Galois' Memoir: Lemma 1

The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Lemma 1: An irreducible equation g(x) cannot have a root in common with a rational equation h(x) without dividing it.

Proof:

(1) Let g(x) be a polynomial with coefficients in a given field K that is irreducible over K.

(2) Let h(x) be a polynomial with coefficients in the field K.

(3) Let r be a root for both h(x),g(x) so that h(r)=0 and g(r)=0

(4) Assume that g(x) does not divide h(x).

(5) Then g(x),h(x) are relatively prime since g(x) is irreducible [see Lemma 1, here]

(6) Then (see Corollary 3.1, here), there exists polynomials A(x), B(x) such that A, B all have coefficients in K and:

1 = A(x)g(x) + B(x)h(x)

(7) Since g(r)=0 and h(r)=0, it follows that:

1 = A(r)*0 + B(r)*0

which is impossible.

(8) So we have a contradiction and we reject our assumption in step #4.

QED

References

• Harold M. Edwards, Galois Theory, Springer, 1984
• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

Galois' Memoir on the Solvability of Equations

Evariste Galois died in 1832 when he was just 20. He had made many attempts to gain attention to his theory of equations but each time had failed.

In 1829 when he was 17, he presented his findings on the the solvability of equations to the Paris Academy. Augustin-Louis Cauchy was appointed referee. There is a popular legend that Cauchy did not appreciate the work or somehow lost it but this does not seem to be the case. It is believed that Cauchy presented detailed comments to Galois and suggested that he resubmit his work. For more information on this, see this article. It was at this time that tragedy struck and Galois' father committed suicide. Galois' submission of his work was delayed.

It is believed that being given encouragement from Cauchy, he rewrote his memoir. He resubmitted his memoir to the Paris Academy in 1830. Jean-Baptiste Joseph Fourier was appointed referee. Unfortunately, Fourier's health took a turn for the worse and he died a few weeks after being appointed referee. No assessment was made and the paper was lost among Fourier's other papers.

In 1831, Galois was asked by Simeon Poisson to resubmit his memoir. After reviewing, Poisson rejected the paper as not fully developed. It is interested to note that Poisson was quite impressed by the quality of work but was not able to verify that they were correct.

Here is what Poisson wrote (see reference below for source):

We have made ever effort to comprehend M. Galois's proof. His arguments are neither sufficiently clear nor developed for us to judge their rigor, and we are not in a position to even give an idea of them in this report...

The author claims that the proposition which is the subject of his memoir is part of a general theory rich in application. Often, different parts of a theory are mutually clarifying, and it is easier to understand them together than in isolation. One should rather wait for the author to publish his work in entirety before forming a definite opinion.

Galois took all this very hard as can be imagined. There is another myth at this point that on the night before Galois' duel where he would die, he wrote up his theory in a single night. While he did write a letter to his friend about the nature of his work, it is clear that he had been working on the ideas since he was 17 and the fullest expression of them up to this point had been the paper that was rejected by Poisson.

Galois would die on May 30, 1832. His funeral was on June 2. His good friend Auguste Chevalier and his brother Alfred collected all his papers in the effort to get notice for his work. They were submitted to the most famous mathematicians of the day: Carl Gauss, Jacobi, and others. By 1843, they had made their way to Joseph Liouville.

Finally, they got the attention they deserved when Liouville published Galois's memoir with additional comments in 1846.

In the next set of blogs, I will review the famous memoir by Galois that had been rejected by Poisson and which was modified the night before his death. For an English translation of Galois' memoir, see Harold Edwards' Galois Theory.

References

• Tony Rothman, "Genius and Biographers: The Fictionalization of Evariste Galois", American Mathematics Monthly, 1982.
  
• Harold M. Edwards, Galois Theory, Springer, 1984.
• "Evariste Galois," Wikipedia.org
  

Joseph Liouville

Joseph Liouville was born on March 24, 1809 in Saint-Omer, France. His father was an officer in Napoleon's army so for his earliest years, he lived with his uncle. Only when Napoleon was defeated did his father return and the family live together in Toul, France.

In Paris, Liouville studied mathematics at the College St. Louis. Already at this time, he showed interest and talent in advanced mathematical topics.

In 1825, when he was 16, he entered the Ecole Polytechnique. There, he attended lectures by Andre Marie Ampere and Dominique Francois Jean Arago. While he did not attend any lectures by Augustin Louis Cauchy, he was greatly influenced by him. Among his examiners when he graduated were Gaspard de Pony and Simeon Denis Poisson.

He graduated in 1827 and entered the Ecole des Ponts et Chaussees with the intention of becoming an engineer. Engineering projects in those days were physically demanding and Liouville found his health severely affected. He took some time off by returning to Toul. He got married to Marie-Louise Balland and decided to resign from the Ecole des Ponts et Chaussees which he did in 1830.

In 1831, he accepted an academic position at the Ecole Polytechnique. He was assistant to Claude Louis Mathieu and the role carried with it responsibilities of 35-40 hours of lectures per week. In doing this, Liouville developed a reputation for focusing on advanced topics and for being difficult to follow.

Many times, Liouville attempted to improve his position at the Ecole Polytechnique without success. He was also frustrated by the quality of math journals in France at the time. In 1836, he started his own math journal, Journal de Mathematiques Pures et Appliquees which was also known as the Journal de Liouville. By this time, he had developed an international reputation based on his papers which he published in Crelle's Journal. The journal was a success and published many significant papers by French mathematicians.

By 1838, he became Professor of Analysis and Mechanics at the Ecole Polytechnique. Many honors followed. In 1840, he was elected to the Academie des Sciences in Astronomy and he was also elected to the Bureau des Longitudes.

Liouville had become close friends with Arago who become the head of the Republican Party in France. Liouville was encouraged to run for office. In 1848, Liouville was elected to the Constituting Assembly. Unfortunately, his political career did not last long for he was voted out of office the following year. This had a big impact on his spirits as noted by one of his biographers:

The political defeat changed Liouville's personality. In earlier letters, he was often depressed because of illness, and could vent his anger towards his enemies such as Libri, but he always fought for what he believed was right. After the election in 1849, he resigned and became bitter, even towards his old friends. When he sat down at his desk, he did not only work, ... he also pondered his ill fate. ... his mathematical notes were interrupted with quotes from poets and philosophers...

In 1850, the mathematical chair at the College de France opened. Up for consideration were Liouville and Cauchy. After a heated contest, the position went to Liouville in 1851.

Liouville's mathematical output was phenomenal writing over 400 mathematical papers. Over 200 of papers were on number theory. Other papers covered a range of topics including mathematical physics, astronomy, as well as pure mathematics.

He introduced the fractional calculus as part of his analysis of electromagnetism. He was also to the first to prove the existence of transcendental numbers, numbers that is not algebraic (that is, it cannot be a solution to an equation of a nonconstant polynomial with rational coefficients). He did very significant work on the boundary value problems with differential equations in what is today called Sturm-Liouville Theory. He did important work in statistical mechanics and measure theory.

Perhaps, his most important impact in mathematics was his discovery of the memoir by Evariste Galois. In 1843, he announced to the Paris Academy that he had discovered very brilliant insights by Galois. Galois's memoir was then published in 1846 which would introduce group theory and place Galois among the most celebrated mathematicians in the history of the subject.

Liouville died on September 8, 1882. Many historians consider him the greatest mathematician of his day. The Liouville Crater on the Moon is named in his honor.

References

• "Joseph Liouville", MacTutor Web Site
• "Joseph Liouville", Wikipedia.org
• "Transcendental Number", Wikipedia.org
• Ioan James, Remarkable Mathematicians: From Euler to Von Neumann, Cambridge University Press, 2003
  

Kronecker's Theorem: An Example

In a previous blog, I presented Kronecker's Theorem.

Today, I will show how it can be used to prove that an equation is not algebraically soluble.

The content in today's blog is taken directly from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Example: x⁵ - ax - b = 0

Let's assume the following:

(1) a,b are positive integers divisible by a prime p

(2) b is not divisible by p²

(3) 4⁴a⁵ is greater than 5⁵b⁴

Here's the analysis:

(1) Using Eisenstein's Criteria, the equation is irreducible over the set of rational numbers. [see Theorem 1, here].

(2) From Sturm's Theorem [see Theorem, here], it is clear that it possesses three real roots and two complex roots since:

(a) We build the following Sturm Chain (see here for details on Sturm Chains):

P₀ = x⁵ - ax - b [The equation itself]

P₁ = 5x⁴ - a [The first derivative, see here for review if needed]

P₂ = 4ax + 5b [The remainder from P₀ and P₁, see here for view if needed]

P₃ = 4⁴a⁵ - 5⁵b⁴ [The remainder from P₁ and P₂]

(b) We know that there are 5 roots from the Fundamental Theorem of Algebra [see here for proof]

(c) From an analysis, I did earlier (see Example 2, here), we know that there are three real roots.

(3) Assume that the equation is algebraically soluble.

(4) Then, from Kronecker's Theorem [see Theorem 4, here], it either has only one real root or all real roots.

(5) But this is not the case from Sturm's Theorem so we have a contradiction.

(6) Thereofore, we reject our assumption in step #3 and conclude that the equation is not algebraically soluble.

References

• Heinrich Dorrie (Translated by David Antin), 100 Great Problems of Elementary Mathematics (Dover, 1965)

Kronecker's Theorem: The Proof

The content in today's blog is taken directly from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma 1:

Let f be an irreducible polynomial over a field F.

Let f be reducible over a field F(λ) where:

λ = K^(1/l) and l is an odd, prime and K ∈ F but λ is not in F.

Let g(x,λ) be a polynomial in F(λ) which divides f.

Let α be the nth root of unity.

Then:

g(x,λαⁱ) divides f.

Proof:

(1) Since g(x,λ) divides f(x), there exists h(x,λ) such that:

f(x) = g(x,λ)*h(x,λ)

(2) Let r be any element of K.

[It is clear that f(r) = g(r,λ)*h(r,λ)]

(3) We can define a function u(x) in F(λ) such that:

u(x) = f(r) - g(r,x)h(r,x)

(4) It is clear that u(λ) = 0 since

f(x) - g(x,λ)h(x,λ)=0 for all x.

(5) Let us also define a function v(x) such that:

v(x) = K^1/l

(6) It is clear that v(x) is irreducible in F. [see Lemma 2, here]

(7) It is also clear that the roots of v(x) are (from the definition of the roots of unity, see here):

λ,
λα
...
λα^l-1

(8) So, from step #4 above, it follows that each of these roots is also a root of u(x) [see Theorem 3, here]

(9) This means that for all these roots:

f(r) - g(r,λαⁱ)h(r,λαⁱ) = 0

(10) But since r can be any element of K (from step #2 above), it follows that:

f(x) - g(x,λαⁱ)h(x,λαⁱ) = 0

QED


Lemma 2:

Let:

ψ(x,λ_v) =u(x,λ_v)v(x,λ_v)

for some v where λ is an nth root of unity

and x ∈ a field F

Then:

ψ(x,λ) =u(x,λ)v(x,λ)

Proof:

(1) Let t(x) = ψ(r,x) - u(r,x)v(r,x)

where r ∈ a field F

[in fact, all r will work since: ψ(x,λ_v) =u(x,λ_v)v(x,λ_v) for all x]

(2) Since t(λ_v) = 0, λ_v is a root of t(x)

(3) Now λ, λ_v, etc. are all roots of unity so they are roots to the equation:

xⁿ - 1 = 0

(4) The polynomial in step #3 above is irreducible in F. [see Lemma 2, here]

(5) So λ, λ₁, etc. are all roots of t(x) [see Theorem 3, here]

(6) So, λ is a root of t(x) and we have for any r [see step #1 above]:

u(λ) = ψ(r,λ) - u(r,x)v(r,λ) = 0

(7) Since it is true for any r, we also have:

f(x) = ψ(x,λ) - u(x,λ)v(x,λ) = 0 for all.

(8) But then it follows that:

ψ(x,λ) = u(x,λ)v(x,λ)

QED


Lemma 3:

Let f(x) be a function irreducible in a field F.

Let Let λ be a number such that:

λ = K^1/l where K ∈ F and l is an odd prime

Let f(x) be reducible in F(λ) such that:

f(x) = ψ(x,λ)φ(x,λ)*ξ(x,λ)*...

where ψ, φ, ξ, ... are irreducible factors in F(λ)

Then:

No two of the ψ(x,λ_i) are equal. That is, if λ_i ≠ λ_j, then ψ(x,λ_i) ≠ ψ(x,λ_j)

Proof:

(1) Assume that λ_i ≠ λ_j, but ψ(x,λ_i) = ψ(x,λ_j)

(2) So that:

ψ(x,λ^μ) = ψ(x,λ^ν)

(3) Let H = the root of unity η^{ν - μ}

(4) So that we have:

ψ(x,λ) = ψ(x,λH)

(5) Hence, we can replace λ with λH to get:

ψ(x,λH) = ψ(x,λH²)

(6) And further that:

ψ(x,λH²) = ψ(x,λH³)

(7) So that we get:

ψ(x,λ) = ψ(x,λH) = ψ(x,λH²) = ψ(x,λH³) ...

(8) Adding the n such equations together we get:

ψ(x,λ) = (1/n)*(ψ(x,λ) + ψ(x,λH) + ψ(x,λH²) + ψ(x,λH³) + ... + ψ(x,λH^n-1)

(9) Now:

(1/n)*(ψ(x,λ) + ψ(x,λH) + ψ(x,λH²) + ψ(x,λH³) + ... + ψ(x,λH^n-1) is a symmetric function [see Definition 1, here].

(10) Further:

λ*λH*...*λH^n-1 = K where K ∈ F

(11) Therefore ψ(x,λ) ∈ F [See Thereom 4, here]

(12) But this is impossible since f is irreducible in F.

(13) So we reject our assumption in step #1.

QED


Theorem 4: Kronecker's Theorem

An algebraically soluble equation of an odd degree that is a prime and which is irreducible over rationals possesses either only one real root or only real roots.

Proof:

(1) Let f be an irreducible polynomial over a field F[x] that is algebraically soluble and has an odd, prime degree n.

(2) Let λ be a number such that:

λ = K^1/l where K ∈ F and l is an odd prime

and

f can be divided into factors over the field F(λ)[x]

(3) Since both l and n are prime numbers and l divides n (see Lemma 2, here), it follows that l=n.

(4) The equation x^l = K is irreducible in F (see Lemma 2, here).

(5) It has the following roots (this derives from step #2 and the definition of the roots of unity, see here):

λ₀ = λ

λ₁ = λ*α

...

λ_n-1 = λ*α^n-1

where α is an nth root of unity.

(6) From step #2 (since f(x) is reducible in F(λ)[x], we can divide up f(x) into irreducible factors:

f(x) = ψ(x,λ)φ(x,λ)*ξ(x,λ)*...

where ψ, φ, ξ, ... represent these factors

(7) Since ψ(x,λ) is a divisor of f(x), it follows that all ψ(x,λ_v) are also factors of f(x). [see Lemma 1 above]

(12) Everyone of the n functions ψ(x,λ_v) is irreducible in F(λ) since:

(a) ψ(x,λ) is irreducible in F(λ) [see step #6 above]

(b) Assume that ψ(x,λ_v) is not irreducible in F(λ)

(c) Then, there exists u(x,λ_v), v(x,λ_v) where:

ψ(x,λ_v) =u(x,λ_v)v(x,λ_v)

(d) But then (from Lemma 2 above):

ψ(x,λ) =u(x,λ)v(x,λ)

(e) Which contradicts (a) and we reject our assumption in (b)

(13) No two of the n functions ψ(x,λ_v) are equal. [see Lemma 3 above]

(14) It follows that f(x) is divisible by the product Ψ(x) of the n different factors ψ(x,λ), ψ(x,λμ), .., ψ(x,λμ^n-1) [from step #13 above and step #12 above]

(15) So that we have:

f(x) = Ψ(x)*U(x)

(16) Since Ψ is a symmetrical function of the roots of xⁿ=K, it follows that Ψ(x) is in F [see Theorem 4, here].

(17) But then U(x) = 1 since f(x) is irreducible in F and we have:

f(x) = Ψ(x) = ψ(x,λ)*ψ(x,λμ)*...*ψ(x,λμ^n-1)

(18) Since f(x) is reducible in F(λ), it follows that if ω, ω₁, ..., ω_n-1 are the roots, then:

x - ω, x - ω₁, ..., x - ω_n-1 are the linear factors of f(x) [see Girard's Theorem, here]

(19) Then (from step #17 above):

x - ω = ψ(x,λ)

x - ω₁ = ψ(x,λμ)

....

x - ω_n-1 = ψ(x,λμ^n-1)

(20) So that we get [from the definition of ψ(x,λ)]:

ω = K₀ + K₁λ + K₂λ² + ... + K_n-1λ^n-1


ω₁ = K₀ + K₁λ₁ + K₂λ₁² + ... + K_n-1λ₁^n-1

....

ω_n-1 = K₀ + K₁λ_n-1 + K₂λ_n-1² + ... + K_n-1λ_n-1^n-1


(21) Now, the equation f(x)=0 has at least one real root since it is an odd degree [see Theorem 3, here]

(22) Let this real root be:

ω = K₀ + K₁λ + K₂λ² + ... + K_n-1λ^n-1

where K₀ ∈ F(α) where α is a primitive nth root of unity.

(23) There are three possibilities that we need to consider:

Case I: λ is a real.

Case II: λ is not real and f(x) is irreducible over F[norm(λ)]

Case III: λ is not real and f(x) is reducible over F[norm(λ)]

(24) If Case I, then all the other roots are not real so it only has one real root. [see Lemma 2, here]

(25) If Case II, then all the roots are real. [see Lemma 3, here]

(26) If Case III, then let Λ = norm(λ) and we can restate step #22 as:

ω = K₀ + K₁Λ + K₂Λ² + ... + K_n-1Λ^n-1

(27) Since Λ is real, then all the other roots are not real and the equation has only one real root. [see Lemma 2, here]

QED

References

• 100 Great Problems of Elementary Mathematics (Dover, 1965)

Kronecker's Theorem: Lemmas on Complex Conjugates

The content in today's blog is taken directly from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma 1:

Let η be a primitive nth root of unity.

Let f(x) be an nth degree polynomial that irreducible over Q where n is an odd prime number and Q is the set of rational numbers extended with η. That is: Q(η).

Let K be a rational number such that λ = K^(1/l) and λ is not a rational number and l is a prime and λ is a real number

Let f(x) be reducible over Q(λ) such that:

f(x) = g(x,λ)*g(x,λη)*...*g(x,λη^n-1)

Then:

Then the roots of f(x) have the following form:

ω₀ = a₀ + a₁λ₀ + ... + a_n-1λ₀^n-1

ω₁ = a₀ + a₁λ₁ + ... + a_n-1λ₁^n-1

...

ω_n-1 = a₀ + a₁λ_n-1 + ... + a_n-1λ_n-1^n-1

where a_i ∈ Q and λ_i = ληⁱ

Proof:

(1) Since n is odd, we know that f(x) has at least one real root. [see Theorem 3, here]

(2) Let us use λ_i to denote the different parameters of g(x,y) so that we have:

λ₀ = λη⁰ = λ

λ₁ = λη¹ = λη

...

λ_n-1 = λη^n-1

(3) Let:

g(x,λ_i) =a₀ + a₁λ_i + ... + a_n-1λ_i^n-1

where a_i ∈ Q

(4) From f(x) = g(x,λ)*g(x,λη)*...*g(x,λη^n-1), we can see that f(x) has n roots:

ω₀ = a₀ + a₁λ₀ + ... + a_n-1λ₀^n-1

ω₁ = a₀ + a₁λ₁ + ... + a_n-1λ₁^n-1

...

ω_n-1 = a₀ + a₁λ_n-1 + ... + a_n-1λ_n-1^n-1

QED


Lemma 2:

Let η be a primitive nth root of unity.

Let f(x) be an nth degree polynomial that irreducible over Q where n is an odd prime number and Q is the set of rational numbers extended with η. That is: Q(η).

Let K be a rational number such that λ = K^(1/l) and λ is not a rational number and l is a prime and λ is a real number

Let f(x) be reducible over Q(λ) such that:

f(x) = g(x,λ)*g(x,λη)*...*g(x,λη^n-1)

Then:

All the roots of f(x) but one are complex and not real.

Proof:

(1) Using Lemma 1 above, we know that f(x) has the following roots:

ω₀ = a₀ + a₁λ₀ + ... + a_n-1λ₀^n-1

ω₁ = a₀ + a₁λ₁ + ... + a_n-1λ₁^n-1

...

ω_n-1 = a₀ + a₁λ_n-1 + ... + a_n-1λ_n-1^n-1

where a_i ∈ Q and λ_i = ληⁱ

(2) Since n is odd, we know that f(x) has at least one real root. [see Theorem 3, here]

(3) Let ω be the real root so that we have:

ω = a₀ + a₁λ + ... + a_n-1λ^n-1

where a_i ∈ Q

(4) We assume that ω = ω₀. (We can make the same argument regardless of which ω_i is real.)

(5) Since ω is real, we know that ω = ω [see Theorem 1, here] which means that:

a₀ + a₁λ + ... + a_n-1λ^n-1 =a₀ + a₁λ + ... + a_n-1λ^n-1

so that:

(a₀ - a₀) + ... + (a_n-1 - a_n-1) = 0

(6) So, that for all i, a_i = a_i

(7) This means that all a_i are real numbers.

(8) Since λ is real, it follows that ληⁱ is not real because i ≠ 0.

(9) λ_v and λ_-v are complex conjugates since (see Theorem 2, here):

λ_v = λ*η^v = λη^-v = λ_-v

(10) So it follows that we have the following (n-1)/2 complex conjugate pairs:

ω_n-1 and ω₁

ω_n-2 and ω₂
...

(11) This shows that all but one of the roots of f(x) are not real.

QED


Lemma 3:

Let η be a primitive nth root of unity.

Let f(x) be an nth degree polynomial that irreducible over Q where n is an odd prime number and Q is the set of rational numbers extended with η. That is: Q(η).

Let K be a rational number such that λ = K^(1/l) and λ is not a rational number and l is a prime and λ is not a real number and f(x) is not reducible over norm(λ)

Let f(x) be reducible over Q(λ) such that:

f(x) = g(x,λ)*g(x,λη)*...*g(x,λη^n-1)

Then:

All the roots of f(x) are real.

Proof:

(1) Using Lemma 1 above, we know that f(x) has the following roots:

ω₀ = a₀ + a₁λ₀ + ... + a_n-1λ₀^n-1

ω₁ = a₀ + a₁λ₁ + ... + a_n-1λ₁^n-1

...

ω_n-1 = a₀ + a₁λ_n-1 + ... + a_n-1λ_n-1^n-1

where a_i ∈ Q and λ_i = ληⁱ

(2) Since n is odd, we know that f(x) has at least one real root. [see Theorem 3, here]

(3) Let ω be the real root so that we have:

ω = a₀ + a₁λ + ... + a_n-1λ^n-1

where a_i ∈ Q

(4) We assume that ω = ω₀. (We can make the same argument regardless of which ω_i is real.)

(5) Since ω is real, we know that ω = ω [see Thereom 1, here] which means that:

a₀ + a₁λ + ... + a_n-1λ^n-1 =a₀ + a₁λ + ... + a_n-1λ^n-1

(6) Let Λ = the norm(λ) so that Λ = λ*λ which then is a real number. [see Lemma 1, here]

(7) So that we have:

a₀ + a₁λ + ... + a_n-1λ^n-1 =a₀ + a₁(Λ/λ) + ... + a_n-1(Λ/λ)^n-1

(8) Now we can define an equation h(x) such that:

h(x) = a₀ + a₁x + ... + a_n-1x^n-1 - a₀ + a₁(Λ/x) + ... + a_n-1(Λ/x)^n-1

(9) From step #7 above, it is clear that λ is a root of this equation.

(10) But λ is also a root of xⁿ = K which is irreducible over Q(Λ) [see Lemma 1, here]

(11) So, by Abel's Lemma [see Theorem 3, here], all the roots of xⁿ = K are also solutions to the equation in step #8.

(12) Now, for every λ_v, we have:

Λ/λ_v = Λ/(λη^v) = λ/(η^v) = ληⁿ = λ_v

(13) Thus, for all λ_v, we have [from step #11 above]:

h(x) = a₀ + a₁x + ... + a_n-1x^n-1 - a₀ + a₁(Λ/x) + ... + a_n-1(Λ/x)^n-1

which means:

a₀ + a₁λ_v + ... + a_n-1λ_v^n-1 =a₀ + a₁(Λ/λ_v) + ... + a_n-1(Λ/λ_v)^n-1

which means:

a₀ + a₁λ_v + ... + a_n-1λ_v^n-1 =a₀ + a₁λ_v + ... + a_n-1λ_v^n-1

so that for all ω_v:

ω_v = ω_v

(14) But this can only be true if all the roots are real [see Theorem 1, here]

QED


References

• 100 Great Problems of Elementary Mathematics (Dover, 1965)

Kronecker's Theorem: Some Lemmas on Irreducible Polynomials

Leopold Kronecker's Theorem is based on Abel's famous theorem. Today, I will present some lemmas that I will use in establishing Kronecker's Theorem.

The content in today's blog is taken from 100 Great Problems of Elementary Mathematics by Heinrich Dorrie.

Definition 1: algebraically soluble

An equation of the nth degree f(x) = 0 is called algebraically soluble when it is soluble by a series of radicals. [See here for a more detailed explanation]

Lemma 1:

Every number of a field F(α), where α is a root of an irreducible equation of the nth degree in F, can be represented as a polynomial of the (n-1)th degree of α with coefficients that are of F and there is only one such way to represent it.

Proof:

(1) Let f(x) be an irreducible equation of the nth degree whose root is α such that:

f(x) = xⁿ + a₁x^n-1 + ... + a_n

and

f(α) = αⁿ + a₁α^n-1 + ... + a_n = 0

(2) Let ζ be a number of field F(α) where α is a root of an irreducible equation of the nth degree in F.

(3) Then, there exists functions Ψ and Φ such that Ψ, Φ are functions in F and:

ζ = Ψ(α)/Φ(α)

(4) It follows from step #1 that:

αⁿ = -a₁α^n-1 - ... - a_n

(5) We can therefore rewrite Ψ and Φ as polynomials of (n-1)th degree.

(6) Since f(x) is irreducible, it follows that f(x) and Φ(x) possess no common divisors.

(7) Using the Bezout Identity for Polynomials (see Corollary 3.1, here), it follows that there exists polynomials u(x) and v(x) such that:

u(x)Φ(x) + v(x)f(x) = 1.

(8) If we set x = α, then we get:

u(α)Φ(α) + v(α)(0) = u(α)Φ(α) = 1.

(9) Now, if we multiply ζ to both sides, we get:

ζ*1 = ζ*u(α)Φ(α) = [Ψ(α)/Φ(α)]*u(α)Φ(α) = Ψ(α)u(α)

(10) If we multiply this out and use step #5, we get the following equation form:

ζ = c₀ + c₁α + ... c_n-1α^n-1

(11) To complete this proof, we need to show that there is only one way to represent this number.

(12) Assume that:

c₀ + c₁α + ... c_n-1α^n-1 = C₀ + C₁α + ... C_n-1α^n-1

(13) If we let d_i = C_i - c_i, then we have:

d₀ + d₁α + ... d_n-1α^n-1 = 0

(14) But then, if we have a root of an irreducible equation of higher degree, then for n-1, it follows all d_i = 0 [See Corollary 3.1, here]

(15) Thus, c_i = C_i for all i.

QED

Lemma 2:

An irreducible equation of the prime number degree p in a field F can become reducible through substitution of a root of another irreducible equation in this group only when p is a divisor of the degree of the latter equation.

Proof:

(1) Let f(x) be an irreducible equation of the pth degree in the field F such that:

f(x) = x^p + a₁x^p-1 + ... + a_p

and p is both odd and prime.

(2) Let g(x) be an irreducible equation of the qth degree in the field F such that:
g(x) = x^q + b₁x^q-1 + ... + b_q

and

g(α) = 0

(3) Assume that f(x) can be reduced in F(α) such that it is the product of two polynomials:

ψ(x,α) and φ(x,α) where ψ is an mth degree polynomial and and φ is an nth degree polynomial

(4) Let us define a function u(x) such that:

u(x) = f(r) - ψ(r,x)φ(r,x)

where r is some rational number.

(5) Now, it is clear that α is a root for u(x).

(6) Let α, α', α'', etc. be the q roots of g(x). [We know that it has q roots from the Fundamental Theorem of Algebra, see here]

(7) We know that α, α', α'', etc. are all roots of u(x). [see Theorem 3, here]

(8) But from step #3 above, we know that:

f(x) - ψ(x,α)φ(x,α) = 0

(9) So that:

u(α) = f(r) - ψ(r,α)φ(r,α) = 0 for all r.

(10) But then from step #7 above, it follows that all for all α', α'', etc.:

f(x) - ψ(x,α')φ(x,α') = 0

[The reasoning here comes from step #9 above. We can define u(x) based on any r. From step #8 above, u(α) = 0 for that r. And then for any r, we can apply step #7 above. If it is true for any r, it is true for all x.]

(11) Thus for all roots α, α', α'' etc. we have:

f(x) = ψ(x,α')φ(x,α')

(12) If we multiply all q of these equations together, we get:

f(x)^q = Ψ(x)Φ(x) where:

Ψ(x) = ψ(x,α)*ψ(x,α')*ψ(x,α'')...

and

Φ(x) = φ(x,α)*φ(x,α')*φ(x,α'')...

(13) Since Ψ(x), Φ(x) are symmetric functions [see Definition 1, here for a definition of symmetric functions], we can represent each as a rational function of the elementary symmetric polynomials [see Theorem 4, here]

(14) Which means that we can restate them as rational functions of the coefficients of g(x) [see Theorem 1, here]

(15) Any root of φ(x,α) is necessarily a root of f(x) and any root of ψ(x,α) is a root of f(x). [see step #11 above].

(16) So it is clear that f(x) shares common roots with both φ(x,α) and ψ(x,α)

(17) Which means that f(x) shares common roots with Φ(x) and Ψ(x)

(18) So, Φ(x) and Ψ(x) are divisible by f(x) without remainder. [see Theoreom 3, here]

(19) It can be shown that both Φ(x) and Ψ(x) can be expressed as powers of f(x) since:

(a) f(x)^q = Ψ(x)Φ(x)

(b) f(x) divides both Ψ(x) and Φ(x) [step #18 above]

(c) Assume that there exists a irreducible polynomial h(x) such that:

Ψ(x) = f(x)^ah(x)

and h(x) is not divisible by f(x)

(d) But then h(x) divides f(x)^q which is impossible since f(x) is irreducible.

(e) Therefore we reject our assumption in (c).

(f) We can make the same argument for Φ(x)

(20) So, there exists μ, ν such that:

Ψ(x) = f(x)^μ

and

Φ(x) = f(x)^ν

and

μ + ν = q

(21) Comparing the degrees of the right and left sides, we obtain (see step #3 above):

mq = μp

and

nq = νp

(22) Since m,n are smaller than p, it follows that p is a divisor of q (since p is prime, p divides m or q [see Euclid's Lemma, here] but p doesn't divide m or n, so p divides q).

QED

References

• 100 Great Problems of Elementary Mathematics (Dover, 1965)

Waring's Method

In today's blog, I will show Edward Waring's method for expressing any symmetric polynomial in terms of the elementary symmetric polynomials (s₁, ..., s_n). For review of the elementary symmetric polynomials, see here. For review of polynomials, see here. For review of a field, see here.

The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Symmetric Polynomial

A polynomial P(x₁, ..., x_n) in n indeterminates is symmetric if and only if it is not altered when the indeterminates are arbitarily permuted among themselves.

That is, for every permutation σ of 1, ..., n, we have:

P(x_σ(1), ..., x_σ(n)) = P(x₁, ..., x_n)

Definition 2: Symmetric Rational Fraction

A rational fraction P/Q in n indeterminates is symmetric if it is not altered when the indeterminates are permuted; i.e. for every permutation σ of 1,..., n:

P(x_σ(1), ..., x_σ(n)) /Q(x_σ(1), ..., x_σ(n)) = P(x₁, ..., x_n)/Q(x₁, ..., x_n)

Note: This is not mean that P,Q are necessarily symmetric. Still, it will be seen later that every symmetric rational fraction can be represented as the quotient of symmetric polynomials.

Definition 3: ∑ x₁^i₁x₂^i₂*...*x_n^i_n

We can use this notation to characterize a symmetric polynomial. In this case, each monomial has the form expressed in the sum.

In using this notation, it is important to specify the value of n and the number of indeterminates. Otherwise, it is not clear how it is expressed.

For example for a symmetric polynomial in two variables:

∑ x₁²x₂ = x₁²x₂ + x₁x₂²

For a symmetric polynomial in three variables, we have:

∑ x₁²x₂ = x₁²x₂ + x₁x₂²+ x₁²x₃ + x₁x₃² + x₂²x₃ + x₂x₃2

We can also use this notation to represent the elementary symmetric polynomials:

s₁ = ∑ x₁

s₂ = ∑ x₁x₂

...

s_n-1= ∑ x₁*...*x_n-1

s_n = ∑ x₁*...*x_n


Definition 4: deg (∑ x₁^i₁x₂^i₂*...*x_n^i_n)= (i₁, ..., i_n)

Using the notation ∑ x₁^i₁x₂^i₂*...*x_n^i_n, by degree, I mean the set of i₁, ..., i_n that make up the symmetric polynomial.

For any non-zero polynomial P = P(x₁, ..., x_n) in n indeterminates x₁, ..., x_n over a field, the degree of P is defined as the largest n-tuple (i₁, ..., i_n) for which the coefficient x₁^i₁*...*x_n^i_n in P is nonzero.

For purposes of comparison, we assume that i₁, ..., i_n are ordered such that i₁ ≥ i₂ ≥ ... ≥ ... i_n

Here are some examples:

deg s₁ = (1, 0, 0, ..., 0)

deg s₂ = (1,1,0,....,0)

...

deg s_n-1 = (1,1,...1,0)

deg ∑ x₁²x₂ = (2,1,0, ..., 0)


Definition 5: Nⁿ

Let Nⁿ be the set of n-tuples of integers of the form of Definition 4.

So that:

Nⁿ = { (i₁, ..., i_n), (j₁, ..., j_n), (k₁, ..., k_n), ... }


Definition 6: Ordering of Nⁿ: (i₁, ..., i_n) ≥ (j₁, ..., j_n)

(i₁, ..., i_n) ≥ (j₁, ..., j_n) if and only if for all values:

i₁ is greater than j₁ or

i₁ = j₁ and i₂ is greater than j₂ or

for all u, i_u = j_u or

there exists v, such that i_v is greater than j_v and for all u less than v, i_u = j_u


Lemma 1: deg(P+Q) ≤ max(deg P, deg Q)

Proof:

(1) Let P = ∑ x₁^i₁x₂^i₂*...*x_n^i_n

(2) Let Q = ∑ x₁^j₁x₂^j₂*...*x_n^j_n

(3) deg P = (i₁, ..., i_n)

(4) deg Q = (j₁, ..., j_n)

(5) Assume that (i₁, ..., i_n) is greater than (j₁, ..., j_n)

(6) So, max(deg P,deg Q) = (i₁, ..., i_n)

(7) If none of the resulting coefficients changes to zero, then:

deg(P+Q) = (i₁, ..., i_n) [This follows from definition 4 above]

(8) If at least one of the resulting coefficients changes to zero, then:

deg(P+Q) is less than (i₁, ..., i_n)

(9) We know that deg(P+Q) cannot be higher because addition may change the coefficient but it cannot change the power of any of the terms.

QED


Lemma 2: deg(PQ) = deg P + deg Q

Proof:

(1) Let P = ∑ x₁^i₁x₂^i₂*...*x_n^i_n

(2) Let Q = ∑ x₁^j₁x₂^j₂*...*x_n^j_n

(3) deg P = (i₁, ..., i_n)

(4) deg Q = (j₁, ..., j_n)

(5) deg P + deg Q = (i₁ + j₁, ..., i_n+j_n)

(6) P*Q = (∑ x₁^i₁x₂^i₂*...*x_n^i_n)*(∑ x₁^j₁x₂^j₂*...*x_n^j_n) = ∑ (x₁^i₁+j₁x₂^i₂+j₂*...*x_n^i_n+j_n)

(7) deg(P*Q) = (i₁ + j₁, ..., i_n+j_n)

QED

Corollary 2.1: deg a^x = x*deg(a)

Proof:

deg a^x = deg (a*a*...*a) = deg(a) + deg(a) + ... + deg(a) = x*deg(a)

QED


Lemma 3: Nⁿ does not contain any infinite strictly decreasing sequence of elements.

That is if we take any x ∈ Nⁿ, we can only decrease it a finite amount of times.

Proof:

(1) This is clearly true for n=1.

(2) So, we can assume that this is true up to n-1.

(3) Assume that we have an infinite strictly decreasing sequence in Nⁿ such that:

(i₁₁, i₁₂, ..., i_1n) is greater than (i₂₁, i₂₂, ..., i_2n) which is greater than ... which is greater than (i_m1, i_m2, ..., i_mn) is greater than ....

(4) Using Definition 6 above, we know that:

i₁₁ ≥ i₂₁ ≥ ... ≥ i_m1 ≥ ....

(5) Since i₁₁ is finite, it follows that the only way that this can be infinite is if this sequence is eventually constant.

(6) Let us assume that it becomes constant starting with i_M1 so that for all m ≥ M, i_m1 = i_M1

(7) By our assumption in step #3, it follows that the following sequence must also be infinite:

(i_M1, i_M2, ..., i_Mn) is greater than (i_(M+1)1, i_(M+1)2, ...., i_(M+1)n) is greater than ... and so on.

(8) Since all of the first elements are equal and from definition 6 above, we can remove the first element in all cases to get the following infinite sequence:

(i_M2, ..., i_Mn) is greater than (i_(M+1)2, ...., i_(M+1)n) is greater than ... and so on.

(9) But now we have a contradiction. Since we assumed in step #2 that there are no infinite strictly decreasing sequence of elements in N^n-1

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

QED

Theorem 4: Waring's Method (Fundamental Theorem of Symmetric Polynomials)

A polynomial in n indeterminates x₁, ..., x_n over a field F can be expressed as a polynomial in s₁, ..., s_n if and only if it is symmetric.

In other words, for any symmetric polynomial, there exists a function g such that:

P(x₁, ..., x_n) = g(s₁, ..., s_n)

where s₁, s₂, ..., s_n are the elementary symmetric polynomials.

Proof:

(1) Let P ∈ F[x₁, ..., x_n] be a non-zero symmetric polynomial.

(2) Let deg P = (i₁, ..., i_n) ∈ Nⁿ where i₁ ≥ i₂ ≥ ... ≥ i_n

(3) I will now show that P can be expressed as a function of the elementary symmetric polynomials.

(4) Let us define the following polynomial:

f = s₁^{i₁ - i₂}s₂^i₂-i₃*...*s_n-1^i_n-1-i_ns_n^i_n

(5) Using Lemma 2 above, we have:

deg f = deg(s₁^{i₁ - i₂}) + deg(s₂^{i₂ - i₃}) + ... + deg(s_n^i_n)

(6) Using Corollary 2.1 above, we have:

deg f = (i₁ - i₂)deg(s₁) + (i₂ - i₃)deg(s₂) + ... + i_ndeg(s_n)

(7) Based on the definition of the elementary symmetric polynomials (see here), we have:

deg f = (i₁ - i₂,0,...,0) + (i₂ - i₃, i₂ - i₃,0,...,0) + ... + (i_n, ..., i_n) = (i₁, i₂, ..., i_n)

(8) Also from the definition of the elementary symmetric polynomials, we know that the leading coefficient of f is 1.

(9) So, we can restate f as:

f = x₁^i₁*...*x_n^i_n + (terms of lower degree)

(10) Let a ∈ F^x be the leading coefficient of P such that:

P = ax₁^i₁*...*x_n^i_n + (terms of lower degree)

(11) Let:

P₁ = P - af

(12) We can see that P₁ has the following properties:

(a) deg P₁ is less than deg P [See definition 4 above]

(b) P₁ is symmetric since P and f are symmetric

(c) We can assume that P₁ is nonzero. [If it were 0, we would be done with the proof. We only need to handle the case when P₁ is nonzero to finish the proof]

(13) Now since P₁ is a symmetric polynomial, we can repeat step #12 such that we define a polynomial P₂

(14) In this way, we can continue to reduce P_i until we P_i - af = 0.

(15) We know that this process will eventually complete from Lemma 3 above.

QED

Example 4.1: S = ∑ x₁⁴x₂x₃ + ∑ x₁³x₂³

(1) Let:

S = ∑ x₁⁴x₂x₃ + ∑ x₁³x₂³

(so that: S = x₁⁴x₂x₃ + x₁x₂⁴x₃ + x₁x₂x₃⁴ + x₁³x₂³ + x₁³x₃³ + x₂³x₃³)

[See Definition 3 above for details if needed]

(2) Since (4,1,1) is greater than (3,3,0) [see Definition 6 above], it follows that [see Definition 4 above]:

deg(S) = (4,1,1)

(3) Let f = s₁^4-1s₂^1-1s₃¹ = s₁³s₃

(4) Using the definitions for s₁, ..., s₃, we have:

s₁³s₃ = (∑ x₁)³(x₁x₂x₃) = ∑ x₁⁴x₂x₃ + 3∑ x₁³x₂²x₃ + 6x₁²x₂²x₃²

(5) If we let S₁ = S - f, then we get:

S₁ = ∑ x₁³x₂³ - 3∑ x₁³x₂²x₃ - 6x₁²x₂²x₃²

(6) deg S₁ = (3,3,0)

(7) Let f₁ = s₁^3-3s₂^3-0s₃⁰ = s₂³

(8) Using the definitions for s₁, ..., s₃, we have:

s₂³ = (∑ x₁x₂)³ = ∑ x₁³x₂³ + 3∑ x₁³x₂²x₃ + 6x₁²x₂²x₃²

(9) If we let S₂ = S₁ - f₁, then we get:

S₂ = - 6∑ x₁³x₂²x₃ - 12x₁²x₂²x₃²

(10) deg S₂ = (3,2,1)

(11) Let f₂ = s₁^3-2s₂^2-1s₁¹ = s₁s₂s₃

(12) Using the definitions for s₁, ..., s₃, we have:

s₁s₂s₃ = (∑ x₁)(∑ x₁x₂)(∑ x₁x₂x₃) = ∑x₁³x₂²x₃ + 3x₁²x₂²x₃²

(13) If we let S₃ = S₂ + 6f₂, then we get:

S₃ = 6x₁²x₂²x₃²

(14) deg(S₃) = (2,2,2)

(15) Let f₃ = s₁^2-2s₂^2-2s₃² = s₃²

(16) Using the definitions for s₁, ..., s₃, we have:

s₃² = (∑ x₁x₂x₃)² = x₁²x₂²x₃²

(17) We can see that S₃ - 6f₃ = 0 so we are done.

(18) The resulting function in terms of the elementary symmetric polynomials is:

S = s₁³s₃ + s₂³ - 6s₁s₂s₃ + 6s₃²


Lemma 5:

Let P,Q be polynomials such that P/Q is symmetric

Then:

P is symmetric if and only Q is symmetric

Proof:

(1) Assume that P is symmetric

(2) Assume that Q is not symmetric such that Q' is a permutation of Q and Q' ≠ Q.

(3) Let P' be the same permutation as Q'.

(4) Since P is symmetric, P' = P

(5) But P'/Q' = P/Q' ≠ P/Q

(6) But this is impossible since we assumed that P/Q is symmetric.

(7) Therefore, we reject our assumption in step #2.

(8) We can make the exact same argument if we assume that Q is symmetric and P is not.

QED

Theorem 6:

A rational fraction in n indeterminates x₁, ..., x_n over a field F can be expressed as a rational fraction in s₁, ..., s_n if it is symmetric

Proof:

(1) Let P,Q be polynomials in n indeterminates x₁, ..., x_n such that the rational fraction P/Q is symmetric.

(2) We can assume that P,Q are not symmetric.

If P is symmetric, then Q is too (from Lemma 5 above) and we can use Theorem 4 above to get our result. So, to prove the theorem, we need only handle the case where both P,Q are not symmetric.

(3) Since Q is not symmetric, let Q₁, ..., Q_r be the distinct polynomials (other than Q) obtained from Q through permutations of the indeterminates.

(4) The product QQ₁*...*Q_r is symmetric since any permutation of the indeterminates simply permutes the factors.

(5) Since P/Q is symmetric, it follows that P/Q = P*(Q₁*...*Q_r)/[Q*(Q₁*...*Q_r)] is symmetric too.

(6) It further follows that P*(Q₁*...*Q_r) is symmetric from Lemma 5 above.

(7) Using Theorem 4 above, we know that there exists functions f,g such that:

P*(Q₁*...*Q_r) = f(s₁, ..., s_n)

and

Q*(Q₁*...*Q_r) = g(s₁, ..., s_n)

(8) Thus,

P/Q = f(s₁, ..., s_n)/g(s₁, ..., s_n)

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001