## Wedderburn’s little theorem (1)

Posted: September 24, 2011 in Division Rings, Noncommutative Ring Theory Notes
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In this two-part note I’m going to prove that every finite division ring is a field. This result is called Wedderburn’s little theorem. The proof we are going to give is due to Ernst Witt and that’s probably the best proof available. But before getting into the proof, we need to know a little bit about cyclotomic polynomials.

Notation. For any integer $n \geq 1$ we have the $n$-th root of unity $\zeta_n = e^{2 \pi i/n}.$

Definition. The $n$-th cyclotomic polynomial is defined by $\Phi_n(x) = \prod_{1 \leq k \leq n, \ \gcd(k,n)=1}(x - \zeta_n^k), \ n \geq 1.$

Lemma 1. $x^n - 1 = \prod_{d \mid n} \Phi_d(x).$ In particular, $\Phi_n(x) \mid x^n -1.$

Proof. We have $x^n-1 = \prod_{j=1}^n (x - \zeta_n^j)=\prod_{d \mid n} \prod_{\gcd(j,n)=d}(x-\zeta_n^j).$ But

$\prod_{\gcd(j,n)=d} (x - \zeta_n^j) = \prod_{\gcd(k, n/d)=1}(x - \zeta_n^{kd})$

and obviously $\zeta_n^d = \zeta_{n/d}.$ Hence

$x^n-1 = \prod_{d \mid n} \prod_{\gcd(k,n/d)=1}(x - \zeta_{n/d}^k)=\prod_{d \mid n} \Phi_{n/d}(x)=\prod_{d \mid n} \Phi_d(x). \ \Box$

Corollary. For every $n \geq 1 : \ \Phi_n(x) \in \mathbb{Z}[x].$

Proof. By induction over $n.$ There is nothing to prove if $n=1$ because $\Phi_1(x)=x-1.$ Now let $n \geq 2$ and suppose the $\Phi_m(x) \in \mathbb{Z}[x]$ for all $m < n.$ Note that cyclotomic polynomials are all monic. Thus, by Lemma 1

$x^n-1 = g(x) \Phi_n(x), \ \ \ \ \ \ \ (*)$

for some monic polynomial $g(x) \in \mathbb{Z}[x].$ Since $x^n-1$ is monic too, it follows from $(*)$ that $\Phi_n(x) \in \mathbb{Z}[x]. \ \Box$

Lemma 2. If $1 \leq d < n$ and $d \mid n,$ then $\displaystyle \Phi_n(x) \mid \frac{x^n - 1}{x^d - 1}.$

Proof. By Lemma 1 we have

$x^n - 1 = \prod_{m \mid n} \Phi_m(x)=\Phi_n(x) \prod_{m

Therefore, again by Lemma 1, $x^n-1 = \Phi_n(x) (x^d -1) f(x),$ where $f(x) =\prod_{m \nmid d, \ m \mid n} \Phi_m(x). \ \Box$

We will continue our discussion in part (2).