tex]x^2+x+1[/tex] has a root in [tex]F_p[/tex] if and only if [tex]p\equiv1 (mod 3)[/tex].

    I thought i could show this in two steps, by showing that:
    a) a solves [tex]x^2+x=-1(mod p)[/tex] if and only if a is an element of order 3 in [tex]F^x_p[/tex].
    b)[tex]F^x_p[/tex] contains an element of order 3 if and only if [tex]p\equiv1 (mod 3)[/tex].

    I've proved part b, but i can't seem to get a hold of a.
    Please help :cry:
     
  2. jcsd
  3. Jan 16, 2005 #2

    Hurkyl

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    Well, first I think you need to review the problem and definitions... the things you've said don't seem to connect to one another.


    Anyways, I think this might help: note that

    x^3 - 1 = (x - 1) (x^2 + x + 1)
     
  4. Jan 16, 2005 #3
    Hmm....sorry, i don't see what you mean.
     
  5. Jan 16, 2005 #4

    Hurkyl

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    It means that if a is a root of x^2 + x + 1, then it is also a root of x^3 - 1. (i.e. it is a cube root of 1)
     
  6. Jan 16, 2005 #5
    ah, so a^3=1, and a has order 3. We can apply the argument backwards, and that will prove a). I see, thanks Hurkyl! :smile:
    There's also a second part to this problem, which says (p) is prime ideal in R if and only if p=-1 (mod 3)
    Apparently the first part of this problem applies, but i'll have to think about this more.
     
  7. Jan 16, 2005 #6

    Hurkyl

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    Ah, so that's why you mentioned R.

    (Incidentally, I think you meant algebraic integers, not Gaussian integers)

    I don't know if it will help, but note that if Fp has a root of x^2 + x + 1, then there is a homomorphism from R onto Fp.
     
  8. Jan 26, 2005 #7
    I was looking at a somewhat similar problem, https://www.physicsforums.com/showthread.php?t=60863

    You use quadratic reciprocity on [tex] X^2\equiv-3 [/tex] Mod p to discover that p is of the form 3k+1. Thus, this prime splits over a field with[tex] \sqrt{-3}, [/tex]so it would not generate a prime ideal.

    Example: [tex]7=2^2+2+1=(2-\zeta)(2-\zeta^2) [/tex]

    By the way, as Hurkyl points out, this is an algebratic number theory problem.
     
    Last edited: Jan 27, 2005
  9. Feb 12, 2005 #8

    mathwonk

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    this is elementary. look at the group homomorphism from the multiplicative group Fp - {0} to itself defined by cubing. then if there is a primitive cube root of 1, the map is 3 to 1, and has image of order 1/3 the order of the group, i.e. then 3 divides p-1. on the other hand if 3 divides the order of the group, it is elementary group theory that there exists an element of order 3.
     
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