Definition A Euclidean domain is an integral domain R, together with a map $f: R \backslash \{0\} \rightarrow \mathbb{Z}_{>=0},$ such that $\forall a, b\in R$ with $b\neq 0:\ \exists q,r \in R:\ a = bq+r$ and either $r = 0$ or $f(r) < f(b)$.
Claim $\mathbb{Z}[\sigma]$ is an Euclidean domain, where $\sigma = \frac{\sqrt{5}-1}{2}$.
Proof attempt For $\alpha = a+b\sigma \in \mathbb{Z}[\sigma]\backslash\{0\}$, let $f(\alpha) = a^2 - ab - b^2$. Notice that $f$ commutes with products. Now take an arbitrary $\beta = c+d\sigma \in \mathbb{Z}[\sigma]$. We will express $\beta = \alpha \gamma + \theta$ for $\gamma \in \mathbb{Z}[\sigma]$ s.t. either $\theta=0$ or $f(\theta) < f(\alpha)$. So consider $\beta/\alpha = r + s \sigma \in \mathbb{Q}[\sigma]$ for $r,s \in \mathbb{Q}$.
Let $k, l\in \mathbb{Z}$ be closest integers to $r,s$ respectively, so that $|k-r|\leq\frac{1}{2}$ and $|l-s|\leq\frac{1}{2}$. Additionally, let $\gamma = k+l\sigma,\ m = r-k,\ n =s-l,\ \theta=(a+b\sigma)(m+n\sigma)$.
Then by construction we have $$\begin{align}c + d\sigma &= (a+b\sigma)(r+s\sigma)\\ &= (a+b\sigma)(k+l\sigma)+(a+b\sigma)(m+n\sigma)\end{align}$$ Ie. $\beta = \alpha\gamma + \theta$. Now assume $\theta \neq 0$. We have that $$\begin{align} f(\theta) &= f(a+b\sigma)\cdot f(m+n\sigma)\\ &= (a^2 - ab - b^2) (m^2 - mn - n^2)\\ &\leq (a^2-ab-b^2)(1/4+1/4+1/4)\\ &\leq \frac{3}{4}(a^2 -ab -b^2)\\ &\leq f(\alpha)\end{align}$$
Questions
- Is the above proof attempt OK?
- If so, why is it that $\mathbb{Z}[\sqrt{5}]$ is not a PID?
One reason $\mathbb{Z}[\sqrt{5}]$ is not a PID is that it is not integrally closed. (*)
Indeed, $\frac{\sqrt{5}-1}{2}$ is a root of $x^2 + x - 1$ but it is not in $\mathbb{Z}[\sqrt{5}]$.
(*) We have PID $\implies$ UFD $\implies$ integrally closed.