## Friday, September 25, 2009

### Kronecker's Theorem: The Proof

The content in today's blog is taken directly from David Antin's translation of Heinrich Dorrie's 100 Great Problems of Elementary Mathematics.

Lemma 1:

Let f be an irreducible polynomial over a field F.

Let f be reducible over a field F(λ) where:

λ = K(1/l) and l is an odd, prime and K ∈ F but λ is not in F.

Let g(x,λ) be a polynomial in F(λ) which divides f.

Let α be the nth root of unity.

Then:

g(x,λαi) divides f.

Proof:

(1) Since g(x,λ) divides f(x), there exists h(x,λ) such that:

f(x) = g(x,λ)*h(x,λ)

(2) Let r be any element of K.

[It is clear that f(r) = g(r,λ)*h(r,λ)]

(3) We can define a function u(x) in F(λ) such that:

u(x) = f(r) - g(r,x)h(r,x)

(4) It is clear that u(λ) = 0 since

f(x) - g(x,λ)h(x,λ)=0 for all x.

(5) Let us also define a function v(x) such that:

v(x) = K1/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
...
λαl-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,λv) =u(x,λv)v(x,λv)

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,λv) =u(x,λv)v(x,λv) for all x]

(2) Since t(λv) = 0, λv is a root of t(x)

(3) Now λ, λv, 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 = K1/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,λi) are equal. That is, if λi ≠ λj, then ψ(x,λi) ≠ ψ(x,λj)

Proof:

(1) Assume that λi ≠ λj, but ψ(x,λi) = ψ(x,λj)

(2) So that:

ψ(x,λμ) = ψ(x,λν)

(3) Let H = the root of unity ην - 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,λHn-1)

(9) Now:

(1/n)*(ψ(x,λ) + ψ(x,λH) + ψ(x,λH2) + ψ(x,λH3) + ... + ψ(x,λHn-1) is a symmetric function [see Definition 1, here].

(10) Further:

λ*λH*...*λHn-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 = K1/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 xl = 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

...

λn-1 = λ*αn-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,λv) are also factors of f(x). [see Lemma 1 above]

(12) Everyone of the n functions ψ(x,λv) is irreducible in F(λ) since:

(a) ψ(x,λ) is irreducible in F(λ) [see step #6 above]

(b) Assume that ψ(x,λv) is not irreducible in F(λ)

(c) Then, there exists u(x,λv), v(x,λv) where:

ψ(x,λv) =u(x,λv)v(x,λv)

(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,λv) 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,λμn-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,λμn-1)

(18) Since f(x) is reducible in F(λ), it follows that if ω, ω1, ..., ωn-1 are the roots, then:

x - ω, x - ω1, ..., x - ωn-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 - ωn-1 = ψ(x,λμn-1)

(20) So that we get [from the definition of ψ(x,λ)]:

ω = K0 + K1λ + K2λ2 + ... + Kn-1λn-1

ω1 = K0 + K1λ1 + K2λ12 + ... + Kn-1λ1n-1

....

ωn-1 = K0 + K1λn-1 + K2λn-12 + ... + Kn-1λn-1n-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 + ... + Kn-1λn-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 + ... + Kn-1Λn-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