## Sunday, March 02, 2008

### 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(μk)(μp) which is defined in Definition 6, here. For definition of Kf, see Definition 5, here.

Lemma 1:

Let ζ be a p-th period of unity.

Let ef=p-1

Let ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i

Then:

σei) = ηi

Proof:

σei) = σei + ζe+i + ζ2e+i + ... + ζe(f-1)+i) =
e+i + ζ2e+i + ζ3e+i + ... + ζef+i =
= ζe+i + ζ2e+i + ... + ζp-1+i =
= ζe+i + ζ2e+i + ... + ζe(f-1)+i + ζ0+i = ηi

QED

Corollary 1.1:

Let ζ be a p-th period of unity.

Let ef=p-1

Let ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i

Let a ∈ Q

Then:

σe(aηi) = aηi

Proof:

σe(aηi) = σe(a[ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i ]) =

= σe(aζi + aζe+i + aζ2e+i + ... + aζe(f-1)+i ) =

= aζe+i + aζ2e+i + aζ3e+i + ... + aζef+i =

= aζe+i + aζ2e+i + aζ3e+i + ... + aζp-1+i =

= aζe+i + aζ2e+i + ... + aζe(f-1)+i + aζ0+i = aηi

QED

Lemma 2:

Let ζ be a p-th period of unity.

Let ef=p-1

Let ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i

Then if ai ∈ Q:

σe(a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1) = a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1

Proof:

(1) σe(a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1) = σe(a0η0) + σe(a1η1) + σe(a2η2) + ... + σe(ae-1ηe-1) [From Definition 5, here]

(2) From Corollary 1.1 above, we have:

σe(a0η0) + σe(a1η1) + σe(a2η2) + ... + σe(ae-1ηe-1) = a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1

QED

Corollary 2.1:

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

Proof:

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

f(ζ) = a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1

g(ζ) = b0η0 + b1η1 + b2η2 + ... + be-1ηe-1

(2) σe(f(ζ)g(ζ)) = σe(f(ζ))σe(g(ζ)) [See Lemma 6, here]

(3) σe(f(ζ))σe(g(ζ)) = σe(a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1e(b0η0 + b1η1 + b2η2 + ... + be-1ηe-1) = (a0η0 + a1η1 + a2η2 + ... + ae-1ηe-1)(b0η0 + b1η1 + b2η2 + ... + be-1ηe-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 ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i

Let ω be a k-th root of unity

Then:

0 + ωηh + ... + ωk-1ηh(k-1))k = a0η0 + ... + ae-1ηe-1

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

Proof:

(1) I will show that 0 + ωηh + ... + ωk-1ηh(k-1))k can be expressed as

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

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

(3) Assume that our hypothesis in step #2 is true up to n such that k ≤ n, then the hypothesis holds for 0 + ωηh + ... + ωk-1ηh(k-1))k

(4) Then 0 + ωηh + ... + ωk-1ηh(k-1))n+1= (η0 + ωηh + ... + ωk-1ηh(k-1))n0 + ωηh + ... + ωk-1ηh(k-1))= (a0η0 + ... + ae-1ηe-1)(η0 + ωηh + ω2η2h + ... + ωk-1ηh(k-1))

(5) (a0η0 + ... + ae-1ηe-1)(η0 + ωηh + ω2η2h + ... + ωk-1ηh(k-1)) = (a0η0)(η0 + ωηh + ω2η2h + ... + ωk-1ηh(k-1)) + ... + (ae-1ηe-1)(η0 + ωηh + ω2η2h + ... + ωk-1ηh(k-1)) = a0η0ω0η0 + ... + ae-1ηe-1ωk-1ηh(k-1)

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

(7) aiηiωjηj*h = (aiωj)(ηiηj*h)

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

(aiωj)(b0η0 + ... + be-1ηe-1) where bi ∈ 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 ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i

Let ω be a k-th root of unity

Then:

0 + ωηh + ... + ωk-1ηh(k-1))k = = a0η0 + ... + ah-1ηh-1 + ahηh + ... + a2h-1η2h-1 + + ... + + ah(k-1)ηh(k-1) + ... + ae-1ηe-1

Proof:

(1) From Lemma 3 above, we have:

0 + ωηh + ... + ωk-1ηh(k-1))k = a0η0 + ... + ae-1ηe-1

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

(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:

0 + ωηh + ... + ωk-1ηh(k-1))k = = a0η0 + ... + ah-1ηh-1 + ahηh + ... + a2h-1η2h-1 + + ... + + ah(k-1)ηh(k-1) + ... + ae-1ηe-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 ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i
Let ξj = ζj + ζh+j + ζ2h+j + ... + ζh(g-1)+j

Then:

ηi + ηh+i + ... + ηh(k-1)+i = ξi for i = 0, ..., h-1

Proof:

(1) ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i [From the definition above]

(2) ηi + ηh+i + ... + ηh(k-1)+i =
= [ ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i] + [ ζh+i + ζh+e+i + ζh+2e+i + ... + ζh+e(f-1)+i] + ...
+ [ ζh(k-1)+i + ζh(k-1)+e+i + ζh(k-1)+2e+i + ... + ζh(k-1)+e(f-1)+i] =

= [ ζi + ζh + i + ... + ζh(k-1)+i ] + [ ζe + i + ζh + e + i + ... + ζh(k-1)+e+i] + ...
+ [ζ
e(f-1)+i + ζh+e(f-1)+i + ... + ζh(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:

[ ζi + ζh + i + ... + ζh(k-1)+i ] + [ ζe + i + ζh + e + i + ... + ζh(k-1)+e+i] + ...
+ [ζ
e(f-1)+i + ζh+e(f-1)+i + ... + ζh(k-1)+e(f-1)+i]=

[ ζi + ζh + i + ... + ζh(k-1)+i ] + [ ζhk + i + ζh + hk + i + ... + ζh(k-1)+hk+i] + ...
+ [ζ
hk(f-1)+i + ζh+hk(f-1)+i + ... + ζh(k-1)+hk(f-1)+i] =

[ ζi + ζh+i + ... + ζh(k-1) ] + [ ζhk+i + ζh(k+1)+i + ... + ζh(2k-1)+i] + ... + [ζh[kf-k] + i + ζh[kf-k+1] + i + ... + ζh[kf-1]+i]

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

[ ζi + ζh+i + ... + ζh(k-1) ] + [ ζhk+i + ζh(k+1)+i + ... + ζh(2k-1)+i] + ... + [ζh[kf-k] + i + ζh[kf-k+1] + i + ... + ζh[kf-1]+i] =

[ ζi + ζh+i + ... + ζh(k-1) ] + [ ζhk+i + ζh(k+1)+i + ... + ζh(2k-1)+i] + ... + [ζh[g-k] + i + ζh[g-k+1] + i + ... + ζh[g-1]+i] = ξi

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 ηi = ζi + ζe+i + ζ2e+i + ... + ζe(f-1)+i
Let ξj = ζj + ζh+j + ζ2h+j + ... + ζh(g-1)+j

Let ω be a k-th root of unity

Let t(ω) = η0 + ωηh + ω2η2h + ... + ωk-1ηh(k-1)

Then:

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

Proof:

(1) By Corollary 3.1 above, we have:

t(ω)k = (η0 + ωηh + ω2η2h + ... + ωk-1ηh(k-1))k = = a0η0 + ... + ah-1ηh-1 + ahηh + ... + a2h-1η2h-1 + + ... + + ah(k-1)ηh(k-1) + ... + ae-1ηe-1

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

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

h + ωη2h + ... + ωk-1ηh(k))k = (ηh + ωη2h + ... + ωk-1η0)k = = a0ηh + ... + ah-1η2h-1 + + ahη2h + ... + a2h-1η3h-1 + + ... + + ah(k-1)η0 + ... + ae-1ηh-1

(5) We note that ω-1(t(ω)) = ηh + ωη2h + ... + ωk-1.

(6) We further note that:

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

(7) So that it is clear that t(ω)k = h + ωη2h + ... + ωk-1η0)k = = a0ηh + ... + ah-1η2h-1 + + ahη2h + ... + a2h-1η3h-1 + + ... + + ah(k-1)η0 + ... + ae-1ηh-1

(8) Since we can make the same argument for ω2t(ω) with σ2h applied, ω3t(ω) with σ3h, ..., ωk-1t(ω) with σh(k-1), we can make the following observation:

kt(ω)k = (a0 + ... + ah(k-1))(η0 + ... + ηh(k-1)) + + (a1 + ... + ah(k-1)+1)(η1 + ... + ηh(k-1)+1) + + ... + + (ah-1 + ... + ae-1)(ηh-1 + ... + ηe-1).

(9) Using Lemma 4 above that ηi + ηh+i + ... + ηh(k-1)+i = ξi for i = 0, ..., h-1, it follows that t(ω)k is rationally expressed in terms of ω and ξ0, ..., ξh-1 as:

t(ω)k = 1/k((a0 + ... + ah(k-1)0 + ... + (ah-1 + ... + ae-1h-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 + ... + ωn-2ζn-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
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