Original proof of Chebotarev's density theorem

Question

As the title suggests, I am currently trying to understand Chebotarev'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.

Edit: Thanks to alpoge's excellent answer, I now understand all the details of the abelian extension case, so all that remains is the counting argument to reduce from the general case to the abelian case. I thought I should add that I looked at the reference to Lang's Algebraic Number Theory, but their nod towards an actual counting argument amounts to "there is a bijection between the two sets in question" which didn't really help.

Answer 1

Collecting comments into a community wiki answer:

Regarding why $L(ζ)=K(ζ)$: $L(ζ)⊆K(ζ)$ is the compositum of the subextensions $L⊆K(ζ)$ and $F(ζ)⊆K(ζ)$. Now (changing notation) when $F⊆K$ and $K'⊆L$ are subextensions of a Galois $L/F$, the compositum $K⋅K'⊆L$ has $\operatorname{Gal}(L/K⋅K')=\operatorname{Gal}(L/K)∩\operatorname{Gal}(L/K')$ (an element of $\operatorname{Gal}(L/F)$ fixes $K⋅K'$ pointwise if and only if it fixes both $K$ and $K'$ pointwise).

Note. Lenstra and Stevenhagen's proof is not Chebotarev's original proof. Lenstra and Stevenhagen say their proof "follows his original strategy, if not his tactics." How is it not his tactics? The counting argument they cite from Lang's book, which reduces the task to a cyclic extension, is due to Deuring 10 years after Chebotarev's proof. Chebotarev's proof only treats base field $\mathbb{Q}$, and Deuring's reduction is incompatible with that since it usually increases the base field. For Chebotarev's proof in German, see "Die Bestimmung der Dichtigkeit einer Menge von Primzahlen, welche zu einer gegebenen Substitutionsklasse gehören", Mathematische Annalen 95 (1926), 191–228. For the Russian version, see "Détermination de la densité des nombres premiers appartenants à une classe donnée de substitutions," Part I and Part II, Bulletin de l'Académie des Sciences de Russie VI série, 17:1–18 (1923), 205–250. Click the "PDF file" link.