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

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.

• For $L/K$ abelian, global class field theory shows that the Frobenius elements generate all of $\operatorname{Gal}(L/K)$. See for example J Milne's notes on class field theory, Chapter VII.4. Aug 7, 2020 at 15:55
• @LukasKofler, Thank you for pointing it out. I should have mentioned in my post that the answer is known to me in the abelian case.
– user814559
Aug 7, 2020 at 16:03
• So may I ask if Frobenius elements are extensions of prime ideals (which is extension of prime numbers in ring) in Galois field? Aug 7, 2020 at 17:20

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.
• Thanks! What if we ask for generators, not topological generators in the case of infinite extensions $K_S$? Is it possible to describe the non-Frobenius elements in a minimal generating set? Similarly, are any minimal generating sets of $Gal(\bar{K}/K)$ itself known, topological or not?