Radical Tower implies Solvability by Radicals First, I define a radical tower as I have it in front of me (in case our definitions differ):
By a radical tower over a field F we mean a sequence of finite extensions $F=F_0{\subset}F_1{\subset}..{\subset}F_r$ having the property that there exist positive integers $d_i$, elements $a_i$ in $F_i$ and $\alpha_i$ with ${\alpha_i}^{d_i}=a_i$ such that $F_{i+1}=F_i(\alpha_i)$. We say that E is contained in a radical tower if there exists a radical tower above such that $E{\subset}F_r$.
Anyway, now onto the main problem: Let E be a finite extension of F (F characteristic 0) and suppose E is contained in a radical tower. Show that there exists a radical tower $F{\subset}E_0{\subset}E_1{\subset}...{\subset}E_m$ such that:
a) $E_m$ is Galois over F and $E{\subset}E_m$
b) $E_0=F(\zeta)$ where $\zeta$ is a primitive nth root of unity
c) For each i, $E_{i+1}=E_i(\alpha_i)$ where ${\alpha_i}^{d_i}=a_i$ $\in$$E_i$, and $d_i$|n.
I am rather stuck on this problem and have not been able to make much headway. Would anyone be able to give any hints/help me solve this problem?
Thank you
 A: So, we have a radical tower $F=F_0\subset F_1\subset\cdots\subset F_r$ with $F_{i+1}=F_i(\alpha_i)$, $\alpha_i^{d_i}=a_i$, $a_i$ in $F_i$, and $E\subset F_r$. Let $n$ be the least common multiple of $d_1,\dots,d_r$. Now consider the tower $F\subset E_0\subset E_1\subset\cdots\subset E_m$ given by $E_0=F(\zeta)$ where $\zeta$ is a primitive $n$th root of unity, and $E_{i+1}=E_i(\alpha_i)$. 
This is clearly a radical tower. 
Condition b) is clearly met. 
Condition c) is met since $n$ is the least common multiple of the $d_i$. 
$E$ is contained in $F_r$, and $F_r$ is contained in $E_m$, so $E\subset E_m$. 
So all that remains is to show that $E_m/F$ is Galois. 
Now $E_m=F(\zeta,\alpha_1,\dots,\alpha_r)$, so $E_m/F$ is finite. It is also normal, since it contains all the conjugates of all of its generators. So, it's Galois. 
A: Say $A$ is an algebraic closure of $F$. Any finite extensions of $F$ are regarded as subfields of $A$. We fix the radical tower of $F$ as specified in your statement.
We first observe the following observations. (1) Let $\phi:F_r\to A$ be an embedding. Then we have a corresponding radical tower, where we start from $\phi(F)$ and each time we adjoint $\phi(\alpha_i)$ to the previous field. (2) Let $L$ be a finite extension of $F$. Then $L$ admits a corresponding radical tower, where $L_{i+1} = L_i(\alpha_i)$ and $\alpha_i^{d_i} = a_i\in L_i$.
Let $n = d_0d_1\cdots d_{r-1}$, let $\zeta$ be a primitive $n$'th root of unity, and let $E = F(\zeta)$. So we have a corresponding radical tower $E = E_0 \subseteq \cdots \subseteq E_r$. Let $\{\Phi^{(j)}:1\leq j\leq N\}$ be the collection of all $F$-embeddings from $E_r$ into $A$, where $\Phi^{(1)}$ is the identity map. Then for each $\Phi^{(j)}$ we get a corresponding radical tower. Note that since $E/F$ is a normal extension, we have $\Phi^{(j)}(E) = E$.
Now we form our desired radical tower by induction. The first step is just to form the tower $E = E_0\subseteq \cdots \subseteq E_r$, which corresponds to the radical tower of $F$ and the finite extension $E/F$. Assume for some $1\leq J$ we have formed radical towers over $E$, such that the terminal field equals $H_J:=E_r^{(1)}E_r^{(2)}\cdots E_r^{(J)}$, where $E^{(j)} = \Phi^{(j)}(E_r)$. We know that corresponding to the tower $E = E_0\subseteq \cdots \subseteq E_r$ and the embedding $\Phi^{(J+1)}$ there is a tower starting from $E=\Phi^{(J+1)}(E)$ and terminates with $E_r^{(J+1)}$. Since $H_J$ is a finite extension of $E$, we get a corresponding tower starting from $H_J$ and terminates with $H_JE_r^{(J+1)}$. This finishes the inductive construction. The final field in the constructed tower would be $E_r^{(1)}E_r^{(2)}\cdots E_r^{(N)}$, which shows that it is a normal extension of $F$.
Note that in the above, $n$ can be chosen to be any common multiple of the $d_i$'s.
