In today's blog, I present some initial lemmas that I will use to establish Step 1 of the proof using field extensions. For the proof using the ideas that Niels Abel originally presented, see here.
Lemma 1:
Let F be a field containing a primitive n-th root of unity.
If a ∈ F is not an nth power and q is prime, then xq - a is irreducible over F. (see Definition 4, here).
Proof
This follows from Lemma 2, here.
QED
For Lemma 2 below, I will use the notation F(u) which is defined in Definition 3, here.
Lemma 2:
Let:
α be a root of xq - a = 0
Then:
every γ ∈ F(α) can be written in the form:
γ = a0 + a1α + ... + aq-1αq-1
where ai are in F.
Proof:
This follows directly from the definition of F(u) [see Definition 3, here.]
QED
Lemma 3:
Let:
q be prime
and
m,k be integers such that 1 ≤ m,k ≤ q-1
Then:
there exists integers r,s such that:
rq + sk = m
and
0 ≤ s ≤ q-1
Proof:
(1) Since q is prime and k ≤ q-1, it follows that gcd(k,q)=1. [That is, they are relatively prime]
(2) It follows that 0*k, 1*k, ..., (q-1)*k forms a complete residue system modulo q. [See Lemma 3, here]
(3) This means that for any integer m, there exists an integer s such that sk ≡ m (mod q) and 0 ≤ s ≤ q-1. [See Definition 1, here for review of a complete residue system]
(4) So that m - sk ≡ 0 (mod q). [See here for review of modular arithmetic if needed]
(5) Since q divides m -sk, it follows that there exists an integer r such that rq = m - sk
(6) Then, it follows that rq + sk = m.
QED
For Corollary 3.1 below, you will need to know that F[x] refers to the set of polynomials in field F. (See here for review of polynomials and the meaning of F[x] ). I will also use the notation F(u) which is defined in Definition 3, here.
Corollary 3.1:
Assume that xq - a ∈ F[x] is irreducible and that α is a root (see here for review of roots of a polynomial).
Let γ be a nonzero element of F(α) with γ not in F.
Then:
There is an element β ∈ F(α) such that βq ∈ F
and there exists b0, b2, ..., bq-1 ∈ F such that:
γ = b0 + β + b2β2 + ... + bq-1βq-1
Proof:
(1) From Lemma 2 above, we know that there exists a0, a1, ..., aq-1 where:
γ = a0 + a1α + ... + aq-1αq-1
(2) Since γ is nonzero, we know that there exists a smallest integer k where 1 ≤ k ≤ q-1 and ak ≠ 0.
(3) Let β = akαk
(4) We know that βq ∈ F since:
(a) ak ∈ F [from Lemma 2 above]
(b) Since α is a root, we know that αq - a = 0 which implies that αq = a
(c) Since a ∈ F, it follows that αq ∈ F.
(5) Now for each k+1 ≤ m ≤ q-1, using Lemma 3 above, we know that:
there exists r,s such that:
rq + sk = m
where 0 ≤ s ≤ q-1
(6) Then there exists cs such that:
αm = (αq)r(αk)s = csβs
where cs ∈ F.
(7) Thus, if we set c0 = a0, we have:
γ = c0 + β + ck+1βk+1 + ... + cq-1βq-1
where each ci ∈ F.
QED
Lemma 4:
Let q be prime and and ζ be a primitive qth root of unity.
Then:
For each integer i:
1 + ζi + ζ2i + ... + ζ(q-1)i =
Proof:
(1) Assume that q divides i.
(2) Then there exists an integer x such that qx = i.
(3) So then:
1 + ζi + ζ2i + ... + ζ(q-1)i = 1 + (ζq)x + ... + (ζq)x*(q-1) = 1 + 1x + ... + 1x*(q-1) = q
(4) Assume q does not divide i.
(5) Then gcd(i,q)=1 and 0*i, 1*i, ..., (q-1)*i form a complete residue system modulo q. [See Lemma 3, here]
(6) Now, since ζ is a primitive q-th root of unity, we know that:
1 + ζ + ζ2 + ... + ζq-1 = 0 [See Lemma 1, here]
(7) Now, since 0*i, 1*i, ... , (q-1)*i is a complete residue system modulo q, we know that each value is uniquely congruent to a value in the complete residue system {0, 1, 2, ..., q-1}
(8) Further, we know that if i ≡ j (mod q), then ζi = ζj since ζ is a primitive qth root of unity [See Lemma 2, here]
(9) But then we have:
1 + ζi + ζ2i + ... + ζ(q-1)i = 1 + ζ + ζ2 + ... + ζq-1 = 0
QED
Lemma 5:
Let y be an a root of an irreducible polynomial f(x) over K.
Let L be a field extension of K such that y ∈ L
Then:
L is a splitting field for f(x)
Proof:
(1) If one root y is an element of L, then all the roots are elements of L. [See Theorem 3, here]
(2) From the Fundamental Theorem of Algebra (see Theorem, here), we know that if y1, ..., yn are the roots, then:
f(x) = (x - y1)*...*(x - yn)
(3) It therefore follows that L is a splitting field for f(x). [See Definition 3, here]
QED
References
- Michael I. Rosen, "Niels Hendrik Abel and Equations of the Fifth Degree", The American Mathematical Monthly, Vol. 102, No. 6 (Jun. - Jul., 1995), pp 495-595.
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