Let $K$ be any field, $\overline K$ is the separable closure of $K$ and $K^{ab}$ is the maximal abelian extension of $K$. I want to prove the following relation

$$G(\overline K|K)^{ab}=G(K^{ab}|K)$$

Remember that if $G$ is a group $G^{ab}$ is its abelianization.

How can I prove this result? I'm quite sure that I have to use the following isomorphism given by the fundamental theorem of Galois theory (for infinite Galois extensions): $$G(\overline{K}|K)\big/G(\overline K|k^{ab})\cong G(K^{ab}|K)$$

From here I'm trying to see if $G(K^{ab}|K)$ respects the usual universal property of the abelianization...


2 Answers 2


The statement is true, assuming that by the abelianization $G^{ab}$ of a profinite group $G$ we mean its abelianization as a profinite group: $G^{ab} = G/[G,G]$, where $[G,G]$ is the closed subgroup generated by all commutators in $G$.

Let's write $G = G(\overline{K}\mid K)$. We want to show that $G^{ab}$ is the Galois group of $K^{ab}$ over $K$. Of course, since $G^{ab}$ is a quotient of $G$, the usual Galois theory for fields tells us that $G^{ab}$ is the Galois group of a well-defined field $L$ with $K \subseteq L \subseteq \overline K$, namely the fixed field in $\overline K$ of the commutator subgroup $[G, G]$ of $G$. So we want to show that $K^{ab} = L$.

For this, first notice that $L$ is an abelian extension of $K$, since by construction its Galois group $G^{ab}$ is abelian. Since $K^{ab}$ is the maximal abelian extension of $K$, this means that $L \subseteq K^{ab}$. Therefore, by the Galois correspondence, we get the reverse inclusion on the side of groups, namely $G(\overline K\mid K^{ab}) \subseteq G(\overline K \mid L) = [G, G]$.

On the other hand, since $K^{ab}$ is an abelian extension, any commutator in $[G, G]$ must fix $K^{ab}$ element-wise, so we also get $[G, G] \subseteq G(\overline K \mid K^{ab})$. Combined with the previous inclusions this shows that $G(\overline K\mid K^{ab}) = G(\overline K \mid L)$, so $K^{ab}$ and $L$ correspond to the same subgroup of $G$, and again using the Galois correspondence then shows that $K^{ab} = L$.

  • $\begingroup$ Since $G$ is infinite, don't we need, in order to apply the Galois correspondence, that $[G,G]$ is a closed subgroup of $G$ with respect to its profinite topology? Is this always true? $\endgroup$
    – Symòn
    Jan 2, 2018 at 14:36
  • 1
    $\begingroup$ @Symòn In the context of profinite groups, $[G,G]$ denotes the closed subgroup generated by all commutators, so it is closed by definition. The reason for this is that otherwise the quotient will not be a profinite group. I will edit my answer to clarify. $\endgroup$
    – Marc
    Jan 2, 2018 at 15:29

It's immediate conclusion following from the fact that $G(\overline{K}/K^{ab})$ is closure of a commutator subgroup of $G_K$:$G(\overline{K}/K^{ab})=\overline{[G_K:G_K]}$.

In fact by the basic formula $G(L/K_1\cdot K_2)=G(L/K_1)\cap G(L/K_2)$, we have

$G(\overline{K}/K^{ab})$ $=G(\overline{K}/\biguplus_{\text{$L_i/K$ is abelian}} L_i)$ $=\bigcap_{L_i/K\ is\ abelian} G(\overline{K}/L_i)$ $=\bigcap_{G_K/H_i\ is\ abelian} H_i$ $=\overline{[G_K:G_K]}$,

where $\biguplus$ denotes compositum of fields.

So, $G(K^{ab}/K)\cong G_K/G(\overline{K}/K^{ab}) =G_K/\overline{[G_K:G_K]} =G_K^{ab}$.


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