# Wedderburn's little theorem

In mathematics, Wedderburn's little theorem states that every finite domain is a field. In other words, for finite rings, there is no distinction between domains, skew-fields and fields.

The Artin–Zorn theorem generalizes the theorem to alternative rings: every finite simple alternative ring is a field.

## History

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 Template:Harv, 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, $\operatorname {Br} (k)=H^{2}(k^{\text{al}}/k)$ coincides with $H^{1}(k^{\text{al}}/k)$ , 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

$a\mapsto ax,a\mapsto xa:A\to A$ 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

$q^{n}-1=q-1+\sum {q^{n}-1 \over q^{d}-1}$ 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

$\Phi _{f}(q)$ .

From the polynomial identities

$x^{n}-1=\prod _{m|n}\Phi _{m}(x)$ and $x^{d}-1=\prod _{m|d}\Phi _{m}(x)$ ,

we set x = q to see that

$\Phi _{n}(q)$ divides both qn−1 and ${q^{n}-1 \over q^{d}-1}$ ,

so by the above class equation $\Phi _{n}(q)$ must divide q−1, and therefore

$|\Phi _{n}(q)|\leq q-1$ .

To see that this forces n to be 1, we will show

$|\Phi _{n}(q)|>q-1$ for n > 1 using factorization over the complex numbers. In the polynomial identity

$\Phi _{n}(x)=\prod (x-\zeta )$ ,

where ζ runs over the primitive n-th roots of unity, set x to be q and then take absolute values

$|\Phi _{n}(q)|=\prod |q-\zeta |$ .

For n > 1,

$|q-\zeta |>|q-1|$ by looking the location of q, 1, and ζ in the complex plane. Thus

$|\Phi _{n}(q)|>q-1$ .