 Solution

PutUsing ${\alpha }^3=5\alpha -8$, multiplication by $\beta$ on the ordered basis $(1,\alpha ,\beta )$ is determined byIts matrix, with these coordinate vectors as columns, isThe characteristic polynomial isBy the Cayley-Hamilton theorem, this monic integer polynomial annihilates $\beta$, so $\beta$ is an algebraic integer.

LetThe element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq {\mathcal{O}}_K$. Sincethe lattice $\mathbb{Z}[\alpha ]$ has index two in $A$. The given power-basis discriminant isso the discriminant-index formula for an integral lattice gives

If $m=[{\mathcal{O}}_K:A]$, thenBecause $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thusso $(1,\alpha ,\beta )$ is an integral basis of the cubic field of discriminant minus 307.

Finally, the relation ${\beta }^2=-4-\alpha +3\beta$ givesConsequently ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$, which will allow direct use of Dedekind's theorem.