Let $K/k$ be a finite unramified Galois extension of number fields, let $K'$ be the Hilbert class field of $K$, let $H:=\mathrm{Gal}(K'/k)$, $\mathrm{Cl}_k$ denotes the ideal class group of $k$.

How to prove that $\mathrm{Cl_k}\simeq H/[H,H]$? The isomorphism is induced by the Artin map.

It's welcome to apply any big theorem in Class field theory which doesn't use this fact.

  • $\begingroup$ @reuns Will $K'$ contain $ k'$ (Hilbert class field of k) in my case? $\endgroup$
    – CYC
    Commented Jul 17, 2019 at 12:20

1 Answer 1

  • For a finite extension $L/k$ of number field for each prime ideal of $O_k$ we have the unique factorization in prime ideals $P O_L = \prod_j Q_j^{e_{P,j}}$. Then $L/k$ is said unramified at the finite places if $\forall P,j, e_{P,j} = 1$.

    $L/k$ is said unramified at the infinite places if for any non-real embedding $\sigma:L \to \Bbb{C}$ then $\sigma(k) \not \subset \Bbb{R}$.

  • If $F/L,L/k$ are unramified then $P O_F = PO_L O_F = \prod_j Q_j O_F = \prod_j \prod_i q_{j,i}$ with the $q_{j,i}$ distinct prime ideals of $O_F$, thus $F/k$ is unramified. Conversely with the same argument if $L/k$ is ramified then $F/k$ is ramified.

  • Thus it makes sense to consider $H_k$ the maximal unramified abelian extension of $k$ (assume we proved it is always a finite extension). Let $K/k$ be Galois unramified.

  • $H_k K/ K$ is unramified : not sure how to prove it, I claim it because $L/k$ unramified at $P$ means for every prime $Q$ above $P$ (with $L_w,k_v$ the $Q$-adic completions) $L_w = k_v(\zeta_{N(Q)-1})$, and this property stays true for compositum.

    Since $H_k K/K$ is also abelian then $H_kK \subset H_K$ and $H_k \subset H_K$.

    If $\sigma \in Gal(\overline{k}/k)$ then $H_K^\sigma/K^\sigma = H_K^\sigma/K$ is abelian unramified, thus $ H_K^\sigma \subset H_K$ and $H_K/k$ is Galois.

    Let $G = Gal(H_K/k)$ then $H_K^{[G,G]}/k$ is the maximal abelian subextension of $H_K/k$ thus $H_k \subset H_K^{[G,G]}$,

    and $H_K/k$ Galois unramified implies $H_K^{[G,G]}/k$ is abelian unramified thus $H_K^{[G,G]} \subset H_k$ and $H_k = H_K^{[G,G]}$.

All this has nothing to do with the main theorem of CFT which is that there is an isomorphism $Cl_k \to Gal(H_k/k)$ given by the Artin map.


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