Fermat's Last Theorem

2020-05-05T01:58:00.001-07:00

Hi Everyone,

It's been almost 11 years since I worked on this blog! Wow.

I have recently started a new blog called All About Primes. If you are interested, come check it out. I will try to posts new content there at least once a week.

Cheers,

-Larry

Galois' Memoir: Corollaries to Proposition I

2009-10-12T01:28:00.001-07:00

Although Evariste Galois does not state it directly, there are important implications to his Proposition I which I will present.

The content that follows is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equation.

Here is the statement of Proposition I from Galois' Memoir without proof. For a proof, see here.

Proposition I:

Let f(x₁, ..., x_n) be a rational fraction in n indeterminates x₁, ..., x_n with coefficients in a field F.

For σ ∈ Gal(P/F):

f(σ(r₁), ..., σ(r_n)) is defined whenever f(r₁, ..., r_n) is defined.

and

f(r₁, ..., r_n) ∈ F if and only if f(σ(r₁),..., σ(r_n)) = f(r₁, ..., r_n) for all σ ∈ Gal(P/F).

Following from this proposition are the following corollaries.

Definition 1: Automorphism

An automorphism is a bijective homomorphism from a group to itself.

For details on the meaning of this definition (including the definition of bijective and the definition of a homomorphism, see here.)

Corollary 1.1:

Each permutation σ ∈ Gal(P/F) can be extended to an automorphism of F(r₁, ..., r_n) which leaves every element in F invariant by setting

σ(f(r₁, ..., r_n)) = f(σ(r₁), ..., σ(r_n)) for any rational fraction f(x₁, ..., x_n) for which
f(r₁, ..., r_n) is defined.

Proof:

(1) Assume that f(r₁, ..., r_n) is defined.

(2) Using Proposition I (see here), we know that:

f(σ(r₁), ..., σ(r_n)) ∈ F(r₁, ..., r_n) is defined.

(3) Let σ(f(r₁,..., r_n)) = f(σ(r₁), ..., σ(r_n))

(4) First, we need to show that σ(f(r₁, ..., r_n)) leaves every element in F invariant.

(5) Assume that f(r₁, ...., r_n) = g(r₁, ..., r_n) for two rational fractions f,g ∈ F(x₁, ..., x_n)

(6) Let us define a function H such that H(x₁, ..., x_n) = f(x₁, ..., x_n) - g(x₁, ..., x_n)

(7) By step #5 above, we know that H(r₁, ..., r_n) = 0

(8) So it follows that H(r₁, ..., r_n) ∈ F since 0 ∈ F.

(9) Using Proposition I (see here), we have:

H(r₁, ..., r_n) = H(σ(r₁), ..., σ(r_n))

(10) So:

f(σ(r₁), ... σ(r_n) ) - g(σ(r₁), ..., σ(r_n)) = 0

and further:

f(σ(r₁), ..., σ(r_n)) = g(σ(r₁), ..., σ(r_n))

(11) This proves that σ(f)=σ(g) = f(σ(r₁), ..., σ(r_n)) = g(σ(r₁), ..., σ(r_n)) if and only if
f(r₁, ..., r_n) = g(r₁, ..., r_n)

(12) To complete the proof, we show that σ is an automorphism [see Definition 1 above].

(13) We know it is bijective because σ is bijective.

(14) We know that it is a homomorphism since:

σ(f + g) = f(σ(r₁), ..., σ(r_n)) + g(σ(r₁), ..., σ(r_n)) = σ(f) + σ(g)

σ(fg) = f(σ(r₁), ..., σ(r_n))*g(σ(r₁), ..., σ(r_n)) = σ(f)*σ(g)

QED

Corollary 1.2:

The set Gal(P/F) does not depend on the choice of the Galois resolvent V.

Proof:

(1) Let V' ∈ F(r₁, ..., r_n) be another Galois resolvent (see Definition 2, here) for p(x)=0,

(2) Let π' be its minimum polynomial over F.

(3) Let f_i' ∈ F(x) for i=1,...,n be a rational fraction such that:

r_i = f_i'(V') for i = 1, ..., n

(4) Using Corollary 1.1 above and using step #3 above, we have:

σ(r_i) = f_i'(σ(V'))

since σ leaves every element in F invariant and since it maps each r_i to a unique value.

(5) Likewise, we can apply σ to both sides of:

π'(V') = 0

to get:

σ(π(V')) = π'(σ(V')) = σ(0) = 0

which gives us that:

π(σ(V')) = 0

(6) This shows that σ(V') is a root of π'.

(7) Thus we have shown that every element of Gal(P/F) as defined earlier with V is also an element of Gal(P/G) as defined with V'.

(8) We can also use the same argument to show the reverse which demonstrates that Gal(P/F) for V = Gal(P/F) for V'.

QED

Corollary 1.3:

Gal(P/F) is a subgroup of the group of all permutations of r₁, ..., r_n

Proof:

(1) Gal(P/F) is by definition a subset of all the permutations of r₁, ..., r_n. [see Definition 1, here, for definition of Gal(P/F)]

(2) So, to complete this proof (see Definition 6, here and Definition 2, here), I need to show that Gal(P/F) has closure, associativity, inversion, and identity on the operation of permutations.

(3) It has closure since for any element τ ∈ Gal(P/F), the composition τ * σ ∈ Gal(P/F) since:

(a) Let τ be the map: f_i(V_j) → f_i(V_k) for some k=1,..., m

(b) Let σ be the map: r_i → f_i(V_j)

(b) It follows that τ * σ : r_i → f_i(V_k)

(c) Therefore, τ * σ ∈ Gal(P/F)

(4) It has identity since the identity map is in Gal(P/F) is clear since the map is σ₁ in the definition of the Galois Group since r₁ = f₁(V₁), r₂ = f₂(V₁), ..., and so on.

(5) It has inversion since:

(a) Let σ ∈ Gal(P/F)

(b) By definition of Gal(P/F), σ(r_i) = f_i(V_j)

(c) We know that V_j is also a Resolvent of P(X)=0. [see Lemma 4, here]

(d) So, we can define Gal(P/F) according to V_j and it will be the same. [see Corollary 1.2 above]

(e) So, it follows that the map:

f_i(V_j) → f_i(V₁)

is an element of Gal(P/F)

(6) It has associativity since:

(a) Let's assume that we have three elements of Gal(P/F) so that we have:

σ, τ, φ

(b) Let's define the following maps:

φ: f_i(V_k) → f_i(V_l)

τ: f_i(V_j) → f_i(V_k)

σ: r_i → f_i(V_j)

(c) (φ * τ ) : f_i(V_j) → f_i(V_l)

(d) (φ * τ ) * σ: r_i → f_i(V_l)

(e) (τ * σ) : r_i → f_i(V_k)

(f) φ*(τ * σ): r_i → f_i(V_l)

QED

References

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

Galois' Memoir: Proposition 1 (Galois Group)

2009-10-06T23:45:00.000-07:00

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.

Definition 1: Galois Group Gal(P/F)

Let r_i, ..., r_n be the roots of a polynomial P with coefficients in F

Let V₁, ..., V_m be the roots of the minimum polynomial for a given Galois Resolvent where the Galois Resolvent V = V₁

Let σ_j be a map: σ_j: r_i → f_i(V_j) for i=1, ..., n

The set of all possible permutations { σ₁, ..., σ_m } is called the Galois group of P(X)=0 over F and is denoted Gal(P/F).

Theorem: Galois' Proposition I

Let an equation be given whose m roots are a, b, c, ...

These will always be a group of permutations of the letters a,b,c,... which will have the following property:

1. That each function invariant under the substitution of this group will be known rationally

2. Conversely, that every function of the roots which can be determined rationally will be invariant under these substitutions

In other words:

Let f(x₁, ..., x_n) be a rational fraction in n indeterminates x₁, ..., x_n with coefficients in F.

Then:

f(r₁, ..., r_n) ∈ F if and only if for all σ ∈ Gal(P/F),

f(r₁, ..., r_n) = f(σ(r₁), ..., σ(r_n))

Proof:

(1) Let f = φ/ψ where φ, ψ ∈ F[x₁, ..., x_n]

(2) First, we need to prove that f is a rational function which comes down to showing that:

ψ(x₁, ..., x_n) ≠ 0 → ψ(σ(x₁), ..., σ(x_n)) ≠ 0 where σ ∈ Gal(P/F)

(3) We can show this since:

(a) Using Lemma 3 before (see Lemma 3, here), we know that there exists functions f₁, ..., f_n such that:

ψ(r₁, ..., r_n) = ψ(f₁(V), ..., f_n(V))

where V is the Galois Resolvent (see Definition 1, here) and f_i ∈ F[X]

(b) But since each f_i ∈ F[X] and ψ ∈ F[X], it follows that there exists a function g ∈ F[X] such that:

g(V) = ψ(f₁(V), ..., f_n(V))

(c) Let σ be a permutation in Gal(P/F)

(d) From Lemma 4 before (see Lemma 4, here), we know that each permutation corresponds to a root of the minimal irreducible polynomial of the Galois Resolvent so that there exists V' such that:

V' is a root of the minimal irreducible polynomial of the Galois Resolvent

and

σ: r_i → f_i(V')

(e) So that we have:

ψ(σ(r₁), ..., σ(r_n)) = ψ(f₁(V'), ..., f_n(V')) = g(V')

(f) It is clear that if ψ(σ(r₁), ..., σ(r_n)) = 0, then g(V')= 0

(g) So that when ψ is 0, V' is a root.

(h) Using Theorem 1, here, it clear that:

g(V') =0 if and only if g(V) is 0.

(i) So further, ψ(σ(r₁), ..., σ(r_n)) = 0 if and only if ψ(r₁, ..., r_n) = 0. [from step #3a above]

(4) This shows that f(σ(r₁), ..., σ(r_n)) is defined when f(r₁, ..., r_n) is defined.

(5) Assume that f(r₁, ..., r_n) ∈ F

(6) From step #3a above, we have:

f(r₁, ..., r_n) = f(f₁(V), ..., f_n(V))

(7) But since each f_i ∈ F[X] and f ∈ F[X], it follows that there exists a function h ∈ F[X] such that:

h(X) = f(f₁(X), ..., f_n(X))

(8) So that:

h(V) = f(f₁(V), ..., f_n(V))

and:

h(X) - f(f₁(X), ..., f_n(X)) = 0

(9) If we define a function H such that:

H(X) = h(X) - f(f₁(X), ..., f_n(X))

It is clear that H(V)=0

(10) Using Theorem 1, here, it follows that all m roots of the minimal irreducible polynomial of the Galois Resolvent are also roots of H so that it's roots include: V₁, ...., V_m

(11) So it follows for all j = 1, ..., m that:

h(V_j) = f(f₁(V_j), ..., f_n(V_j))

(12) Now since f(r₁, ..., r_n) ∈ F [by our assumption in step #5], it follows that we can define a function G such that:

G(X) = h(X) - f(r₁, ..., r_n)

(13) From step #6 above, we note that V is a root of G(X) and by the same logic as step #10, we conclude that all V₁, ..., V_m are also roots of G(X).

(14) So that we have for all j= 1, ..., m:

h(V_j) = f(r₁, ..., r_n)

(15) But if this is true for all V_j, then it follows that it is true for all σ ∈ Gal(F/G) [by Lemma 4, here] so that we have:

f(σ_j(r₁, ..., σ_j(r_n)) = f(r₁, ..., r_n) for j = 1, ..., m

(16) Assume that f(σ_j(r₁, ..., σ_j(r_n)) = f(r₁, ..., r_n) for j = 1, ..., m

(17) From step #3a above, we know that:

f(r₁, ..., r_n) = f(f₁(V), f₂(V), ..., f_n(V)) where f, f_i ∈ F[X]

(18) So, we can define a function g(x) such that:

g(x) = f(f₁(x),f₂(x), ..., f_n(x))

so that we have:

g(V) = f(r₁, ..., r_n)

(19) Since for all j, we have f(σ_j(r₁, ..., σ_j(r_n)) = f(r₁, ..., r_n), it follows that:

for all j = 1..m, g(V_j) = f(r₁, ..., r_n)

(20) Further we note that:

m*f(r₁, ..., r_n) = g(V₁) + g(V₂) + ... + g(V_m)

so that we have:

f(r₁, ..., r_n) = (1/m)*(g(V₁) + ... + g(V_m))

(21) We can define the following function H such that:

H(x₁, ..., x_m) = (1/m)*g(x₁) + ... + g(x_m)

(22) H is clearly a symmetric function so it can be expressed as a rational fraction in the elementary symmetric polynomials s₁, ..., s_m [see Theorem 6, here]

(23) Now the elementary symmetric polynomials correspond to the coefficients of a polynomial. [see Theorem 1, here]

(24) This shows that if we substitute V₁, ..., V_m for x₁, ..., x_m the elementary symmetric polynomial will correspond to the coefficients of the minimal irreducible polynomial for the Galois Resolvent whose coefficients are in F.

(25) Therefore H(V₁, ..., V_m) ∈ F

(26) But H(V₁, ..., V_m) = (1/m)*[g(V₁) + g(V₂) +... + g(V_m)] = f(r₁, ..., r_n)

(26) So we have shown that f(r₁, ..., r_n) ∈ F

QED

References

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

Galois' Memoir: Lemma 4

2009-10-05T02:05:00.000-07:00

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 4:

Suppose one has formed the equation for V and that one has taken one of its irreducible factors so that V is the root of an irreducible equation.

Let V, V', V'', ... be the roots of this irreducible equation .

If a = f(V) is one of the roots of the given equation, f(V') will also be a root of the given equation.

In fact, we can show that:

r₁ = f₁(V), r₂ = f₂(V), ..., r_n = f_n(V)

Then it follows that for any root V' or V'' or ... the complete set of distinct roots is:

f₁(V'), f₂(V'), ..., f_n(V')

Proof:

(1) Let V be a Galois Resolvent of P(X) = 0. [see Lemma 2, here]

(2) Let r₁, ..., r_n be the roots of P(X).

(3) We know that for each r_i, there exists f_i [see Lemma 3, here] such that:

r_i = f_i(V)

with f_i(X) ∈ F(X)

(4) Let V₁ = V, V₂, ..., V_m be the roots of the minimum polynomial of V over F [see Theorem 1, here] which are in F(r₁, ..., r_n) [see Corollary 1.1, here]

(5) Since r_i = f_i(V), we have P(f_i(V)) = 0.

(6) If we view P(f_i(X)) as a polynomial, then, we have P(f_i(V_j)) = 0 for j = 1, ..., m since P(f_i(X)) and the minimal polynomial of V over F share at least one root V. [see Theorem, here]

(7) This shows that each f_i(V_j) must correspond to a root r_i since P(X)=0 if and only if X is a root.

(8) Now, I will end this proof by showing that for any root V_i, that the n roots are:

f₁(V_i), f₂(V_i), ..., f_n(V_i)

(9) Now we know that for V₁, f₁(V₁) ≠ f₂(V₁) ≠ ... ≠ f_n(V₁) [from step #3 above]

(10) Assume that f_u(V_i) = f_v(V_i)

where i ≠ 1

(11) Then V_j is a root of the polynomial f_u - f_v.

(12) But then all V₁, ..., V_m must likewise be roots of f_u - f_v. [see see Theorem, here]

(13) But this also means that V₁ is a root of f_u - f_v

(14) So it follows that f_u(V₁) = f_v(V₁)

(15) But from step #9 above this is only possible if u = v.

(16) Hence, we have shown that for any V_i,

f₁(V_i), ..., f_n(V_i) must be the n distinct roots [since they cannot be equal and each one is equal to a root.]

QED

References

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

Galois' Memoir: Lemma 4: Preliminary Results (Existence of Minimum Polynomial)

2009-10-04T12:30:00.001-07:00

Before jumping into Lemma 4 by Evariste Galois, I will prove the existence of minimum polynomials.

The theorem in today's blog comes from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Definition 1: Minimum Polynomial of u over F

For a nonzero polynomials in F[X] which have u as a root, the minimum polynomial is the polynomial of least degree that divides all other nonzero polynomials and is unique, monic, and irreducible.

Theorem 1: Existence of Minimum Polynomials

(a) Every element u ∈ F(r₁, ..., r_n) has a polynomial expression in r₁, ..., r_n such that:

u = φ(r₁, ..., r_n)

for some polynomial φ ∈ F[x₁, ..., x_n]

(b) For every element u ∈ F(r₁, ..., r_n), there is a unique monic irreducible polynomial π ∈ F[X] such that π(u) = 0.

This polynomial π splits into a product of linear factors over F(r₁, ..., r_n)

Proof:

(1) Let u ∈ F(r₁, ..., r_n) be such that:

u = φ(r₁, ..., r_n)

for some polynomial φ ∈ F[x₁, ..., x_n]

(2) Let Θ(X,x₁, ..., x_n) = ∏(σ) [X - φ(σ(x₁), ..., σ(x_n))]

where σ runs over the set of permutations of x₁, ..., x_n

(3) Since Θ is symmetric in x₁, ..., x_n, we can write Θ as a polynomial in X and the elementary symmetric polynomials s₁, ..., s_n in x₁, ..., x_n [see Theorem 4, here]

(4) Let Θ(X,x₁, ..., x_n) = Ψ(X,s₁, ..., s_n)

for some polynomial Ψ with coefficients in F.

(5) Substituting r₁, ..., r_n into Θ, we get:

Θ(X,r₁, ..., r_n)= Ψ(X,a₁, ..., a_n) ∈ F[X]

where a_i are the coefficients of the polynomial where r₁, ..., r_n are the roots. [see Theorem 1, here]

(6) From the definition of Θ in step #2 above, we have:

Θ(u,r₁, ..., r_n) = 0

since in the permutation σ that leaves x_i unchanged, the result will be 0.

(7) It follows that Ψ(X,a₁, ..., a_n) is a polynomial in F[X] which has u as a root. [see step #5 above]

(8) We also note that Θ(X,r₁, ..., r_n) is a product of linear factors. [see step #2 above]

(9) Thus, we have shown that Ψ(X,a₁,...,a_n) splits into a product of linear factors over F(r₁, ..., r_n) [see step #4 above]

(10) We know that we can break down Ψ(X,a₁, ..., a_n) into a product of monic irreducible factors of Ψ [see Theorem 3, here] so that:

Ψ = cP₁*...*P_r

(11) Since u is a root, it follows that u must be the root of one of these monic irreducible factors.

(12) So that there exists i such that:

P_i is a monic, irreducible polynomial over F[X] and u is a root of P_i and P_i divides Ψ.

(13) Since Ψ splits into a product of linear factors over F(r₁, ..., r_n), it follows that P_i must also split into a product of linear factors over F(r₁, ..., r_n).

(14) Finally, we note that there is only one monic irreducible polynomial P_i which has u as a root since:

(a) Assume that Q ∈ F[X] is another polynomial with the same properties.

(b) First, we note that P must divide Q. [see Lemma 2, here]

(c) But, by the same argument Q must divide P.

(d) So, then it follows that P=Q.

(15) This completes part(b) of the proof.

(16) Let V be the Galois Resolvent (see Definition 1, here) such that:

V ∈ F(r₁, ..., r_n)

(17) We know that r₁, ..., r_n are rational fractions in V. [see Lemma 3, here]

(18) Since u is a rational fraction of F(r₁, ..., r_n), it follows that u is also rational fraction in V and further u ∈ F(V).

(19) So, u can be expressed as a polynomial in V and:

u = Q(V)

for some polynomial Q ∈ F[X].

(20) Substituting f(r₁, ..., r_n) for V, we obtain:

u = Q(f(r₁, ..., r_n))

(21) This is a polynomial expression in r₁, ..., r_n since Q and f are polynomials.

(22) This completes part(a) of the proof.

QED

Corollary 1.1:

Let V be any Galois Resolvent of P(X) = 0 over F and let V₁, ..., V_m be the roots of its minimum polynomial over F (among which V lies)

Then:

F(r₁, ..., r_n) = F(V) = F(V₁, ..., V_m)

Proof:

(1) r₁, ..., r_n are rational fractions in V [see Lemma 3, here]

(2) So, we have:

F(r₁, ..., r_n) ⊂ F(V)

(3) We know that the roots V₁, ..., V_m of the minimum polynomial of V are in F(r₁, ..., r_n) since:

(a) V is a polynomial g(x₁, ..., x_n) ∈ F[X] [see Definition 1, here]

(b) Applying part(b) of Theorem 1 above, we note that each V_i has g(V_i)=0 and can be expressed over linear factors of F(r₁, ..., r_n)

(4) So, we have:

F(V₁, ..., V_m) ⊂ F(r₁, ..., r_n)

(5) Since V₁, ..., V_m are roots of V, it follows that:

F(V) ⊂ F(V₁, ..., V_m)

(6) Therefore we have:

F(r₁, ..., r_n) = F(V) = F(V₁, ..., V_m)

QED

References

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

Galois' Memoir: Lemma 3

2009-10-02T08:54:00.000-07:00

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 3:

When a function V is chosen as satisfies Lemma 2 (see here), it will have the property that all the roots of the given equation can be expressed as rational functions of V.

That is:

Let P be a function of degree n with the roots: r₁, ..., r_n where each root is distinct.

Let V(r₁, ..., r_n) be the Galois resolvent where each distinct permutation has a distinct value.

Then for all roots r_i ∈ F(V).

Proof:

(1) Let V be a Galois Resolvent [see Definition 2, here]

(2) Let r₁ be any root of an equation P

Since r₁ is any root, the proof is done if we can show that r_i ∈ F(V).

(3) Let:

g(x₁, ..., x_n) = ∏ (for each permutation σ) [V - f(x₁, σ(x₂), ..., σ(x_n)) ] ∈ F(V)[x₁, ..., x_n]

where σ runs over all permutations of x₂, ..., x_n

(4) Since g is symmetric in x₂, ..., x_n, g can be written as a polynomial in x₁ and the elementary polynomials s₁, ..., s_n-1. (see Lemma, here)

(5) Let:

g(x₁, x₂, ..., x_n) = h(x₁, ..., s_n-1)

for some polynomial h with coefficients in F(V).

(6) Substituting in various ways the roots r₁, ..., r_n of P for the indeterminates x₁, ..., x_n which has the effect of substituting a₁, ..., a_n-1 ∈ F for s₁, ..., s_n-1 [see Theorem 1, here, for details on the mapping between elementary symmetric polynomials and the coefficents of a polynomial], we obtain:

g(r₁, r₂, ..., r_n) = h(r₁, a₁, ..., a_n-1)

and generalizing this, we get:

g(r_i, r₁, ...., r_i-1, r_i+1, ..., r_n) = h(r_i, a₁, ..., a_n-1)

(7) Since V is a Galois Resolvent for a function f such that V = f(r₁, ..., r_n), we have:

V ≠ f(r_i, σ(r₁), σ(r₂), ..., σ(r_i-1), σ(r_i+1), ..., σ(r_n))

for i ≠ 1 and for any permutation σ of { r₁, ..., r_i-1, r_i+1, ..., r_n }. [see Lemma 2, here]

(8) Therefore:

g(r_i, r₁, r₂, ..., r_i-1, r_i+1, ...., r_n) ≠ 0 for i ≠ 1.

(9) On the other hand, the definition of g and V show that [see definition of g in step #3 above]:

g(r₁, ..., r_n) = 0

(10) From step #9 above and step #6 above, we know that:

h(X,a₁, ..., a_n-1) ∈ F(V)[X]

vanishes for X = r₁ but not for X = r_i with i ≠ 1.

(11) Therefore, it is divisible by X - r₁ but not by X - r_i for i ≠ 1. [This follows from Girard's Theorem, see here]

(12) We then consider the monic greatest common divisor D(X) of P(X) and h(X,a₁, ..., a_n-1) in F(V)[X]. [see Theorem 1, here for proof of the existence of a greatest common divisor for polynomials]

(13) Since P(X) = (X - r₁)*...*(X - r_n), we know that X - r₁ divides P(X).

(14) From step #11 above, we know that X - r₁ divides h.

(15) Since x - r₁ divides both P(X) and h(X,a₁, ..., a_n-1), it divides D.

(15) On the other hand, h(X,a₁, ..., a_n-1) is not divisible by X - r_i for i ≠ 1. [see step #10 above]

(16) Hence, D has no other factor than X - r₁.

(17) Thus, D = X - r₁ whence r₁ ∈ F(V) since D ∈ F(V)[X].

QED

References

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

Galois' Memoir: Lemma 2 (Galois Resolvent)

2009-09-28T23:16:00.000-07:00

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.

Definition 1: Galois Resolvent Function

For any equation f(x) with distinct roots, the Galois Resolvent Function is a function g(x₁, ..., x_n) of the roots that no matter how the roots are permuted on the function, no two of the values are equal.

Definition 2: Galois Resolvent

The Galois Resolvent is a value of the Galois Resolvent Function where the roots of the equation f(x) are passed in as parameters.

Lemma 2: Galois Resolvent Function Exists

Given any equation f(x) with distinct roots a,b,c,... one can always form a function V of the roots such that no two of the values one obtains by permuting the roots in this function are equal.

For example, one can take:

V = Aa + Bb + Cc + ...

A, B, C, ... being suitably chosen whole numbers.

Proof:

(1) Let the n distinct roots of f(x) be denoted a,b,c, ...

(2) Since these roots are distinct, the discriminant (a - b)²(a - c)²(b - c)²*... = D is not zero [For review of the discriminant, see here].

(3) What needs to be shown is that n integers A,B,C, ... can be chosen so that the n! numbers AS(a) + BS(b) + CS(c) + ... + are distinct where S ranges over all n! permutations of the roots a,b,c,...

[for details on why count(n permutations) = n!, see Corollary 1.1, here]

(4) Let P be the product of the squares of the differences of these n! numbers that is:

P = ∏ (S,T) [ A(S(a) - T(a)) + B(S(b) - T(b)) + ... ]²

where the product is all over n!(n! - 1)/2 pairs (unordered) of permutations S and T in which S≠ T.

Note: The purpose of this equation is to verify that all n! numbers are distinct. P is the product of all possible differences between any two permutations.

We know that there are n! possible permutations (see step #3 above).

We pick one of these permutations (1 out of n!) and call it S. Then, we pick a second one (1 out of n! - 1) and call it T. Since ordering doesn't matter and there are two ways to pick the same combination, we only need to deal with n!*(n!-1)/2 comparisons between S and T.

(5) To complete the proof, we only need to show that we can pick A,B,C, ... etc. such that P is nonzero.

If any of the permutations are not distinct, then the difference between S and T will be 0. If any of the differences are 0, then P = 0. So if P ≠ 0, it follows that we have found values for A,B,C,... such that all permutations are distinct.

(6) Let A, B, C, ..., be regarded at first as variables.

(7) Then P is a polynomial in variables A, B, C,... whose coefficients are polynomials in the roots a,b,c,...

(8) P is symmetric in the roots (this follows directly from the definition of P and the definition of symmetric polynomials, see Definition 1, here).

(9) Since P is symmetric, the coefficients are symmetric in the roots so using Waring's Method (see Theorem 4, here), P is a polynomial in A, B, C, ... with the coefficients symmetric in roots.

(10) So, we can determine the coefficients of P based on the elementary symmetric polynomials using the roots [see Theorem 1, here].

(11) We can therefore assume that the coefficients are known since we are assuming that the roots are known.

(12) Since P is a product of nonzero polynomials, it is nonzero. [see Theorem, here]

(13) Therefore once can assign integer values to A,B,C,... in such a way as to make P ≠ 0 [see Theorem, here].

QED

References

Harold M. Edwards, Galois Theory, Springer, 1984

Galois' Memoir: Lemma 1

2009-09-28T02:11:00.000-07:00

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

2009-09-27T22:55:00.000-07:00

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

2009-09-26T20:44:00.001-07:00

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

2009-09-25T23:26:00.001-07:00

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

2009-09-25T12:49:00.000-07:00

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

2009-09-24T21:32:00.001-07:00

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

2009-09-13T02:57:00.000-07:00

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

2009-09-09T07:49:00.000-07:00

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

Edward Waring

2009-09-03T23:35:00.001-07:00

Edward Waring, despite not being very well known even today, was one of the most talented mathematicians of his time. He was cursed by the inability to properly communicate his ideas coupled with a fascination for esoteric mathematical topics. One of his biographers wrote:

Waring was one of the profoundest mathematicians of the eighteenth century; but the inelegance and obscurity of his writings prevented him from obtaining that reputation to which he was entitled.

Waring was born in Shropshire, England in 1736. His father was a successful farmer and he attended school in Shrewsbury. In 1753, he entered Magdalene College at Cambridge. Originally, he entered as a sizar which meant that he paid a reduced admission fee but had to take on extra duties at the school.

His mathematical abilities soon drew the attention of his teachers. He graduated with top honors in 1757. One year after graduation, he was elected as a fellow to Magdalene College. In 1759, his name was put forward as the Cambridge Lucasian Chair of mathematics even though he was only two years past graduation.

William Powell, one of the professors at St. John's College challenged this nomination. Powell wrote a pamphlet questioning Waring's mathematical knowledge. Waring responded to this with his own pamphlet and Powell wrote a rebuttal. The debate finally ended when a famous mathematician of this time, John Wilson, intervened on Waring's behalf. In 1760, Edward Waring became Lucasian Professor of Mathematics. He had not yet turned 24.

In 1762, Waring published his most famous work: Meditationes Algebraicae. The work shows his thoughts on topics in equations, number theory, and geometry. The work was well received and he was elected to the Royal Society in 1763. He would later extend this work into three separate volumes.

Despite the book's high praise by many top mathematicians, the book was not widely read among mathematicians. In 1764, an influential math book claimed that there were no first rate mathematicians in England. Waring was alarmed at being overlooked but admitted:

... never could hear of any reader in England, out of Cambridge, who took pains to read and understand it ...

Even though he was the Lucasian Professor of Mathematics, he decided to also study medicine. He received a medical degree in 1767. His medical career did not go as well as he had hoped and he gave up medicine by 1770. In 1776, he married Mary Oswell.

After quitting medicine, he expanded his original mathematical work: releasing a volume on geometry in 1770 and a volume on number theory and equations in 1772.. In these works, he did significant work with symmetric functions and also the cyclotomic equation. His ideas were a precursor to what later became group theory. In number theory, he presented a problem that was later solved by David Hilbert in 1909.

Despite being the Lucasian Professor, he did not lecture much. In fact, he did not correspond very often with the mathematicians of his day. His works were not systematic and most of his ideas were hundreds of years ahead of their time.

As he grew older, he struggled with poverty. In 1795, three years before his death, he resigned from the Royal Society because he could not afford its dues. He died on August 15, 1798.

References

"Biography of Edward Waring", O'Connor, J. J. & Robertson, E. F., MacTutor
"Edward Waring," Wikipedia.org
"Edward Waring", LucasianChair.org

Girard's Theorem

2009-08-29T00:09:00.000-07:00

The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations. Although Albert Girard was the first to propose this theorem, he never was able to provide a correct proof. The proof presented today is based on the work done by Leopold Kronecker.

For review of polynomials, see here. For review of irreducible polynomials, see here. For review of fields, see here. For review of rings, see here. For review of ideals, see here.

Lemma 1:

Let P be a polynomial ∈ F[X] where F is a field.

Let (P) be the set of multiples of P such that (P) = {PQ where Q ∈ F[X]}

Let P be irreducible over F[X]

Then:

F[X]/(P) is a field containing F and a root of P.

Proof:

(1) Since (P) is an ideal of F[X] (see Lemma 1, here), it follows that F[X]/(P) is a ring (see Lemma 3, here)

(2) To prove that F[X]/(P) is a field, we need to show that every nonzero element Q + (P) in F[X]/(P) is invertible and that F[X]/P has unity. [See Definition 3, here for definition of a field]

(3) Since 1 ∈ F[X], it follows that 1 + (P) ∈ F[X]/P and this shows that F[X]/P has unity since (see here for review of the operations of a quotient ring):

[x + (P)]*[1 + (P)] = (x*1) + (P) = x + (P)

(4) Assume that Q + (P) in F[X]/(P) is a nonzero element.

(5) It follows that Q is not in (P) [since if it was Q + (P) = 0 + (P), see Lemma 2, here, but by assumption Q + (P) is nonzero]

(6) So, Q is not divisible by P. [From the definition of (P)]

(7) Since P is irreducible over F[X], P,Q are relatively prime. [see Definition 1, here]

(8) Therefore, there exists P₁, Q₁ ∈ F[X] such that (See Corollary 3.1, here):

PP₁ + QQ₁ = 1

(9) Since -PP₁ = QQ₁ - 1, it follows that:

P divides QQ₁ - 1

(10) Since P divides QQ₁ - 1, it follows that:

QQ₁ - 1 ∈ (P) [See Definition of (P) above]

(11) But then [see here]:

QQ₁ + (P) = 1 + (P)

(10) From a previous result (see Lemma 1, here), this shows that:

QQ₁ + (P) = 1 + (P)

(11) Since (Q + (P))*(Q₁ + (P)) = QQ₁ + (P) [see here for review of the operations of a quotient ring], it follows that:

(Q+(P))*(Q₁+(P)) = 1 + (P)

(12) This shows that Q₁ + (P) is the inverse of Q + (P) in F[X]/(P)

(13) Assume that x ∈ F and x is nonzero

(14) Then, x + (P) is nonzero since no nonzero is divisible by P. [See here for review of polynomials and divisibility]

(15) Thus, F is a subfield of F[X]/(P) since there is a clear mapping from F[X] to F[X]/(P).

(16) Using the operations defined for quotient rings, we get:

P(X + (P)) = P(X) + (P) since:

(a) Let P(X) = aXⁿ + bX^n-1 + ... + cX + d

(b) P(X + (P)) = a(X + (P))ⁿ + b(X + (P))^n-1 + ... + c(X + (P)) + d =

= a[Xⁿ + (P)] + b[X^n-1 + (P)] + ... + cX + (P) + d =

= (aXⁿ + bX^n-1 + ... + cX + d) + (P)

(17) Since P(X) ∈ (P), it follows that P(X) + (P) = 0 + (P) [See Lemma 2, here]

(18) Combining step #16 and step #17, gives us:

P(X + (P)) = 0

(19) This shows that X+(P) is a root of P in F[X]/(P).

QED

Theorem: Girard's Theorem

For any nonconstant polynomial P ∈ F[X], there is a field K containing F such that P splits over K into a product of linear factors:

P = a(X - x₁)*...*(X - x_n) in K[X]

Proof:

(1) Let P be a polynomial such that P ∈ a field F.

(2) By the nature of polynomials (see Theorem 3, here), we can break up P into a product of irreducible factors in F[X] such that:

P = P₁*...*P_r

(3) Let s be the number of linear factors in P so that s is between 0 and r.

(4) If (deg P) - s = 0, then each of the factors P₁, ..., P_r is linear and K = F.

(5) If (deg P) - s ≥ 1, then at least one of the factors P₁, ..., P_r has degreee greater than or equal to 2.

(6) Assume that deg P₁ ≥ 2.

(7) Let F₁ = F[X]/(P₁)

(8) From Lemma 1 above, we know that P₁ has a root in F₁

(9) So, we can decompose P₁ over F₁ with at least one linear factor. [See Theorem, here]

(10) The decomposition of P into irreducible factors over F₁ is then at least s+1 .

(11) We can repeat this same sequence for all factors of P₁ and all factors of P.

(12) In this way, we can our field using Lemma 1 above and find a field K where P can be decomposed into linear factors.

QED

References

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

Leopold Kronecker

2009-08-01T16:46:00.000-07:00

Leopold Kronecker was born on December 7, 1823 in the city of Liegnitz which is today part of Poland. At the time of his birth, it was part of the Kingdom of Prussia. His father was a wealthy businessman and his mother had come from a wealthy family. As he was growing up, his education was handled by private tutors.

He entered the Gymnasium at Liegnitz. There, he grew interested in mathematics after attending lectures by the mathematician Ernst Kummer.

In 1841, he enrolled at the Berlin University. There, he studied under Johann Dirichlet and Jakob Steiner. He also became quite interested in the philosophies of Rene Descartes, Wilhelm Leibniz, Benedict Spinoza, and Georg Hegel. He attended one semester at the University of Bonn to study astronomy and one year at the University of Breslau to study under Kummer who had recently been appointed the chair of mathematics.

His doctoral thesis was on algebraic number theory and was very well received. He soon became friends with the mathematicians Carl Gustav Jacobi and Ferdinand Eisenstein. These mathematicians would have a great effect on his thinking about mathematics.

In 1848, he married the daughter of his uncle. He helped to manage the family estate and by this time, he had come to his share of the family fortune. He studied mathematics solely for his own enjoyment.

In 1855, he returned to Berlin in order to continue his work in mathematics among the top mathematicians of his day. Kummer had recently transfered to Berlin to take over a position left open by Dirichlet. Carl Borchardt was also in Berlin at this time as he had recently become the editor of Crelle's Journal. Karl Weierstrass came to Berlin in 1856.

Despite the fact that Kronecker was not a professor, he still wrote numerous well-received mathematical papers. His topics included number theory, elliptic functions, algebra, the theory of determinants, the theory of integrals, and the interrelations between these topics. In 1861, he was elected to the Berlin Academy.

Even though he was not a professor, being a member of the Berlin Academy entitled him to lecture at the university. His lectures were hard to follow and not very popular with the students. Still, between his papers and his lectures, his mathematical reputation shined. He was offered the mathematics chair of the University of Gottingen which he declined because he prefered to stay in Berlin. He was elected as a member of the Paris Academy and in 1883, he became math chair of the University of Berlin. In 1884, he was elected a member of the Royal Society of London.

Kronecker had long had certain radical views on the nature of mathematics. He once said:

God created the integers, all else is the work of man.

By this, he meant that in his view, mathematics should deal only with finite numbers and functions that involved a finite number of operations on those numbers. He did not approve the use of irrational numbers, upper and lower limits, transcendental numbers or any other concept which could not be derived in a finite way. Kronecker opposed the publication of Heinrich Heine's work on trigonometric series and the set theory work done by Georg Cantor. This had significance since Kronecker was on the editorial staff of Crelle's Journal and later, in 1880, became editor of the influential math journal. In 1883, he became one of the codirectors of the mathematical seminar in Berlin.

He made his views public when criticized the theory of irrational numbers in 1886:

... the introduction of various concepts by the help of which it has frequently been attempted in recent times (but first by Heine) to conceive and establish the "irrationals" in general. Even the concept of an infinite series, for example one which increases according to definite powers of variables, is in my opinion only permissible with the reservation that in every special case, on the basis of the arithmetic laws of constructing terms (or coefficients), ... certain assumptions must be shown to hold which are applicable to the series like finite expressions, and which thus make the extension beyond the concept of a finite series really unnecessary.

Kronecker was easily offended and often broke contact with mathematicians who's ideas he did not agree with. For example, he did not get along with Weierstrass, Dedekind, and Cantor. When he became math chair of the University of Berlin, Weierstrass planned to move to Switzerland but changed his mind when he decided that someone needed to oppose the views of Kronecker.

Kronecker died on December 29, 1891. The majority of mathematicians of his day accepted the theory of irrational numbers. Today, his strong rejection of the work by Cantor and Dedekind seems quite eccentric. Still, it is important to remember that his thoughts had great impact on Jules Poincare and Luitzen Brouwer in their work on Intuitionism which exerts a strong influence to this day.

References

"Leopold Kronecker", MathTutor

Sturm's Theorem: Examples

2009-02-07T22:48:00.000-08:00

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

2009-02-06T23:32:00.000-08:00

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

2009-02-03T09:07:00.000-08:00

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

2009-01-17T21:27:00.000-08:00

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

2008-10-29T22:20:00.000-07:00

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

2008-10-27T17:21:00.000-07:00

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

2008-10-25T16:42:00.000-07:00

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.