# Euclidean norms in quadratic fields

I'm currently reading a set of lecture notes of Number Theory, and there's a small part I'm having trouble understanding.

A norm $N: R \rightarrow \mathbb{N}$ is Euclidean if it satisfies: for all $a \in R, >b \in R$, there exists $q, r \in R$ such that $a = qb + r$ and $N(r) < N(b)$.

Consider the norm given by $N(a + b \sqrt{D}) = |a^2 - Db^2|$

Now, we consider the quadratic field $\mathbb{Z}[\sqrt{-1}]$: to show that this is a Euclidean norm, we must show that for all $\alpha \in \mathbb{Q}(\sqrt{-1})$ there exists $\beta \in \mathbb{Z}(\sqrt{-1})$ with $N( \alpha - \beta ) < 1$

My question is: Why is $\alpha \in \mathbb{Q}(\sqrt{-1})$ if we are considering $\mathbb{Z}[\sqrt{-1}]?$

For the notes I'm using, see: http://math.uga.edu/~pete/4400quadrings.pdf page 5.

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Is that really all your notes say? I'd expect a bit more proof here. – Chris Eagle Feb 27 '13 at 14:46
@ChrisEagle I added a link. – user64219 Feb 27 '13 at 14:50

You are trying to do Euclidean division in $\Bbb{Z}[i]$. So you take $a, b \in \Bbb{Z}[i]$, with $b \ne 0$, and consider $\alpha = a b^{-1} \in \Bbb{Q}(i)$, as $\Bbb{Q}(i)$ is a field.
If you can find a $\beta \in \Bbb{Z}[i]$ such that $N(\alpha - \beta) < 1$ (PS and you can indeed), then $$\alpha = a b^{-1} = \beta + (\alpha - \beta)$$ so that, multiplying by $b$, $$a = b \beta + r \tag{div}$$ where $r = (\alpha - \beta) b = a - b \beta \in \Bbb{Z}[i]$, as $a, b, \beta \in \Bbb{Z}[i]$. Moreover, $$N(r) = N((\alpha - \beta) b) = N(\alpha - \beta) \cdot N(b) < N(b),$$ so $r$ is the remainder, and in (div) you've just done Euclidean division.