Wedderburn's Theorem

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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).


This needs considerable tedious hard slog to complete it.
In particular: Invoke Normalizer of Conjugate is Conjugate of Normalizer to formalise the independence of representative choice
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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 $D$.


This article, or a section of it, needs explaining.
In particular: We cannot take $D$ in the first place, since $D$ is not a group under multiplication. Doesn't it make sense to start with $\map U D$ directly? --Wandynsky (talk) 16:51, 30 July 2021 (UTC)
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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:

\(\ds \order {\map U D}\) \(=\) \(\ds \order {\map Z {\map U D} } - 1 + \sum_{j \mathop = 1}^l \frac {\order D} {\order {\map {N_D} {u_j} } }\)
\(\ds \) \(=\) \(\ds p^n - 1 + \sum_{\alpha_i} \frac {\paren {p^n}^m - 1} {\paren {p^n}^{\alpha_i} - 1}\)


This article is incomplete.
In particular: Clean the following up. This bit is due to Ernst Witt.
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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.


This page or section has statements made on it that ought to be extracted and proved in a Theorem page.
In particular: The above two results need to be proved, on their own pages.
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Aiming for a contradiction, suppose $m > 1$.

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.

$\blacksquare$


Also known as

Wedderburn's Theorem is also known as Wedderburn's little theorem.


Source of Name

This entry was named for Joseph Henry Maclagan Wedderburn.


Historical Note

Joseph Henry Maclagan Wedderburn first published the result now known as Wedderburn's Theorem in $1905$.

However, his proof had a gap in it.

The first complete proof was supplied by Leonard Eugene Dickson.


Sources

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