= Solution
The <Kronecker–Weber theorem> says that \b[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 $d\equiv1\pmod4$, and $4d$ otherwise. A nontrivial quadratic <discriminant> supported on the odd prime $\ell$ must therefore be
$$
\ell^*=(-1)^{(\ell-1)/2}\ell.
$$
Indeed $d=\pm1$ cannot give a nontrivial extension unramified at two, whereas exactly one of $\ell,-\ell$ is congruent to one modulo four. Thus
$$
F_\ell=\mathbb Q(\sqrt{\ell^*})\subseteq\mathbb Q(\zeta_\ell).
$$
This 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 $\prod_{\ell\mid d,\ \ell\text{ odd}}\ell^*$. 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 $\ell^*$ removes all possible odd-prime <ramification>; the remaining quadratic factor is contained in
$$
\mathbb Q(\zeta_8)=\mathbb Q(i,\sqrt2),\qquad \sqrt2=\zeta_8+\zeta_8^{-1}.
$$
The 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 $F$, put
$$
D=D_2\prod_{\substack{\ell\mid D\\\ell\text{ odd}}}\ell^*,\qquad D_2\in\{1,-4,8,-8\}.
$$
The possibilities for $D_2$ follow immediately from the quadratic <discriminant> formula. The corresponding fields are trivial, $\mathbb Q(i)$, $\mathbb Q(\sqrt2)$ 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
$$
\boxed{\mathbb Q(\sqrt D)\subseteq\mathbb Q(\zeta_{|D|}).}
$$
Back to article page
