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Homework Help: Prove that Associates Have the Same Norm

  1. Aug 9, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to prove that any two quadratic integers that are associates must also have the same norm.

    2. Relevant equations

    If α = a + b√d, the norm of α is N(α) = a^2 - b^2*d.

    If two quadratic integers α and β are associates, α divides β, β divides α, and α/β and β/α both equal some unit, although each may be equal to a different unit.

    N(unit) = ±1

    3. The attempt at a solution

    This is what I've done so far:

    α/β = ε
    N(α/β) = N(ε)
    N(α/β) = ±1
    N(α)/N(β) = ±1
    N(α) = ±N(β)

    From here, I guess I need to show that it is impossible to have N(α) = -N(β), but I'm not sure how to do that. Any ideas?
     
    Last edited: Aug 9, 2007
  2. jcsd
  3. Sep 16, 2007 #2
    Sorry to go gravedigging here, but I just had a new idea. Is it even necessary that associates have the same norm?

    α and β are defined as associates if α = β*ε

    Thus, we can take any quadratic integer α in any quadratic field that has units that have negative norms, and multiply α by some unit ε with a negative norm to get β. Then when we take the norm we have:

    N(α) = N(β*ε)
    N(α) = N(β)*N(ε)
    N(α) = -N(β)

    Is all of my logic here correct? Because if so, it seems I may have been trying to prove a statement that is actually false.
     
  4. Sep 16, 2007 #3

    HallsofIvy

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    Can you actually find two such associates?

    By the way, have you actually proved either [itex]N(\alpha /\beta)= N(\alpha)/N(\beta)[/itex] or [itex]N(\alpha\beta)= N(\alpha)N(\beta)[/itex]?
     
  5. Sep 19, 2007 #4
    I have proved all of your statements about norms.

    In Q[sqrt(2)], 7 - 5*sqrt(2) has norm -1, which makes it a unit. I chose α = 2 + 2*sqrt(2) to test my initial assumption:

    β = (2 + 2*sqrt(2))(7 - 5*sqrt(2))
    β = -6 + 4*sqrt(2)

    N(α) = -4
    N(β) = 4

    Did I do something wrong, or did I prove that my original assumption was incorrect?
     
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