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