_{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 x

^{n}- 1 [See Definition 1, here for definition of Φ

_{n}], then, there exists a polynomial g such that:

x

^{n}- 1 = fg

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

(5) Since f(ω) = 0, it follows that ω

^{n}= 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 ω

^{n}= 1

(6) We also know that (ω

^{p})

^{n}= 1 since:

(ω

^{p})

^{n}= ω

^{pn}= (ω

^{n})

^{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: x

^{n}- 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:

x

^{n}- 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) = a

_{0}+ a

_{1}x + ... + a

_{r-1}x

^{r-1}+ x

^{r}

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

g(x) = a

_{0}

^{p}+ a

_{1}

^{p}x + ... + a

_{r-1}

^{p}x

^{r-1}+ x

^{r}

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

g(x

^{p}) = a

_{0}

^{p}+ a

_{1}

^{p}x

^{p}+ ... + a

_{r-1}

^{p}x

^{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}) = (a

_{0}+ a

_{1}x + ... + a

_{r-1}x

^{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:

x

^{n}- 1 = f(x)g(x) = (f' φ)(g'φ ) in F

_{p}[x]

(26) Let ψ = f '*g'

(27) Then we have:

x

^{n}- 1 = φ

^{2}ψ 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 x

^{n}- 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 = p

_{1}*...*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 a

_{0}+ a

_{1}α + a

_{2}α

^{2}+ ... + a

_{n-1}α

^{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:

a

_{0}+ a

_{1}η + a

_{2}η

^{2}+ ... + a

_{n-1}η

^{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

_{0,1}η + ... + a

_{0,n-1}η

^{n-1}) + (a

_{1,0}+ a

_{1,1}η + ... + a

_{1,n-1}η

^{n-1})x + ... + (a

_{n-1,0}+ a

_{n-1,1}η + ... + a

_{n-1,n-1}η

^{n-1})x

^{n-1}

where φ(y,x) = (a

_{0,0}+ a

_{0,1}y + ... + a

_{0,n-1}y

^{n-1}) + (a

_{1,0}+ a

_{1,1}y + ... + a

_{1,n-1}y

^{n-1})x + ... + (a

_{n-1,0}+ a

_{n-1,1}y + ... + a

_{n-1,n-1}y

^{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 ζ

^{n}= 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}= (ζ

^{n})

^{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:

a

_{1}ζ + a

_{2}ζ

^{2}+ ... + a

_{p-1}ζ

^{p-1}

for some a

_{1}, ..., 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 = ζ

^{i}.

(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 = a

_{0}+ a

_{1}ζ + a

_{2}ζ

^{2}+ ... + a

_{p-2}ζ

^{p-2}

with a

_{i}∈ Q(μ

_{k})

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

Φ

_{p}(ζ) = 1 + ζ + ζ

^{2}+ ... + ζ

^{p-1}= 0

which means that:

ζ + ζ

^{2}+ ... + ζ

^{p-1}= -1

(8) This gives us that:

a

_{0}= -a

_{0}(ζ + ζ

^{2}+ ... + ζ

^{p-1})

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

a = (a

_{1}- a

_{0})ζ + (a

_{2}- a

_{0})ζ

^{2}+ ... + (a

_{p-2}- a

_{0})ζ

^{p-2}+ (-a

_{0})ζ

^{p-1}

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

a

_{1}ζ + ... + a

_{p-1}ζ

^{p-1}= b

_{1}ζ + ... + b

_{p-1}ζ

^{p-1}

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

ζ

^{p-1 }= -1 - ζ - ζ

^{2}+ ... -ζ

^{p-2}

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

a

_{1}ζ + ... + a

_{p-1}(-1 - ζ - ζ

^{2}+ ... -ζ

^{p-2}) = b

_{1}ζ + ... + b

_{p-1}(-1 - ζ - ζ

^{2}+ ... -ζ

^{p-2})

which reduces to:

-a

_{p-1}+ (a

_{1}- a

_{p-1})ζ + (a

_{2}- a

_{p-1})ζ

^{2}+ ... + (a

_{p-2}- a

_{p-1})ζ

^{p-2}

^{ }= -b

_{p-1}+ (b

_{1}- b

_{p-1})ζ + (b

_{2}- b

_{p-1})ζ

^{2}+ ... + (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:

a

_{1}= b

_{1}

...

a

_{p-2}= b

_{p-2}

QED

References

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