Wedderburn's Theorem

Theorem
Every finite division ring $D$ is a field.

Proof
Let $D$ be a finite division ring. If $D$ is shown commutative then $D$ is a field.

Denote $Z(D) = \{z \in D: zd = dz \forall d \in D \}$ called the center of the ring. Then $Z(D)$ is a finite field since the center of a division ring is a field. Thus the characteristic of $Z(D)$ is a prime number $p$; in otherwords, $Z(D)$ is isomorphic to $\Z/(p)$.

Now, we can consider two modules. $Z(D)$ is a module over $\Z/(p)$ and $D$ is a module over $Z(D)$ of dimension $n$ and $m$ respectively. Thus $Z(D)$ has $p^n$ elements and $D$ has $(p^n)^m$ elements.

Now the idea behind the rest of the proof is as follows. We want to show $D$ is commutative. We know $Z(D)$ is commutative, and if we can show $D=Z(D)$ we are done. Now if we can show $|D| = |Z(D)|$ then $D=Z(D)$, and again, we are done. Now by considering $Z(D)$ and $D$ as modules as we have that if $m=1$ then $|D| = |Z(D)|$, and we are done. Thus it remains to show that $m=1$.

Now in a finite group, let $x_j$ be a representative of the conjugacy class $(x_j)$ (the representative does not matter), and let there be $l$ (distinct) non-singleton conjugacy classes. Let $N_D(x)$ be the normalizer of $x$ with respect to $D$. Then we know by the conjugacy class equation theorem that $\displaystyle |D| = |Z(D)| + \sum_{j=0}^{l-1} [D:N_D(x_j)]$ which by the theorem of Lagrange is $\displaystyle |Z(D)| + \sum_{j=1}^l \frac{|D|}{|N_D(x_j)|}$.

Now we specialize just a bit. We consider the group of multiplicative units $U(D)$ in $D$.

Consider what the above equation tells if we start with $U(D)$ instead of $D$. Now, if we centralize a multiplicative unit that is in the center we get a singleton class. And the above sum only considers non-singleton classes.

Thus choose some element $u$ not in the center, so $N_D(u)$ is not $D$; however, $Z(D) \subset N_D(u)$ since any element in the center commutes with everything in $D$ including $u$. Then $\left|{N_D(u)}\right| = (p^n)^m$ for $r < m$.

Suppose there $l$ such $u$.

Then:
 * $\displaystyle |U(D)| = |Z(U(D))| - 1 + \sum_{j=1}^l \frac{|D|}{|N_D(u_j)|} = p^n - 1 +\sum_{\alpha_i}\frac{(p^n)^m-1}{(p^n)^{\alpha_i}-1}$

We need two results to finish.
 * 1) ) If $p^k-1 \backslash p^j-1$ then $k \backslash j$ (where $\backslash$ denotes divides)
 * 2) ) If $j \backslash k$ then the $n$th cyclotomic polynomial (IS THERE A PAGE FOR THIS?) denoted $\Phi_n \backslash \dfrac{x^j-1}{x^k-1}$

Now we argue by contradiction to show that $m=1$.

Assume $m>1$. Then let $\gamma_i$ be an $m$th primitive root of unity. Then the above used conjugacy class theorem tells us how to compute size of $U(D)$ using non-central elements $u_j$.

However, in doing so, we have that $ (q^n)^{\alpha_i} - 1 \backslash (q^n)^m - 1$. Thus by the first result $ \alpha_i \backslash m$. Thus $\Phi_m \backslash \dfrac{x^m-1}{x^{\alpha_i}-1}$.

However, $|p^n-\gamma_i| > p^n-1$, and so the division is impossible, contradicting our assumption that $n>1$.

He first published it in 1905. However, his proof had a gap in it.

The first complete proof was supplied by Leonard Dickson.

It is also known as Wedderburn's Little Theorem.