The original proof was given by Joseph Wedderburn in 1905,[2] who went on to prove the theorem in 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.[2] Witt's proof is sketched below. Alternatively, the theorem is a consequence of the Skolem–Noether theorem by the following argument.[3] Let be a finite division algebra with center. Let and denote the cardinality of . Every maximal subfield of has elements; so they are isomorphic and thus are conjugate by Skolem–Noether. But a finite group (the multiplicative group of in our case) cannot be a union of conjugates of a proper subgroup; hence, .
A later "group-theoretic" proof was given by Ted Kaczynski in 1964.[4] This proof, Kaczynski's first published piece of mathematical writing, was a short, two-page note which also acknowledged the earlier historical proofs.
Relationship to the Brauer group of a finite field
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.
The triviality of the Brauer group can also be obtained by direct computation, as follows. Let and let be a finite extension of degree so that Then is a cyclic group of order and the standard method of computing cohomology of finite cyclic groups shows that
where the norm map is given by
Taking to be a generator of the cyclic group we find that has order and therefore it must be a generator of . This implies that is surjective, and therefore is trivial.
Proof
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 elementary group theory[5] that the nonzero elements of form a group under multiplication. Thus, is a division ring.
Since the center of is a field, is a vector space over with finite dimension . Our objective is then to show . If is the order of , then has order . Note that because contains the distinct elements and , . For each in that is not in the center, the centralizer of is a vector space over , hence it has order where is less than . Viewing , , and as groups under multiplication, we can write the class equation
where the sum is taken over the conjugacy classes not contained within , and the are defined so that for each conjugacy class, the order of for any in the class is . In particular, the fact that is a subgroup of implies that divides , whence divides by elementary algebra.
Parshall, K. H. (1983). "In pursuit of the finite division algebra theorem and beyond: Joseph H M Wedderburn, Leonard Dickson, and Oswald Veblen". Archives of International History of Science. 33: 274–99.