State Dedekind's theorem on the factorisation of rational primes into prime ideals.

A rational prime is said to ramify totally in a field with degree $n$ if it is the $n$-th power of a prime ideal in the field. Show that, in the quadratic field $\mathbb{Q}(\sqrt{d})$ with $d$ a squarefree integer, a rational prime ramifies totally if and only if it divides the discriminant of the field.

Verify that the same holds in the cyclotomic field $\mathbb{Q}(\zeta)$, where $\zeta=e^{2 \pi i / q}$ with $q$ an odd prime, and also in the cubic field $\mathbb{Q}(\sqrt[3]{2})$.

$[$ The cases $d \equiv 2,3(\bmod 4)$ and $d \equiv 1(\bmod 4)$ for the quadratic field should be carefully distinguished. It can be assumed that $\mathbb{Q}(\zeta)$ has a basis $1, \zeta, \ldots, \zeta^{q-2}$ and discriminant $(-1)^{(q-1) / 2} q^{q-1}$, and that $\mathbb{Q}(\sqrt[3]{2})$ has a basis $1, \sqrt[3]{2},(\sqrt[3]{2})^{2}$ and discriminant $\left.-108 .\right]$