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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?

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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 $\overline{i}=-i$.

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$char(K)=0$ 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 $K=F(i)$.

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  • $\begingroup$ How do you know automatically that $[K:F]=2$? In my question above, I had to assume $\overline{i}=-i$ to show that $K=F(i)$. $\endgroup$
    – user829347
    Dec 6, 2021 at 20:32
  • $\begingroup$ @thewonderfulwizardofoz The fixed field of an automorphism of order $2$ is automatically such that $[K:F]=2$. My answer says that $K$ algebraically closed of characteristic not $2$ and $[K:F]=2$ implies that $K=F(i)$. $\endgroup$
    – reuns
    Dec 6, 2021 at 21:40

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