 Solution

The factorization form of Kummer-Dedekind theorem is as follows. If $K=\mathbb{Q}(\theta )$ with $\theta$ integral, minimal polynomial $f$, and $p$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\theta ]]$, factor $\bar{f}={\prod }_i{\bar{f}}_i^{e_i}$ into distinct monic irreducibles over ${\mathbb{F}}_p$. ThenThe index condition is essential. All integral bases used below make the index one, so it applies to every prime.

For a squarefree quadratic $d\equiv 2,3(\mathrm{mod}4)$, take $\theta =\sqrt{d}$, $f=X^2-d$ and field discriminant $D=4d$. At an odd prime, a repeated factor occurs exactly when $p\mid d$, in which case $\bar{f}=X^2$ and $p{\mathcal{O}}_K=(p,\sqrt{d}{)}^2$. At $p=2$, reduction is $X^2$ if $d$ is even and $(X+1{)}^2$ if $d$ is odd, so $2$ is totally ramified in both cases. Thus the totally ramified primes are exactly those dividing $4d$.

For $d\equiv 1(\mathrm{mod}4)$, take $\theta =(1+\sqrt{d})/2$, with polynomial $X^2-X+(1-d)/4$ and discriminant $D=d$. At an odd prime it has a double root exactly when $p\mid d$, giving a single linear factor squared and total ramification. At $p=2$, its derivative is $1$, so no repeated factor occurs. Hence again

For the prime cyclotomic field, use ${\Phi }_q(X)=1+X+\cdots +X^{q-1}$. Modulo $q$ this is $(X-1{)}^{q-1}$, so $q{\mathcal{O}}_K=(q,\zeta -1{)}^{q-1}$, proving total ramification. For $p\ne q$, the discriminant has no prime factor $p$, so the reduction is squarefree and $p$ is unramified. There is an actual exponent error in the printed discriminant: the correct value is $(-1{)}^{(q-1)/2}{q}^{q-2}$, not $q^{q-1}$. To check it, at a nontrivial root ${\zeta }^j$,and multiplying these derivatives and using ${\prod }_{j=1}^{q-1}(1-{\zeta }^j)=q$ gives the exponent $q-2$. The prime support is still just $\{q\}$, so the requested ramification conclusion is unaffected.

For $\theta =\sqrt[3]{2}$, the integral basis gives $f=X^3-2$ and $D=-108$. Modulo $2$, $\bar{f}=X^3$; modulo $3$, $\bar{f}=(X+1{)}^3$. Dedekind factorization gives $(2)=(2,\theta {)}^3$ and $(3)=(3,\theta +1{)}^3$. For other primes the polynomial is squarefree because its discriminant is not divisible by the prime. Therefore the equivalence also holds in this cubic field. This is a conclusion for these specific fields, not the false general assertion that every ramified prime in every higher-degree number field is totally ramified.