 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 $N_{L/K}{C}_L$, andEquivalently, 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 ofThe 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 satisfiesIts 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 $K_v^{\ast }$, rather than only the positive real factors used in a narrow ray class group.