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Thread: integral closure of Z[2i]?

  1. #1
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    integral closure of Z[2i]?

    Find the quotient field $\displaystyle K$ of $\displaystyle \mathbb{Z}[2i]$ and the integral closure of $\displaystyle \mathbb{Z}[2i]$ in $\displaystyle K$.
    ($\displaystyle i=\sqrt{-1}$)
    Last edited by mr fantastic; Mar 21st 2011 at 01:32 AM. Reason: Title.
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  2. #2
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    Quote Originally Posted by KaKa View Post
    Find the quotient field $\displaystyle K$ of $\displaystyle \mathbb{Z}[2i]$ and the integral closure of $\displaystyle \mathbb{Z}[2i]$ in $\displaystyle K$.
    ($\displaystyle i=\sqrt{-1}$)
    It is easy to see that K=Q(i). Now, is $\displaystyle i \notin \mathbb{Z}[2i]$ integral over $\displaystyle \mathbb{Z}[2i]$? Think about $\displaystyle x^2-2ix-1 \in \mathbb{Z}[2i][x]$. Can you see that the integral closure of $\displaystyle \mathbb{Z}[2i]$ in $\displaystyle K$ is simply $\displaystyle \mathbb{Z}[i]$? .
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