Questions about the proof of constructible $n$-polygon I have several questions about the proof of the following theorem.
(In the second paragraph, the converse part)$\def\Gal{\operatorname{Gal}} \def\Q{\Bbb Q}$


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*Is it true that $H=\Gal(\Q(w)/\Q(\cos(2\pi/n)))$ is also Galois since $\Gal(\Q(w)/\Q)$ is Galois? I mean, if $L_1\subseteq L_2\subseteq L_3$ and $L_3/L_1$ finite and Galois, then can we deduce that $L_3/L_2$ also Galois? I think yes.

*Why $|H_{i+1}:H_i|=2$ implies $|L_i:L_{i+1}|=2$? 
I think the '$L_i=F(H_i)$' should be fixed field of $H_i$ in $H_{i+1}$. Apply the fundamental theorem of Galois on $H_i/H_0$ for each $i$.

*Why $L_i=L_{i+1}(\sqrt{u_i})$ for some $u_i$? And why each $L_i$ is contained in $\Q(\cos(2\pi/n))$?


 A: 1. Yes, this is trivial over fields of characteristic $0$ as, over these fields, finite Galois extensions are precisely splitting fields (any splitting fields). 
So, if $F$ is of characteristic $0$ and $E$ is a finite Galois extension of $F$, then $E$ is a splitting field of, say $f(x)\in F[X]$, over $F$. Then, if$K$ is a field lying between $E$ and $F$, $E$ is a splitting field over $K$ too and of exactly the same polynomial, $f(x)$, making $E/K$ Galois. In your question, $F=\mathbb Q$, $E=\mathbb Q(w),$ and $K=\mathbb Q(\operatorname{cos(\frac{2\pi}{n})})$.
2. The author seems to have made a mistake in the definition. It looks like what the author really meant to say was that $H_r=Gal(\mathbb Q(w):\mathbb Q)$, $H_0=Gal(\mathbb Q(w):\mathbb Q(\operatorname{cos(\frac{2\pi}{n})}))$ instead of the other way around.
Then the $H_i$s ascend from $Gal(\mathbb Q(w):\mathbb Q(\operatorname{cos(\frac{2\pi}{n})}))$ to $H_r=Gal(\mathbb Q(w):\mathbb Q)$. (The error becomes more obvious when you note that $|Gal(\mathbb Q(w):\mathbb Q(\operatorname{cos(\frac{2\pi}{n})}))|=[\mathbb Q(w):\mathbb Q(\operatorname{cos(\frac{2\pi}{n})})]=2$, so $Gal(\mathbb Q(w):\mathbb Q(\operatorname{cos(\frac{2\pi}{n})}))$ has only a trivial proper subgroup.) 
And yes, each $L_i$ seems to be defined as the fixed field of each $H_i$. Then, as you say, by the fundamental theorem of Galois theory each $H_i=Gal(\mathbb Q(w):L_i)$, so $$|H_{i+1}:H_i|=|Gal(\mathbb Q(w):L_{i+1}):Gal(\mathbb Q(w):L_i)|=\frac{|Gal(\mathbb Q(w):L_{i+1})|}{|Gal(\mathbb Q(w):L_i)|}=\frac{[\mathbb Q(w):L_{i+1}]}{[\mathbb Q(w):L_i]}=[L_i:L_{i+1}]$$
where the second equality holds due to Lagrange's theorem, the third due to the basic result of Galois theory on the number of automorphisms of a Galois extension, and the fourth holds due to the tower theorem.
3. Since, by 2., we have $|L_i:L_{i+1}|=2$ always, we may write, for each $i$, $L_i=L_{i+1}(v_i)$ for some $v_i\in L_i$ that is of algebraic degree $2$ over $L_{i+1}$. This means $v_i$ is the root of some polynomial of the form $a_ix^2+b_ix+c_i$ where $a_i,b_i,c_i\in L_{i+1}$ and $a_i$ is $\neq 0$. We can apply the quadratic formula to get $$v_i=\frac{-b_i\pm\sqrt{b_i^2-4a_ic_i}}{2a_i}$$ This gives us that $$L_i=L_{i+1}\left(\frac{-b_i\pm\sqrt{b_i^2-4a_ic_i}}{2a_i}\right)$$ But since $a_i$ and $b_i$ lie in $L_{i+1}$, this extension equals $$L_{i+1}\left(\sqrt{b_i^2-4a_ic_i}\right)$$ Now, if we set $u_i=b_i^2-4a_ic_i$, we get the desired result. 
As for why each $L_i$ is contained in $\mathbb Q(\operatorname{cos(\frac{2\pi}{n})})$ - this uses the fact that, for any normal extension, $E/F$ and any fields $K_1$ and $K_2$ between $E$ and $F$, we have $Gal(E:K_1)\subset Gal(E:K_2)\iff K_1\supset K_2$. 
Assuming that the correction I pointed out in 2. agreed with what the author meant, for each $i$, $Gal(\mathbb Q(w):L_i)=H_i\supset H_0=Gal(\mathbb Q(w):\mathbb Q(\operatorname{cos(\frac{2\pi}{n})}))$, so it follows that $L_i\subset \mathbb Q(\operatorname{cos(\frac{2\pi}{n})})$ always.
