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$$E_i, and d_i|n. I am rather stuck on this problem and have not been able to make much headway. Would anyone be able to give any hints/help me solve this problem? Thank you • @hardmath oh yes sorry my fault completely when I was formatting the question I accidentally left it out. Thank you very much for point it out to me! – Daniele1234 Aug 17 '17 at 0:41 • @hardmath Do you have an answer to the question because no one seems to be answering? – Daniele1234 Aug 17 '17 at 8:13 • It seems to me that (b) and (c) give you a road map on how to reach (a), which concerns the last field E_m in the tower. Note that symbol n appears in (b) and (c), presumably with the same value (but did not appear previously in the problem statement). Do you know some results about a finite extension E that might help us here? – hardmath Aug 17 '17 at 17:08 • Interesting observation. I have proved already that the normal closure of E must also be contained in a radical tower. I also proved that if E' is the conjugate of E over F, then E' is contained in a radical tower as well. – Daniele1234 Aug 17 '17 at 17:24 • @hardmath About your comment on the tower, it is an alpha in E_{i+1}=E_i(\alpha_i) if you look closely enough, and an 'a' (english letter) in a_i \in$$E_i$ – Daniele1234 Aug 17 '17 at 22:25

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.

• Myers Thank you for your answer, it is explained in a clear and simple way so easy to follow thank you. Just one thing: How do you know that $E_m$ contains all the conjugates of all of its generators? – Daniele1234 Aug 18 '17 at 0:46
• The conjugates of $\alpha_i$ are of the form $\eta^j\alpha_i$ where $\eta^{d_i}=1$, but that makes $\eta$ a power of $\zeta$. – Gerry Myerson Aug 18 '17 at 13:15
• How do you claim: "The conjugates of $\alpha_i$ are of the form $\eta^j$$\alpha_i$ where $\eta$^$d_i$=1." – Daniele1234 Aug 18 '17 at 17:17
• Well, you've got a point there. The conjugates of $\alpha_i$ over $F_i$ are certainly of that form, as those are the roots of $x^{d_i}=a_i$. But what I really need is the conjugates of $\alpha_i$ over $E$, so I'll have to work a little harder. – Gerry Myerson Aug 19 '17 at 4:46

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.