Wedderburn's little theorem

Theorem. Every finite division ring $K$ is a field.

This classical result was first proven by Joseph Wedderburn in 1905. It is actually a rather unexpected fact that the finiteness of elements in $K$ imposes such a strong condition as the commutativity of multiplication. The finiteness assumption is clearly an essential one, because there exist infinite, non commutative division rings, such as the ring $\mathbb{H} $ of real quaternions.

The proof of Wedderburn's theorem usually given in modern textbooks (as the ones cited in the references below) is due to Ernst Witt (1931) and goes as follows.

Proof. Let $Z:=Z(K)$ be the center of $K$.  By assumption, $Z$ is a finite field and then it has $q=p^s$ elements, where $p$ is the characteristic and $s= [Z : \mathbb{F}_p]$ is the dimension of $Z$ as a $\mathbb{F}_p$-vector space.  We want to show that $Z=K$.

Since $K$ is a finite-dimensional vector space over $Z = \mathbb{F}_q$, it has $q^t$ elements for some $t \geq 1$. For all $x \in K$, let us denote by $C_x$ the centralizer of $x$ in $K$, namely $$C_x :=\{y \in K \, | \, xy=yx \}.$$ Then $C_x$ is a $Z$-vector space, and we call $q^{d(x)}$  its dimension.  Then $d(x)$ divides $t$ and $d(x)=t$ if and only if $x \in Z$.

Setting $G:=K^*$ for the multiplicative group of $K$, we see that $G$ is a finite group whose center is $Z^*$; moreover the centralizer in $G$ of an element $x$ is $C_x^*$. Then the equation of conjugate classes for $G$ yields
\begin{equation} \label{eq:classes}
q^t-1 = |G| = |Z^*| + \sum \frac{|G|}{|{C_x}^*|} = q-1 + \sum \frac{q^t-1}{q^{d(x)}-1},
\end{equation} where the sum is extended to a complete system of representatives for the conjugacy classes of $G \setminus Z^*= K \setminus Z$.

We want now to show that the only possibility is $d(x)=t=1$, namely $K=Z$. In order to do this, we exploit the theory of cyclotomic polynomials. In fact, for any positive divisor $d$ of $t$ such that $1 < d < t$, we have $$X^t -1 = \prod_{k | t} \Phi_k(X) =\Phi_t(X)(X^d-1) \left(\prod_{k<t, \, k |t, \, k \nmid d } \Phi_k(X) \right).$$ This implies that $$\frac{X^t-1}{X^d-1}$$ is a polynomial with integer coefficients, which is divided by $\Phi_t(X)$. In particular, setting $X=q$ and $d=d(x)$, we deduce that $\Phi_t(q)$ divides $(q^t-1)/(q^{d(x)}-1)$, hence by  using \eqref{eq:classes} we infer that $\Phi_t(q)$ divides $q-1$.

We now obtain the desired contradiction: indeed, if $\zeta$ is a primitive $t^{\mathrm{th}}$ root of unity, since $t >1$ and $q \in \mathbb{N}$  we have  $|q - \zeta| > q-1,$ so  $$\Phi_t(q) = \prod_{(k, \, t)=1} |q - \zeta^k| > q-1, $$ which is absurd.   $\; \Box$

Wedderburn's theorem is a purely algebraic result; however, it has unexpected consequences in projective geometry. In fact, it is known that a projective plane $\mathbb{P}$ is Desarguesian if and only if it is of the form $\mathbb{P}^2(K)$ for some division ring  $K$, and that in that case it is Pappian if and only if $K$ is a field. So we obtain the following 

Corollary. A finite projective plane $\mathbb{P}^2(K)$ is Desarguesian if and only if it is Pappian.

It is remarkable that this is the only known proof of the implication $$\textrm{Desarguesian} \Longrightarrow \textrm{Pappian}$$ for a finite projective plane $\mathbb{P}^2(K).$

References

• E. Artin: Algèbre Géométrique, Gauthier-Villars, p. 37.
• S. Gabelli: Teoria delle equazioni e teoria di Galois, Springer, p. 171.
• I. N. Herstein: Algebra, Editori Riuniti, p. 397.
• https://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem