Minkowski's bound

If a number field of degree $n$ and signature $(r_1,r_2)$ has discriminant $d_K$, every ideal class contains an integral ideal $I$ with

For ${\alpha }^3=6$, the ring of integers is $\mathbb{Z}[\alpha ]$ and its field discriminant is $-972$. The Minkowski bound for ideal classes is $16\sqrt{3}/\pi <9$. Prime ideals of norm at most eight lie above $2,3,5,7$. The primes above two, three and the degree-one prime above five are generated by $\alpha -2$, $3+2\alpha +{\alpha }^2$ and $\alpha -1$, of norm magnitudes $2,3,5$. At seven the residue roots are $3,5,6$; the primes at roots six and three are generated by $\alpha +1$ and ${\alpha }^2+\alpha -5$, both of norm seven. Their product with the remaining prime is $(7)$, making that prime principal too. Every possible Minkowski representative is therefore principal.

Since every nonzero integral ideal has norm at least one, the Minkowski bound for ideal classes impliesfor every number field of degree $n$ and signature $(r_1,r_2)$.

For $K=\mathbb{Q}(\sqrt[3]{3})$ with ${\mathcal{O}}_K=\mathbb{Z}[\sqrt[3]{3}]$, the field discriminant is $-243$ and the Minkowski bound is $8\sqrt{3}/\pi <5$. The prime ideals above $2$ and $3$ are principal, so every ideal class has a principal representative and

The imaginary quadratic field $K=\mathbb{Q}(\sqrt{-163})$ has discriminant $-163$, and its Minkowski bound for ideal classes isThe polynomial $X^2-X+41$ defining its ring of integers of a quadratic field has no root modulo $2,3,5,$ or $7$, so these rational primes are inert. Factoring an integral ideal of norm at most eight into prime ideals then shows that its only possible nontrivial factor is the principal ideal $(2)$, of norm four. Every ideal class is therefore principal, and

The field has ring of integers $\mathbb{Z}[\sqrt{-33}]$, discriminant $-132$, and Minkowski bound below eight. The nonprincipal ramified ideals above two and three generate four classes, and the split primes above seven introduce no new class. Hence

The ring of integers is $\mathbb{Z}[\sqrt{-22}]$, its discriminant is $-88$, and its Minkowski bound for ideal classes is $4\sqrt{22}/\pi <6$. The primes three and five are inert, whileThe ideal ${\mathfrak{p}}_2$ is nonprincipal because $a^2+22b^2=2$ has no integer solution. Every class is represented by an ideal of norm at most five, so $[{\mathfrak{p}}_2]$ is the only possible nontrivial class and the class group is $C_2$.

The equation $y^3=x^2+22$ has no integer solutions. Factoring it in $\mathbb{Z}[\sqrt{-22}]$, the two conjugate factors are coprime. Unique factorization of ideals in a number field and the absence of three-torsion in the class group force $x+\sqrt{-22}$ to be a cube up to a unit. Writing it as $(a+b\sqrt{-22}{)}^3$ makes its $\sqrt{-22}$ coefficientwhich has no integer solution.