= Integral basis of the cubic field of discriminant minus 140
{title2=$\mathcal O_{\mathbb Q(\theta)}=\mathbb Z[\theta],\quad\theta^3+2\theta+2=0$}
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.
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