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.
== Also known as ==
== Historical Note ==
== Sources ==
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Category:Wedderburn's Theorem
Category:Division Rings
Category:Galois Fields