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.