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)