Division Rings, Finite Division Ring is a Field

Center of a Division Ring

The center of a division ring K is the set of elements that commute with all of K. If x and y are two such elements then c*(x+y) = cx+cy = xc+yc = (x+y)*c. In other words, the center is closed under addition. The same holds for multiplication. Finally, start with cx = xc and multiply by x inverse on the left and the right to show the inverse of x lies in the center. Thus the center of K is a field. It may not be the largest field however, as shown by the complex numbers in the quaternions.

Finite Division Ring is a Field

Let K be a finite division ring and let F be the center, a field of characteristic p. Suppose K is larger than F. Thus K is an F vector space of dimension n > 1.

The multiplicative group K^* is a finite nonabelian group. Consider its class equation. In the following equation, a ranges over the nontrivial orbits, the nontrivial conjugacy classes of K. Here c(a) is the size of that conjugacy class.

|K^*| = |F^*| + ∑c(a)

The size of an orbit is the index of its stabilizer, hence c(a) = |K^*| divided by the size of the centralizer of a.

Let q = |F|, the size of the central field. Obviously |F^*| = q-1. Since K is an F vector space, |K^*| = qⁿ-1.

For a fixed element a, consider all the elements that commute with a. Certainly F commutes with a. If x and y commute with a then so does x+y. Fill in the details, and the centralizer of a is an F vector space. Since a is not in F, the vector space has dimension > 1.

Say its dimension is r. Setting zero aside, the centralizer has size q^r-1, and c(a) = qⁿ-1 over q^r-1.

If x and y commute with a then so does xy. Fill in the details, and the centralizer of a is a division ring E inside K. Now K is an E vector space, and E is an F vector space, and K is an F vector space. Combine dimensions, and r divides n. We already showed q^r-1 goes into qⁿ-1; and now we know that r goes into n.

In 1903 Wedderburn (biography) developed a complicated theorem to finish the proof, but the cyclotomic polynomials provide a simpler conclusion.

We are interested in the quotient qⁿ-1 over q^r-1. We'll insert the value of q later. For now, step back and write xⁿ-1 as a product of cyclotomic polynomials. The primitive n^th roots produce a polynomial of degree φ(n). Roots of lesser order combine to form their own cyclotomic polynomials. The root 1, with order 1, forms its own polynomial, x-1. Multiply these together to get xⁿ-1.

Do the same thing for x^r-1. Since r divides n, every cyclotomic polynomial in x^r-1 appears in xⁿ-1. divide these out, and we are left with a subset of cyclotomic polynomials drawn from xⁿ-1. Multiply them together and replace x with q to get c(a).

Since n is strictly larger than r, there are no primitive n^th roots in x^r-1. This means ζ_n(x) remains. Other cyclotomics may remain as well, but we know ζ_n(q) is part of c(a).

Let t = ζ_n(q). Now t divides each term c(a), and it also divides qⁿ-1, which is the left side of the class equation. Therefore t divides |F^*|, or q-1.

Move to the complex plane, and write ζ_n(q) as a product of factors q-y, where y ranges over the primitive n^th roots of 1. This is an equivalent formula for t.

Take the absolute value of the product, i.e. its distance from the origin. This is the product of the absolute values of the individual factors q-y. We know that q is at least 2, and each y lies on the unit circle. Thus q-y has absolute value at least 1. Each factor makes |t| a little larger. since 1 is not a primitive n^th root, we don't have to worry about q-1. All the factors q-y are larger, in absolute value, than q-1. Multiply them together and |t| exceeds q-1. Therefore t cannot divide q-1.

This is a contradiction, thus K is a field.

Earlier in ring theory we showed that a finite domain is a division ring. Combine that result with this theorem, and every finite domain is a field.