A PID that is not Euclidean

Such a ring is \[ R=\mathbb Z\left[\frac{1+\sqrt{-19}}2\right], \] the ring of integers of \(\mathbb Q(\sqrt{-19})\).

First, \(R\) is a PID. The discriminant is \(-19\), so Minkowski’s bound for ideal classes in this imaginary quadratic field is \[ \frac{2}{\pi}\sqrt{19}<3. \] Hence every ideal class contains an integral ideal of norm \(1\) or \(2\). But \(2\) is inert: the defining polynomial \[ x^2-x+5\equiv x^2+x+1\pmod2 \] has no root in \(\mathbb F_2\). Thus there is no ideal of norm \(2\). Every ideal class therefore contains an ideal of norm \(1\), hence the unit ideal; the class group is trivial, so \(R\) is a PID.

It is not Euclidean for any Euclidean function. Indeed, in any Euclidean domain that is not a field, choose a nonzero nonunit \(b\) of minimal Euclidean value. For every \(a\), Euclidean division gives \[ a=qb+r,\qquad r=0\text{ or }\delta(r)<\delta(b). \] By minimality of \(\delta(b)\) among nonzero nonunits, every nonzero remainder \(r\) must be a unit. Hence every residue class modulo \((b)\) is represented by \(0\) or by a unit.

For the present ring, the norm is \[ N(a+b\theta)=a^2+ab+5b^2, \qquad \theta=\frac{1+\sqrt{-19}}2. \] The only units are \(\pm1\). Therefore such a quotient \(R/(b)\) would have at most three residue classes, so \(|R/(b)|=|N(b)|\) would have to be \(2\) or \(3\). But \[ a^2+ab+5b^2 =\left(a+\frac b2\right)^2+\frac{19}{4}b^2 \] never equals \(2\) or \(3\) for integers \(a,b\). This contradiction shows that \(R\) admits no Euclidean function.