Normalized root tower and primitive roots of unity

\mathbf{Problem 1:} Let p be a prime unequal to the characteristic of the field F. Show that, if a\in F, then x^p-a is either irreducible in F[x] or it has a root in F.

\mathbf{Proof:} Let w be a p-th primitive root of unity, K=F(w). Since F has characteristic 0, K contains p distinct pth root of unity.

Let U=\{\text{pth roots of unity contained in } K\}. U is a multiplicative subgroup of K.

Suppose E is a splitting field of x^p-a over K and r is one root of x^p-a=0 in E. All the roots of x^p-a are  zr, z\in U. So E=K(r). \forall\, \xi\in \text{Gal }E/K, \xi(r)=zr, z\in U. We can verify that this induces a monomorphism

\text{Gal }E/K\mapsto U

\xi\to z

Since U is cyclic of order p. G which is a subgroup of U can only be the trivial group \{1\} or the whole group U.

If G=\{1\}, E=K=F(w). And x^p-a splits over F(w). Then it must have a root in F.

If G=U, then [K(r):K]=[E:K]=|G|=p, x^p-a must be irreducible in K[x], hence also in F[x].

\text{Q.E.D}\hfill \square

\mathbf{Problem 2:} Let E/F be the cyclotomic field of the pth 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 over F such that the integers n_i are primes and [F_{i+1}:F_i]=n_i. Call such a root tower normalized.

\mathbf{Proof:} Since  E is the cyclotomic field over F of characteristic 0, G=\text{Gal }E/F is abelian by lemma 1 in page 252.

Then G is a finite solvable group. So G has a composition series

\displaystyle G=G_1\rhd G_2\rhd \cdots\rhd G_{r+1}=\{1\} 

whose composition factor G_i/G_{i+1} is cyclic of prime order p_i. By Galois corresponding theorem, we have an increasing chain of subfields

F=F_1\subset F_2\subset\cdots\subset F_{r+1}=E

such that G_i=\text{Gal }E/F_i and \displaystyle G_i/G_{i+1}=\text{Gal }F_{i+1}/F_i. So \displaystyle [F_{i+1}:F_i]=p_i is prime. This is a normalized root tower.

\text{Q.E.D}\hfill \square

\mathbf{Problem3:} Obtain normalized root tower over \mathbb{Q} of the cyclotomic fields of 5th and 7th roots of unity.

\mathbf{Proof:} Suppose w is a primitive 5th roots of unity. [\mathbb{Q}(w):\mathbb{Q}]=4 and  G=\text{Gal }\mathbb{Q}(w)/\mathbb{Q} is multiplicative group of \mathbb{Z}_5, because \forall\,\zeta\in G maps \{w,w^2,w^3,w^4,w^5=1\} to itself. The automorphism \zeta:w\to w^3 is a generator of G=\{\zeta,\zeta^2,\zeta^3,\zeta^4=1\}.

G=\langle \zeta\rangle\rhd \langle \zeta^2\rangle\rhd=\{1\}

the corresponding root tower is

\mathbb{Q}\subset \mathbb{Q}(w+w^4)\subset \mathbb Q(w)

It is easy to verify this is a normalized root tower.

For 7th roots of unity, suppose w is a primitive 7th root of unity,

\mathbb{Q}\subset \mathbb{Q}(w+w^2+w^4)\subset \mathbb Q(w)

is a normalized tower where [\mathbb{Q}(w+w^2+w^4):\mathbb{Q}]=2.

\text{Q.E.D}\hfill \square

\mathbf{Problem4:} Prove that, if f(x)=0 has a solvable Galois group over a field F of characteristic 0. Then its splitting field can be imbedded in an extension field which has a normalized tower over F.

Suppose splitting field is E and [E:F]=n, the proof relies on adjoin nth roots of unity to E.

\mathbf{Remark:} Jacobson p256.

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