For notations and the results we have already proved, see part (1).

Lemma 3.  Let $q \geq 1, \ n \geq 2$ be integers. Then $|\Phi_n(q)| > q-1.$

Proof. By definition of $\Phi_n,$ it suffices to prove that $|q - \alpha| > q- 1$ for any $n$-th root of unity $\alpha \neq 1.$ So if $\alpha = \cos \theta + i \sin \theta,$ then $\cos \theta < 1$ and thus

$|q - \alpha|^2=q^2-(2\cos \theta)q + 1 > (q-1)^2 \geq q-1. \ \Box$

Wedderburn’s Little Theorem. (J. M. Wedderburn, 1905) Every finite division ring is a field.

Proof. Let $D$ be a finite division ring with the center $Z.$ Then $Z$ is a (finite) field and $D$ is a finite dimensional vector space over $Z.$ So if $|Z|=q$ and $\dim_Z D=n,$ then $|D|=q^n.$ If $n = 1,$ then $D=Z$ and we are done. So we will assume that $n \geq 2$ and we will get a contradiction. Let $D^{\times}=D \setminus \{0\},$ as usual, be the multiplicative group of $D.$ Clearly $Z^{\times}=Z \setminus \{0\}$ is the center of $D^{\times}.$ Also, for any $a \in D,$ let $C(a)$ be the centralizer of $a$ in $D.$ Then $C(a)$ is also a finite division ring and thus a finite dimensional vector space over $Z.$ Let $\dim_Z C(a)=n_a.$ Then $|C(a)|=q^{n_a}.$ It is clear that the centralizer of $a$ in $D^{\times}$ is $C(a)^{\times}=C(a) \setminus \{0\}.$ So the class equation of $D^{\times}$ gives us

$\displaystyle q^n-1=|D^{\times}|=|Z^{\times}| + \sum_{a}[D^{\times}:C(a)^{\times}]=q-1 + \sum_{a} \frac{q^n-1}{q^{n_a}-1}. \ \ \ \ \ \ (\dagger)$

By Lemma 1 and Lemma 2 in part (1), $|\Phi_n(q)|$ divides both $q^n-1$ and $\sum_a \displaystyle \frac{q^n-1}{q^{n_a}-1}.$ So $|\Phi_n(q)| \mid q-1,$ by $(\dagger ).$ Thus $|\Phi_n(q)| \leq q-1,$ contradicting Lemma 3. $\Box$

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