Question: 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

Answer: 
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.
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.