Class number one for Q of cube root six ($h_{\mathbb{Q}(\sqrt[3]{6})}=1$)

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 $16\sqrt{3}/\pi <9$. 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.