Fermat Numbers

While of all Pierre de Fermat's proposed theorems really turned out to be theorems, not all of his conjectures turned out to be true.

In all fairness to Fermat, these theorems and conjectures were published after his death and were based on the notes he kept.

Pierre de Fermat conjectured that the F_n = 2²ⁿ + 1 is prime for every integer n. These values F_n are today called Fermat Numbers.

The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

Now, for n=0, 1, 2, 3, 4 the results are F₀=3, F₁=5, F₂=17, F₃=257, and F₄=65537 which are all prime.

In 1732, Leonhard Euler showed that F₅ = 641*6,700,417.

Since then, it has been shown that for 5 ≤ n ≤ 16, F_n is not prime.

Despite all these efforts, no other Fermat number has been found to be prime and no proof has been offered that would resolve this issue.

I will end today's blog with the following proof:

Lemma 1: If a prime number p has the form 2^m + 1, then m is a power of 2.

Proof:

(1) Assume that m is not a power of 2.

(2) Then it is divisible by some odd integer k.

(3) Now, we know that:

x^k + 1 = (x + 1)(x^k-1 - x^k-2 + ... - x + 1) [See Lemma 4, here where a=x, b=1]

(4) Let x= 2^m/k

(5) This then gives us:

(2^m/k)^k + 1= (2^m/k + 1)([2^m/k]^k-1 - [2^m/k]^k-2 + ... - [2^m/k] + 1)

(6) Since (2^m/k)^k + 1 = 2^m + 1, it follows that 2^m + 1 cannot be a prime since it is divisible by (2^m/k + 1).

(7) Therefore, we have a contradiction and we reject our assumption in step #1.

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

Wantzel: Constructibility Criterion

The criterion presented in today's blog was known to Carl Friedrich Gauss in 1796 when he presented his criterion for the constructibility of regular polygons. Even so, the proof of the criteroin was first presented in 1837 by Pierre Laurent Wantzel.

The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

To simplify the proof, I offer the following postulates.

Postulate 1:

It is possible to draw a line between any two points that are already determined.

Posulate 2:

It is possible to draw a circle with a center at a point already determined and with the radius equal to the distance between any two points already determined.

Postulate 3:

If two lines are not parallel and not colinear, then they intersect at exactly one point.

Postulate 4:

If a circle's center is at a point on a line then the circle intersects the extended line at two points.

I will now use these postulates to present some lemmas:

Lemma 1: (0,v) is constructible if and only if (v,0) is constructible

Proof:








This follows from Postulate 2 above setting center A at (0,0) and radius = v. If C at (v,0) is constructible, then we can obtain B at (0,v). If B at (0,v) is constructible, then we can obtain C at (v,0).

QED

Lemma 2: (u,0) is constructible and (0,v) is constructible if and only if (u,v) is constructible















Proof:

(1) Assume that B at (u,0) and D at (0,v) are constructible.

(2) If B at (u,0) is constructible and D at (0,v) is constructible, then F at (u,v) is constructible since:

(a) Let BG be a line that intersects B at (u,0) and is parallel to the y-axis. [Euclid: Book I, Prop 12, see here]

(b) Let DE be a line that insersects D at (0,v) and is parallel to the x-axis. [Euclid: Book I, Prop 12, see here]

(c) Then F at (u,v) is constructible since it is the point where BG,DE intersect [See Postulate 3 above]

(4) Assume that F at (u,v) is constructible.

(5) Since F at (u,v) is constructible, we know that B at (u,0) is constructible and D at (0,v) is constructible since:

(a) Let FG be a line parallel to the y-axis that passes through F at (u,v) [Euclid: Book I, Prop 12, see here]

(b) Then B at (u,0) is the point where FG and the x-axis intersect. [See Postulate 3 above]

(c) Let EF be a line parallel to the x-axis that passes through F at (u,v) [Euclid: Book I, Prop 12, see here]

(d) Then D at (0,v) is the point where EF and the y-axis intersect. [See Postulate 3 above]

QED

Lemma 3:

If a line passes through two points (a₁,b₁) and (a₂,b₂), then it has the equation of form:

αX + βY = γ

where

α = b₂ - b₁
β = a₁ - a₂
γ = a₁b₂ - b₁a₂

Proof:

(1) This lemma is proved if it can be shown that both points satisfy the equation given.

That is, both equations, lead to γ = a₁b₂ - b₁a₂

(2) Case I: (a₁,b₁)

α(a₁) + β(b₁) = (b₂ - b₁)(a₁) + (a₁ - a₂)(b₁) = a₁b₂ - b₁a₂

(3) Case II: (a₂,b₂)

α(a₂) + β(b₂) = (b₂ - b₁)(a₂) + (a₁ - a₂)(b₂) = a₁b₂ - b₁a₂

QED

Lemma 4:

If we construct a circle with center (a₁,b₁) and radius equal to the distance between (a₂,b₂) and (a₃,b₃), then it's equation has the form:

X² + Y² = αX + βY + γ

where:

α, β, γ are rational expressions of a₁, a₂, a₃, b₁, b₂, b₃

Proof:

(1) The equation for this circle is (see Theorem 1, here):

(X - a₁)² + (Y - b₁)² = (a₂ - a₃)² + (b₂ - b₃)²

(2) Multiplying out the squares involving X,Y gives us:

X² - 2a₁X + a₁² + Y² - 2b₁Y - b₁²

(3) Moving everything but X²,Y² to the other side gives us:

X² + Y² = 2a₁X + 2b₁Y + [-a₁² -b₁² + (a₂ - a₃)² + (b₂ - b₃)²]

QED

Theorem 5: Criterion of Constructibility

Let O be a circle that with center at (0,0) and radius = 1 unit.

The point (u,v) is constructible from O if and only if (u,v) can be obtained from (1,0) by a sequence of operations of the following types:

(i) rational operations
(ii) extraction of square roots

Proof:

(1) I will prove the first part of the theorem using induction.

(2) For the base case, we assume that u=a or 0 and v=a or 0.

(3) To prove the base case, we have four cases to prove:

(a) (a,0),(0,0) [Constructible from the given]

(b) (0,a) [Constructible from (a,0) and Proposition 2 above if we set center at (0,0) and radius=a].

(c) (a,a) [Constructible from (a,0) and Lemma 1 above]

(4) So we assume that all points (u',v') are constructible.

(5) To complete the first half of the proof, we need to show that if (u',v') is constructible and (u,v) is obtained in a single operation from (u',v'), then (u,v) is constructible.

(6) Using Lemma 2 above, step #5 is achieved if we show that the following points are constructible:

(u'+v',0)
(u' -v',0)
(u'v',0)
(u'v'^-1,0) [assuming v ≠ 0 ]
(√u',0) [assuming u ≥ 0]

(7) (u'+v',0) and (u' -v',0) are constructible [From Postulate 2 above where the center is (u',0) and the radius=v and Postulate 4 above where (u'-v',0) and (u'+v',0) are the intersection of the circle and the x-axis.]

(8) (u'v',0) is constructible since:















(a) We are given that F at (0,v') is constructible.

(b) We are given that C at (u',0) is constructible.

(c) Let E at (0,b) be a point where b is less than v' and b is greater than 0 such that we consider b our unit of measure (i.e. b is length = 1 unit)

(d) Draw a line EC connecting E at (0,b) and C at (u',0) [See Posulate 1 above]

(e) Draw a line called FB parallel to EC and passing through F at (0,v') [Euclid: Book I, Prop 12, see here]

(f) Let B at (z,0) be the point where FB intersects with the x-axis. [See Postulate 3 above]

(g) Triangles EAC and FAB are similar since:

∠ EAC and ∠ FAB are the same.

∠ AEC is congruent to ∠ AFB and ∠ ACE is congruent to ∠ ABF since FB and EC are parallel [Euclid: Book I, Prop 29]

(h) So, we know that:

AC/AE = AB/AF [see Lemma 3, here]

(i) Since in step #8c above we assumed that AE = 1 unit, we have:

AC = AB/AF

or equivalently:

AB = AF*AC

(j) Since the length of AB is z [since A is at (0,0) and B is at (z,0)], the length of AF is v' [since A is at (0,0) and F is at (0,v')] and the length of AC is u' [since A is at (0,0) and C is at (u',0)], it follows that:

z = u'*v'

(9) (u'v'^-1,0) is constructible since:

(a) Using the same diagram as step #8 above, we are given that F at (0,v') is constructible.

(b) Let us assume that we are given B at (u',0) is constructible.

(c) Let E at (0,b) be a point where b is less than v' and b is greater than 0 such that we consider b our unit of measure (i.e. b is length = 1 unit)

(d) Draw a line EC connecting E at (0,b) with a point C at (c,0) on the x-axis. [Euclid: Book I, Prop 12, see here]

(e) We can use the same arguments as #8g above to establish that Triangles EAC and FAB are similar

(f) Using the same argument as #8i, we have:

AC = AB/AF

(g) Since the length of AB is u' [since A is at (0,0) and B is at (u',0)], the length of AF is v' [since A is at (0,0) and F is at (0,v')] and the length of AC is c [since A is at (0,0) and C is at (c,0)], it follows that:

c = u'/v' = u'*v'^-1

(10) Assuming u ≥ 0, then ( √u , 0) is constructible since:













(a) Assume that point B is at (0,0)

(b) Let D be a point on the x-axis at (d,0) such that BD = 1 unit.

(c) Let C be a point on the x-axis at (1+u',0) where 1 = BD. [See step #7 above]

(d) Let us a draw a circle O whose center is at the midpoint M of BC and whose radius is MC. [See Postulate 2 above]

(e) Let A at (1,a) be the intersection of O and a line drawn from D that is perpendicular to BC. [Euclid: Book I, Prop 11, see here]

(f) We can see that triangle BDA and triangle ADC are similar since:

Both ∠ BDA and ∠ ADC are right angles

And, ∠ BAC is a right angle [see Euclid: Book III, Prop 31]

So, ∠ BAD + ∠ DAC = 90 °

Since triangle ADC is a right triangle, we also have ∠ DAC + ∠ DCA = 90 ° [See Lemma 4, here]

So, we have shown that ∠ BAD ≅ ∠ DCA.

(g) Since the two triangles are equiangular, we have [see Lemma 3, here]:

DA/BD = DC/DA

(h) Since the length of DA is c [since A is at (1,c) and D is at (1,0)], the length of BD is 1 [since B is at (0,0) and D is at (1,0)] and the length of DC is u' [since D is at (1,0) and C is at (1+u,0)], it follows that:

c/1 = u'/c

Further, this gives us that:

c² = u'

which means that:

c = √u'

(11) To complete the second half of the proof, I will show that if a point (u,v) can be constructed from (1,0) then it can be obtained from (1,0) through a finite sequence of operations.

(12) To accomplish, I need to show that the following imply points obtained from (1,0):

(a) two lines intersecting
(b) two circles intersecting
(c) line intersecting with a circle

(13) A point at two lines intersecting implies that this new point is obtained from the previous points since:

(a) By Lemma 3 above, the two lines can be represented by:

α₁X + β₁Y = γ₁

α₂X + β₂Y = γ₂

(b) Assuming that α₁ ≠ 0, we multiply the top equation by (-α₂/α₁) and add the two equations together to get:

(-α₂/α₁)β₁Y + β₂Y = γ₁ + γ₂

which gives us that:

Y[(-α₂/α₁)β₁ + β₂] = γ₁ + γ₂

which shows that Y can be obtained from the given points.

Y = [γ₁ + γ₂ ]/[(-α₂/α₁)β₁ + β₂]

(c) Now, all we need to do is show that X can be obtained from the given points which can be shown using the first equation:

α₁X + β₁Y = γ₁

which then shows:

X = (γ₁ - β₁Y)/(α₁)

(14) A point at two circles intersecting implies that this new point is obtained from the previous points since:

(a) By Lemma 4 above, the two circles can be represented by:

X² + Y² = α₁X + β₁Y + γ₁

X² + Y² = α₂X + β₂Y + γ₂

which gives us:

α₁X + β₁Y + γ₁ = α₂X + β₂Y + γ₂

and moving X to one side and Y to the other gives us:

α₁X - α₂X = β₂Y - β₁Y + γ₂ - γ₁

(b) Now, assuming that (α₁ - α₂) ≠ 0, we can state X in terms of Y

X = [(β₂ - β₁)Y + γ₂ - γ₁]/(α₁ - α₂)

(c) Now, we can state Y as a quadratic equation since:

Using:

X² + Y² = α₁X + β₁Y + γ₁

We get:

{[(β₂ - β₁)Y + γ₂ - γ₁]/(α₁ - α₂)}² + Y² = α₁{[(β₂ - β₁)Y + γ₂ - γ₁]/(α₁ - α₂)} + β₁Y + γ₁

(d) Since a quadratic equation is solvable in terms of the coefficients (see Theorem, here if needed), we have shown that in this case, the points are obtained from the previous points.

(15) A point at a circle intersecting with a line is obtained from the previous points since:

(a) By Lemma 3 above, the line can be represented by:

α₁X + β₁Y = γ₁

(b) By Lemma 4 above, the circle can be represented by:

X² + Y² = α₂X + β₂Y + γ₂

(c) Assuming α₁ ≠ 0, we can restate X in terms of Y using step #15a since:

X = (γ₁ - β₁Y)/α₁

(d) Substituting into the equation in #15b gives us the following quadratic equation:

([γ₁ - β₁Y]/α₁)² + Y² = α₂([γ₁ - β₁Y]/α₁) + β₂Y + γ₂

(e) Since a quadratic equation is solvable in terms of the coefficients (see Theorem, here if needed), we have shown that in this case, the points are obtained from the previous points.

QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

Gauss: Proof that all roots of unity are solvable by radicals

After Carl Friedrich Gauss had solved the seventeenth root of unity, he proceed to generalize this result to show that all roots of unity are solvable by radicals. In today's blog, I will give Gauss's proof which was first presented as part of Gauss's classic work Disquisitiones Arithmeticae when Gauss was 21.

Today's content is taken straight from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

In the proofs below, I use σ(f) where f ∈ Q(μ_k)(μ_p) which is defined in Definition 6, here. For definition of K_f, see Definition 5, here.

Lemma 1:

Let ζ be a p-th period of unity.

Let ef=p-1

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i

Then:

σ^e(η_i) = η_i

Proof:

σ^e(η_i) = σ^e(ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i) =
=ζ_e+i + ζ_2e+i + ζ_3e+i + ... + ζ_ef+i =
= ζ_e+i + ζ_2e+i + ... + ζ_p-1+i =
= ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i + ζ_0+i = η_i

QED

Corollary 1.1:

Let ζ be a p-th period of unity.

Let ef=p-1

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i

Let a ∈ Q

Then:

σ^e(aη_i) = aη_i

Proof:

σ^e(aη_i) = σ^e(a[ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i ]) =

= σ^e(aζ_i + aζ_e+i + aζ_2e+i + ... + aζ_e(f-1)+i ) =

= aζ_e+i + aζ_2e+i + aζ_3e+i + ... + aζ_ef+i =

= aζ_e+i + aζ_2e+i + aζ_3e+i + ... + aζ_p-1+i =

= aζ_e+i + aζ_2e+i + ... + aζ_e(f-1)+i + aζ_0+i = aη_i

QED

Lemma 2:

Let ζ be a p-th period of unity.

Let ef=p-1

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i

Then if a_i ∈ Q:

σ^e(a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1η_e-1) = a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1η_e-1

Proof:

(1) σ^e(a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1η_e-1) = σ^e(a₀η₀) + σ^e(a₁η₁) + σ^e(a₂η₂) + ... + σ^e(a_e-1η_e-1) [From Definition 5, here]

(2) From Corollary 1.1 above, we have:

σ^e(a₀η₀) + σ^e(a₁η₁) + σ^e(a₂η₂) + ... + σ^e(a_e-1η_e-1) = a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1ηe-1

QED

Corollary 2.1:

if f(ζ), g(ζ) ∈ K_f, then: f(ζ)*g(ζ) ∈ K_f

Proof:

(1) Using Theorem 2, here, there exists a₀, ..., a_e-1 and b₀, ..., b_e-1 such that:

f(ζ) = a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1η_e-1

g(ζ) = b₀η₀ + b₁η₁ + b₂η₂ + ... + b_e-1η_e-1

(2) σ^e(f(ζ)g(ζ)) = σ^e(f(ζ))σ^e(g(ζ)) [See Lemma 6, here]

(3) σ^e(f(ζ))σ^e(g(ζ)) = σ^e(a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1η_e-1)σ^e(b₀η₀ + b₁η₁ + b₂η₂ + ... + b_e-1η_e-1) = (a₀η₀ + a₁η₁ + a₂η₂ + ... + a_e-1η_e-1)(b₀η₀ + b₁η₁ + b₂η₂ + ... + b_e-1η_e-1) =

= f(ζ)g(ζ) [This follows directly from Lemma 2 above]

QED

Lemma 3

Let ζ be a p-th period of unity.

Let ef=gh=p-1 where f divides g and k=g/f

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i

Let ω be a k-th root of unity

Then:

(η₀ + ωη_h + ... + ω^k-1η_h(k-1))^k = a₀η₀ + ... + a_e-1η_e-1

where the coefficients a₀, ..., a_e-1 are rational expressions in ω over Q [that is, they are rational expressions in Q(μ_k)]

Proof:

(1) I will show that (η₀ + ωη_h + ... + ω^k-1η_h(k-1))^k can be expressed as

a₀η₀ + ... + a_e-1η_e-1 where the coefficients a₀, ..., a_e-1 are rational expressions in ω over Q

(2) Assume that k = 1, then it is clear that a₀ = 1, a₁ = ω, a₂ = ω², ..., a_k-1 = ω^k-1

(3) Assume that our hypothesis in step #2 is true up to n such that k ≤ n, then the hypothesis holds for (η₀ + ωη_h + ... + ω^k-1η_h(k-1))^k

(4) Then (η₀ + ωη_h + ... + ω^k-1η_h(k-1))ⁿ⁺¹= (η₀ + ωη_h + ... + ω^k-1η_h(k-1))ⁿ(η₀ + ωη_h + ... + ω^k-1η_h(k-1))= (a₀η₀ + ... + a_e-1η_e-1)(η₀ + ωη_h + ω²η_2h + ... + ω^k-1η_h(k-1))

(5) (a₀η₀ + ... + a_e-1η_e-1)(η₀ + ωη_h + ω²η_2h + ... + ω^k-1η_h(k-1)) = (a₀η₀)(η₀ + ωη_h + ω²η_2h + ... + ω^k-1η_h(k-1)) + ... + (a_e-1η_e-1)(η₀ + ωη_h + ω²η_2h + ... + ω^k-1η_h(k-1)) = a₀η₀ω⁰η₀ + ... + a_e-1η_e-1ω^k-1η_h(k-1)

(6) It is clear that each of the resulting terms is in the form of a_iη_iω^jη_j*h so if I can prove that each of these form reduces to a rational expression of ω and η_i that I am done.

(7) a_iη_iω^jη_j*h = (a_iω^j)(η_iη_j*h)

(8) Using Corollary 2.1 above and Theorem 2 here, this gives us that each term reduces to:

(a_iω^j)(b₀η₀ + ... + b_e-1η_e-1) where b_i ∈ Q.

QED

Corollary 3.1:

Let ζ be a p-th period of unity.

Let ef=gh=p-1 where f divides g and k=g/f

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i

Let ω be a k-th root of unity

Then:

(η₀ + ωη_h + ... + ω^k-1η_h(k-1))^k = = a₀η₀ + ... + a_h-1η_h-1 + a_hη_h + ... + a_2h-1η_2h-1 + + ... + + a_h(k-1)η_h(k-1) + ... + a_e-1η_e-1

Proof:

(1) From Lemma 3 above, we have:

(η₀ + ωη_h + ... + ω^k-1η_h(k-1))^k = a₀η₀ + ... + a_e-1η_e-1

where the coefficients a₀, ..., a_e-1 are rational expressions in ω over Q [that is, they are rational expressions in Q(μ_k)]

(2) Since f divides g and ef = gh, it follows that g/f = e/h so that h divides e.

(3) This allows us to divide 0 .. e-1 into:

(η₀ + ωη_h + ... + ω^k-1η_h(k-1))^k = = a₀η₀ + ... + a_h-1η_h-1 + a_hη_h + ... + a_2h-1η_2h-1 + + ... + + a_h(k-1)η_h(k-1) + ... + a_e-1η_e-1

QED

Lemma 4:

Let ζ be a p-th period of unity.

Let ef=gh=p-1 where f divides g and k=g/f

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i
Let ξ_j = ζ_j + ζ_h+j + ζ_2h+j + ... + ζ_h(g-1)+j

Then:

η_i + η_h+i + ... + η_h(k-1)+i = ξ_i for i = 0, ..., h-1

Proof:

(1) η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i [From the definition above]

(2) η_i + η_h+i + ... + η_h(k-1)+i =
= [ ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i] + [ ζ_h+i + ζ_h+e+i + ζ_h+2e+i + ... + ζ_h+e(f-1)+i] + ...
+ [ ζ_h(k-1)+i + ζ_h(k-1)+e+i + ζ_h(k-1)+2e+i + ... + ζ_{h(k-1)+e(f-1)+i}] =

= [ ζ_i + ζ_{h + i} + ... + ζ_h(k-1)+i ] + [ ζ_{e + i} + ζ_{h + e + i} + ... + ζ_h(k-1)+e+i] + ...
+ [ζ_e(f-1)+i + ζ_h+e(f-1)+i + ... + ζ_{h(k-1)+e(f-1)+i}]

(3) Since k = g/f and ef=gh, it follows that e = gh/f = hk

(4) This then gives us:

[ ζ_i + ζ_{h + i} + ... + ζ_h(k-1)+i ] + [ ζ_{e + i} + ζ_{h + e + i} + ... + ζ_h(k-1)+e+i] + ...
+ [ζ_e(f-1)+i + ζ_h+e(f-1)+i + ... + ζ_{h(k-1)+e(f-1)+i}]=

[ ζ_i + ζ_{h + i} + ... + ζ_h(k-1)+i ] + [ ζ_{hk + i} + ζ_{h + hk + i} + ... + ζ_h(k-1)+hk+i] + ...
+ [ζ_hk(f-1)+i + ζ_h+hk(f-1)+i + ... + ζ_{h(k-1)+hk(f-1)+i}] =

[ ζ_i + ζ_h+i + ... + ζ_h(k-1) ] + [ ζ_hk+i + ζ_h(k+1)+i + ... + ζ_h(2k-1)+i] + ... + [ζ_{h[kf-k] + i} + ζ_{h[kf-k+1] + i} + ... + ζ_h[kf-1]+i]

(5) Since k = g/f, it follows that kf = g so that we have:

[ ζ_i + ζ_h+i + ... + ζ_h(k-1) ] + [ ζ_hk+i + ζ_h(k+1)+i + ... + ζ_h(2k-1)+i] + ... + [ζ_{h[kf-k] + i} + ζ_{h[kf-k+1] + i} + ... + ζ_h[kf-1]+i] =

[ ζ_i + ζ_h+i + ... + ζ_h(k-1) ] + [ ζ_hk+i + ζ_h(k+1)+i + ... + ζ_h(2k-1)+i] + ... + [ζ_{h[g-k] + i} + ζ_{h[g-k+1] + i} + ... + ζ_h[g-1]+i] = ξ_i

QED

Theorem 5:

Let ζ be a p-th period of unity.

Let ef=gh=p-1 where f divides g and k=g/f

Let η_i = ζ_i + ζ_e+i + ζ_2e+i + ... + ζ_e(f-1)+i
Let ξ_j = ζ_j + ζ_h+j + ζ_2h+j + ... + ζ_h(g-1)+j

Let ω be a k-th root of unity

Let t(ω) = η₀ + ωη_h + ω²η_2h + ... + ω^k-1η_h(k-1)

Then:

t(ω)^k has a rational expression in terms of a₀ξ₀ + ... + a_h-1ξ_h-1 where the coefficients a₀, ..., a_h-1 are rational expressions in ω over Q

Proof:

(1) By Corollary 3.1 above, we have:

t(ω)^k = (η₀ + ωη_h + ω²η_2h + ... + ω^k-1η_h(k-1))^k = = a₀η₀ + ... + a_h-1η_h-1 + a_hη_h + ... + a_2h-1η_2h-1 + + ... + + a_h(k-1)η_h(k-1) + ... + a_e-1η_e-1

where the coefficients a₀, ..., a_e-1 are rational expressions in ω over Q

(2) Apply σ^h to both sides of the equation (see Lemma 7, here) gives us:

(η_h + ωη_2h + ... + ω^k-1η_h(k))^k = (η_h + ωη_2h + ... + ω^k-1η₀)^k = = a₀η_h + ... + a_h-1η_2h-1 + + a_hη_2h + ... + a_2h-1η_3h-1 + + ... + + a_h(k-1)η₀ + ... + a_e-1η_h-1

(5) We note that ω^-1(t(ω)) = η_h + ωη_2h + ... + ω^k-1.

(6) We further note that:

[ω^-1t(ω)]^k = (ω^-1)^kt(ω)^k = (ω^k)^-1t(ω)^k = t(ω)^k

(7) So that it is clear that t(ω)^k = (η_h + ωη_2h + ... + ω^k-1η₀)^k = = a₀η_h + ... + a_h-1η_2h-1 + + a_hη_2h + ... + a_2h-1η_3h-1 + + ... + + a_h(k-1)η₀ + ... + a_e-1η_h-1

(8) Since we can make the same argument for ω²t(ω) with σ^2h applied, ω³t(ω) with σ^3h, ..., ω^k-1t(ω) with σ^h(k-1), we can make the following observation:

kt(ω)^k = (a₀ + ... + a_h(k-1))(η₀ + ... + η_h(k-1)) + + (a₁ + ... + a_h(k-1)+1)(η₁ + ... + η_h(k-1)+1) + + ... + + (a_h-1 + ... + a_e-1)(η_h-1 + ... + η_e-1).

(9) Using Lemma 4 above that η_i + η_h+i + ... + η_h(k-1)+i = ξ_i for i = 0, ..., h-1, it follows that t(ω)^k is rationally expressed in terms of ω and ξ₀, ..., ξ_h-1 as:

t(ω)^k = 1/k((a₀ + ... + a_h(k-1))ξ₀ + ... + (a_h-1 + ... + a_e-1)ξ_h-1)

QED

Corollary 5.1: For every integer n, the n-th roots of unity have expressions by radicals

Proof:

(1) This is clearly true if n=1 or n=2 since the roots of unity are (1) or (1,-1)

(2) We may assume that for every integer k less than n, the k-th roots of unity are expressible by radicals

(3) If n is not prime, then since each of its factors are less than n, we can conclude that n is expressible by radicals (see Theorem 2, here).

(4) So, we can assume that n is prime.

(5) We can then order the n-th roots of unity other than 1 with the aid of a primitive root of n [see Theorem 3, here]

(6) We can then consider the Lagrange resolvent (see Theorem, here):

t(ω) = ζ₀ + ωζ₁ + ... + ω^n-2ζ_n-2

where ω is an (n-1)-st root of unity.

(7) By the induction hypothesis, ω can be expressed by radicals.

(8) By the Theorem 5 above (with k = g = n-1), this shows that t(ω)^n-1 has a rational expression in terms of ω

(9) We can then use Lagrange's formula (see Theorem, here), to get the following formula:



QED

References

• Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001