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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – bsbb4
    Aug 7, 2020 at 15:55
  • $\begingroup$ @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. $\endgroup$
    – user814559
    Aug 7, 2020 at 16:03
  • $\begingroup$ So may I ask if Frobenius elements are extensions of prime ideals (which is extension of prime numbers in ring) in Galois field? $\endgroup$ Aug 7, 2020 at 17:20

1 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.

  • $\begingroup$ 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? $\endgroup$
    – user814559
    Aug 15, 2020 at 18:00

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