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
- 100 Great Problems of Elementary Mathematics (Dover, 1965)
No comments:
Post a Comment