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Math Help - Eisenstein Integers, need help with a proof

  1. #1
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    Eisenstein Integers, need help with a proof

    Hey all, I am having a problem with an exercise from the book "A Classical Introduction to Modern Number Theory". It involves the ring \mathbb{Z}[\omega], where
    <br />
\omega = \frac{-1 + \sqrt{-3}}{2}<br />
    The question is
    For any \alpha \in \mathbb{Z}[\omega], show that \alpha is congruent to 1, -1 or 0 mod 1 - \omega

    I figured the best thing to use was Euclidean division, so I set
    \alpha = a + b\omega
    and worked out that
    \frac{\alpha}{1 - \omega} = \frac{2a - b + (a + b)\omega}{3}

    I can see how it would work, and have tested it a few times on random Eisenstein integers, but cannot work out how to actually prove it!

    Any help would be much appreciated
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  2. #2
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    Quote Originally Posted by Aprshaw View Post
    Hey all, I am having a problem with an exercise from the book "A Classical Introduction to Modern Number Theory". It involves the ring \mathbb{Z}[\omega], where
    <br />
\omega = \frac{-1 + \sqrt{-3}}{2}<br />
    The question is
    For any \alpha \in \mathbb{Z}[\omega], show that \alpha is congruent to 1, -1 or 0 mod 1 - \omega

    I figured the best thing to use was Euclidean division, so I set
    \alpha = a + b\omega
    and worked out that
    \frac{\alpha}{1 - \omega} = \frac{2a - b + (a + b)\omega}{3}

    I can see how it would work, and have tested it a few times on random Eisenstein integers, but cannot work out how to actually prove it!

    Any help would be much appreciated


    What does it mean "to be congruent to" in a general ring??

    Tonio
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  3. #3
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    I believe
    \alpha is congruent to \beta \pmod{1 - \omega}
    is the same as saying
    \alpha = (1 - \omega) \cdot \kappa + \beta
    Which is Euclidean division in the ring \mathbb{Z}[\omega]

    However, I am not completely certain of this, but this is the assumption I have made.
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  4. #4
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    Quote Originally Posted by Aprshaw View Post
    I believe
    \alpha is congruent to \beta \pmod{1 - \omega}
    is the same as saying
    \alpha = (1 - \omega) \cdot \kappa + \beta
    Which is Euclidean division in the ring \mathbb{Z}[\omega]

    However, I am not completely certain of this, but this is the assumption I have made.

    Of course. I just missed the "mod" thingy in your original post...

    1) Check first that a+b\omega=0\!\!\pmod {(1-\omega )}\Longleftrightarrow a+b=0\!\!\mod 3

    2) Now deduce from the above your problem's solution...or

    Directly, show that a+b=i\!\!\pmod {(1-\omega )}\Longleftrightarrow a+b=i\!\!\pmod 3\,,\,\,for\,\,i=-1,0,1

    Tonio
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