Sunday, February 03, 2008

Gauss: σ notation

In today's blog, I present a mapping notation σ(f) that I will use in proofs about periods of cyclotomic equations. I will talk in more detail in my next blog about Gauss's concept of periods which generalize the same method that Alexander-Theophile Vandermonde used to solve the eleventh root of unity.

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'd = 1m'd = ζd

ζn = ζn'*p + d = (ζp)n'd = 1n'd = 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 = (ζgigj)g

(3) There also exists a,b such that:

gi ≡ a (mod p)
gj ≡ b (mod p)

(4) So that ρ*ω = ζab = 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 = (ζgigj)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 Qk) 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 ∈ Qk), 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

1 comment:

dhodg13 said...

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.