= Solution
The <existence theorem of global class field theory>, together with reciprocity, gives an inclusion-reversing bijection between finite <abelian extensions> $L/K$ inside a fixed algebraic closure and open finite-index subgroups of the <idèle class group> $C_K$. The subgroup for $L$ is $N_{L/K}C_L$, and
$$
\boxed{C_K/N_{L/K}C_L\cong\operatorname{Gal}(L/K).}
$$
Equivalently, for each modulus $\mathfrak m$, subgroups between the ray-principal group $P_K(\mathfrak m)$ and the <ideal> group $I_K(\mathfrak m)$ classify <abelian extensions> whose conductor divides $\mathfrak m$. Their quotients are identified with the corresponding <Galois groups> by the <Artin map>. Changing the modulus gives the same extension when the induced congruence subgroup is the same.
The conductor criterion says that a finite place has conductor exponent zero exactly when the extension is unramified there; a real place is absent from the conductor exactly when it stays real. Thus conductor one means unramified at every finite place and no real-to-complex ramification.
Apply the existence theorem to the open subgroup of $C_K$ which is the image of
$$
U=\prod_{v<\infty}\mathcal O_v^*\times\prod_{v\mid\infty}K_v^*.
$$
The idele-to-ideal map identifies the quotient by this subgroup with the ordinary <ideal class group>, which is finite. Therefore its class field $H$ exists and satisfies
$$
\boxed{\operatorname{Gal}(H/K)\cong\operatorname{Cl}(K),\qquad[H:K]=h_K.}
$$
Its conductor is one, so it is everywhere unramified. Conversely any everywhere-unramified <abelian extension> has conductor one; its <Artin map> kills $U$, hence factors through this quotient and corresponds to an extension contained in $H$. This proves that $H$ is exactly the <Hilbert class field>, the maximal abelian everywhere-unramified extension. Here the ordinary <ideal class group> is appropriate because the infinite factors include all $K_v^*$, rather than only the positive real factors used in a narrow <ray class group>.
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