Integral basis of the cubic field of discriminant minus 140 (${\mathcal{O}}_{\mathbb{Q}(\theta )}=\mathbb{Z}[\theta ],{\theta }^3+2\theta +2=0$)

The power basis $1,\theta ,{\theta }^2$ is an integral basis. The defining polynomial is irreducible by Eisenstein's criterion at two and its discriminant is $-140$. The discriminant-index formula for an integral lattice restricts its index in the ring of integers to one or two. An index of two would supply an algebraic integer $\eta =(a+b\theta +c{\theta }^2)/2$ with $a,b,c\in \{0,1\}$ not all zero. Its field trace $(3a-4c)/2$ forces $a=0$. If $b=0$ this gives ${\theta }^2/2$ directly; if $b=1$, the identity $\theta \eta ={\theta }^2/2-c\theta -c$ gives the same integrality. But ${\theta }^2/2=-1-{\theta }^{-1}$, while $N({\theta }^{-1})=-1/2$ excludes integrality by the field norm. Thus the index is one, proving the claim.