proof of Wedderburn’s theorem

We want to show that the multiplicationPlanetmathPlanetmath operationMathworldPlanetmath in a finite division ring is abelianMathworldPlanetmath.

We denote the centralizerMathworldPlanetmath in D of an element x as CD(x).

Lemma. The centralizer is a subring.

0 and 1 are obviously elements of CD(x) and if y and z are, then x(-y)=-(xy)=-(yx)=(-y)x, x(y+z)=xy+xz=yx+zx=(y+z)x and x(yz)=(xy)z=(yx)z=y(xz)=y(zx)=(yz)x, so -y,y+z, and yz are also elements of CD(x). Moreover, for y0, xy=yx implies y-1x=xy-1, so y-1 is also an element of CD(x).

Now we consider the center of D which we’ll call Z(D). This is also a subring and is in fact the intersectionDlmfMathworldPlanetmath of all centralizers.


Z(D) is an abelian subring of D and is thus a field. We can consider D and every CD(x) as vector spaces over Z(D) of dimension n and nx respectively. Since D can be viewed as a module over CD(x) we find that nx divides n. If we put q:=|Z(D)|, we see that q2 since {0,1}Z(D), and that |CD(x)|=qnx and |D|=qn.

It suffices to show that n=1 to prove that multiplication is abelian, since then |Z(D)|=|D| and so Z(D)=D.

We now consider D*:=D-{0} and apply the conjugacy class formula.


which gives



By Zsigmondy’s theorem, there exists a prime p that divides qn-1 but doesn’t divide any of the qm-1 for 0<m<n, except in 2 exceptional cases which will be dealt with separately. Such a prime p will divide qn-1 and each of the qn-1qnx-1. So it will also divide q-1 which can only happen if n=1.

We now deal with the 2 exceptional cases. In the first case n equals 2, which would D is a vector space of dimension 2 over Z(D), with elements of the form a+bα where a,bZ(D). Such elements clearly commute so D=Z(D) which contradicts our assumptionPlanetmathPlanetmath that n=2. In the second case, n=6 and q=2. The class equationMathworldPlanetmath reduces to 64-1=2-1+x26-12nx-1 where nx divides 6. This gives 62=63x+21y+9z with x,y and z integers, which is impossible since the right hand side is divisible by 3 and the left hand side isn’t.

Title proof of Wedderburn’s theorem
Canonical name ProofOfWedderburnsTheorem
Date of creation 2013-03-22 13:10:50
Last modified on 2013-03-22 13:10:50
Owner lieven (1075)
Last modified by lieven (1075)
Numerical id 8
Author lieven (1075)
Entry type Proof
Classification msc 12E15