integral closure of Z[2i]?

**KaKa** · March 20th 2011, 09:37 PM

Find the quotient field $K$ of $\mathbb{Z}[2i]$ and the integral closure of $\mathbb{Z}[2i]$ in $K$ .
( $i=\sqrt{-1}$ )

**TheArtofSymmetry** · March 21st 2011, 02:28 AM

Originally Posted by KaKa

Find the quotient field $K$ of $\mathbb{Z}[2i]$ and the integral closure of $\mathbb{Z}[2i]$ in $K$ .
( $i=\sqrt{-1}$ )

It is easy to see that K=Q(i). Now, is $i \notin \mathbb{Z}[2i]$ integral over $\mathbb{Z}[2i]$ ? Think about $x^2-2ix-1 \in \mathbb{Z}[2i][x]$ . Can you see that the integral closure of $\mathbb{Z}[2i]$ in $K$ is simply $\mathbb{Z}[i]$ ? .

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