= Class number one for Q of cube root six
{title2=$h_{\mathbb Q(\sqrt[3]{6})}=1$}
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 $16\sqrt3/\pi<9$. 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.
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