Euclidean domains and PIDs

Every Euclidean domain is a PID. Indeed, let \(R\) be Euclidean with Euclidean function \(\delta\), and let \(0\ne I\triangleleft R\). Choose \(0\ne a\in I\) with \(\delta(a)\) minimal. For any \(b\in I\), Euclidean division gives \[ b=qa+r, \] with either \(r=0\) or \(\delta(r)<\delta(a)\). Since \(r=b-qa\in I\), minimality forces \(r=0\). Hence every \(b\in I\) lies in \((a)\), so \(I=(a)\).

Thus \[ \text{Euclidean domain}\Longrightarrow\text{PID}\Longrightarrow\text{UFD}. \] Neither implication reverses in general.

For UFD \(\not\Rightarrow\) PID, take \(k[x,y]\). It is a UFD by Gauss’s theorem, but the ideal \((x,y)\) is not principal: if \((x,y)=(d)\), then \(d\) divides both \(x\) and \(y\), so \(d\) is a unit, contradicting \((x,y)\ne k[x,y]\).

For PID \(\not\Rightarrow\) Euclidean, take \[ R=\mathbb Z\!\left[\frac{1+\sqrt{-19}}2\right]. \] Its discriminant is \(-19\), so Minkowski’s bound is \((2/\pi)\sqrt{19}<3\). Thus every ideal class contains an integral ideal of norm \(1\) or \(2\). The polynomial \(x^2-x+5\) is irreducible modulo \(2\), so \(2\) is inert and there is no ideal of norm \(2\). Hence every ideal class is trivial, so \(R\) is a PID.

It is not Euclidean for any Euclidean function. If \(R\) were Euclidean, choose a nonzero nonunit \(b\) of minimal Euclidean value. Division by \(b\) would show that every residue class modulo \((b)\) is represented by \(0\) or a unit. The norm is \[ N(a+c\theta)=a^2+ac+5c^2, \qquad \theta=\frac{1+\sqrt{-19}}2, \] and the only units are \(\pm1\), so \(|R/(b)|=|N(b)|\) would be at most \(3\) and greater than \(1\), hence \(2\) or \(3\). But \[ a^2+ac+5c^2=\left(a+\frac c2\right)^2+\frac{19}{4}c^2 \] never equals \(2\) or \(3\) for integers \(a,c\). This contradiction shows that \(R\) is not Euclidean.