proof of Wedderburn’s theorem

We want to show that the multiplication operation in a finite divisionring is abelian.

We denote the centralizer in $D$ of an element $x$ as $C_D(x)$.

Lemma. The centralizer is a subring.

$0$ and $1$ are obviously elements of $C_D(x)$ 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 alsoelements of $C_D(x)$. Moreover, for $y\ne 0$, $xy=yx$ implies$y^{-1}x=x{y}^{-1}$, so $y^{-1}$ is also an element of $C_D(x)$.

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

$$Z(D)=\bigcap _{x\in D}{C}_D(x)$$

$Z(D)$ is an abelian subring of $D$ and is thus a field. We can consider$D$ and every $C_D(x)$ as vector spaces over $Z(D)$ of dimension $n$and $n_x$ respectively. Since $D$ can be viewed as a module over$C_D(x)$ we find that $n_x$ divides $n$. If we put $q:=|Z(D)|$, we seethat $q\ge 2$ since $\{0,1\}\subset Z(D)$, and that $|C_D(x)|=q^{n_x}$and $|D|=q^n$.

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

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

$$|D^{*}|=|Z(D^{*})|+\sum _x[D^{*}:C_{D^{*}}(x)]$$

which gives

$$q^n-1=q-1+\sum _x\frac{q^n-1}{q^{n_x}-1}$$

.

By Zsigmondy’s theorem, there exists a prime $p$ that divides $q^n-1$but doesn’t divide any of the $q^m-1$ for , except in 2exceptional cases which will be dealt with separately. Such a prime$p$ will divide $q^n-1$ and each of the $\frac{q^n-1}{q^{n_x}-1}$. Soit 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 $a+b\alpha$ where $a,b\in Z(D)$. Such elements clearly commute so $D=Z(D)$ which contradictsour assumption that $n=2$. In the second case, $n=6$ and $q=2$. Theclass equation reduces to $64-1=2-1+{\sum }_x\frac{2^6-1}{2^{n_x}-1}$where $n_x$ divides 6. This gives $62=63x+21y+9z$ with $x,y$ and $z$integers, which is impossible since the right hand side is divisibleby 3 and the left hand side isn’t.