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The Kronecker–Weber theorem says that every finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic field. We prove its quadratic case by identifying the ramification that a quadratic number field must have.

For an odd prime $\ell$, the cyclic Galois group of $\mathbb{Q}({\zeta }_{\ell })/\mathbb{Q}$ has exactly one subgroup of index two. Its fixed field $F_{\ell }$ is quadratic and is unramified at every finite prime other than $\ell$. For a squarefree integer $d$, the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d\equiv 1(\mathrm{mod}4)$, and $4d$ otherwise. A nontrivial quadratic discriminant supported on the odd prime $\ell$ must therefore beIndeed $d=\pm 1$ cannot give a nontrivial extension unramified at two, whereas exactly one of $\ell ,-\ell$ is congruent to one modulo four. ThusThis identifies the quadratic subfield of a prime cyclotomic field using ramification, without a Gaussian-sum computation.

Now let $F=\mathbb{Q}(\sqrt{d})$ with $d$ squarefree. Multiply $d$ by ${\prod }_{\ell \mid d,\ell \text{ odd}}{\ell }^{\ast }$. Every odd prime then occurs with even exponent, so its square class is represented by one of $1,-1,2,-2$. Thus adjoining the square roots of the ${\ell }^{\ast }$ removes all possible odd-prime ramification; the remaining quadratic factor is contained inThe original $\sqrt{d}$ is a rational multiple of a product of these square roots. It belongs to the compositum of $\mathbb{Q}({\zeta }_8)$ and the prime cyclotomic fields just used. This compositum is itself contained in a cyclotomic field, proving the theorem for $F$.

The sharper conductor statement follows from the same ramification calculation. For the fundamental discriminant $D$ of $F$, putThe possibilities for $D_2$ follow immediately from the quadratic discriminant formula. The corresponding fields are trivial, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$, lying in the cyclotomic fields of conductors $1,4,8,8$. Their conductors are coprime to the odd primes in the product. Hence quadratic Kronecker–Weber theorem by ramification gives the explicit conclusion

We use the following standard facts about cyclotomic fields. For an odd prime $\ell$, the unique quadratic subfield of $\mathbb{Q}({\zeta }_{\ell })$ isAlso $\mathbb{Q}({\zeta }_{{\ell }^r})$ has degree $\phi ({\ell }^r)={\ell }^{r-1}(\ell -1)$, and $\ell$ is totally ramified, with a unique prime and residue field ${\mathbb{F}}_{\ell }$. These are the quadratic subfield of a prime cyclotomic field and total ramification in a prime-power cyclotomic field results.

TakeSince $59\equiv 3(\mathrm{mod}4)$, every $E_r$ contains $K=\mathbb{Q}(\sqrt{-59})$, through its subfield $E_1$. The unique prime above $59$ is totally ramified in $E_r/K$: the absolute ramification index is $58\cdot 59^{r-1}$, the index in $K/\mathbb{Q}$ is $2$, and the relative index is thereforeThe field $K$ has no real places. By part (b), $h_K\mid h_{E_r}$, and part (a) showed $3\mid h_K$. ThusThe ring of integers of a number field is a Dedekind domain, and is a principal ideal domain exactly when its ideal class group is trivial. Hence none of these rings of integers is principal. Finally the degrees $58\cdot 59^{r-1}$ strictly increase, so the fields are pairwise distinct. The explicit infinite family of nonprincipal cyclotomic fields of conductor a prime power is

For an odd prime $\ell$, the unique quadratic subfield of a prime cyclotomic field has no finite ramification outside $\ell$. The classification of quadratic discriminants identifies it with $\mathbb{Q}(\sqrt{{\ell }^{\ast }})$, where ${\ell }^{\ast }=(-1{)}^{(\ell -1)/2}\ell$. A fundamental discriminant factors as $D=D_2{\prod }_{\ell \mid D,\ell \text{ odd}}{\ell }^{\ast }$ with $D_2\in \{1,-4,8,-8\}$. The remaining quadratic number field is generated by $i$, $\sqrt{2}$, or $\sqrt{-2}$ and lies in $\mathbb{Q}({\zeta }_8)$. Taking the corresponding compositum proves the indicated quadratic case of the Kronecker–Weber theorem.