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 $Z \left({D}\right)$ be the center of $D$, that is:

$\left\{{z \in D: \forall d \in D: z d = d z}\right\}$

From Center of Division Ring is Subfield it follows that $Z \left({D}\right)$ is a finite field.

Thus from Characteristic of Finite Field is Prime the characteristic of $Z \left({D}\right)$ is a prime number $p$.


Now, we can consider two vector spaces.

From Field of Prime Characteristic has Unique Prime Subfield, the prime subfield of $Z \left({D}\right)$ is isomorphic to $\Z / \left({p}\right)$.

From Division Ring is Vector Space over Prime Subfield, $Z \left({D}\right)$ is thus a vector space over $\Z / \left({p}\right)$.

From Vector Space over Division Subring, $D$ is a vector space over $Z \left({D}\right)$.

Since $Z \left({D}\right)$ 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 $Z \left({D}\right)$ has $p^n$ elements and $D$ has $\left({p^n}\right)^m$ elements.


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

We want to show $D$ is commutative.

We know $Z \left({D}\right)$ is commutative, and if we can show $D = Z \left({D}\right)$ we have finished.

Now if we can show $\left \vert{D}\right \vert = \left \vert{Z \left({D}\right)}\right \vert$ then $D = Z \left({D}\right)$, and again, we have finished.

Now by considering $Z \left({D}\right)$ and $D$ as modules as we have that if $m=1$ then $\left \vert{D}\right \vert = \left \vert{Z \left({D}\right)}\right \vert$, and we have finished.

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


Now in a finite group, let $x_j$ be a representative of the conjugacy class $\left({x_j}\right)$ (the representative does not matter).

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

Let $N_D \left({x}\right)$ be the normalizer of $x$ with respect to $D$.

Then we know by the Conjugacy Class Equation that:

$\displaystyle \left \vert{D}\right \vert = \left \vert{Z \left({D}\right)}\right \vert + \sum_{j=0}^{l-1} \left[{D: N_D \left({x_j}\right)}\right]$

which by Lagrange's theorem is:

$\displaystyle \left \vert{D}\right \vert + \sum_{j=1}^l \frac{\left \vert{D}\right \vert}{\left \vert{N_D(x_j)}\right \vert}$


Now we specialize just a bit.

We consider the group of units $U \left({D}\right)$ in $D$.

Consider what the above equation tells if we start with $U \left({D}\right)$ instead of $D$.

If we centralize a multiplicative unit that is in the center, from Conjugacy Classes of Center Elements are Singletons 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 $N_D \left({u}\right)$ is not $D$.

However, $Z \left({D}\right) \subset N_D \left({u}\right)$ since any element in the center commutes with everything in $D$ including $u$.

Then $\left|{N_D \left({u}\right)}\right| = \left({p^n}\right)^m$ for $r < m$.

Suppose there are $l$ such $u$.

Then:

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


We need two results to finish.

$(1)\quad$ If $p^k-1 \backslash p^j-1$, then $k \backslash j$ (where $\backslash$ denotes divides).
$(2)\quad$ If $j \backslash k$ then $\Phi_n \backslash \dfrac{x^j-1}{x^k-1}$ where $\Phi_n$ denotes the $n$th cyclotomic polynomial .



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 $U \left({D}\right)$ using non-central elements $u_j$.

However, in doing so, we have that $\left({q^n}\right)^{\alpha_i} - 1 \backslash \left({q^n}\right)^m - 1$.

Thus by the first result $\alpha_i \backslash m$.

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

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



Source of Name

This entry was named for Joseph Wedderburn.

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.