Galois extension over a field of characteristic $0$ of order $2^n$ Assume that $E$ is Galois over a field $F$ of characteristic 0 and $Gal(E/F)$ is a group of order $2^n$. Prove that $E$ can be obtained from $F$ by a sequence of simple extensions by roots of polynomials of the form $x^2 - c$
A group of order $2^n$ is nilpotent and a polynomial is irreducible iff $c$ is a non square.  Does extensions of the form $x^2 -c$ mean extensions by the root?  
 A: Let $F$ be a field of characteristic $\neq 2.$
We say $L/F$ is a quadratic extension if $L\cong F[X]/(X^2-c)$ for some $c \in F.$
We say $M/F$ is a quadratic tower if there exists chain of intermediate fields
$$F=F_0\subseteq F_1 \subseteq \ldots \subseteq F_n=M$$
such that $F_i/F_{i-1}$ is a quadratic extension for each $i=1,\ldots,n.$
Claim. Every non-trivial Galois extension of $F$ whose degree is a power of $2$ is a quadratic tower.
Proof. We proceed by induction.

(Base case). Suppose $K/F$ is a Galois extension with $[K:F]=2.$
Choose $\alpha \in K$ such that $\alpha \notin F$ (we can do this since $[K:F]=2$ and so $K\neq F$).
Then $\alpha$ has minimal polynomial over $F$ of the form $X^2+aX+b$ for some $a,b \in F.$
Take $\beta = \alpha+\frac{1}{2}a \in K\,$ and $c=\frac{1}{4}a^2-b \in F$ (we can divide by 2 and by 4 since char$(F)\neq 2$).
Then $\beta \notin F$ (because $\alpha \notin F$) and $\beta$ is a root of $X^2-c.$
Hence $\beta$ has degree $2$ over $F$ and so we must have $K=F(\beta)\cong F[X]/(X^2-c).$

(Induction step). Suppose that the claim is true for all Galois extensions of $F$ of degree $2^m.$
Let $E/F$ be a Galois extension with $[E:F]=2^{m+1}.$
Since Gal$(E/F)$ is a finite $p$-group (with $p=2,$ of course), it has a subgroup of size $2^m$.
So by the fundamental theorem of Galois theory, we have an intermediate field
$$F\subseteq K \subseteq E$$
such that $E/K$ is Galois with $[E:K]=2^m.$
Hence, by induction hypothesis, $E/K$ is a quadratic tower.
Also, since $[K:F]=2,$ we see that $K/F$ is a quadratic extension (by the base case).
Therefore $E/F$ is a quadratic tower and the claim follows by induction.
