The content in today's blog is taken from the essay by Michael I. Rosen entitled "Niels Hendrik Abel and Equations of the Fifth Degree."

In today's blog, I present the key lemma that I will use to establish Step 1 of the proof using field extensions. For the proof using the ideas that

Niels Abel originally presented, see

here.

Lemma 1:Let:

where

q_{i} is prime,

a_{i} ∈ E_{i}and

E_{i}, E_{i+1} are fields that include the roots of unity

L is a field such that

(E_{i} ∩ L) ⊂ (E_{i+1} ∩ L) ⊂ L and includes the roots of unity.

M_{i+1} = E_{i+1} ∩ LM_{i} = E_{i} ∩ Lγ ∈ M_{i+1} but not in

M_{i}Then:

There exist:

γ_{i} = c_{0} + c_{1}ζ^{i-1}β + c_{2}ζ^{2(i-1)}β^{2} + ... + c_{q-1}ζ^{(q-1)(i-1)}β^{q-1}such that:

c_{i} ∈ E_{i}β ∈ E_{i+1}, β^{q} ∈ E_{i}γ_{i} ∈ M_{i+1}ζ is a primitive

qth root of unity

Proof:

(1) From our assumptions, we know that

γ^{q} ∈ M_{i}(2) Let

P(x) be a polynomial such that

P(γ) = 0, that is,

γ is a root for

P(x).

(3) We further know that (see Corollary 3.1,

here) there exists an element

β ∈ E_{i+1} such that

β^{q} ∈ E_{i}and there exists

b_{0}, b_{2}, ..., b_{q-1} ∈ E_{i} such that:

γ = b_{0} + β + b_{2}β^{2} + ... + b_{q-1}β^{q-1}(4) We can define a function

Q(γ) such that

Q(γ) = P(b_{0} + x + b_{2}x^{2} + ... + b_{q-1}x^{q-1})

(5) From this definition, it is clear that

Q(β) = P(γ) = 0 so that

β is a root for

Q(γ).

(6) Let

a = β^{q}(7) It is clear that

β is the root of

Y^{q} - a.

(8) We also know that

Y^{q} - a is irreducible over

M_{i} [See Lemma 1,

here]

(9) But since

β is a root for both

Y^{q} - a and

Q(Y), it follows that

Y^{q} - a divides

Q(Y) and every root of

Y^{q} - a is also a root of

Q(Y). [See Theorem 3,

here]

(10) By the Fundamental Theorem of Algebra [See Theorem,

here], we know that there are

q roots for

Y^{q} - a and they are:

&beta, ζβ, ζ^{2}β, ..., ζ^{q-1}β where

ζ is a primitive

qth root of unity [see

here for review of the primitive roots of unity].

(11) Since

Y^{q} - a divides

Q(Y) and

ζ^{i}β is a root of

Y^{q} - a, it follows that

Q(ζ^{i}β) = 0 for all integers

i.

(12) So, the numbers [see step #5 above]:

γ = γ_{1} = b_{0} + β + b_{2}β^{2} + ... + b_{q-1}β^{q-1}γ_{2} = b_{0} + ζβ + b_{2}ζ^{2}β^{2} + ... + b_{q-1}ζ^{q-1}β^{q-1}...γ_{q} = b_{0} + ζ^{q-1}β + b_{2}ζ^{2(q-1)}β^{2} + ... + b_{q-1}ζ^{(q-1)(q-1)}β^{q-1}are all roots of

P(x). That is,

P(γ_{k}) = 0 for

k = 1, ..., q.

(13) We also know that the numbers

γ_{1}, γ_{2}, ..., γ_{q} are all in

L since:

(a)

γ ∈ M_{i+1} = E_{i+1} ∩ L [from step #1 above]

(b) So,

γ ∈ E_{i+1} and

γ ∈ L.

(c) Then, using Lemma 5,

here, we can see that each

γ_{k} ∈ L.

QED

Lemma 2:Let:

where

q_{i} is prime,

a_{i} ∈ E_{i}and

E_{i}, E_{i+1} are fields that include the roots of unity

L is a field such that

(E_{i} ∩ L) ⊂ (E_{i+1} ∩ L) ⊂ L and includes the roots of unity.

M_{i+1} = E_{i+1} ∩ LM_{i} = E_{i} ∩ LThen:

where

q_{i} is prime,

a_{i} ∈ M_{i}Proof:

(1) Let

y be an element such that

y ∈ M_{i+1} but

y is not an element in

M_{i}We know that there is at least one such element since

M_{i} ⊂ M_{i+1}(2) Using Lemma 1 above, we know that there exists

y_{1}, ..., y_{q} such that each

y_{i} ∈ M_{i+1} such that:

y = y_{1} = b_{0} + β + b_{2}β^{2} + ... + b_{q-1}β^{q-1}y_{2} = b_{0} + ζβ + b_{2}ζ^{2}β^{2} + ... + b_{q-1}ζ^{q-1}β^{q-1}...y_{q} = b_{0} + ζ^{q-1}β + b_{2}ζ^{2(q-1)}β^{2} + ... + b_{q-1}ζ^{(q-1)(q-1)}β^{q-1}where

β ∈ E_{i+1} but not in

E_{i}β^{q} ∈ E_{i}b_{0}, ..., b_{q-1} ∈ E_{i}(3) Now, if we multiply each equation in step #13 above by

ζ^{1-i}, we get:

y = y_{1} = b_{0} + β + b_{2}β^{2} + ... + b_{q-1}β^{q-1}ζ^{-1}y_{2 }= ζ^{q-1}y_{2} = ζ^{q-1}b_{0} + ζ^{0}β + b_{2}ζ^{1}β^{2} + ... + b_{q-1}ζ^{q-2}β^{q-1}...ζ^{1-q}y_{q} = ζ^{1} = ζb_{0} + ζ^{0}β + b_{2}ζ^{2(q-1)+1}β^{2} + ... + b_{q-1}ζ^{(q-1)(q-1)+1}β^{q-1}(4) If we add up each row, then we have the following values for each column [Using Lemma 4,

here]:

b_{0} + ζ^{-1}b_{0} + ... + ζ^{1-q}b_{0} = b_{0}(1 + ζ^{1*(q-1)} + ζ^{2*(q-1)} + ... + ζ^{(q-1)*(q-1)}) = b_{0}*0 = 0. [Since

q doesn't divide

q-1.]

β + ζ^{0}β + ... + ζ^{0}β = qβb_{2}β^{2} + ζ^{1}b_{2}β^{2} + ... + ζ^{q-1}b_{2}β^{2} = b_{2}β^{2}(1 + ζ^{1*1} + ζ^{2*1} + ... + ζ^{(q-1)*1}) = b_{2}β^{2}*0 = 0...

b_{q-1}β^{q-1} + ζ^{q-2}b_{q-1}β^{q-1} + ... + ζ^{2}b_{2}β^{2} = b_{q-1}β^{q-1}(1 + ζ^{1*1} + ζ^{2*1} + ... + ζ^{(q-1)*1}) = b_{q-1}β^{q-1}*0 = 0(5) So, adding each of the columns together gives us the following equation:

β = (1/q)∑ (i=1,q) ζ^{1-i}y_{i} ∈ LWe know that

β ∈ L since all

ζ^{1-i},q are in

L (by the given) and

y_{i} ∈ L by step #2 above.

(6) Thus

β ∈ L ∩ E_{i+1} = M_{i+1} and

β^{q} = b ∈ L ∩ E = M_{i}(7) Let

γ ∈ M_{i+1} but not

M_{i}(8) Then

there exist (see Lemma 1 above using the same value

β as before):

γ_{i} = c_{0} + c_{1}ζ^{i-1}β + c_{2}ζ^{2(i-1)}β^{2} + ... + c_{q-1}ζ^{(q-1)(i-1)}β^{q-1} with

c_{i}∈ E_{i} and each

γ_{i} ∈ M_{i+1}(9) If we multiply each

γ_{i} by

ζ^{k(1-i)} and add up each column in the same way that we did in steps #15 - #17), we get:

c_{k}β^{k} = ∑ (i=1,q) ζ^{k(1-i)}γ_{i} .

(10) Now since

ζ^{k(1-i)}, γ_{i} ∈ M_{i+1}, it follows that each

c_{k}β^{k} ∈ M_{i+1}(11) Since

β ∈ M_{i+1} [See step #6 above], it follows that:

β^{k} ∈ M_{i+1}, and using step #10 above, we get

c_{k} ∈ M_{i+1}(12) But if

c_{k} ∈ M_{i+1} = E_{i+1} ∩ L, it follows that

c_{k} ∈ L.

(13) And since

c_{k} ∈ E_{i} [see step #8 above] and

c_{k} ∈ L [see step #12 above], it follows that

c_{k} ∈ M_{i} = E_{i} ∩ L.

(14) Thus, we have show that:

where

q_{i} is prime,

a_{i} ∈ M_{i}QED

Theorem 2: If

E/K is a radical tower and

L ⊆ E, then it follows that

L/K is also a radical tower.

Proof:

(1) Since

E/K is a radical tower, we have (see Definition 4,

here):

K = E_{0} ⊂ E_{1} ⊂ E_{2} ⊂ ... ⊂ E_{m} = Ewhere for each

E_{i} we have:

such that:

q_{i} is prime,

a_{i} ∈ E_{i}(2) Since

K ⊂ L, it follows that:

K = E_{0} ∩ L(3) Further, since

E_{i} ⊂ E_{i+1}, it follows that:

E_{i} ∩ L ⊆ E_{i+1} ∩ L(4) Since

L ⊆ E, it follows that

L = E ∩ L(5) Putting steps #2, #3, and #4 together gives us:

K = (E_{0} ∩ L) ⊆ (E_{1} ∩ L) ⊆ ... ⊆ (E_{m-1} ∩ L) ⊆ L(6) Now, if remove all cases where

E_{i} ∩ L = E_{i+1} ∩ L and renumber, then we get the following:

K = (E_{0} ∩ L) ⊂ (E_{1} ∩ L) ⊂ ... ⊂ E_{n} = L(7) Using Lemma 2 above, we know that for each

E_{i}, (E_{i+1} ∩ L)/(E_{i} ∩ L) is a field extension.

(8) Thus, we have shown that

L/K is tower of radicals.

QED

References