Relation between separable algebraic closure and the maximal abelian extension

Question

In the book "Class Field Theory" - Milne, the author mentioned: For a filed $K$, let $G:=\operatorname{Gal}(K^{\text{al}}/K)$, then $\operatorname{Gal}(K^{\text{ab}}/K)=G/[G,G]$. However, I can not find the proof of this statement in his appendix as I expected. How can I prove this? Did I miss something in his appendix that directly implies this statement?

Answer 1

For clarity the separable closure is denoted $K^{sep}$.

$G=Gal(K^{sep}/K)$.

• Let $F$ be the subfield fixed by $[G,G]$ (the subgroup generated by the $aba^{-1}b^{-1},a,b\in G$).

  Clearly $K^{ab} \subset F$.

• $[G,G]$ is a normal subgroup so $F/K$ is a normal and Galois extension. For each automorphisms $a',b'\in Gal(F/K)$ we can extend to some automorphisms $a,b$ of $K^{ab}/K$ so that $aba^{-1}b^{-1}$ acts trivially on $F$. Therefore $a'b'=b'a'$ and $Gal(F/K)$ is abelian.

Whence $F=K^{ab}$.

According to this answer $[G,G]$ is usually not closed in the Krull topology, in that case $Gal(K^{ab}/K)$ is not $G/[G,G]$ but $G/\overline{[G,G]}$.