Saturday, October 25, 2008

Ruffini's Proof: Field Extensions

In today's blog, I will present Paolo Ruffini's proof that if n ≥ 5 and F is a splitting field, then F/k is not a radical tower. This constitutes step two of the Abel-Ruffini proof using field extensions. For a review of splitting fields and field extensions, start here.

Below is Pierre Lauren Wantzel's version of Ruffini's proof. Niels Abel independently presented his own version of this version which I covered previously.

Today's content is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Lemma 1:

Let u,a be functions of n parameters in a field F such that up = a for some prime p.

Let n ≥ 5

Let σ be the following permutation:

x1 → x2 → x3 → x1 and xi → xi for i greater than 3.

so that:

σf(x1, x2, x3, ..., xn) = f(x2, x3, x1, ..., xn)

Let τ be the following permutation:

x3 → x4 → x5 → x3 and xi → xi for i=1,2 and i greater than 5

so that:

τf(x1, x2, x3, x4, x5, ..., xn) = f(x1, x2, x4, x5, x3, ..., xn)

Then:

If a is invariant under the permutations σ and τ, then so is u.

Proof:

(1) From the given, we have up = a where u,a are functions of n parameters.

(2) Applying the permutation σ to both sides gives us:

σ(u)p = σ(a)

(3) Since a is invariant with regard to σ, we have:

σ(u)p = σ(a)= a = up

(4) Dividing both sides by up gives us:

[σ(u)/u]p = 1

(5) Taking the p-th root of each side and multiplying by u gives us:

σ(u) = ζσu

where ζσ is some p-th root of unity.

(6) Applying σ to both sides gives us:

σ2(u) = ζσ2u

(7) Applying σ again we get:

σ3(u) = ζσ3u

(8) Now, we also know from its definition that σ3(u) = u since:

σ3f(x1, x2, x3, x4, ..., xn) = f(x1,x2, x3, x4, ..., xn)

(9) So, that ζσ3 = 1

(10) We can make the same line of argument with τ to show that:

τ(u) = ζτu

where ζτ is a pth root of unity and

ζτ3 = 1

(11) Putting these two results together gives us:

σ*τ(u) = σ(τ(u)) = ζσζτu

σ2*τ = σ2(τ(u)) = ζσ2ζτu

(12) Now, representing each of these as permutations we get:

σ*τ:

x1 → x2 → x3 → x4 → x5 → x1 and xi → xi for i greater than 5.

σ2*τ:

x1 → x3 → x4 → x5 → x2 → x1 and xi → xi for i greater than 5

(13) Using the above permutation maps, we get that:

(σ*τ)5(u) = u

(σ2τ)5(u) = u

(14) Using step #5 and step #10 above, we can use step #13 to conclude that:

(ζσζτ)5 = (ζσ2ζτ)5 = 1

(15) Now, we also note that:

ζσ = ζσ6*ζσ-5 = ( ζσ6*ζσ-5)*(ζτ5*ζτ-5)*(ζσ5*ζσ-5) = ζσ6*(ζτ5*ζσ5)*(ζσ-5*ζσ-5*ζτ-5) =

= ζσ6*(ζσζτ)5(ζσ2ζτ)-5

(16) But then using step #14 with step #15, we have:

ζσ = ζσ6*(ζσζτ)5[(ζσ2ζτ)5]-1 = ζσ6*(1)*(1)-1 = ζσ6

(17) Using step #9 above, we then have:

ζσ =(ζσ3)2 = (1)2 = 1

(18) Now, since (ζσζτ)5 =1, we have:

(1*ζτ)5 =1

so that:

ζτ5 = 1

(19) We can now show that ζτ = 1 since:
ζτ = ζτ6*ζτ-5 = (ζτ3)2*(ζτ5)-1

Taking ζτ3 = 1 (step #10) and ζτ5 = 1 (step #18), we get:

ζτ = (ζτ3)2*(ζτ5)-1 = (1)2*(1)-1 = 1

QED


Theorem 2: Ruffini's Theorem

Let P(X) = (X - x1)*...*(X - xn) = Xn - s1Xn-1 + ... + (-1)nsn = 0

where si are coefficients in field k.

Let K be a field such that K = k(s1, ..., sn)

If n ≥ 5 and F is a splitting field for P(X), then F/K is not a radical tower.

Proof:

(1) Since F is a splitting field for P(X), x1 ∈ F.

(2) Since K is defined around the coefficients of P, it is clear that all elements of K are invariant under the permutations of σ and τ. [See Lemma 2, here]

(3) Assume F/K is a tower of radicals such that:

K = F0 ⊂ F1 ⊂ ... ⊂ Fm-1 ⊂ Fm = F

such that for each 0 ≤ i ≤ m:



where pi is a prime and αi ∈ F0 = K

(4) Then, by induction, all elements Fi are also invariant against the permutations σ and τ since:

(a) We know that all elements of F0 = K are invariant under σ and τ (step #2 above) so we can assume that this is true up to some i where i ≥ 0.

(b) Let a = αi, u = a(1/p) so that:

Fi+1 = Fi(u)

and

a ∈ Fi

(c) But since a ∈ Fi, it follows that a is invariant under σ and τ.

(d) Using Lemma 1 above, we note that this implies that u is also invariant under σ and τ.

(e) But then all elements of Fi(u) = Fi+1 are also invariant under σ and τ.

(5) But now we have a contradiction since x1 ∈ F = Fm is not invariant under σ [since σ(x1) = x2]

(6) So, we reject our assumption in step #3.

QED

References
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Pierre Laurent Wantzel

Pierre Larent Wantzel was born on June 5, 1814 in Paris, France. His father was a professor of applied mathematics at the Ecole speciale du Commerce. From an early age, Wantzel showed tremendous ability in mathematics. Ademar Jean Claude Barre de Saint-Venant relates an anecdote that when Wantzel was 9 years old, his teacher would send for him to help with certain difficult surveying problems.

He entered the Ecole des Arts et Metiers when he was 12. After a year, Wantzel decided that the school was not academic enough and switched to the College Charlesmagne in 1828. He would later marry the daughter of his language coach at the College Charlesmagne.

At 15, Wantzel edited a famous mathbook, Antoine-Andre-Louis Reynaud's Treatise on Arithmetic and added a proof for a method of finding square roots that had previously not had a proof. Later, in 1831, he he was awarded first prize in Latin disseration. When he applied in 1832 for the prestigious Ecole Polytechnique and the Ecole Normale, he placed first in both examinations. By 1838, he had become a lecturer in mathematics at the Ecole Polytechnique and in 1841, he also became a professor of applied mathematics at Ecole de Ponts et Chaussees.

In 1837, Wantzel became the first to prove the impossibility of duplicating the cube and trisecting an angle using only ruler and compass. The great Carl Friedrich Gauss had stated that both of these methods were impossible but had never provided proof.

In 1845, Wantzel gave a revised proof of Abel's Theorem on the impossibility of solving all equations of n ≥ 5 by radicals. In this presentation, Wantzel gave a revised proof of the one done by Paolo Ruffini. In all, he wrote over 20 works on a wide range of subjects.

By May 12, 1848, Wantzel's never-ending pattern of very hard work without break and opium-use caught up with him and he died at the very early age of 33.

Saint-Venant writes:
... one could reproach him for having been too rebellious against those counselling prudence. He usually worked during the evening, not going to bed until late in the night, then reading, and got but a few hours of agitated sleep, alternatively abusing coffee and opium, taking his meals, until his marriage, at odd and irregular hours.
One might wonder how why Wantzel doesn't rank with the greatest mathematicians. Without a doubt, he showed tremendous promise as a child and with all of his hard work, one wonders why he did not reach the highest heights in mathematics. Saint-Venant writes:
...I believe that this is mostly due ot the irregular manner in which he worked, to the excessive number of occupations in which he was engaged, to the continual movement and feverishness of his thoughts, and even to the abuse of his own facilities. Wantzel improvised more than he elaborated, he probably did not give himself the leisure nor the calm necessary to linger long on the same subject.
References
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Wednesday, October 22, 2008

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 qi is prime, ai ∈ Ei

and Ei, Ei+1 are fields that include the roots of unity

L is a field such that (Ei ∩ L) ⊂ (Ei+1 ∩ L) ⊂ L and includes the roots of unity.

Mi+1 = Ei+1 ∩ L

Mi = Ei ∩ L

γ ∈ Mi+1 but not in Mi

Then:

There exist:

γi = c0 + c1ζi-1β + c2ζ2(i-1)β2 + ... + cq-1ζ(q-1)(i-1)βq-1

such that:

ci ∈ Ei

β ∈ Ei+1, βq ∈ Ei

γi ∈ Mi+1

ζ is a primitive qth root of unity

Proof:

(1) From our assumptions, we know that γq ∈ Mi

(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 ∈ Ei+1 such that βq ∈ Ei

and there exists b0, b2, ..., bq-1 ∈ Ei such that:

γ = b0 + β + b2β2 + ... + bq-1βq-1

(4) We can define a function Q(γ) such that Q(γ) = P(b0 + x + b2x2 + ... + bq-1xq-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 = βq

(7) It is clear that β is the root of Yq - a.

(8) We also know that Yq - a is irreducible over Mi [See Lemma 1, here]

(9) But since β is a root for both Yq - a and Q(Y), it follows that Yq - a divides Q(Y) and every root of Yq - 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 Yq - a and they are: &beta, ζβ, ζ2β, ..., ζq-1β where ζ is a primitive qth root of unity [see here for review of the primitive roots of unity].

(11) Since Yq - a divides Q(Y) and ζiβ is a root of Yq - a, it follows that Q(ζiβ) = 0 for all integers i.

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

γ = γ1 = b0 + β + b2β2 + ... + bq-1βq-1

γ2 = b0 + ζβ + b2ζ2β2 + ... + bq-1ζq-1βq-1

...

γq = b0 + ζq-1β + b2ζ2(q-1)β2 + ... + bq-1ζ(q-1)(q-1)βq-1

are all roots of P(x). That is, P(γk) = 0 for k = 1, ..., q.

(13) We also know that the numbers γ1, γ2, ..., γq are all in L since:

(a) γ ∈ Mi+1 = Ei+1 ∩ L [from step #1 above]

(b) So, γ ∈ Ei+1 and γ ∈ L.

(c) Then, using Lemma 5, here, we can see that each γk ∈ L.

QED

Lemma 2:

Let:



where qi is prime, ai ∈ Ei

and Ei, Ei+1 are fields that include the roots of unity

L is a field such that (Ei ∩ L) ⊂ (Ei+1 ∩ L) ⊂ L and includes the roots of unity.

Mi+1 = Ei+1 ∩ L

Mi = Ei ∩ L

Then:



where qi is prime, ai ∈ Mi

Proof:

(1) Let y be an element such that y ∈ Mi+1 but y is not an element in Mi

We know that there is at least one such element since Mi ⊂ Mi+1

(2) Using Lemma 1 above, we know that there exists y1, ..., yq such that each yi ∈ Mi+1 such that:

y = y1 = b0 + β + b2β2 + ... + bq-1βq-1

y2 = b0 + ζβ + b2ζ2β2 + ... + bq-1ζq-1βq-1

...

yq = b0 + ζq-1β + b2ζ2(q-1)β2 + ... + bq-1ζ(q-1)(q-1)βq-1

where β ∈ Ei+1 but not in Ei

βq ∈ Ei

b0, ..., bq-1 ∈ Ei

(3) Now, if we multiply each equation in step #13 above by ζ1-i, we get:

y = y1 = b0 + β + b2β2 + ... + bq-1βq-1

ζ-1y2 = ζq-1y2 = ζq-1b0 + ζ0β + b2ζ1β2 + ... + bq-1ζq-2βq-1

...

ζ1-qyq = ζ1 = ζb0 + ζ0β + b2ζ2(q-1)+1β2 + ... + bq-1ζ(q-1)(q-1)+1βq-1

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

b0 + ζ-1b0 + ... + ζ1-qb0 = b0(1 + ζ1*(q-1) + ζ2*(q-1) + ... + ζ(q-1)*(q-1)) = b0*0 = 0. [Since q doesn't divide q-1.]

β + ζ0β + ... + ζ0β = qβ

b2β2 + ζ1b2β2 + ... + ζq-1b2β2 = b2β2(1 + ζ1*1 + ζ2*1 + ... + ζ(q-1)*1) = b2β2*0 = 0

...

bq-1βq-1 + ζq-2bq-1βq-1 + ... + ζ2b2β2 = bq-1βq-1(1 + ζ1*1 + ζ2*1 + ... + ζ(q-1)*1) = bq-1βq-1*0 = 0

(5) So, adding each of the columns together gives us the following equation:

β = (1/q)∑ (i=1,q) ζ1-iyi ∈ L

We know that β ∈ L since all ζ1-i,q are in L (by the given) and yi ∈ L by step #2 above.

(6) Thus β ∈ L ∩ Ei+1 = Mi+1 and βq = b ∈ L ∩ E = Mi

(7) Let γ ∈ Mi+1 but not Mi

(8) Then there exist (see Lemma 1 above using the same value β as before):

γi = c0 + c1ζi-1β + c2ζ2(i-1)β2 + ... + cq-1ζ(q-1)(i-1)βq-1 with ci∈ Ei and each γi ∈ Mi+1

(9) If we multiply each γi by ζk(1-i) and add up each column in the same way that we did in steps #15 - #17), we get:

ckβk = ∑ (i=1,q) ζk(1-i)γi .

(10) Now since ζk(1-i), γi ∈ Mi+1, it follows that each ckβk ∈ Mi+1

(11) Since β ∈ Mi+1 [See step #6 above], it follows that:

βk ∈ Mi+1, and using step #10 above, we get ck ∈ Mi+1

(12) But if ck ∈ Mi+1 = Ei+1 ∩ L, it follows that ck ∈ L.

(13) And since ck ∈ Ei [see step #8 above] and ck ∈ L [see step #12 above], it follows that ck ∈ Mi = Ei ∩ L.

(14) Thus, we have show that:



where qi is prime, ai ∈ Mi

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 ⊂ ... ⊂ Em = E

where for each Ei we have:



such that: qi is prime, ai ∈ Ei

(2) Since K ⊂ L, it follows that:

K = E0 ∩ L

(3) Further, since Ei ⊂ Ei+1, it follows that:

Ei ∩ L ⊆ Ei+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) ⊆ ... ⊆ (Em-1 ∩ L) ⊆ L

(6) Now, if remove all cases where Ei ∩ L = Ei+1 ∩ L and renumber, then we get the following:

K = (E0 ∩ L) ⊂ (E1 ∩ L) ⊂ ... ⊂ En = L

(7) Using Lemma 2 above, we know that for each Ei, (Ei+1 ∩ L)/(Ei ∩ L) is a field extension.

(8) Thus, we have shown that L/K is tower of radicals.

QED

References
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Monday, October 20, 2008

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

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 some initial lemmas 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 F be a field containing a primitive n-th root of unity.

If a ∈ F is not an nth power and q is prime, then xq - a is irreducible over F. (see Definition 4, here).

Proof

This follows from Lemma 2, here.

QED

For Lemma 2 below, I will use the notation F(u) which is defined in Definition 3, here.

Lemma 2:

Let:

α be a root of xq - a = 0

Then:

every γ ∈ F(α) can be written in the form:

γ = a0 + a1α + ... + aq-1αq-1

where ai are in F.

Proof:

This follows directly from the definition of F(u) [see Definition 3, here.]

QED

Lemma 3:

Let:

q be prime

and

m,k be integers such that 1 ≤ m,k ≤ q-1

Then:

there exists integers r,s such that:

rq + sk = m

and

0 ≤ s ≤ q-1

Proof:

(1) Since q is prime and k ≤ q-1, it follows that gcd(k,q)=1. [That is, they are relatively prime]

(2) It follows that 0*k, 1*k, ..., (q-1)*k forms a complete residue system modulo q. [See Lemma 3, here]

(3) This means that for any integer m, there exists an integer s such that sk ≡ m (mod q) and 0 ≤ s ≤ q-1. [See Definition 1, here for review of a complete residue system]

(4) So that m - sk ≡ 0 (mod q). [See here for review of modular arithmetic if needed]

(5) Since q divides m -sk, it follows that there exists an integer r such that rq = m - sk

(6) Then, it follows that rq + sk = m.

QED

For Corollary 3.1 below, you will need to know that F[x] refers to the set of polynomials in field F. (See here for review of polynomials and the meaning of F[x] ). I will also use the notation F(u) which is defined in Definition 3, here.

Corollary 3.1:

Assume that xq - a ∈ F[x] is irreducible and that α is a root (see here for review of roots of a polynomial).

Let γ be a nonzero element of F(α) with γ not in F.

Then:

There is an element β ∈ F(α) such that βq ∈ F

and there exists b0, b2, ..., bq-1 ∈ F such that:

γ = b0 + β + b2β2 + ... + bq-1βq-1

Proof:

(1) From Lemma 2 above, we know that there exists a0, a1, ..., aq-1 where:

γ = a0 + a1α + ... + aq-1αq-1

(2) Since γ is nonzero, we know that there exists a smallest integer k where 1 ≤ k ≤ q-1 and ak ≠ 0.

(3) Let β = akαk

(4) We know that βq ∈ F since:

(a) ak ∈ F [from Lemma 2 above]

(b) Since α is a root, we know that αq - a = 0 which implies that αq = a

(c) Since a ∈ F, it follows that αq ∈ F.

(5) Now for each k+1 ≤ m ≤ q-1, using Lemma 3 above, we know that:

there exists r,s such that:

rq + sk = m

where 0 ≤ s ≤ q-1

(6) Then there exists cs such that:

αm = (αq)r(αk)s = csβs

where cs ∈ F.

(7) Thus, if we set c0 = a0, we have:

γ = c0 + β + ck+1βk+1 + ... + cq-1βq-1

where each ci ∈ F.

QED

Lemma 4:

Let q be prime and and ζ be a primitive qth root of unity.

Then:

For each integer i:

1 + ζi + ζ2i + ... + ζ(q-1)i =



Proof:

(1) Assume that q divides i.

(2) Then there exists an integer x such that qx = i.

(3) So then:

1 + ζi + ζ2i + ... + ζ(q-1)i = 1 + (ζq)x + ... + (ζq)x*(q-1) = 1 + 1x + ... + 1x*(q-1) = q

(4) Assume q does not divide i.

(5) Then gcd(i,q)=1 and 0*i, 1*i, ..., (q-1)*i form a complete residue system modulo q. [See Lemma 3, here]

(6) Now, since ζ is a primitive q-th root of unity, we know that:

1 + ζ + ζ2 + ... + ζq-1 = 0 [See Lemma 1, here]

(7) Now, since 0*i, 1*i, ... , (q-1)*i is a complete residue system modulo q, we know that each value is uniquely congruent to a value in the complete residue system {0, 1, 2, ..., q-1}

(8) Further, we know that if i ≡ j (mod q), then ζi = ζj since ζ is a primitive qth root of unity [See Lemma 2, here]

(9) But then we have:

1 + ζi + ζ2i + ... + ζ(q-1)i = 1 + ζ + ζ2 + ... + ζq-1 = 0

QED

Lemma 5:

Let y be an a root of an irreducible polynomial f(x) over K.

Let L be a field extension of K such that y ∈ L

Then:

L is a splitting field for f(x)

Proof:

(1) If one root y is an element of L, then all the roots are elements of L. [See Theorem 3, here]

(2) From the Fundamental Theorem of Algebra (see Theorem, here), we know that if y1, ..., yn are the roots, then:

f(x) = (x - y1)*...*(x - yn)

(3) It therefore follows that L is a splitting field for f(x). [See Definition 3, here]

QED

References
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Sunday, October 19, 2008

Abel-Ruffini Proof: Field Extensions

When Niels Abel prepared his impossibility proof, he used the vocabulary of the mathematics of his time. As a result of his work and the work done by Evariste Galois, the very language of mathematics changed. In 1995, Michael I. Rosen wrote an article for The American Mathematical Monthly that explained the ideas of Abel's proof in terms of modern ideas.

In today's blog, I will review the reinterpretation of Abel's proof in terms of modern mathematical concepts. The content in today's blog is taken from Michael Rosen's article.

The content in today's blog, assumes that you feel comfortable with the idea of fields. For review of this concept, start here.

Definition 1: subfield

A set B is a subfield of a set A if both A,B are fields and B is a subset of A.

Definition 2: field extension

A field A is a field extension of a field B if and only if B is a subfield of A.

Definition 3: splitting field over k

A field F is a splitting field over k for a polynomial f(x) if and only if F is a field extension of k and f(x) can be split into linear factors over F.

Definition 4: a radical tower over k

E/k is said to be a radical tower over k if there is a series of intermediate fields such that:

k = E0 ⊂ E1 ⊂ ... ⊂ Em-1 ⊂ Em = E

such that for each 0 ≤ i ≤ m:



where pi is a prime and αi ∈ Ei

Definition 5: solvable in radicals

An equation f(x)=0 is solvable by radicals if and only if there is a radical tower E/k such that F is a splitting field over k and F ⊂ E.

We now state Abel's proof:

Theorem: Abel's Impossibility Proof

Let f(x) = xn - a1xn-1 + ... + (-1)nan be the general equation of degree n.

If n ≥ 5, then this equation is not solvable in radical.

Proof:

(1) An equation f(x)=0 is solvable by radicals if and only if there is a radical tower E/k such that F is a splitting field over k and F ⊂ E. [Definition 5 above]

(2) If F is contained in a radical tower over k, then F/k is itself a radical tower. [see Theorem here for proof]

(3) If n ≥ 5 and F is a splitting field, then F/k is not a radical tower. [See Theorem, here for proof]

(4) Therefore, F is not contained in a radical tower over k.

QED

References
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