Root towers

From: Johannes Beyermann (beyermann.johannes_at_gmx.de)
Date: 12/07/04 

Date: Tue, 07 Dec 2004 14:30:48 +0100

Hello,

i want to solve the following exercise:
Let E/F be the cyclotomic field of the p-th roots of unity over the field F
of characteristic 0. Show that E can be imbedded in a field K which has a
root tower F = F_1 c F_2 c ... c F_r = K, where F_(i+1)=F_i(d_i) and
d_i^(n_i)=a_i e F_i, and n_i are primes and [F_(i+1):F_i]=n_i.

I have the following problem: I dont see, why I cannot assume, that is a
splitting field of a polynomial over F with a solvable Galois group. Why
must E/F be cyclotomic?
When I look at the proof of Galois-Criterion for solvability of an equation
by radicals, there is already constructed such a root tower for every
solvable Galois group.
Here is the proof ("Jacobson Basic Algebra I" same book as the exercise):
"... Conversely, assume that the Galois group G of f over F is solvable. Let
n=|G|=[E:F] where E is a splitting field over F of f. Let F_1=F, F_2=F(z)
where z is a primitive nth root of unity and let K=E(z). By Lemma xx, the
Galois group of K/F_2 is isomorphic to a subgroup H of G. Hence H is
solvable and it has a composition series H=H_1 |> H_2 ... |> H_(r+1)=1
whose composition factor H_i/H_(i+1) is cyclic of prime order p_i ...
Correspondingly, we have an increasing chain of subfields F_2 c F_3 c ... c
F_(r+2)=K where H_i=Gal K/F_(i+1). Hence F_(i+1) is normal over F_i with
cyclic Galois group of prime order p_i. Since p_i | n and F_i contains p_i
p_i-th roots of 1, so by Lemma xx F_(i+1)=F_i(d_i), and d_i^(p_i) e F_i."

Why i'm wrong? I don't see, what is to prove. I think all is said in the
proof?

Many thanks in advance
Johannes

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