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 $$|D| = |Z(D)| + \sum_{j=0}^{l-1} [D:N_D(x_j)]$$ which by the theorem of Lagrange is $$|Z(D)| + \sum_{j=1}^l  \frac{|D|}{|N_D(x_j)|}$$.

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

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 $$\Phi_n \backslash \frac{x^j-1}{x^k-1}$$

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 $$ \Phi_m \backslash \frac{x^m-1}{x^{\alpha_i}-1}$$.

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.