## Monday, February 04, 2008

### Gauss: Periods of Cyclotomic Equations

In today's blog, I show some major results from Carl Friedrich Gauss in his analysis of periods of cyclotomic equations. These results represent a very small subset of Gauss's work in his classic Disquisitiones Arithmeticae which he wrote when he was 21. These properties of periods of cyclotomic equations are later used to demonstrate Gauss's proof that all cyclotomic polynomials are solvable by radicals.

Today's content is taken straight from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations which covers the history of Galois Theory from a mathematical perspective.

Definition 1: μp

Let μp denote the set of p-th roots of unity.

Examples:

μ1 = {1}
μ2 = {1, -1}
μ3 = {1, (1/2)[-1 + √-3], 1/2)[-1 - √-3])
μ4 = {1, -1, i, -i}

I will use μp in the following context:

Definition 2: Q(μp)

Let Q(μp) denote the set of complex numbers are rational expressions in these p-th roots of unity.

So that: Definition 3: period of f terms of a p-th root of unity

For any two positive integers: e,f where ef = p-1, the periods of f terms are:

η0 = ζ0 + ζe + ζ2e + ... + ζe(f-1)

η1 = ζ1 + ζe+1 + ζ2e+1 + ... + ζe(f-1)+1

η2 = ζ2 + ζe+2 + ζ2e+2 + ... + ζe(f-1)+2

...

ηe-1 = ζe-1 + ζ2e-1 + ζ3e-1 + ... + ζe(f-1)+(e-1)

If you review Alexander-Theophile Vandermonde's solution of the eleventh root of unity, it is clear that Carl Friedrich Gauss's theory of periods is a generalization of his solution. Interestingly, it is not clear if Gauss derived his solution from the work of Vandermonde or if he came upon it independently as part of his solution of the seventeenth root of unity.

Definition 4: σi(f)

If we number each time that we apply σ such that σ(σ(f)) = σ12(f)), then:
σi(f) = σ12(...(σi(f))...))

We will now use the equation ef=p-1 (see Definition 3 above) to define a set whose elements are invariant under σe.

Definition 5: Kf

By Kf, let us denote the set of all Q(μp) which are invariant under σe where ef = p-1.

Examples of Kf:

(1) All rational numbers

u ∈ Q → u ∈ Kf [This is clear from Lemma 6, here]

(2) All periods of cyclotomic equations

This is clear from definition 3 above.

Now, we can use these definitions to identify some properties which we will use later.

Theorem 1: Kf is a vector space

Proof:

The proof follows from Definition 2, here since:

(1) Kf is nonempty [See Lemma 6, here since Q is nonempty]

(2) Kf is closed on addition. [See Lemma 5, here]

(3) Kf is closed on scalar multiplication. [See Lemma 7, here]

(4) Kf addition is associative. [See Lemma 5, here]

(5) 0 ∈ Kf [See Lemma 6, here since 0 ∈ Q]

(6) Kf has negative elements [See Lemma 7, here since (-1)*σ(x) = σ(-x)]

(7) Kf addition is commutative. [See Lemma 5, here]

(8) All elements of Kf are distributive since:

σ(a[b + c]) = σ(ab + ac)

(9) Scalar multiplication is associative [See Lemma 7, here]

(10) Existence of 1 [See Lemma 6, here since 1 ∈ Q]

QED

Theorem 2: Every element in Kf can be written in a unique way as a linear combination with rational coefficients of the e periods of f terms.

(1) Let a be an arbitrary element in Kf [See Definition 5 above]

(2) We can write a as follows: [See Definition 2 above and See Definition 1, here, for ζi]

a = a0ζ0 + a1ζ1 + ... + ae-1ζe-1 +
+ aeζe + ae+1ζe+1 + ... + a2e-1ζ2e-1 +
+ ... +
+ ae(f-1)ζe(f-1) + ae(f-1)+1ζe(f-1)+1 + ... + ap-2ζp-2

(3) By the definition of σi [See Definition 4 above], we have:

σe(a) = a0ζe + a1ζe+1 + ... + ae-1ζ2e-1 +
+ a2e + ae+1ζ2e+1 + ... + a2e-1ζ3e-1 +
+ ... +
+ ae(f-1)ζ0 + ae(f-1)+1ζ1 + ... + ap-2ζe-1.

(4) Since a ∈ Kf, we know that:

σe(a) = a

(5) Thus:

a0 = ae = a2e = ... = ae(f-1)

a1 = ae+1 = a2e+1 = ... = ae(f-1)+1

...

ae-1 = a2e-1 = a3e-1 = ... = ap-2

(6) Therefore:

a = a00 + ζe + ... + ζe(f-1)) +
+ a11 + ζe+1 + ... + ζe(f-1)+1) +
+ ... +
+ ae-1e-1 + ζ2e-1 + ... + ζp-2).

(7) This proves that a is a linear combination of the periods, since the expressions between the brackets are the periods of f terms. [See Definition 3 above]

(8) Further, this expression is unique. [See Theorem 4, here]

QED

Corollary 2.1: 1, η, η2, ..., ηe-1 is a basis for Kf

Proof:

This follows directly from Theorem 1 above, Theorem 2 above and Lemma 1, here.

QED

Theorem 3:

1, η, η2, ..., ηe-1 is a basis for the vector space Kf

Proof:

(1) 1, η, η2, ..., ηe-1 are linearly independent [see Definition 1, here for definition of linearly independent if needed] since:

(a) Assume that a0 + a1η + ... + ae-1ηe-1 = 0 for some rational numbers a0, ..., ae-1

(b) Then η is the root of the polynomial p(x) where:

p(x) = a0 + a1x + ... + ae-1xe-1 (from step #1a)

(c) Now if a0 + a1η + ... + ae-1ηe-1 = 0, it follows that:

σ(a0 + a1η + ... + ae-1ηe-1 ) = σ(0) = 0

σ2(a0 + a1η + ... + ae-1ηe-1 ) = σ2(0) = 0

σ3(a0 + a1η + ... + ae-1ηe-1 ) = σ3(0) = 0

...

σe-1(a0 + a1η + ... + ae-1ηe-1 ) = σe-1(0) = 0

(d) So, σ(η), σ2(η), ..., σe-1(η) are all roots of p(x) in step #1b

(e) Now, each of η, σ(η), etc. are the e periods of f terms which are pairwise distinct [See Definition 3 above]

(f) Since by the Fundamental Theorem of Algebra (see Theorem, here), the polynomial p(x) has degree at most e -1, it cannot have as roots the e periods of the f terms unless it is the zero polynomial.

(g) Therefore, a0 = ... = ae-1 = 0

(h) This then proves 1, η, η2, ..., ηe-1 are linearly independent. [See Definition 1, here]

(2) From Corollary 2.1 above, we know that dim Kf = e. [See Theorem 1, here and Definition 2, here]

(3) But then using the fact 1, η, η2, ..., ηe-1 are linearly independent and Lemma 2, here, we can conclude that:

1, η, η2, ..., ηe-1 is a basis for Kf.

QED

Corollary 3.1:

If η, η' are periods of f terms, then:

η' = a0 + a1η + ... + ae-1ηe-1

for some rational numbers a0, ..., ae-1

Proof:

This follows from Theorem 3 above since η' ∈ Kf and 1, η, η2, ..., ηe-1 is a basis for the vector space Kf.

QED

Lemma 4:

if gh=ef=p-1 and f divides g, then it follows that:

Kg ⊂ Kf

Proof:

(1) Since gh=ef and f divides g, there exists an integer k such that:

k = g/f = e/h

(2) Therefore e = hk which gives us that:

σe = (σh)k

(3) This means that every element that is invariant under σh is also invariant under σe since:

(a) Assume that an element a is invariant under σh such that:

σh(a) = a

(b) Further:

σh1h2(...(σhk(a)...))) = a

(4) Since h*k = e, it follows from definition 4 above that:

σh1h2(...(σhk(a)...))) = σe(a)

(5) And it follows that:

σe(a) = a

(6) Since σh(a) = a → a ∈ Kg and σe(a) = a → a ∈ Kf, it follows that:

Kg ⊂ Kf

QED

Lemma 5:

Let f,g be divisors of p-1.

If f divides g, then every element in Kf is a root of a polynomial of degree g/f with coefficients in Kg

Proof:

(1) Let a be an element of Kf

(2) Let us define k such that:

k = g/f

Since ef = gh, it follows that:

k = g/f = e/h

(3) Let use define P(x) such that:

P(x) = (x - a)(x - σh(a))(x - σ2h(a))*...*(x - σh(k-1)(a))

(4) P(x) has degree hk/h = k = g/f

(5) It is also clear that a is a root of P(x). [Since if x=a, then P(x)=0]

(6) We note that:

σhh(k-1)(a)) = σhk(a) = (σh(a))k

(7) Since k = e/h, it follows that e=hk and:

h(a))k = σe(a) = a

(8) Step #3 and step #6 and step #7 give us that:

σh(P(x)) = P(x)

(9) Therefore, we conclude that P(x) has coefficients in Kg.

QED

Corollary 5.1:

Let f,g be divisors of p-1 and let η, ξ be periods of f and g terms respectively.

If f divides g, then η is a root of a polynomial of degree g/f whose coefficients are rational expressions of ξ

Proof:

(1) ξ ∈ Kg, and η ∈ Kf

(2) Using Lemma 5 above, we know that η is a root of a polynomial P(x) of degree g/f with coefficients in Kg

(3) Using Theorem 3 above, it follows that:

P(x) has coefficients which are rational expressions of ξ.

QED

References

## 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