Algebraically closed field of characteristic zero I have a question about a proof which looks similar to the algebraically closedness of $\mathbb{C}$:
Let $F$ be a field of characteristic $0$, such that every odd polynomial over $F$ has at least one root in $F$. Assume that $E$ is a degree $2$ extension of $F$
such that for each $a\in E$, the polynomial $x^2-a$ has a root in $E$. Prove that $E$ is algebraically closed.
 A: Sketching this (standard) solution:


*

*Because $[E:F]=2$, and we are in characteristic zero, $E/F$ is Galois. Let's use the notation familiar from the case $F=\Bbb{R}$, $E=\Bbb{C}$ for the non-trivial automorphism, namely denote the $F$-conjugate of $z\in E$ by $\overline{z}$. We extend that conjugation operation to an automorphism of the polynomial ring $E[x]$ by conjugating all the coefficients.

*Let $p(x)\in E[x]$ be an arbitrary polynomial. Then $r(x)=p(x)\overline{p}(x)\in F[x]$. Let $K$ be a splitting field of $r(x)$ over $F$, and let $G=Gal(K/F)$ be the Galois group.

*The group $G$ is finite, so we can write its order as $|G|=2^nm$, where $m$ is an odd natural number. Let $P\le G$ be a Sylow $2$-subgroup, so $[G:P]=m$. Let $L=K^P$ be the fixed field of $P$. By Galois theory $[L:F]=m$ is odd.

*The extension $L/F$ is finite and separable. Hence there exists an element $\alpha\in L$ such that $L=F[\alpha]$. The minimal polynomial $m(x)$
of $\alpha$ is irreducible over $F$ and of odd degree $m$. Therefore we must have $m=1$.

*Ok, so $P=G$. We claim that $|P|\le 2$. Assume contariwise that $n\ge2$. Because $P$ is  nilpotent it has a normal series
$$\cdots P_2\unlhd P_1\unlhd P $$
such that $[P:P_1]=[P_1:P_2]=2$.

*Show that the fixed field of $P_1$ must be (up to isomorphism) $E$, and that the fixed field of $P_2$ must be a quadratic extension of $E$. Fields of the latter kind don't exist so we must have $K=E$ and we are done. 

