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Math Help - Showing that Z[Sqrt[-3]] is a Euclidean Domain

  1. #1
    Senior Member roninpro's Avatar
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    Showing that Z[Sqrt[-3]] is a Euclidean Domain

    Hello everyone,

    I'm having a bit of trouble showing that \mathbb{Z}[\sqrt{-3}] is a Euclidean domain. My attempt is:

    For this ring, define the norm N(a+b\sqrt{-3})=a^2+3b^2. We need to show that there is a division algorithm, so let \alpha=a+b\sqrt{-3} and \beta=c+d\sqrt{-3}. We will compute \alpha/\beta as complex numbers and fish for an appropriate quotient and remainder. We have
    \displaystyle \frac{\alpha}{\beta}=\frac{ac+3bd}{c^2+3d^2}+\frac  {bc-ad}{c^2+3d^2}\sqrt{-3}
    Now choose a quotient q\in \mathbb{Z}[\sqrt{-3}] such that |Re(q)-\frac{ac+3bd}{c^2+3d^2}|\leq 1/2 and |Im(q)-\frac{bc-ad}{c^2+3d^2}|\leq \sqrt{3}/2. Then we have \alpha/\beta=q+r/\beta, where r/\beta is the leftover "fractional" term. From the choice of q, we have N(r/\beta)\leq 1, and N(r)\leq N(\beta).

    At this point, I can't shake the possible equality. I would appreciate any suggestions on this issue.
    Last edited by roninpro; November 22nd 2010 at 11:07 PM.
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  2. #2
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    Quote Originally Posted by roninpro View Post
    Hello everyone,

    I'm having a bit of trouble showing that \mathbb{Z}[\sqrt{-3}] is a Euclidean domain.


    No wonder: it is not. For example, check 4=2\cdot 2=(1+\sqrt{-3})(1-\sqrt{-3}) ...

    Tonio



    My attempt is:

    For this ring, define the norm N(a+b\sqrt{-3})=a^2+3b^2. We need to show that there is a division algorithm, so let \alpha=a+b\sqrt{-3} and \beta=c+d\sqrt{-3}. We will compute \alpha/\beta as complex numbers and fish for an appropriate quotient and remainder. We have
    \displaystyle \frac{\alpha}{\beta}=\frac{ac+3bd}{c^2+3d^2}+\frac  {bc-ad}{c^2+3d^2}\sqrt{-3}
    Now choose a quotient q\in \mathbb{Z}[\sqrt{-3}] such that |Re(q)-\frac{ac+3bd}{c^2+3d^2}|\leq 1/2 and |Im(q)-\frac{bc-ad}{c^2+3d^2}|\leq \sqrt{3}/2. Then we have \alpha/\beta=q+r/\beta, where r/\beta is the leftover "fractional" term. From the choice of q, we have N(r/\beta)\leq 1, and N(r)\leq N(\beta).

    At this point, I can't shake the possible equality. I would appreciate any suggestions on this issue.
    .
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  3. #3
    Senior Member roninpro's Avatar
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    Ah, yes - the statement of the problem in the book was a little bit confusing. I think that they meant to put \alpha=\frac{1+\sqrt{-3}}{2} and to show \mathbb{Z}[\alpha] is a Euclidean domain. I'll go back and look at it again.
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