# integral closure of Z[2i]?

• Mar 20th 2011, 09:37 PM
KaKa
integral closure of Z[2i]?
Find the quotient field $K$ of $\mathbb{Z}[2i]$ and the integral closure of $\mathbb{Z}[2i]$ in $K$.
( $i=\sqrt{-1}$)
• Mar 21st 2011, 02:28 AM
TheArtofSymmetry
Quote:

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 $\mathbb{Z}[i]$? .