The content in today's blog is taken straight from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.
Lemma 1:
for any prime p, if m ≡ n (mod p), and ζ is a p-th root of unity
then:
ζm = ζn
Proof:
(1) Assume m ≡ n (mod p) [See here for a review of modular arithmetic if needed]
(2) Then, there exists an integer d such that:
0 ≤ d ≤ p-1
and
m ≡ d (mod p)
n ≡ d (mod p)
(3) So there exists m' and n' such that:
m = m'*p + d
n = n'*p + d
(4) Since ζp = 1 (see here for review of roots of unity if needed), this gives us that:
ζm = ζm'*p + d = (ζp)m'*ζd = 1m'*ζd = ζd
ζn = ζn'*p + d = (ζp)n'*ζd = 1n'*ζd = ζd
QED
Definition 1: ζi
Let ζi = ζgi where g is a primitive root of a prime p.
Lemma 2:
ζp-1 = ζ0
ζp = ζ1
Proof:
(1) Since g is a primitive root, gp-1 ≡ 1 (mod p) [By Fermat's Little Theorem, see here].
(2) So using Lemma 1 above, it follows that:
ζp-1 = ζgp-1 = ζ1 = ζg0 = ζ0
and
ζp = ζg(p-1)+1 = ζgp-1*g1 = (ζgp-1)g = (ζ1)g = ζg1 = ζ1
QED
Definition 2: μp
Let μp denote the set of p-th roots of unity so that:
μp = { 1, ζ0, ζ1, ..., ζp-2 }
Example:
μ1 = {1}
μ2 = {1, -1}
μ3 = {1, (1/2)[-1 + √-3], 1/2)[-1 - √-3])
μ4 = {1, -1, i, -i}
Definition 3: σ(ζ) where ζ ∈ μp
Let σ be a map that changes f(ζ) to f(ζg)
Lemma 3: σ(ζi) = ζi+1
Proof:
(1) From the definition of ζi [See Definition 1 above]
σ(ζi) = σ(ζgi)
(2) From the definition of σ [See Definition 2 above]
σ(ζgi) = ζgi+1 = ζi+1
QED
Lemma 4:
if ρ, ω ∈ μp
then:
σ(ρω) = σ(ρ)σ(ω)
Proof:
(1) Since ρ, ω ∈ μp, there exists i,j such that:
ρ = ζi
ω = ζj
(2) σ(ρ)σ(ω) = (ζgi+1)*(ζgj+1) = (ζgi)g*(ζgj)g = (ζgi*ζgj)g
(3) There also exists a,b such that:
gi ≡ a (mod p)
gj ≡ b (mod p)
(4) So that ρ*ω = ζa*ζb = ζa+b
(5) There exists d such that a+b ≡ d (mod p) and 0 ≤ d ≤ p-1 so it follows that ζd ∈ μp
(6) There exists k such that gk ≡ d (mod p) so we have:
σ(ρ*ω) = σ(ζgk) = ζgk+1 = (ζgk)g = (ζa + b)g = (ζgi + gj)g = (ζgi*ζgj)g
QED
Definition 4: Q(μp)
Let Q(μp) denote the set of complex numbers that are rational expressions in these p-th roots of unity.
This gives us that:
Definition 5: σ(f) where f ∈ Q(μp)
σ(a0ζ0 + ... + ap-2ζp-2) = a0σ(ζ0) + ... + ap-2σ(ζp-2)
where ai ∈ Q and ζi ∈ μp
Lemma 5: σ is well-defined on the whole of Q(μp)
Proof:
This follows from Definition 5 above and Theorem 4, here.
QED
Lemma 6:
The map σ is a field automorphism of Q(μp) which leaves every element of Q invariant.
Proof:
(1) σ is bijective . [See Definition 1, here for definition of bijective; see Definition 5 above]
(2) σ(ua + vb) = uσ(a) + vσ(b) for a,b ∈ Q(μp) and u,v ∈ Q. [see Definition 5 above]
(3) If a ∈ Q, then using Corollary 1.1, here, we have:
a = (-a)ζ + (-a)ζ2 + ... + (-a)ζp-1
where ζ is a primitive p-th root of unity
(4) Since each of these ζi corresponds to a different p-th root of unity (see Theorem 3, here), this implies that:
a = (-a)ζ0 + (-a)ζ1 + ... + (-a)ζp-2
(5) This shows that every rational number is invariant under σ.
(6) Finally: σ(ab) = σ(a)σ(b) where a,b ∈ Q(μp) since:
(a) We can define a,b, and ab as summations:
a = ∑ (i=0, p-2) aiζi
b = ∑ (j=0, p-2) bjζj
ab = ∑ (i,j =0, p-2) aibjζiζj
(b) From Definition 5 above, we have:
σ(ab) = ∑ (i,j=0, p-2) σ(aibjζiζj)
(c) Since ai, bj ∈ Q, we have:
σ(ab) = ∑ (i,j=0, p-2) aibjσ(ζiζj)
(d) Since ζi, ζj ∈ μp, using Lemma 4 above, we have:
σ(ab) = ∑ (i,j=0, p-2) aibjσ(ζi)σ(ζj)
(e) Also, using Definition 5 above, we have:
σ(a)σ(b) = [∑ (i=0, p-2) σ(aiζi)][∑ (j=0, p-2) σ(bjζj)]
(f) Since ai, bj ∈ Q, we have:
σ(a)σ(b) = [∑ (i=0, p-2) aiσ(ζi)][∑ (j=0, p-2) bjσ(ζj)] =
= ∑ (i,j=0, p-2) aibjσ(ζi)σ(ζj)
(7) This shows that σ is a field automorphism of Q(μp) [See Definition 6, here, for definition of field automorphism]
QED
Definition 6: σ(f) where f ∈ Q(μk)(μp) where k divides p-1.
σ(a0ζ0 + ... + ap-2ζp-2) = a0σ(ζ0) + ... + ap-2σ(ζp-2)
where ai ∈ Q(μk) and ζi ∈ μp
Lemma 7: σ is well-defined on the whole of Q(μk)(μp)
Proof:
This follows from Definition 6 above and Corollary 3.1, here.
QED
Lemma 8:
The map σ is a field automorphism of Q(μk)(μp) which leaves every element of Q(μk) invariant.
Proof:
(1) σ is bijective . [See Definition 1, here for definition of bijective; see Definition 6 above]
(2) σ(ua + vb) = uσ(a) + vσ(b) for a,b ∈ Q(μk)(μp) and u,v ∈ Q(μk). [see Definition 6 above]
(3) If a ∈ Q(μk), then using Corollary 1.1, here, we have:
a = (-a)ζ + (-a)ζ2 + ... + (-a)ζp-1
where ζ is a primitive p-th root of unity
(4) Since each of these ζi corresponds to a different p-th root of unity (see Theorem 3, here), this implies that:
a = (-a)ζ0 + (-a)ζ1 + ... + (-a)ζp-2
(5) This shows that every number a ∈ Q(μk) is invariant under σ.
(6) Finally: σ(ab) = σ(a)σ(b) where a,b ∈ Q(μk)(μp) since:
(a) We can define a,b, and ab as summations:
a = ∑ (i=0, p-2) aiζi
b = ∑ (j=0, p-2) bjζj
ab = ∑ (i,j =0, p-2) aibjζiζj
(b) From Definition 6 above, we have:
σ(ab) = ∑ (i,j=0, p-2) σ(aibjζiζj)
(c) Since ai, bj ∈ Q, we have:
σ(ab) = ∑ (i,j=0, p-2) aibjσ(ζiζj)
(d) Since ζi, ζj ∈ μp, using Lemma 4 above, we have:
σ(ab) = ∑ (i,j=0, p-2) aibjσ(ζi)σ(ζj)
(e) Also, using Definition 6 above, we have:
σ(a)σ(b) = [∑ (i=0, p-2) σ(aiζi)][∑ (j=0, p-2) σ(bjζj)]
(f) Since ai, bj ∈ Q(μk), we have:
σ(a)σ(b) = [∑ (i=0, p-2) aiσ(ζi)][∑ (j=0, p-2) bjσ(ζj)] =
= ∑ (i,j=0, p-2) aibjσ(ζi)σ(ζj)
(7) This shows that σ is a field automorphism of Q(μk)(μp) [See Definition 6, here, for definition of field automorphism]
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001
1 comment:
Thank you so much for creating this blog. I had to form an opinion Pierre de Fermat's works. To do that I had to largely understand the proof of the theorem itself.
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