2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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! $\endgroup$ – Daniele1234 Aug 17 '17 at 0:41
  • $\begingroup$ @hardmath Do you have an answer to the question because no one seems to be answering? $\endgroup$ – Daniele1234 Aug 17 '17 at 8:13
  • $\begingroup$ 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? $\endgroup$ – hardmath Aug 17 '17 at 17:08
  • $\begingroup$ 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. $\endgroup$ – Daniele1234 Aug 17 '17 at 17:24
  • 1
    $\begingroup$ @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$ $\endgroup$ – Daniele1234 Aug 17 '17 at 22:25
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Daniele1234 Aug 18 '17 at 0:46
  • $\begingroup$ 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 $\zeta$. $\endgroup$ – Gerry Myerson Aug 18 '17 at 13:15
  • $\begingroup$ How do you claim: "The conjugates of $\alpha_i$ are of the form $\eta^j$$\alpha_i$ where $\eta$^$d_i$=1." $\endgroup$ – Daniele1234 Aug 18 '17 at 17:17
  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Aug 19 '17 at 4:46
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.