Wedderburn's theorem

A ring R with unit element 1 is called a skew-field or division ring iff, any given nonzero element a is invertible, that is, there exists an element in R denoted by a−1 such that the relation aa−1 = a−1a = 1, holds tue (we do not require multiplication to be commutative). Thus a commutative division ring is a field.

For example, the sets ℚ, ℝ and ℂ are fields, whereas the set ℍ of quaternions (introduce by Hamilton) is only a skew-field:

ℍ = {ai + bj + ck + d where a,b,c,d ∈ ℝ, i2 = j2 = k2 = ijk = − 1, ij = −ji = k, jk = −kj = i, ki = −ik = j}

Lemma 8.6.1. If x is an integer greater than 1, xm − 1 divides xn − 1 iff m | n.

Proof. If we divide xn − 1 by xm − 1, we get the following

xn − 1 = (xm − 1) (xn−m + xn−2m + xn−3m + ... + xn−km) + (xn−km − 1)

where k is the largest positive integer sucht that n − km ≥ 0. From this we see that xm − 1 divides xn − 1 iff n = km.  □

8.6.2 Wedderburn's theorem. A finite skew field is commutative (i.e., a field).

Proof. We shall show that the center Z(K) = {k ∈ K | kx = xk ∀x ∈ K} coincides with K.
Let |Z(K)| = q. Since Z(K) is a subfield of K, K is a vector space over Z(K), and thereby has qn elements. We shall thus prove that n = 1. Let a ∈ K, and let C(a) its Centralizer: it is a sub-skew-field of K containing the center, thus its order is m which divides n. Let C(a)* = C(a) \{0} the multiplicative subgroup K*, of K, hence by Lagrange's theorem qm − 1 is a divisor of qn −1, thus m | n (Lemma 8.6.1). The class equation for the group K* can be written as

q n − 1 = q − 1 + ∑ m | n , m ≠ n q n − 1 q m − 1   8.4.1

which is a relation among integers. The condition m ≠ n is equivalent to a ∉ Z(K*). We shall show that if n > 1, this relation cannot hold. Suppose there exists an integer dividing (qn − 1)/(qm −1) in Eq. 8.4.1 then it must divide q − 1 as well. If this does not happen Eq. 8.4.1 could be verified only for n = 1. To determine this integer we resort to the nth cyclotomic polynomial, Φn(x) a polynomial in ℤ[x]. The relation

x n − 1 = Φ n ( x ) ( x d − 1 ) ∏ k | n k ∤ d k ≠ n Φ k ( x )

a relation of the form

xn − 1 = Φn(x)(xd − 1)f(x),    f(x) ∈ ℤ[x]

which holds for every x. In particular if q is an integer

Φ n ( q ) | q n − 1 q m − 1 ∀ m | n , m ≠ n

Since Φn(q) divides qn − 1, based on Eq. 8.4.1, it must divide q − 1 as well. We must have |Φn(q)| ≤ qn − 1. For n > 1 this would be impossible: Φ(q) = Π(q − θ), with θ variable among the nth primitive roots of unity, and |q − θ| > |q| − |θ| = q − 1, that is |Φn(q)| > q − 1.  □

«Finite Fields Index Normal Extensions»