Kronecker's Theorem: Some Lemmas on Irreducible Polynomials

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

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

Definition 1: algebraically soluble

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

Lemma 1:

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

Proof:

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

f(x) = xⁿ + a₁x^n-1 + ... + a_n

and

f(α) = αⁿ + a₁α^n-1 + ... + a_n = 0

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

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

ζ = Ψ(α)/Φ(α)

(4) It follows from step #1 that:

αⁿ = -a₁α^n-1 - ... - a_n

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

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

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

u(x)Φ(x) + v(x)f(x) = 1.

(8) If we set x = α, then we get:

u(α)Φ(α) + v(α)(0) = u(α)Φ(α) = 1.

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

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

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

ζ = c₀ + c₁α + ... c_n-1α^n-1

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

(12) Assume that:

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

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

d₀ + d₁α + ... d_n-1α^n-1 = 0

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

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

QED

Lemma 2:

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

Proof:

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

f(x) = x^p + a₁x^p-1 + ... + a_p

and p is both odd and prime.

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

and

g(α) = 0

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

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

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

u(x) = f(r) - ψ(r,x)φ(r,x)

where r is some rational number.

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

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

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

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

f(x) - ψ(x,α)φ(x,α) = 0

(9) So that:

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

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

f(x) - ψ(x,α')φ(x,α') = 0

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

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

f(x) = ψ(x,α')φ(x,α')

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

f(x)^q = Ψ(x)Φ(x) where:

Ψ(x) = ψ(x,α)*ψ(x,α')*ψ(x,α'')...

and

Φ(x) = φ(x,α)*φ(x,α')*φ(x,α'')...

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

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

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

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

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

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

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

(a) f(x)^q = Ψ(x)Φ(x)

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

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

Ψ(x) = f(x)^ah(x)

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

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

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

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

(20) So, there exists μ, ν such that:

Ψ(x) = f(x)^μ

and

Φ(x) = f(x)^ν

and

μ + ν = q

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

mq = μp

and

nq = νp

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

QED

References

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

Waring's Method

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

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

Definition 1: Symmetric Polynomial

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

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

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

Definition 2: Symmetric Rational Fraction

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

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

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

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

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

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

For example for a symmetric polynomial in two variables:

∑ x₁²x₂ = x₁²x₂ + x₁x₂²

For a symmetric polynomial in three variables, we have:

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

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

s₁ = ∑ x₁

s₂ = ∑ x₁x₂

...

s_n-1= ∑ x₁*...*x_n-1

s_n = ∑ x₁*...*x_n


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

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

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

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

Here are some examples:

deg s₁ = (1, 0, 0, ..., 0)

deg s₂ = (1,1,0,....,0)

...

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

deg ∑ x₁²x₂ = (2,1,0, ..., 0)


Definition 5: Nⁿ

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

So that:

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


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

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

i₁ is greater than j₁ or

i₁ = j₁ and i₂ is greater than j₂ or

for all u, i_u = j_u or

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


Lemma 1: deg(P+Q) ≤ max(deg P, deg Q)

Proof:

(1) Let P = ∑ x₁^i₁x₂^i₂*...*x_n^i_n

(2) Let Q = ∑ x₁^j₁x₂^j₂*...*x_n^j_n

(3) deg P = (i₁, ..., i_n)

(4) deg Q = (j₁, ..., j_n)

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

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

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

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

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

deg(P+Q) is less than (i₁, ..., i_n)

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

QED


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

Proof:

(1) Let P = ∑ x₁^i₁x₂^i₂*...*x_n^i_n

(2) Let Q = ∑ x₁^j₁x₂^j₂*...*x_n^j_n

(3) deg P = (i₁, ..., i_n)

(4) deg Q = (j₁, ..., j_n)

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

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

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

QED

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

Proof:

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

QED


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

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

Proof:

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

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

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

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

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

i₁₁ ≥ i₂₁ ≥ ... ≥ i_m1 ≥ ....

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

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

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

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

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

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

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

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

QED

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

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

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

P(x₁, ..., x_n) = g(s₁, ..., s_n)

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

Proof:

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

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

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

(4) Let us define the following polynomial:

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

(5) Using Lemma 2 above, we have:

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

(6) Using Corollary 2.1 above, we have:

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

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

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

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

(9) So, we can restate f as:

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

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

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

(11) Let:

P₁ = P - af

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

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

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

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

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

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

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

QED

Example 4.1: S = ∑ x₁⁴x₂x₃ + ∑ x₁³x₂³

(1) Let:

S = ∑ x₁⁴x₂x₃ + ∑ x₁³x₂³

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

[See Definition 3 above for details if needed]

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

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

(3) Let f = s₁^4-1s₂^1-1s₃¹ = s₁³s₃

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

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

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

S₁ = ∑ x₁³x₂³ - 3∑ x₁³x₂²x₃ - 6x₁²x₂²x₃²

(6) deg S₁ = (3,3,0)

(7) Let f₁ = s₁^3-3s₂^3-0s₃⁰ = s₂³

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

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

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

S₂ = - 6∑ x₁³x₂²x₃ - 12x₁²x₂²x₃²

(10) deg S₂ = (3,2,1)

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

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

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

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

S₃ = 6x₁²x₂²x₃²

(14) deg(S₃) = (2,2,2)

(15) Let f₃ = s₁^2-2s₂^2-2s₃² = s₃²

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

s₃² = (∑ x₁x₂x₃)² = x₁²x₂²x₃²

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

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

S = s₁³s₃ + s₂³ - 6s₁s₂s₃ + 6s₃²


Lemma 5:

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

Then:

P is symmetric if and only Q is symmetric

Proof:

(1) Assume that P is symmetric

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

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

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

(5) But P'/Q' = P/Q' ≠ P/Q

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

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

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

QED

Theorem 6:

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

Proof:

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

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

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

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

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

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

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

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

P*(Q₁*...*Q_r) = f(s₁, ..., s_n)

and

Q*(Q₁*...*Q_r) = g(s₁, ..., s_n)

(8) Thus,

P/Q = f(s₁, ..., s_n)/g(s₁, ..., s_n)

QED

References

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

Edward Waring

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

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

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