# Why $\mathrm{Cl_k}\simeq H/[H,H]$?

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.

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

• 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.