 Solution

Let $\theta =\sqrt[3]{3}$. The power-basis discriminant isIf $\mathbb{Z}[\theta ]$ were a proper sublattice of ${\mathcal{O}}_K$, part (a) would produce a prime $p$ whose square divides $243$ and a new integral element with denominator $p$. The only possible prime is $3$. But $x^3-3$ is Eisenstein at $3$, and part (b) says that every integral element of $(1/3)\mathbb{Z}[\theta ]$ already belongs to $\mathbb{Z}[\theta ]$. This is a contradiction. Hence