Sturm's Theorem: Examples

In a previous entry, I posted the proof of Sturm's Theorem. This is a method for determining the number of real roots in a given interval.

Today, I will show some examples of its use.

Example 1: f(x) = x⁵ - 3x - 1 in the interval [-2, +2]

The Sturm Chain for this polynomial is:

f₀ = x⁵ - 3x - 1

f₁ = 5x⁴ - 3

f₂ = 12x + 5

f₃ = 1

For x=-2, there are 3 sign changes.

For x=-1, there are 2 sign changes

For x=0, there is 1 sign change

For x=1, there is 1 sign change

For x=2, there is 0 sign changes

So, between -2 and -1, there is 3-2=1 real zero

Between -1 and 0, there is 2-1=1 real zero

Between 0 and 1, there are no 1-1=0 real zeros.

Between 1 and 2, there is 1-0=1 real zero.

In summary, between -2 and +2, there are 3-0=3 real zeros.

Example 2: x⁵ -ax -b when a,b are positive and 4⁴a⁵ is greater than 5⁵b⁴

The Sturm Chain for this polynomial is:

f₀ = x⁵ -ax -b

f₁ = 5x⁴ - a

f₂ = 4ax + 5b

f₃ = 4⁴a⁵ - 5⁵b⁴

For this example, let's look at the interval between -∞ and +∞

For -∞, there are 3 sign changes.

For +∞, there are 0 sign changes.

So, all equations of this form have 3-0=3 real roots.

References

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

Sturm's Theorem: The Proof

In a previous blog entry, I talked about the properties of Sturm Chains. In today's blog entry, I will show how they can be used to establish Sturm's Theorem.

Definition 1: Number of Sign Changes Across a Sturm Chain at x

Let f₀, f₁, ..., f_s be a Sturm Chain where for all 0 ≤ i ≤ s, f_i ≠ 0. The number of sign changes across a Sturm Chain at x is the number of times that a neighboring function changes sign: the number of times when f_i(x) is negative and f_i+1(x) is positive or when f_i(x) is negative and f_i+1(x) is positive.

Example:

For the function f(x) = x⁵ - 3x - 1, the Sturm Chain is (see here for details on how this Sturm Chain is constructed):

f₀(x) = x⁵ - 3x - 1

f₁(x) = 5x⁴ - 3

f₂(x) = 12x + 5

f₃(x) = 1

For x=-2, we see that:

f₀ = -

f₁ = +

f₂ = -

f₃ = +

So, the number of sign changes across the Sturm Chain at x=-2 is 3.

Definition 2: Open and Closed Intervals

An interval is open on a,b if for all x in a,b, a is less than x and is less than b. An open interval is represented (a,b). [It is called open because there is no minimum or maximum value]

An interval is closed on a,b if for all x in a,b, a ≤ x ≤ b. A closed interval is represented [a,b] (It is called closed because it has both a minimum and a maximum value)

We can also mix the two notations so that x in [a,b) means that a ≤ x and x is less than b.

x in (a,b] means that a is less than x ≤ b.

I use this notation below.

Lemma 1:

Let f be a continuous function such that f(a) and f(b) have different signs,

Then:

There exists x such that x in (a,b) and f(x)=0

Proof:

This follows directly from the Intermediate Value Theorem (see Theorem, here).

QED

Corollary 1.1:

If f is continuous on [a,b] and for all x in [a,b]:

f(x) ≠ 0

Then:

for all x in [a,b]:

f(x) has the same sign.

Proof:

(1) Assume that f changes sign over [a,b]

(2) Then by Lemma 1 above there exists x such that f(x)=0

(3) But this is not true so we reject our assumption at step #1.

QED

Corollary 1.2:

Let f₀, f₁, ..., f_s be a Sturm Chain

Let k be the only zero for any f_i and it is a zero where i ≥ 1.

Let a,b be an interval such that k is in [a,b]

Then:

There are no sign changes across the Sturm Chain for x in [a,b]

Proof:

(1) For all x in [a,b]:

f_i-1(x) ≠ 0 and f_i+1(x) ≠ 0 and if x ≠ k, then f_i(k) ≠ 0.

(2) Since f_i(k)=0, we know that f_i-1(k) and f_i+1(k) have opposite signs. (see for properties of Sturm Chains, see Definition 1, here)

(3) Since f_i-1 and f_i+1 are nonzero on [a,b], we know from Corollary 1.1 above that f_i-1 and f_i+1 have the same opposite signs for all x in [a,b].

(4) But this means that any sign change on f_i doesn't affect the total number of sign changes from f_i-1 to f_i to f_i+1 since:

Before k:

There are the following possibilities for all f_i-1, f_i, f_i+1

+, +, - = 1 sign change
+, -, - = 1 sign change
-, +, + = 1 sign change
-, -, + = 1 sign change

After k:

There are the following possibilities for f_i-1, f_i, f_i+1

+, -, - = 1 sign change
+, +, - = 1 sign change
-, -, + = 1 sign change
-,+,+ = 1 sign change

(5) The above property is true even if there are multiple Sturm functions that are zero at k. The key point is that by the properties of Sturm Chains, we know that no neighboring functions are both 0 (see here for details on the properties of a Sturm Chain).

(6) It is possible that f_i, f_i+2, etc. are all 0 on k. But even in this case, the above logic holds and there are no sign changes for any of the Sturm functions.

(5) Because in all other cases, the Sturm functions are nonzero, it follows from Corollary 1.1 above that they don't change sign and we are done.

QED

Corollary 1.3:

Let f₀, f₁, ..., f_s be a Sturm Chain

Let [a,b] be an interval such that for all x in [a,b]:

f₀(x) ≠ 0 (but the other Sturm functions may have a zero).

Let Z(x) be the number of sign changes that occur across a Sturm Chain for a given x (see Definition 1 above)

Then:

There are no sign changes over the Sturm Chain in [a,b]. That is, Z(a) - Z(b) = 0.

Proof:

(1) Let f₀, f₁, ..., f_s be a Sturm Chain

(2) Assume that k is the only zero in [a,b] for any function in the Sturm Chain where i ≥ 1.

(3) In this case, the Theorem holds using Corollary 1.2 above.

(4) Assume that we order all zeros in [a,b] for any function in the Sturm Chain and up to the nth zero k, there is no sign change across the Sturm Chain.

(7) Let k be the nth zero in [a,b] and k' be the n+1th zero in [a,b] and k'' be the n+2th zero. Let i,i',i'' be the function that is zero so that we have: f_i(k)=0, f_i'(k')=0, and f_i''(k'') = 0.

(8) Since k' is the only zero on (k,k''), we can use Corollary 1.2 above to complete the inductive proof.

QED

Lemma 2:

Let f₀, f₁, ..., f_s be a Sturm Chain

Let k be the only zero for f₀ in [a,b]

Let Z(x) be the number of sign changes that occur across a Sturm Chain for a given x.

Then:

Z(a) - Z(b) = 1.

Proof:

(1) Let h be a point before k and l be a pointer after k such that f'(h) has the same sign as f'(k) and as f'(l). (See Property 3 of Sturm Chains in Definition 1, here)

(4) We have the following cases to consider:

Case I: f(h) is positive and f(l) is negative

(a) For all x in (h,l):

f(x) is decreasing, so its derivative f'(x) is negative. [See Lemma here if needed]

(b) So at h:

f₀(h)=f(h) is positive and f₁(h)=f'(h) is negative.

(c) At l:

f₀(l) is negative and f₁(l) is negative.

(d) So, Z(h) - Z(l) = 1.

Case II: f(h) is negative and f(l) is positive

(a) For all x in (h,l):

f(x) is increasing, so its derivative f'(x) is positive. [See Lemma here if needed]

(b) So at h:

f₀(h) is negative and f₁(h) is positive.

(c) At l:

f₀(l) is positive and f₁(l) is positive.

(d) So, again, Z(h) - Z(l) = 1.

QED

Theorem: Sturm's Theorem

Let f(x) be an algebraic equation with real coefficients and with only simple roots.

Let (a,b) be an interval such that f(a) ≠ 0 and f(b) ≠ 0 and a is less than b.

Let Z(x) be the number of sign changes that occur across a Sturm Chain for a given x (see Definition 1 above if needed).

Then:

The number of real roots occurring in (a,b) is equal to Z(a) - Z(b)

Proof:

(1) Let f₀, f₁, ..., f_s be a Sturm Chain for f(x) where f₀ = f. [See Lemma 1, here]

(2) Let [a,b] be an interval such that a is less than b and a,b are real numbers.

(3) Let Z(x) be the number of the sign changes across the Sturm Chain for a given x.

(4) There are three cases that we need to consider to prove this theorem.

Case I: No zeros for any function in the Sturm Chain

(a) To prove the theorem for this case, we need to show that Z(a) - Z(b) = 0.

(b) For all x in [a,b], for all 0 ≤ i ≤ s, f_i(x) ≠ 0

(c) But then by Corollary 1.1 above, for f_i, there is no sign changes for any function in the Sturm Chain.

(d) So, Z(a) - Z(b)=0

Case II: No zeros in (a,b) for f₀(x) but at least one zero for the other functions in the Sturm Chain.

This case is equivalent to Corollary 1.3 above.

Case III: At least one zero in (a,b) for f(x)

(a) ∃x such that a ≤ x ≤ b and f₀(x) = 0

(b) To prove this, I will use induction.

(c) Assume that there is only one such zero in [a,b]

(d) Using Lemma 2 above, we know that in this case, Z(a) - Z(b) = 1.

(e) Assume that the theorem is true up to the nth zero of f in [a,b]

(f) Let k be the nth zero, k' be the n+1th zero and k'' be the n+2th zero.

(g) Then there is only zero in (k,k'') which is located at k'.

(h) Let l be a point in (k,k') and l' be a point in (k',k'').

(i) Then, by Lemma 2 above Z(l) - Z(l') = 1.

(j) From step e above, we have Z(a) - Z(l) = n and from step i above we have Z(a) - Z(l') = n+1.

QED

References

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

Cauchy's Bound for Real Roots

Augustin-Louis Cauchy came up with a useful bound for real roots in a polynomial equation. This works well with Sturm's Theorem.

Theorem: Cauchy's Bound for Real Roots

Let f(x) = a_nxⁿ + a_n-1x^n-1 + ... + a₀

Let c be a root of f(x) such that f(c)=0

Then

abs(c) ≤ 1 + {abs(a_n-1 + ... + abs(a₀)}/abs(a_n)

Proof:

(1) Assume that abs(c) is greater than 1. [The alternative is true since 1 + abs(x) ≥ 1]

(2) Since c is a root, then a_ncⁿ + a_n-1c^n-1 + ... + a₀ = 0, and we have:

a_ncⁿ = -a_n-1c^n-1 + .... + -a₀

(3) Using a basic inequality (see Lemma 2, here), we have:

abs(a_n)*abs(c)ⁿ ≤ abs(a_n-1)*abs(c)^n-1 + ... + abs(a₀)

(4) Let H = max{abs(a_n-1), ..., abs(a₀) }

(5) So,

abs(a_n)*abs(c)ⁿ ≤ H(abs(c)^n-1 + ... + 1)

(6) Since (abs(c)-1)*(abs(c)^n-1 + ... + 1) = abs(c)ⁿ - 1 and since abs(c) is greater than 1, it follows that:

H(abs(c)^n-1 + ... + 1) = H[abs(c)ⁿ - 1]/[abs(c) - 1]

and therefore:

abs(a_n)*abs(c)ⁿ ≤ H[abs(c)ⁿ]/[abs(c) - 1]

(7) So, we can state:

abs(a_n) ≤ H/[abs(c) - 1]

(8) Since abs(a_n) and abs(c)-1 are positive, we can also state:

abs(c) - 1 ≤ H/abs(a_n)

and finally,

abs(c) ≤ 1 + H/abs(a_n)

QED

Sturm's Theorem: Sturm Chains

Before proceeding to Sturm's Theorem, let's talk about Sturm Chains.

Definition 1: Sturm Chain

A Sturm Chain is a series of polynomials P₀, ...., P_m

such that:

(1) For any value of i where i ≥ 1, if P_i(x)=0, then P_i-1(x) = -P_i+1(x)

(2) If P_i(x) = 0 then P_i+1(x) ≠ 0 and if i ≥ 1, then P_i-1(x) ≠ 0.

(3) For a sufficiently small area surrounding a zero point of P₀(x), P₁(x) is everywhere greater than 0 or everywhere smaller than 0.

Lemma 1: Constructibility of a Sturm Chain from a Polynomial

If a polynomial P is differentiable and has only simple roots, then it is possible to construct a Sturm Chain based on it.

Proof:

(1) Let P₀ be the polynomial and P₁ be the first derivative of P₀.

(2) We can now use an alternate form of Euclid's Algorithm for Greatest Common Divisor of Polynomials (see Theorem 1, here) to get the following:

P₀ = Q₁P₁ - P₂
...

P_m-2 = Q_m-1P_m-1 - P_m

where all deg P_i is less than deg P_i-1 and deg P_m = 0.

(3) Since P₀ is simple, we know that P₀ and P₁ are relatively prime [See Lemma 2, here].

(4) Therefore, P_m is degree 0. [See definition 3, here for the definition of relatively prime polynomials]

(5) Now, we can show that P₀, P₁, ..., P_m form a Sturm Chain. Here's the reasoning.

(6) Using step #2 above, we know for i, we have the following equation:

P_i-1 = Q_iP_i - P_i+1

(7) Assume that P_i(x) = 0.

(8) Then it follows that:

P_i-1 = -P_i+1

(9) Assume that P_i(x) = 0 and P_i+1(x) = 0

(10) Then it follows that P_i+2(x) = 0 and so on.

(12) So that P_m(x) = 0. But this is impossible since P_m is a constant (a polynomial of degree 0)

(13) Therefore, we reject our assumption in step #7 and conclude that P_i+1(x) ≠ 0.

(14) In the same way, we can show that if P_i(x) = 0, then it follows that P_i-1 ≠ 0.

(15) Finally, since P₀=P is a polynomial with only simple roots and P₁ = P' is its derivative, we know that (see Lemma, here), for a sufficiently small area surrounding a zero point of P₀(x), P₁(x) is everywhere negative or everywhere positive.

QED

References

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

Sturm's Theorem: An Initial Lemma

Here's a major problem that Jacques Charles Francois Sturm solved:

Find the number of real roots of an algebraic equation with real coefficients over a given interval.

There are two possible cases:

(I) The real roots of the equation in question are all simple over the given interval.

(II) The equation also possesses multiple real roots over the interval.

Before proceeding to his solution, let's start with a lemma.

Lemma:

Any equation of multiple real roots can be broken down into a set of equations with only simple real roots.

Proof:

(1) Let the F(x) = 0 have the distinct roots α, β, γ, ...

(2) Let the root α be a-fold, the root β be b-fold, γ c-fold, etc where a,b,c,... are not necessarily 1.

(3) Using the Fundamental Theorem of Algebra (see Thereom, here), we have:

F(x) = (x - α)^a(x - β)^b(x - γ)^c*...

(4) Using some basic properties of the derivative (see Corollary 2.2, here), we have:

F'(x)/F(x) = a/(x - α) + b/(x - β) + c/(x - γ) + ... =

= [a(x - β)(x - γ)(x - λ)*... + b(x - α)(x - γ)(x - λ)*... + ...]/[(x - α)(x - β)(x - γ)*...]

(5) Let:

p(x) = [a(x - β)(x - γ)(x - λ)*... + b(x - α)(x - γ)(x - λ)*... + ...]

q(x) = [(x - α)(x - β)(x - γ)*...]

so that we have:

F'(x)/F(x) = p(x)/q(x)

(6) We note that p(x) and q(x) do not have any common divisors since for each factor of q(x), we are left with a remainder of the form c/(x - d) where c,d are constants.

(7) Let G(x) = F(x)/q(x)

(8) Then:

F(x) = G(x)*q(x)

F'(x) = G(x)*p(x) [since F'(x)/F(x)=p(x)/q(x) → F'(x) = F(x)*p(x)/q(x) = G(x)*p(x)]

(9) Since p(x) and q(x) do not have any common divisors, it follows that G(x) is the greatest common divisor of F(x) and F'(x)

(10) Since we can always figure out the greatest common divisor of two equations (see Theorem 1, here for the greatest common divisor of polynomials), it follows that G(x) is obtainable based solely on F(x) and F'(x).

(11) Now since F(x)=G(x)*q(x), it follows that F(x)=0 divides into two equations:

G(x)=0

q(x)=0

(12) Since q(x) = (x - α)*(x - β)*(x - γ)*..., it is clear that it consists only of simple roots [from step#1]

(13) Since we can apply the same logic to G(x), it follows that we can always break down an equation of multiple real roots into a set of equations with only simple real roots.

QED

References

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

Jacques Charles Francois Sturm

Jacques Charles Francois Sturm was born on September 29, 1803 in Geneva, Switzerland. His father was a math teacher. When Sturm was 16, his father died and his family fell into a difficult financial situation.

At the Geneva Academy, Sturm's strong mathematical ability was recognized by his instructors. One of his teachers, Jean-Jacques Schaub, arranged financial support for young Sturm so he could attend school full time. At the Geneva Academy, Sturm met Daniel Colladon whose friendship and collaboration was an important part of his early work in mathematics.

When Sturm graduated from the Academy, he accepted a position as the tutor to Madame de Stael's youngest son in 1823. Madam de Stael had been a very successful and famous French writer who had died in 1817.

The family spent six months each year in Paris and Sturm was able to join them. Through the family, he was able to meet many of the intellectual luminaries of French society including Dominique Francois Jean Arago, Pierre-Simon Laplace, Simeon Denis Poisson, Jean Baptiste Joseph Fourier, Joseph Louis Gay-Lussac, and Andre Marie Ampere among others.

In 1824, Sturm and Colladon attempted to win a prize offered by the Paris Academy on the compressibility of water. The results were not as expected and Colladon severely injured his hand. They tried again in 1825. This time Sturm got a job tutoring Arago's son and was able to use Ampere's laboratory and received support and advice from Fourier. With all this new help, even if they did not win, they had made significant improvement from the previous year.

The next year, both Sturm and Colladon worked as assistants to Fourier. Additionally, they continued their experiments on the compressability of water and this time, they won the Grand Prix of the Academies de Sciences. The prize money was enough that they could stay in Paris and devote themselves to their research.

In 1829, Sturm published what would become one of his most famous papers: Mémoire sur la résolution des équations numériques. In it, he presented a major simplification of a method discovered by Cauchy to identify the number of real roots that an equation had over a specified interval. His method was largely based on methods from Fourier but the result was undeniably impressive. Her is Hermite's response:

Sturm's theorem had the good fortune of immediately becoming a classic and of finding a place in teaching that it will hold forever. His demonstration, which utilises only the most elementary considerations, is a rare example of simplicity and elegance.

Despite the well-received paper, Sturm had trouble finding work until the revolution of 1830. With the help of Arago, Sturm became professor of mathematics at the College Rollin. Three years later, he became a French citizen and three years after that he was admitted to the Academie des Sciences.

He would make significant contributions to differential equations relating to Poisson's theory of heat. Today, this work along with with the work done by Liouville form what is known as Sturm-Liouville Theory. Later in his career, he was professor at the Ecole Polytechnique. He made contributions to infinitesimal geometry, projective geometry, differential geometry, and geometric optics.

He died on December 18, 1855 in Paris.

References

• "Jacques Charles Francois Sturm", MacTutor
  

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 u^p = a for some prime p.

Let n ≥ 5

Let σ be the following permutation:

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

so that:

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

Let τ be the following permutation:

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

so that:

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

Then:

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

Proof:

(1) From the given, we have u^p = 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 = u^p

(4) Dividing both sides by u^p gives us:

[σ(u)/u]^p = 1

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

σ(u) = ζ_σu

where ζ_σ is some p-th root of unity.

(6) Applying σ to both sides gives us:

σ²(u) = ζ_σ²u

(7) Applying σ again we get:

σ³(u) = ζ_σ³u

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

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

(9) So, that ζ_σ³ = 1

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

τ(u) = ζ_τu

where ζ_τ is a pth root of unity and

ζ_τ³ = 1

(11) Putting these two results together gives us:

σ*τ(u) = σ(τ(u)) = ζ_σζ_τu

σ²*τ = σ²(τ(u)) = ζ_σ²ζ_τu

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

σ*τ:

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

σ²*τ:

x₁ → x₃ → x₄ → x₅ → x₂ → x₁ and x_i → x_i 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:

(ζ_σζ_τ)⁵ = (ζ_σ²ζ_τ)⁵ = 1

(15) Now, we also note that:

ζ_σ = ζ_σ⁶*ζ_σ^-5 = ( ζ_σ⁶*ζ_σ^-5)*(ζ_τ⁵*ζ_τ^-5)*(ζ_σ⁵*ζ_σ^-5) = ζ_σ⁶*(ζ_τ⁵*ζ_σ⁵)*(ζ_σ^-5*ζ_σ^-5*ζ_τ^-5) =

= ζ_σ⁶*(ζ_σζ_τ)⁵(ζ_σ²ζ_τ)^-5

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

ζ_σ = ζ_σ⁶*(ζ_σζ_τ)⁵[(ζ_σ²ζ_τ)⁵]^-1 = ζ_σ⁶*(1)*(1)^-1 = ζ_σ⁶

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

ζ_σ =(ζ_σ³)² = (1)² = 1

(18) Now, since (ζ_σζ_τ)⁵ =1, we have:

(1*ζ_τ)⁵ =1

so that:

ζ_τ⁵ = 1

(19) We can now show that ζ_τ = 1 since:
ζ_τ = ζ_τ⁶*ζ_τ^-5 = (ζ_τ³)²*(ζ_τ⁵)^-1

Taking ζ_τ³ = 1 (step #10) and ζ_τ⁵ = 1 (step #18), we get:

ζ_τ = (ζ_τ³)²*(ζ_τ⁵)^-1 = (1)²*(1)^-1 = 1

QED


Theorem 2: Ruffini's Theorem

Let P(X) = (X - x₁)*...*(X - x_n) = Xⁿ - s₁X^n-1 + ... + (-1)ⁿs_n = 0

where s_i are coefficients in field k.

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

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 ⊂ ... ⊂ F_m-1 ⊂ F_m = F

such that for each 0 ≤ i ≤ m:



where p_i is a prime and α_i ∈ F₀ = K

(4) Then, by induction, all elements F_i 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 = α_i, u = a^(1/p) so that:

F_i+1 = F_i(u)

and

a ∈ F_i

(c) But since a ∈ F_i, 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 F_i(u) = F_i+1 are also invariant under σ and τ.

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

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

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
• Jorg Bewersdorff, Galois Theory for Beginners, American Mathematical Society, 2006.
  

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

• "Pierre Laurent Wantzel", MacTutor