Dedekind: irreducibility of Φp over Q(μk)

Carl Friedrich Gauss made an assumption about the irreducibility of Φ_p over Q(μ_k). This assumption is correct but the proof of this assumption was first given by Leopold Kronecker in 1854.

The proof presented in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations and is based on "some ideas of [Richard] Dedekind".

For review if needed, see here for review of a monic polynomials. See Definition 1, here for definition of an irreducible factor of a polynomial.

Lemma 1:

Let f be a monic irreducible factor of Φ_n in Q[X]

Let p be a prime number which does not divide n.

If ω ∈ C is a root of f

Then:

ω^p is also a root of f and:

f(ω) = 0 → f(ω^p) = 0

Proof:

(1) Assume that f(ω) = 0

(2) Assume that f(ω^p) ≠ 0

(3) Since f divides Φ_n and Φ_n divides xⁿ - 1 [See Definition 1, here for definition of Φ_n], then, there exists a polynomial g such that:

xⁿ - 1 = fg

(4) We know that g is monic. [See Lemma 1, here]

(5) Since f(ω) = 0, it follows that ωⁿ = 1 since:

(a) (ω)ⁿ - 1 = f(ω)g(ω) = 0*g(ω) = 0

(b) Adding 1 to both sides of the equation gives us ωⁿ = 1

(6) We also know that (ω^p)ⁿ = 1 since:

(ω^p)ⁿ = ω^pn = (ωⁿ)^p = (1)^p = 1
(7) So, we have: (ω^p)ⁿ - 1 = f(ω^p)g(ω^p) = 0

(8) But f(ω^p) ≠ 0 from step #2 above so we can conclude that g(ω^p) = 0

(9) This shows that ω is a root of g(x^p)

(10) Since f is irreducible, we can use a previous result (see Lemma 2, here) to conclude that f(x) divides g(x^p)

(11) Thus, there exists a polynomial h(x) such that:

g(x^p) = f(x)h(x)

(12) Further, we can conclude that f,g, and h all have integer coefficients since:

(a) Step #3 above shows that f,g have integer coefficients. [See Corollary 2.1, here] since:

f,g are monic [f is monic from the given; g is monic from step #4 above] and they both divide a monic polynomial with integral coefficients: xⁿ - 1.

(b) Step #11 above shows that h has integer coefficients since: g is a monic polynomial with integral coefficients [step #12a above] and h is monic since f,g are monic [see Lemma 1, here]

(13) So, we can consider the polynomials: f, g, and h whose coefficients are the congruence classes modulo p of the coefficients of f,g, and h. [See here for review of congruence classes modulo p]

(14) From step #3 above, we can conclude:

xⁿ - 1 = f(x)g(x) in F_p[x]

where F_p[x] is a polynomial whose coefficients are the set of congruence classes modulo p Z/pZ.

(15) From step #11 above, we can conclude:

g(x^p) = f(x)h(x) in F_p[x]

(16) Using Fermat's Little Theorem (see Theorem, here), we know that:

a^p-1 ≡ 1 (mod p)

which further implies that:

a^p ≡ a (mod p)

(17) Assume that:

g(x) = a₀ + a₁x + ... + a_r-1x^r-1 + x^r

(18) Then using step #16 we also have:

g(x) = a₀^p + a₁^px + ... + a_r-1^px^r-1 + x^r

(19) But from step #17, we also have:

g(x^p) = a₀^p + a₁^px^p + ... + a_r-1^px^p(r-1) + x^pr

(20) Using the result (see Lemma, here) that (a + b)^p ≡ a^p + b^p (mod p), we further have:

g(x^p) = (a₀ + a₁x + ... + a_r-1x^r-1 + x^r)^p = g(x)^p in F_p[x]

(21) From step #20 and step #15, we can conclude that:

g^p = fh

(22) But then if f divides g^p, it is clear that f and g must have a common factor.

(23) Let φ(x) be the the nonconstant common factor of f and g .

(24) Then, there exists polynomials f ' and g ' such that:

f = f ' φ g = g' φ

(25) Using step #14, we have:

xⁿ - 1 = f(x)g(x) = (f' φ)(g'φ ) in F_p[x]

(26) Let ψ = f '*g'

(27) Then we have:

xⁿ - 1 = φ²ψ in F_p[x]

(28) Taking the derivatives from both sides (see here for linear combination rule and product rule):

nx^n-1 = φ*(2dφ*ψ + φ*dψ)

(29) But this now gives us a contradiction since by step #27:

φ divides xⁿ - 1

and by step #28:

φ divides nx^n-1

which is impossible

(30) So, we reject our assumption in step #2.

QED

Theorem 2:

For every integer n ≥ 1, the cyclotomic polynomial Φ_n is irreducible over Q

Proof:

(1) Let f be a monic irreducible factor of Φ_n in Q[x]

(2) To establish this result, I will show that every root of Φ_n in C is a root of f and finally that f = Φ_n.

(3) Let ζ be a root of f.

(4) Then ζ is a root of Φ_n since:

(a) Since f is a factor of Φ_n, there exists a polynomial g such that:

Φ_n = fg

(b) Φ_n(ζ) = f(ζ)g(ζ) = 0*g(ζ) = 0

(5) From step #4, we can conclude that ζ is a primitive n-th root of unity since:
Φ_n only includes primitive n-th roots of unity as roots [See Definition 1, here]

(6) Since ζ is a primitive n-th root of unity, all other primitive n-th roots of unity (if they exist) have the form ζ^k where k is an integer relatively prime to n between 1 and n-1. [See Theorem 3, here]

(7) Factoring k in prime factors we get (see Theorem 3, here):

k = p₁*...*p_s

(8) We can now use Lemma 1 above to show that any root of Φ_n is a root of f since if f(ζ) = 0, then also:

f(ζ^p₁) = 0 f(ζ^p₁p₂) = 0 ... f(ζ^k) = 0

(9) Thus, f has as its root every primitive n-th root of unity and every root of Φ_n

(10) Using a previous result (see Corollary 2.2, here), we can further conclude that Φ_n divides f.

(11) So if Φ_n divides f and f divides Φ_n, it follows that Φ_n = f

QED

Theorem 3:

If m and n are relatively prime integers, then Φ_n is irreducible over Q(μ_m)

Proof:

(1) Let f be a monic irreducible factor of Φ_n in Q(μ_m)[x], that is, the coefficients of f are in Q(η_m)

(2) Let ζ ∈ C be a root of f so that ζ is a primitive n-th root of unity [see step #5 in Theorem 2 above].

(3) Let η be a primitive m-th root of unity.

(4) Since all roots of unity are powers of the primitive m-th root of unity (see Theorem 3, here), we have:

Q(μ_m) = Q(η)

(5) From an earlier result (see Lemma 3, here), we can conclude that every coefficient of f is a polynomial expression in η with rational coefficients since:

(a) ζ ∈ C is a root of f ∈ Q[x] which is irreducible and is of degree less than n.

(b) Let Q(α) be the set of elements in C which are rational expressions of α with coefficients in Q where α ∈ C.

(c) Q(α) = u(α)/v(α) ∈ C such that u,v ∈ Q[x] and v(α) ≠ 0.

(d) So using Lemma 3, here, we can conclude that every element in Q(α) can be uniquely written in the form a₀ + a₁α + a₂α² + ... + a_n-1α^n-1 with a_i ∈ Q.

(e) From 5d, it is clear that we can define a polynomial

(6) Therefore f(x) = φ(η,x) for some polynomial φ(y,x) ∈ Q[y,x]

(a) f(x) ∈ Q(μ_n)[x] [See step #1 above]

(b) Q(μ_n) = Q(η) [See step #4 above]

(c) Using step #5d, it is clear that all elements of Q(η) can be expressed as:

a₀ + a₁η + a₂η² + ... + a_n-1η^n-1 with a_i ∈ Q

(d) From step #6c, it is clear that we can define a polynomial φ(η,x) such that:

f(x) = (a_0,0 + a_0,1η + ... + a_0,n-1η^n-1) + (a_1,0 + a_1,1η + ... + a_1,n-1η^n-1)x + ... + (a_n-1,0 + a_n-1,1η + ... + a_n-1,n-1η^n-1)x^n-1

where φ(y,x) = (a_0,0 + a_0,1y + ... + a_0,n-1y^n-1) + (a_1,0 + a_1,1y + ... + a_1,n-1y^n-1)x + ... + (a_n-1,0 + a_n-1,1y + ... + a_n-1,n-1y^n-1)x^n-1 and φ(y,x) ∈ Q[y,x]

(7) Let ρ = ζη.

(8) Since m,n are relatively prime, it follows from a previous result (see Theorem 2, here) that:

ρ is a primitive mn-th root of unity.

(9) Since m,n are relatively prime, there exists integers r,s such that (see Lemma 1, here):

mr + ns = 1.

(10) Since ζⁿ = 1 and η^m = 1, we have:

ζ = ζ^mr = ζ^mr*(1)^r =ζ^mr*(η^m)^r =ρ^mr

and

η = η^ns = (1)^s*η^ns= (ζⁿ)^s*η^ns =ρ^ns

(11) Since f(ζ) = 0, we have [see step #6 above]:

φ(η,ζ) = 0

and therefore [from step #10 above],

φ(ρ^ns,ρ^mr) = 0

(12) From an earlier result (see Lemma 2, here), we can conclude that Φ_mn(x) divides φ(x^ns,x^mr) since:

(a) Φ_mn(x) and φ(x^ns,x^mr) are polynomials with coefficients in Q(μ_m).

(b) Since φ(x^ns,x^mr) = f(x) [see step #6 above] and f(x) is irreducible in Q(μ_m), it follows that:

φ(x^ns,x^mr) is irreducible in Q(μ_m)

(c) Finally, Φ_mn(x) and φ(x^ns,x^mr) have a common root ρ in field C which contains the field Q(μ_m) [See step #8 above and step #11 above]

(d) Therefore, we can use Lemma 2, here to conclude that Φ_mn(x) divides φ(x^ns,x^mr)

(13) It follows then that: φ(ω^ns,ω^mr) = 0 for every mn-th root of unity ω since:

if Φ_mn(x) divides φ(x^ns,x^mr), then there exists a polynomial g such that:

φ(x^ns,x^mr) = Φ_mn(x)g(x)

so that if Φ_mn(a)=0, it follows that φ(a^ns,a^mr) =0.

(14) Let k be an integer relatively prime to n such that 1 ≤ k ≤ n-1.

(15) Let l = kmr + ns [Derived from the equation in step #9 above]

(16) Since mr + ns=1, we have mr ≡ 1 (mod n) and ns ≡ 1 ( mod m)

(17) We further have:

l ≡ k (mod n)

and

l ≡ 1 (mod m)

(18) It follows that ζ^l = ζ^k and η^l = η [See Lemma 1, here]

(19) Since we already have observed (see step #10 above) that ζ = ρ^mr and η = ρ^ns, we have:

ρ^lmr = ρ^kmr = (ρ^mr)^k= ζ^k [Since ζ is an primitive n-th root of unity]

and

ρ^lns =ρ^ns = η [Since η is a primitive m-th root of unity]

(20) On the other hand, the congruences in step #17 also show that l is relatively prime to mn.

(21) Therefore, ρ^l is a primitive mn-th root of unity [See Definition 2, here].

(22) step #11 combined with Lemma 1 above gives us:

φ(ρ^lns,ρ^lmr) = 0

since:

φ(ρ^ns, ρ^mr) = 0 and Lemma 1 above implies that:

φ([ρ^l]^ns,[ρ^l]^mr) = 0.

(23) So since φ([ρ^l]^ns,[ρ^l]^mr) = φ(η,ζ^k) [see step #19 above], we have:

φ(η,ζ^k) = 0

(24) Since we have φ(η,ζ^k) = f(ζ^k) [see step #6 above], we can conclude:

f(ζ^k) = 0.

(25) To complete the proof, we use the same reasoning as in Theorem 2 above since:

(a) We have shown that every root of Φ_n in C is a root of f [see step #14 since for every root r, there exists k such that r = ζ^k]

(b) Using a previous result (see Corollary 2.2, here), we can further conclude that Φ_n divides f.

(c) So if Φ_n divides f and f divides Φ_n, it follows that Φ_n = f

QED

Corollary 3.1:

Let p be a prime number and let k be an integer which divides p-1.

Let ζ ∈ C be a primitive p-th root of unity

Then:

Every element in Q(μ_k)(μ_p) can be uniquely written in the form:

a₁ζ + a₂ζ² + ... + a_p-1ζ^p-1

for some a₁, ..., a_p-1 in Q(μ_k)

Proof:

(1) The hypothesis on k ensures that k is relatively prime to p since k ≡ 1 (mod p).

(2) Hence, Φ_p is irreducible over Q(μ_k) by Theorem 3 above.

(3) Let ζ be a primitive p-th root of unity.

(4) Then, Q(μ_k)(μ_p) = Q(μ_k)(ζ) since:

Using Theorem 3, here, we can conclude for all x ∈ μ_p, there exists an integer i such that x = ζⁱ.

(5) We make the following observations:

(a) ζ ∈ C is a root of Φ_p [See step #3 above]

(b) Φ_p is irreducible over Q(μ_k) [see step #2 above]

(c) Φ_p has degree p-1. [See Lemma 1, here]

(6) Using Lemma 3, here, we can conclude that every element in Q(μ_k)(ζ) can be uniquely written in the form:

a = a₀ + a₁ζ + a₂ζ² + ... + a_p-2ζ^p-2

with a_i ∈ Q(μ_k)

(7) Using the cyclotomic equation (see Lemma 1, here), we have:

Φ_p(ζ) = 1 + ζ + ζ² + ... + ζ^p-1 = 0

which means that:

ζ + ζ² + ... + ζ^p-1 = -1

(8) This gives us that:

a₀ = -a₀(ζ + ζ² + ... + ζ^p-1)

(9) Combining step #6 with step #8 gives us:

a = (a₁ - a₀)ζ + (a₂ - a₀)ζ² + ... + (a_p-2 - a₀)ζ^p-2 + (-a₀)ζ^p-1

(10) To prove uniqueness, let's assume that:

a₁ζ + ... + a_p-1ζ^p-1 = b₁ζ + ... + b_p-1ζ^p-1

(11) From step #7, we also have:

ζ^p-1 = -1 - ζ - ζ² + ... -ζ^p-2

(12) Putting this into step #10 gives us:

a₁ζ + ... + a_p-1(-1 - ζ - ζ² + ... -ζ^p-2) = b₁ζ + ... + b_p-1(-1 - ζ - ζ² + ... -ζ^p-2)

which reduces to:

-a_p-1 + (a₁ - a_p-1)ζ + (a₂ - a_p-1)ζ² + ... + (a_p-2 - a_p-1)ζ^p-2 = -b_p-1 + (b₁ - b_p-1)ζ + (b₂ - b_p-1)ζ² + ... + (b_p-2 - b_p-1)ζ^p-2

(13) From Lemma 3 here, we can conclude that the coefficients on both sides are equal which gives us:

a_p-1 = b_p-1

which then gives us:

a₁ = b₁
...
a_p-2 = b_p-2

QED

References

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