If $K$ is algebraically closed, is the fixed field of an involution real-closed?

Question

Let $K$ be an algebraically closed field of characteristic zero equipped with an involution $x\mapsto\overline{x}$ (that is, an order-2 field automorphism). Its fixed field $F:=\lbrace x\in K:\overline{x}=x\rbrace$ is then a proper subfield of $K$ (if it were not proper, then the involution would be trivial).

In the case $K=\mathbb{C}$ with the involution given by complex conjugation, we have that $F=\mathbb{R}$, which is real-closed. But is $F$ always real-closed?

Since $K$ is algebraically closed and has characteristic zero, the polynomial $x^2+1$ has two distinct roots $i$,$-i$ and clearly $\overline{i}=\pm i$. If $\overline{i}=-i$, then each $x\in K$ can be expressed as $$\frac{x+\overline{x}}{2}+\frac{x-\overline{x}}{2i}i\in F(i),$$ so that $F(i)=K$ is algebraically closed, hence $F$ is real-closed. I might have done something silly here but I can't quite rule out the possibility that $\overline{i}=i$ or find a counterexample. Have I missed something?

Answer 1

$char(K)=0$ and $[K:F]=2$ so $K = F(\sqrt{d})$ for some $d\in F$.

Let $\sigma$ be the non-trivial automorphism of $K/F$.

$$\sigma(d^{1/4})^2 = \sigma(d^{1/2})=-d^{1/2}$$ so that $$\sigma(d^{1/4})= \pm i d^{1/4}$$

If $i\in F$ then $\sigma$ has order $4$, impossible.

Whence $i\not \in F$ and $K=F(i)$.

Answer 2

Figured it out now: If $\overline{i}=i$, then $x^2+1$ is reducible over $F$, so $F$ is not real-closed. So $F$ is real-closed precisely when $\overline{i}=-i$.