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, by definition, $D$ is a field.

Let $\map Z D$ be the center of $D$, that is:


 * $\map Z D := \set {z \in D: \forall d \in D: z d = d z}$

From Center of Division Ring is Subfield it follows that $\map Z D$ is a Galois field.

Thus from Characteristic of Galois Field is Prime the characteristic of $\map Z D$ is a prime number $p$.

Let $\Z / \ideal p$ denote the quotient ring over the principal ideal $\ideal p$ of $\Z$.

From Field of Prime Characteristic has Unique Prime Subfield, the prime subfield of $\map Z D$ is isomorphic to $\Z / \ideal p$.

From Division Ring is Vector Space over Prime Subfield, $\map Z D$ is thus a vector space over $\Z / \ideal p$.

From Vector Space over Division Subring is Vector Space, $D$ is a vector space over $\map Z D$.

Since $\map Z D$ and $D$ are finite, both vector spaces are of finite dimension.

Let $n$ and $m$ be the dimension of the two vector spaces respectively.

It now follows from Cardinality of Finite Vector Space that $\map Z D$ has $p^n$ elements and $D$ has $\paren {p^n}^m$ elements.

Now the idea behind the rest of the proof is as follows.

We want to show $D$ is commutative.

By definition, $\map Z D$ is commutative.

Hence it is to be shown that $D = \map Z D$.

It is shown that:
 * $\order D = \order {\map Z D}$

Hence $D = \map Z D$, and the proof is complete.

$\map Z D$ and $D$ are considered as modules.

We have that if $m = 1$ then:
 * $\order D = \order {\map Z D}$

and the result then follows.

Thus it remains to show that $m = 1$.

In a finite group, let $x_j$ be a representative of the conjugacy class $\tuple {x_j}$ (the representative does not matter).

Let there be $l$ (distinct) non-singleton conjugacy classes.

Let $\map {N_D} x$ denote the normalizer of $x$ with respect to $D$.

Then we know by the Conjugacy Class Equation that:
 * $\ds \order D = \order {\map Z D} + \sum_{j \mathop = 0}^{l - 1} \index D {\map {N_D} {x_j} }$

which by Lagrange's theorem is:
 * $\ds \order D + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {x_j} } }$

Consider the group of units $\map U D$ in $D$.

Consider what the above equation tells if we start with $\map U D$ instead of $D$.

If we centralize a multiplicative unit that is in the center, from Conjugacy Class of Element of Center is Singleton we get a singleton conjugacy class.

Bear in mind that the above sum only considers non-singleton classes.

Thus choose some element $u$ not in the center, so $\map {N_D} u$ is not $D$.

However, $\map Z D \subset \map {N_D} u$ because any element in the center commutes with everything in $D$ including $u$.

Then:
 * $\order {\map {N_D} u} = \paren {p^n}^m$

for $r < m$.

Suppose there are $l$ such $u$.

Then:

We need two results to finish.
 * $(1):\quad$ If $p^k - 1 \divides p^j - 1$, then $k \divides j$

where $\divides$ denotes divisibility.


 * $(2)\quad$ If $j \divides k$ then $\Phi_n \divides \dfrac{x^j - 1} {x^k - 1}$

where $\Phi_n$ denotes the $n$th cyclotomic polynomial.

$m > 1$.

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 $\map U D$ using non-central elements $u_j$.

However, in doing so, we have that:
 * $\paren {q^n}^{\alpha_i} - 1 \divides \paren {q^n}^m - 1$

Thus by the first result:
 * $\alpha_i \divides m$

Thus:
 * $\Phi_m \divides \dfrac {x^m - 1} {x^{\alpha_i} - 1}$

However:
 * $\size {p^n - \gamma_i} > p^n - 1$

Thus the division is impossible.

This contradicts our assumption that $m > 1$.

Hence $m = 1$ and the result follows, as determined above.