= 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 $\mathbb F_p$. Then
$$
p\mathcal O_K=\prod_i\mathfrak p_i^{e_i},\qquad
\mathfrak p_i=(p,f_i(\theta)),\qquad
[\mathcal O_K/\mathfrak p_i:\mathbb F_p]=\deg f_i.
$$
The index condition is essential. All integral bases used below make the index one, so it applies to every <prime>.
For a <squarefree> quadratic $d\equiv2,3\pmod4$, 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 $d\equiv1\pmod4$, 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
$$
\boxed{p\text{ is totally ramified }\Longleftrightarrow p\mid D}.
$$
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 $\zeta^j$,
$$
\Phi_q'(\zeta^j)=\frac{q(\zeta^j)^{q-1}}{\zeta^j-1},
$$
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.
Back to article page
