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Math Help - Congruences classes, Quadratic Field

  1. #1
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    Congruences classes, Quadratic Field

    Find the congruence classes (\bmod{(3+\sqrt{-3}})/2) in \mathbb{Q}(\sqrt{-3}).

    I'm having trouble on this one. I don't know where to start.
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    Quote Originally Posted by Pn0yS0ld13r View Post
    Find the congruence classes (\bmod{(3+\sqrt{-3}})/2) in \mathbb{Q}(\sqrt{-3}).

    I'm having trouble on this one. I don't know where to start.
    Do you perhaps mean \mathbb{Z}(\sqrt{-3})?

    Because \mathbb{Q}(\sqrt{-3}) is a field and so any two elements are divisible by any non-zero number. Which would imply the congruence class is the entire field itself. And in general, the congruence class mod any unit is always the entire ring itself.
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    Quote Originally Posted by ThePerfectHacker View Post
    Do you perhaps mean \mathbb{Z}(\sqrt{-3})?

    Because \mathbb{Q}(\sqrt{-3}) is a field and so any two elements are divisible by any non-zero number. Which would imply the congruence class is the entire field itself. And in general, the congruence class mod any unit is always the entire ring itself.
    I don't think its a typo, but I copied it exactly how the problem is written. I'm not sure if it makes any difference, but I think the problem assumes that we're only dealing with quadratic integers in \mathbb{Q}(\sqrt{-3}).

    My book gives the following definition for divisibility in \mathbb{Q}(\sqrt{d}):

    If \alpha and \beta are integers in \mathbb{Q}(\sqrt{d}), with \alpha \neq 0, we say that \alpha divides \beta if there is an integer \gamma in \mathbb{Q}(\sqrt{d}) such that \beta = \alpha\gamma.
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    Quote Originally Posted by Pn0yS0ld13r View Post
    My book gives the following definition for divisibility in \mathbb{Q}(\sqrt{d}):

    If \alpha and \beta are integers in \mathbb{Q}(\sqrt{d}), with \alpha \neq 0, we say that \alpha divides \beta if there is an integer \gamma in \mathbb{Q}(\sqrt{d}) such that \beta = \alpha\gamma.
    But \frac{3}{2} + \frac{\sqrt{3}i}{2} is not an integer in \mathbb{Q}[\sqrt{-3}]
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    Quote Originally Posted by ThePerfectHacker View Post
    But \frac{3}{2} + \frac{\sqrt{3}i}{2} is not an integer in \mathbb{Q}[\sqrt{-3}]
    Yes it is, because if d \equiv 1 \bmod{4}, then the integers of \mathbb{Q}(\sqrt{d}) are those numbers of the form \dfrac{a+b\sqrt{d}}{2} where a,b \in \mathbb{Z} and a and b are both even or both odd.

    So \dfrac{3+\sqrt{-3}}{2} is a quadratic integer in \mathbb{Q}(\sqrt{-3}).
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