# analytic number theory – Original Proof of Chebatorev’s Density Theorem

As the title suggests, I am currently trying to understand Chebatorev’s original proof of his density theorem, based on the proof in the appendix here. I am fully on-board with the cyclotomic extension case (which is essentially just a slightly more generalised form of Dirichlet’s theorem). However, I am having difficulties with the reductions.

Specifically, in the case of an abelian extension, we pick some prime $$m nmid Delta_{K/mathbb{Q}}$$ and a primitive $$m^{th}$$ root of unity. The Galois group of the extension $$K(zeta)/F$$ then decomposes as
$$text{Gal}(K(zeta)/F) cong text{Gal}(K/F) times text{Gal}(F(zeta)/F) =: G times H$$
by restricting the automorphisms of $$K(zeta)$$ to the respective subfields. If we now consider a Frobenius substitution (a.k.a. Artin symbol) $$(sigma, tau) in G times H$$ and add the condition that $$(K : F) mid text{ord}(tau)$$ then we find that $$langle (sigma, tau) rangle cap G times {1} = {(1,1)}$$. This is because if $$(sigma, tau)^k = (sigma^k, 1)$$ then it means that $$text{ord}(tau) mid k$$. By transitivity, it follows that $$(K : F) mid k$$ as well and so $$sigma^k=1$$, as claimed.

Now we arrive at my problem. Stevenhagen and Lenstra claim that this intersection property means that the fixed field $$L=K(zeta)^{langle (sigma, tau)rangle}$$ satisfies $$L(zeta) = K(zeta)$$, so that $$K(zeta) / L$$ is an intermediate cyclotomic extension. I’m sure I’m missing something obvious but I simply don’t see why this is true.

Question: Why does $$L(zeta) = K(zeta)$$?

If anyone is familiar with this proof, I would also love to better understand the counting argument which reduces the general extension case to the cyclic extension case. I know a proof using representation theory of finite groups, but I feel that this combinatorial argument is likely more elementary.

Posted on Categories Articles