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)