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

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?

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 $$\overline{i}=-i$$.

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

• How do you know automatically that $[K:F]=2$? In my question above, I had to assume $\overline{i}=-i$ to show that $K=F(i)$. Dec 6, 2021 at 20:32
• @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 $K=F(i)$. Dec 6, 2021 at 21:40