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Let $\overline{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$. The absolute galois group $G_\mathbb{Q}$ of $\mathbb{Q}$ acts on the set of real-closed subfields of $\overline{\mathbb{Q}}$.

Does it act transitively?

The real-closed subfields are in bijection with the involutions of $G_\mathbb{Q}$ under the Galois correspondence, so another way to ask the question would be,

Do the involutions of $G_{\mathbb{Q}}$ form a single conjugacy class?

I am asking out of curiosity. I have been unable to locate the answer in any of my texts on real fields, or via internet search, but please forgive me if it is well known. Because the order on a real-closed field is unique, and $\mathbb{Q}$ is order-dense in its real closure, the real-closed subfields $K$ of $\overline{\mathbb{Q}}$ have trivial automorphism group, and it follows that the stabilizer of each $K$ for the action of $G_\mathbb{Q}$ is just the involution fixing $K$ pointwise; thus the action is almost free, and this made me curious if it is transitive.

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    $\begingroup$ $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ is probably beyond apprehension on its own, but we know a lot of its quotients. If this is false, maybe there's some easy quotient that witnesses its falsity? $\endgroup$
    – LSpice
    Commented Sep 4, 2018 at 21:14
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    $\begingroup$ Oh, wait. An involution in a quotient of $G_{\mathbb Q}$ doesn't obviously lift to an involution in $G_{\mathbb Q}$, so maybe it's subtler than that. $\endgroup$
    – LSpice
    Commented Sep 4, 2018 at 21:17
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    $\begingroup$ This answer by Matt E on math.SE says that all involutions in the absolute Galois group are conjugate: math.stackexchange.com/a/622935/448 $\endgroup$
    – David E Speyer
    Commented Sep 4, 2018 at 21:18
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    $\begingroup$ I think Matt E is saying that all the involutions are "complex conjugations" i.e. they are involutions fixing a real-closed field, but I don't believe he is asserting that they are conjugate in $G_\mathbb{Q}$... $\endgroup$
    – benblumsmith
    Commented Sep 4, 2018 at 21:20
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    $\begingroup$ This fact is mentioned - with some references - at page 5 of this notes by Keith Conrad. $\endgroup$
    – Jarek Kuben
    Commented Sep 4, 2018 at 21:55

1 Answer 1

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Any real-closed subfield $R\subseteq\overline{\mathbb Q}$ is a real closure of $\mathbb Q$ (being real closed and algebraic over $\mathbb Q$). Thus, by uniqueness of real closures, any two such fields are isomorphic, and an isomorphism of $R$ to $R'$ extends to an isomorphism of $R(i)=\overline{\mathbb Q}$ to $R'(i)=\overline{\mathbb Q}$, i.e., an automorphism of $\overline{\mathbb Q}$.

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    $\begingroup$ That's it! The argument I gave above in comments is essentially this, only muddied up by the embeddings in $\mathbb{R}$ and $\mathbb{C}$. $\endgroup$
    – benblumsmith
    Commented Sep 4, 2018 at 21:43
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    $\begingroup$ The key property of base field $\mathbf Q$ is that $\mathbf Q$ has just one ordering. If we use a number field $K$ in place of $\mathbf Q$ and ask about conjugacy classes of involutions in ${\rm Gal}(\overline{K}/K)$, those are in bijection with the orderings on $K$ (equivalently, they are bijection with the embeddings of $K$ into $\mathbf R$), so ${\rm Gal}(\overline{K}/K)$ has involutions and they are all conjugate if and only if $K$ has exactly one real embedding. $\endgroup$
    – KConrad
    Commented Sep 4, 2018 at 23:46

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