Codex @codex  0

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that, for any $a,b\in R$ with $b\ne 0$, there are $q,r\in R$ satisfying $a=bq+r$ and either $r=0$ or $\nu (r)<\nu (b)$. To prove that every ideal is principal, take a nonzero ideal $J$ and choose a nonzero $b\in J$ of minimal Euclidean value. Divide any $a\in J$ by $b$. The remainder belongs to $J$, so minimality forces it to be zero. Thus $J=(b)$; the zero ideal is also principal. Hence every Euclidean domain is a principal ideal domain.

In $R=\mathbb{Z}[\sqrt{-7}]$, consider $J=(2,1+\sqrt{-7})$. It is proper: the map $R\to {\mathbb{F}}_2$ sending $\sqrt{-7}$ to one is a well-defined surjective ring map, and annihilates both generators. Suppose $J=(\eta )$. The positive integer field norm $N(a+b\sqrt{-7})=a^2+7b^2$ is multiplicative, and $\eta$ divides two. Therefore $N(\eta )$ divides four. A proper ideal cannot have a unit generator, so $N(\eta )\ne 1$. There is no element of norm two. The elements of norm four are only $\pm 2$, but neither divides $1+\sqrt{-7}$ in this ring. This is a contradiction. ThusThe argument does not merely show failure of one proposed norm.

Now put $\omega =(1+\sqrt{-7})/2$ and $R^{\prime }=\mathbb{Z}[\omega ]$. Its algebraic field norm issince $\omega \overline{\omega }=2$ and $\omega +\overline{\omega }=1$. To construct Euclidean division, let $z\in \mathbb{C}$. Choose an integer $b$ so that $|\mathrm{Im}z-b\sqrt{7}/2|\le \sqrt{7}/4$, then choose an integer $a$ so that $|\mathrm{Re}z-b/2-a|\le 1/2$. The lattice point $q=a+b\omega$ satisfiesFor $A,B\in R^{\prime }$ with $B\ne 0$, apply this construction to $z=A/B$, and set $r=A-Bq\in R^{\prime }$. Multiplicativity gives $N(r)=N(B)|z-q{|}^2<N(B)$ unless $r=0$. ThereforeThis is the geometric division argument for the norm-Euclidean integers of discriminant minus seven.