Subgroup generated by Frobenius elements in Galois group of a number field extension

Question

I have an elementary question.

Let $K$ be a number field.

Suppose $L$ is a finite Galois extension of $K$ with Galois group $\text{Gal}(L/K)$.

What is the (normal) subgroup generated by all the Frobenius elements in $\text{Gal}(L/K)$?

I'd like to ask the same question about $\text{Gal}(\bar{K}/K)$, where $\bar{K}$ is an algebraic closure of $K$, but I wonder if it makes sense in this case.

Answer 1

If the Frobenius at a prime $\mathfrak{P}$ above $\mathfrak{p}$ is $g$, then the Frobenius at a conjugate prime $\sigma \mathfrak{P}$ is $\sigma g \sigma^{-1}$, so every element in the conjugacy class will occur.

Is is a theorem of Cebotarev (https://en.wikipedia.org/wiki/Chebotarev%27s_density_theorem) that every conjugacy class of $G$ occurs as the conjugacy class of Frobenius elements of primes $\mathfrak{P}$ above $\mathfrak{p}$ for infinitely many $\mathfrak{p}$, so certainly every element of the group occurs (infinitely often) as a Frobenius element, and so they generate the entire group $G$.

If you take $K_{S}$ to be the maximal extension of $K$ unramified outside $S$ for a finite $S$, then as a consequence you deduce that every element occurs (infinitely often) as Frobenius for any finite quotient of $\mathrm{Gal}(K_S/K)$. It follows that the Frobenius elements inside $\mathrm{Gal}(K_S/K)$ are topologically dense (with the natural inverse limit topology), and so they certainly generate the group topologically. It doesn't quite make sense to take $S$ to be all primes because then Frobenius elements are not defined.

You do have to take the topological closure to get everything; if you take the extension $L = \mathbf{Q}(\zeta_{p^{\infty}})$ then $\mathrm{Gal}(L/\mathbf{Q}) = \mathbf{Z}^{\times}_p$, and the Frobenius elements all have the form $q$ for some prime $q \ne p$. This literally only generates the countable subgroup $\mathbf{Q}^{\times} \cap \mathbf{Z}^{\times}_p$, but it is topologically dense.