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 Φn in Q[X]
Let p be a prime number which does not divide n.
If ω ∈ C is a root of f
Then:
ωp is also a root of f and:
f(ω) = 0 → f(ωp) = 0
Proof:
(1) Assume that f(ω) = 0
(2) Assume that f(ωp) ≠ 0
(3) Since f divides Φn and Φn divides xn - 1 [See Definition 1, here for definition of Φn], 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 (ωp)n = 1 since:
(ωp)n = ωpn = (ωn)p = (1)p = 1
(7) So, we have: (ωp)n - 1 = f(ωp)g(ωp) = 0
(8) But f(ωp) ≠ 0 from step #2 above so we can conclude that g(ωp) = 0
(9) This shows that ω is a root of g(xp)
(10) Since f is irreducible, we can use a previous result (see Lemma 2, here) to conclude that f(x) divides g(xp)
(11) Thus, there exists a polynomial h(x) such that:
g(xp) = 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 Fp[x]
where Fp[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(xp) = f(x)h(x) in Fp[x]
(16) Using Fermat's Little Theorem (see Theorem, here), we know that:
ap-1 ≡ 1 (mod p)
which further implies that:
ap ≡ a (mod p)
(17) Assume that:
g(x) = a0 + a1x + ... + ar-1xr-1 + xr
(18) Then using step #16 we also have:
g(x) = a0p + a1px + ... + ar-1pxr-1 + xr
(19) But from step #17, we also have:
g(xp) = a0p + a1pxp + ... + ar-1pxp(r-1) + xpr
(20) Using the result (see Lemma, here) that (a + b)p ≡ ap + bp (mod p), we further have:
g(xp) = (a0 + a1x + ... + ar-1xr-1 + xr)p = g(x)p in Fp[x]
(21) From step #20 and step #15, we can conclude that:
gp = fh
(22) But then if f divides gp, 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 Fp[x]
(26) Let ψ = f '*g'
(27) Then we have:
xn - 1 = φ2ψ in Fp[x]
(28) Taking the derivatives from both sides (see here for linear combination rule and product rule):
nxn-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 nxn-1
which is impossible
(30) So, we reject our assumption in step #2.
QED
Theorem 2:
For every integer n ≥ 1, the cyclotomic polynomial Φn is irreducible over Q
Proof:
(1) Let f be a monic irreducible factor of Φn in Q[x]
(2) To establish this result, I will show that every root of Φn in C is a root of f and finally that f = Φn.
(3) Let ζ be a root of f.
(4) Then ζ is a root of Φn since:
(a) Since f is a factor of Φn, there exists a polynomial g such that:
Φn = fg
(b) Φn(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:
Φn 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 ζk 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*...*ps
(8) We can now use Lemma 1 above to show that any root of Φn is a root of f since if f(ζ) = 0, then also:
f(ζp1) = 0 f(ζp1p2) = 0 ... f(ζk) = 0
(9) Thus, f has as its root every primitive n-th root of unity and every root of Φn
(10) Using a previous result (see Corollary 2.2, here), we can further conclude that Φn divides f.
(11) So if Φn divides f and f divides Φn, it follows that Φn = f
QED
Theorem 3:
If m and n are relatively prime integers, then Φn is irreducible over Q(μm)
Proof:
(1) Let f be a monic irreducible factor of Φn in Q(μm)[x], that is, the coefficients of f are in Q(ηm)
(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(μm) = 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 + ... + an-1αn-1 with ai ∈ 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(μn)[x] [See step #1 above]
(b) Q(μn) = 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 + ... + an-1ηn-1 with ai ∈ Q
(d) From step #6c, it is clear that we can define a polynomial φ(η,x) such that:
f(x) = (a0,0 + a0,1η + ... + a0,n-1ηn-1) + (a1,0 + a1,1η + ... + a1,n-1ηn-1)x + ... + (an-1,0 + an-1,1η + ... + an-1,n-1ηn-1)xn-1
where φ(y,x) = (a0,0 + a0,1y + ... + a0,n-1yn-1) + (a1,0 + a1,1y + ... + a1,n-1yn-1)x + ... + (an-1,0 + an-1,1y + ... + an-1,n-1yn-1)xn-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 ηm = 1, we have:
ζ = ζmr = ζmr*(1)r =ζmr*(ηm)r =ρmr
and
η = ηns = (1)s*ηns= (ζn)s*ηns =ρns
(11) Since f(ζ) = 0, we have [see step #6 above]:
φ(η,ζ) = 0
and therefore [from step #10 above],
φ(ρns,ρmr) = 0
(12) From an earlier result (see Lemma 2, here), we can conclude that Φmn(x) divides φ(xns,xmr) since:
(a) Φmn(x) and φ(xns,xmr) are polynomials with coefficients in Q(μm).
(b) Since φ(xns,xmr) = f(x) [see step #6 above] and f(x) is irreducible in Q(μm), it follows that:
φ(xns,xmr) is irreducible in Q(μm)
(c) Finally, Φmn(x) and φ(xns,xmr) have a common root ρ in field C which contains the field Q(μm) [See step #8 above and step #11 above]
(d) Therefore, we can use Lemma 2, here to conclude that Φmn(x) divides φ(xns,xmr)
(13) It follows then that: φ(ωns,ωmr) = 0 for every mn-th root of unity ω since:
if Φmn(x) divides φ(xns,xmr), then there exists a polynomial g such that:
φ(xns,xmr) = Φmn(x)g(x)
so that if Φmn(a)=0, it follows that φ(ans,amr) =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 ζl = ζk and ηl = η [See Lemma 1, here]
(19) Since we already have observed (see step #10 above) that ζ = ρmr and η = ρns, we have:
ρlmr = ρkmr = (ρmr)k= ζk [Since ζ is an primitive n-th root of unity]
and
ρlns =ρns = 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, ρl is a primitive mn-th root of unity [See Definition 2, here].
(22) step #11 combined with Lemma 1 above gives us:
φ(ρlns,ρlmr) = 0
since:
φ(ρns, ρmr) = 0 and Lemma 1 above implies that:
φ([ρl]ns,[ρl]mr) = 0.
(23) So since φ([ρl]ns,[ρl]mr) = φ(η,ζk) [see step #19 above], we have:
φ(η,ζk) = 0
(24) Since we have φ(η,ζk) = f(ζk) [see step #6 above], we can conclude:
f(ζk) = 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 Φn in C is a root of f [see step #14 since for every root r, there exists k such that r = ζk]
(b) Using a previous result (see Corollary 2.2, here), we can further conclude that Φn divides f.
(c) So if Φn divides f and f divides Φn, it follows that Φn = 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(μk)(μp) can be uniquely written in the form:
a1ζ + a2ζ2 + ... + ap-1ζp-1
for some a1, ..., ap-1 in Q(μk)
Proof:
(1) The hypothesis on k ensures that k is relatively prime to p since k ≡ 1 (mod p).
(2) Hence, Φp is irreducible over Q(μk) by Theorem 3 above.
(3) Let ζ be a primitive p-th root of unity.
(4) Then, Q(μk)(μp) = Q(μk)(ζ) since:
Using Theorem 3, here, we can conclude for all x ∈ μp, there exists an integer i such that x = ζi.
(5) We make the following observations:
(a) ζ ∈ C is a root of Φp [See step #3 above]
(b) Φp is irreducible over Q(μk) [see step #2 above]
(c) Φp has degree p-1. [See Lemma 1, here]
(6) Using Lemma 3, here, we can conclude that every element in Q(μk)(ζ) can be uniquely written in the form:
a = a0 + a1ζ + a2ζ2 + ... + ap-2ζp-2
with ai ∈ Q(μk)
(7) Using the cyclotomic equation (see Lemma 1, here), we have:
Φp(ζ) = 1 + ζ + ζ2 + ... + ζp-1 = 0
which means that:
ζ + ζ2 + ... + ζp-1 = -1
(8) This gives us that:
a0 = -a0(ζ + ζ2 + ... + ζp-1)
(9) Combining step #6 with step #8 gives us:
a = (a1 - a0)ζ + (a2 - a0)ζ2 + ... + (ap-2 - a0)ζp-2 + (-a0)ζp-1
(10) To prove uniqueness, let's assume that:
a1ζ + ... + ap-1ζp-1 = b1ζ + ... + bp-1ζp-1
(11) From step #7, we also have:
ζp-1 = -1 - ζ - ζ2 + ... -ζp-2
(12) Putting this into step #10 gives us:
a1ζ + ... + ap-1(-1 - ζ - ζ2 + ... -ζp-2) = b1ζ + ... + bp-1(-1 - ζ - ζ2 + ... -ζp-2)
which reduces to:
-ap-1 + (a1 - ap-1)ζ + (a2 - ap-1)ζ2 + ... + (ap-2 - ap-1)ζp-2 = -bp-1 + (b1 - bp-1)ζ + (b2 - bp-1)ζ2 + ... + (bp-2 - bp-1)ζp-2
(13) From Lemma 3 here, we can conclude that the coefficients on both sides are equal which gives us:
ap-1 = bp-1
which then gives us:
a1 = b1
...
ap-2 = bp-2
QED
References
- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001
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