The original proof was given by Joseph Wedderburn in 1905, who went on to prove it two other ways. Another proof was given by Leonard Eugene Dickson shortly after Wedderburn's original proof, and Dickson acknowledged Wedderburn's priority. However, as noted in (Parshall 1983), Wedderburn's first proof was incorrect – it had a gap – and his subsequent proofs appeared only after he had read Dickson's correct proof. On this basis, Parshall argues that Dickson should be credited with the first correct proof.
A simplified version of the proof was later given by Ernst Witt. Witt's proof is sketched below. Alternatively, the theorem is a consequence of the Skolem–Noether theorem by the following argument. Let D be a finite division algebra with center k. Let [D : k] = n2 and q denote the cardinality of k. Every maximal subfield of D has qn elements; so they are isomorphic and thus are conjugate by Skolem–Noether. But a finite group (the multiplicative group of D in our case) cannot be a union of conjugates of a proper subgroup; hence, n = 1.
A later "group-theoretic" proof was given by Theodore Kaczynski. This proof, Kaczynski's first published piece of mathematical writing, was a short, two-page note which also acknowledged the earlier historical proofs.
The theorem is essentially equivalent to saying that the Brauer group of a finite field is trivial. In fact, this characterization immediately yields a proof of the theorem as follows: let k be a finite field. Since the Herbrand quotient vanishes by finiteness, coincides with , which in turn vanishes by Hilbert 90.
Let A be a finite domain. For each nonzero x in A, the two maps
are injective by the cancellation property, and thus, surjective by counting. It follows from the elementary group theory that the nonzero elements of A form a group under multiplication. Thus, A is a skew-field.
To prove that every finite skew-field is a field, we use strong induction on the size of the skew-field. Thus, let A be a skew-field, and assume that all skew-fields that are proper subsets of A are fields. Since the center Z(A) of A is a field, A is a vector space over Z(A) with finite dimension n. Our objective is then to show n = 1. If q is the order of Z(A), then A has order qn. Note that because Z(A) contains the distinct elements 0 and 1, q>1. For each x in A that is not in the center, the centralizer Zx of x is clearly a skew-field and thus a field, by the induction hypothesis, and because Zx can be viewed as a vector space over Z(A) and A can be viewed as a vector space over Zx, we have that Zx has order qd where d divides n and is less than n. Viewing Z(A)*, A*, and the Z*x as groups under multiplication, we can write the class equation
where the sum is taken over the conjugacy classes not contained within Z(A)*, and the d are defined so that for each conjugacy class, the order of Z*x for any x in the class is qd − 1. qn − 1 and qd − 1 both admit polynomial factorization in terms of cyclotomic polynomials
In the polynomial identities
we set x = q. Because each d is a proper divisor of n,
so by the above class equation must divide q − 1, and therefore
To see that this forces n to be 1, we will show
for n > 1 using factorization over the complex numbers. In the polynomial identity
where ζ runs over the primitive n-th roots of unity, set x to be q and then take absolute values
For n > 1, we see that for each primitive n-th root of unity ζ,
because of the location of q, 1, and ζ in the complex plane. Thus