Abel's Proof: Field Extensions: Step 1: Proof

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 ∩ L

M_i = E_i ∩ L

γ ∈ M_i+1 but not in M_i

Then:

There exist:

γ_i = c₀ + c₁ζ^i-1β + c₂ζ^2(i-1)β² + ... + c_q-1ζ^(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 b₀, b₂, ..., b_q-1 ∈ E_i such that:

γ = b₀ + β + b₂β² + ... + b_q-1β^q-1

(4) We can define a function Q(γ) such that Q(γ) = P(b₀ + x + b₂x² + ... + b_q-1x^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, ζ²β, ..., ζ^q-1β 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 ζⁱβ is a root of Y^q - a, it follows that Q(ζⁱβ) = 0 for all integers i.

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

γ = γ₁ = b₀ + β + b₂β² + ... + b_q-1β^q-1

γ₂ = b₀ + ζβ + b₂ζ²β² + ... + b_q-1ζ^q-1β^q-1

...

γ_q = b₀ + ζ^q-1β + b₂ζ^2(q-1)β² + ... + b_q-1ζ^(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 γ₁, γ₂, ..., 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 ∩ L

M_i = E_i ∩ L

Then:



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 y₁, ..., y_q such that each y_i ∈ M_i+1 such that:

y = y₁ = b₀ + β + b₂β² + ... + b_q-1β^q-1

y₂ = b₀ + ζβ + b₂ζ²β² + ... + b_q-1ζ^q-1β^q-1

...

y_q = b₀ + ζ^q-1β + b₂ζ^2(q-1)β² + ... + b_q-1ζ^(q-1)(q-1)β^q-1

where β ∈ E_i+1 but not in E_i

β^q ∈ E_i

b₀, ..., b_q-1 ∈ E_i

(3) Now, if we multiply each equation in step #13 above by ζ^1-i, we get:

y = y₁ = b₀ + β + b₂β² + ... + b_q-1β^q-1

ζ^-1y₂ = ζ^q-1y₂ = ζ^q-1b₀ + ζ⁰β + b₂ζ¹β² + ... + b_q-1ζ^q-2β^q-1

...

ζ^1-qy_q = ζ¹ = ζb₀ + ζ⁰β + b₂ζ^2(q-1)+1β² + ... + b_q-1ζ^(q-1)(q-1)+1β^q-1

(4) If we add up each row, then we have the following values for each column [Using Lemma 4, here]:

b₀ + ζ^-1b₀ + ... + ζ^1-qb₀ = b₀(1 + ζ^1*(q-1) + ζ^2*(q-1) + ... + ζ^(q-1)*(q-1)) = b₀*0 = 0. [Since q doesn't divide q-1.]

β + ζ⁰β + ... + ζ⁰β = qβ

b₂β² + ζ¹b₂β² + ... + ζ^q-1b₂β² = b₂β²(1 + ζ^1*1 + ζ^2*1 + ... + ζ^(q-1)*1) = b₂β²*0 = 0

...

b_q-1β^q-1 + ζ^q-2b_q-1β^q-1 + ... + ζ²b₂β² = b_q-1β^q-1(1 + ζ^1*1 + ζ^2*1 + ... + ζ^(q-1)*1) = b_q-1β^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-iy_i ∈ L

We 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 = c₀ + c₁ζ^i-1β + c₂ζ^2(i-1)β² + ... + c_q-1ζ^(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_kβ^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_kβ^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 = E₀ ⊂ E₁ ⊂ E₂ ⊂ ... ⊂ E_m = E

where for each E_i we have:



such that: q_i is prime, a_i ∈ E_i

(2) Since K ⊂ L, it follows that:

K = E₀ ∩ 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 = (E₀ ∩ L) ⊆ (E₁ ∩ 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 = (E₀ ∩ L) ⊂ (E₁ ∩ 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

• Michael I. Rosen, "Niels Hendrik Abel and Equations of the Fifth Degree", The American Mathematical Monthly, Vol. 102, No. 6 (Jun. - Jul., 1995), pp 495-595.
• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001