Algebra: Polynomials, Galois Theory and Applications by Frédéric Butin

Author:Frédéric Butin
Language: eng
Format: epub
Publisher: INscribe Digital
Published: 2016-07-15T16:00:00+00:00

6.2 Cyclotomic polynomials

Let K be a field, Ω be an algebraic closure of K, and be an integer that is not divisible by the characteristic of K. Then according to lemma 6.1.2, Ω contains at least a primitive nth root of unity, denoted by ω.

We define the nth cyclotomic polynomial of K by the formula

This is a monic polynomial of degree φ(n), which depends only on the prime subfield of K (see the remark that follows corollary 3.2.6).

Proposition 6.2.1

• Let K be a field. Then .

• We have .

• For every prime number p that does not divide n, (reduction modulo p of ).

Proof:

• The set of nth roots of 1 is the disjoint union of the primitive dth roots of 1 for d that divides n. The first assertion of the proposition is an immediate consequence of this fact.

• By induction on , let us show that . The result is obvious for n = 1. Let us assume that the result is true for 1, . . . , n − 1, and let us prove it for n. Let us set . By the induction hypothesis, P belongs to . Then, as the leading coefficient of P is an invertible element of , the Euclidean division in provides two polynomials Q and R of such that Xn − 1 = PQ + R with ∂°R < ∂°P. Moreover, (first part of the proof), and by uniqueness of the Euclidean division in , we deduce that (and R = 0). Consequently, belongs to .

• By induction on , let us show that for every prime number p that does not divide n, we have . The result is obvious for n = 1. Let us assume that the result is true for 1, . . . , n − 1, and let us prove it for n. Setting , then according to the first two parts of the proof, we have Xn − 1 = , and P and belong to . Therefore, by reducing modulo p, we get . As p does not divide n, p does not divide any divisor of n, thus by the induction hypothesis, we have . Moreover, according to the first part of the proof, we also have in , and by uniqueness of the Euclidean division in , we deduce that .

Theorem 6.2.2 (Wedderburn)

Every finite field is commutative.

Proof:

• Let K be a finite field. Then the characteristic of K is a prime number p, and there is an injective map from to K. The center Z of K is a commutative subfield of K, and we set q := |Z|. Then K is a Z-vector space whose dimension is denoted by d := [K : Z], so that |K| = qd. The set Zx of elements of K that commute with x is a subfield of K that contains Z, and we set dx := [Zx : Z]. By multiplicativity of the degrees, dx divides d.

• The group K× acts by conjugation on itself, and for every x ∈ K×, the stabilizer (K×)x of x is (Zx)×.

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