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I have been trying to come up with a characterization of formally real fields in terms of their finite Galois theory, in order to pique the curiosity of a friend who is a distinguished algebraist, who cares a great deal about finite Galois theory and not at all about formally real fields. I believe the following is such a characterization and want to know if you agree:

Let $K$ be a field.

I think that $K$ is formally real iff it has a quadratic extension $L$ such that, for any finite Galois extension $M/K$ containing $L$, there is an involution of $M/K$ that extends the involution of $L/K$.

Do you agree?

Here is my thinking. $K$ is formally real if and only if it lies in a real-closed subfield of its algebraic closure $\overline K$ (because if it is formally real, its real-closure is such a subfield, and if it lies in a real-closed field, it inherits an order by restriction so it is formally real). In turn, there is a real-closed subfield $K_R$ of $\overline K$ containing $K$ if and only if there is an involution of $\overline K / K$; if there is such an involution, then $K_R$ is its fixed field, and if $K_R\supset K$ exists, then $[K:K_R]=2$, thus there is an involution fixing $K_R$ and therefore $K$. To summarize, $K$ is formally real iff its absolute galois group $G_K := \operatorname{Gal}(\overline K/K)$ contains an involution.

The above is an attempt to render the statement "$G_K$ contains an involution" in finite-Galois-theoretic terms. Since $G_K$ is the inverse limit of the finite Galois groups over $K$, it seems to me that "$G_K$ contains an involution" literally means "there is a Galois field extension $L/K$ with an involution and such that every Galois extension of $K$ containing $L$ contains an involution that restricts to this one." Thus if there is no involution in $G_K$, no such $L$ exists at all. It remains to argue that if there is such an involution, then $L$ can be taken to be quadratic over $K$. I think this because if there is an involution in $G_K$, then let $K_R$ be its fixed field. Then $K_R$ is formally real, so doesn't contain $i = \sqrt{-1}$, so then neither does $K$, and then $L$ can be taken to be $K(i)$.

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Ben Blum-Smith
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This is an immediate result of Artin-Schreier theorem for real closed fields. You can also look here (just replace $\mathbb Q$ with $K$).

Jose
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  • I take it you agree. (Incidentally, the question is really about if I have correctly translated the statement "$K$ is formally real iff its absolute Galois group contains an involution" correctly into finite Galois theory.) – Ben Blum-Smith Sep 07 '18 at 01:48
  • Yes I agree. If the absolute Galois group has an involution then the fixed field of the subgroup generated by this involution is a real closed field (by Artin-Schreier). Furthermore, $K$ lies in this real closed field. Any subfield of a real closed field is real. But nothing beats the natural definition of a real field: It is simply a field $K$ such that for a finite sum of squares $\sum a_i^2 = 0$ one has (iff) $a_i=0$ for all $a_i\in K$ – Jose Sep 07 '18 at 06:43
  • I am not sure. Because you cannot use Galois correspondence the way you use in finite Galois groups. If the absolute Galois group contains an involution, then does it mean that it contains an involution whose induced subgroup is closed in the absolute Galois group? In other words, if you have a subgroup of order 2 in a profinite group then can you also have a closed subgroup of order 2? – quantum Sep 07 '18 at 11:38
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    @quantum - Since a profinite group is $T_1$, in fact Hausdorff, a finite subgroup is automatically closed. – Ben Blum-Smith Sep 07 '18 at 11:41
  • @Jose - I agree, but my friend finds the natural definition uncompelling. I was hoping a Galois-theoretic characterization would be more compelling to him. Incidentally, it seems to me the full strength of Artin-Schreier is more power than needed here. One needs the knowledge that if a field's algebraic closure is quadratic over it, it is real-closed, but not the stronger statement that this is true if its algebraic closure is any finite extension. – Ben Blum-Smith Sep 07 '18 at 15:44
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    Ok you will reduce the proof of Artin-Schreier considerably. But I believe you still need some of the arguments in the original proof: I would show that the field has characeristic that is not 2 (only finally you come to the conclusion it has characteristic 0), then use Kummer Extension theorem. I happened to have compiled the proof of Artin-Schreier from different sources where you can see what I mean in this comment: See here – Jose Sep 08 '18 at 16:35