Wedderburn's little theorem
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.
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.
Sketch of 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 the elementary group theory that the nonzero elements of A form a group under multiplication. Thus, A is a skew-field. 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. For each x in A that is not in the center, the centralizer Zx of x has order qd where d divides n and is less than n. Viewing Z(A)*, Z*x and A* as groups under multiplication, we can write the class equation
where the sum is taken over all representatives x that is not in Z(A) and d are the numbers discussed above. qn−1 and qd−1 both admit factorization in terms of cyclotomic polynomials
From the polynomial identities
- and ,
we set x = q to see that
- divides both qn−1 and ,
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,
by looking the location of q, 1, and ζ in the complex plane. Thus
- Shult, Ernest E. (2011). Points and lines. Characterizing the classical geometries. Universitext. Berlin: Springer-Verlag. p. 123. ISBN 978-3-642-15626-7. Zbl 1213.51001.
- Lam (2001), p. 204
- Theorem 4.1 in Ch. IV of Milne, class field theory, http://www.jmilne.org/math/CourseNotes/cft.html
- e.g., Exercise 1.9 in Milne, group theory, http://www.jmilne.org/math/CourseNotes/GT.pdf
- 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.
- Lam, Tsit-Yuen (2001). A first course in noncommutative rings. Graduate texts in mathematics 131 (2 ed.). Springer. ISBN 0-387-95183-0.