Understanding a proof of the Wedderburn theorem

Question

Wedderburn theorem: Every finite division ring is a field.

The following proof seem easier than other which I manage to understand the theorem. Although some logic(highlighted ones) still seems unclear to me.

proof: Let $D$ be a finite division ring with center $Z$. Suppose $Z$ has $q$ element and $D$ has $q^n$ elements. We claim that $D=Z$ and $n=1$. Since, $D^{\times}=D-\{0\}$ is multiplicative group. we split up the multiplicative group $D^{\times}$ into its conjugate classes. Each element of $Z$ forms a conjugate class by itself. Suppose, there are classes $C_1,\cdots,C_k$ in addition each containing more than one element. We know that conjugate class $C$ consists of $\frac{q^n-1}{q^r-1}$ elements where $r|n$.

"conjugate class $C$ consists of $\frac{q^n-1}{q^r-1}$ elements where $r|n$", I didn't recognize or manage this result.

So corresponding to each class $C_i$ there is a factor $r_i$ of $n$ such that $C_i$ consists of $\frac{q^n-1}{q^{r_i}-1}$ elements and $r_i<n$. Hence, $$D^{\times}=\cup_{x\in Z^{\times}}[x]\bigcup \cup_{x\notin Z^{\times}}[x]\implies O(D^{\times})=O(Z^{\times})+\sum_{x\notin Z^{\times}}|C_i|\implies q^n-1=q-1+\sum_{i=1}^k \frac{q^n-1}{q^{r_i}-1}$$ Now, the nth cyclotomic polynomial $\Phi_n$ is a factor of both the polynomial $X^n-1$ and $\frac{X^n-1}{X^{r_i}-1}$.

The nth cyclotomic polynomial, for any positive integer $n$, is the unique irreducible polynomial with integer coefficients that is a divisor of $x^{n}-1$ and is not a divisor of $x^{k}-1$ for any $k < n$. but here it say also for $\frac{X^n-1}{X^{r_i}-1}$, which I couldn't understand.

Let $a=\Phi_n(q)$ then $a$ divides $q^n-1$ and $\frac{q^n-1}{q^{r_i}-1}$. Hence, $a|q-1$. If $n>1$ then for every primitive nth root of unity $\xi$ in the field of complex numbers $\mathbb C$. we have $|q-\xi|>|q-1|$ hence $|a|=\prod |q-\xi|>q-1$.

I couldn't manage to understand $|a|=\prod |q-\xi|>q-1$. Why the product come in?

And hence $a$ can't be a factor of $q-1$. It follows that there are no classes $C_i$ containing more than one element. Hence, $n=1$ and $D=Z$.

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I couldn't manage to understand the in-between logic I highlight above.

To be honest I couldn't manage to understand this theorem intuitively, like a connected logics to write the proof by own (after understanding once). It will be great help if anyone enlighten me the intuition.

Answer 1

I will enumerate your questions by 1,2,3.

1. This is because of the orbit-stabilizer theorem. Here, $D^\times$ acts on itself by conjugation. The orbits are the conjugacy classes (by their very definition). The stabilizers are the centralizers (by their very definition). Each centralizer, including $0$, is a vector space over the center $Z$, therefore has $q^r$ elements with some $r$. The orbit-stabilizer theorem now implies that the orbit has $(q^n - 1)/(q^r - 1)$ elements. Hence, $q^r - 1 \mid q^n - 1$, and this implies $r \mid n$.

2. We have the factorization $X^n - 1 = \prod_{\large d \mid n} \Phi_d$, and then also $X^{r_i} - 1 = \prod_{\large d \mid r_i} \Phi_d$. Thus, $(X^n - 1)/(X^{r_i} - 1) = \prod_{\large d \mid n, ~ d \not\mid r_i} \Phi_d$, and $\Phi_n$ divides $(X^n - 1)/(X^{r_i}-1)$.

3. The polynomial $\Phi_n$ factors as $\prod_{\zeta} (X - \zeta)$, where $\zeta$ runs through all primitive $n$-th roots of unity. Substitute $X$ by $q$, then we get $a = \Phi_n(q) = \prod_{\zeta} (q-\zeta)$.

Let me mention that all of these questions are also answered by Wikipedia's proof.

Answer 2

"conjugate class $C$ consists of $\frac{q^n-1}{q^r-1}$ elements where $r \mid n$"

Let $a\in C_1$ and $\mathrm{Fix}(a)$ is the elements fixing it under conjugation (admitting $0$): \begin{equation*} \mathrm{Fix}(a)=\{x\in D:\ ax=xa\}. \end{equation*} It is easy to check $\mathrm{Fix}(a)$ is a subring of $D$, and is a division ring: $0,1\in D$, closeness under addition, multiplication, negative elemnts are easy to check, and as for inverse of non-zero members, \begin{equation*} xa=ax\implies xax^{-1}=a\implies ax^{-1}=x^{-1}a. \end{equation*} It is immediate that $D\supseteq Z$. Recall every module over a division ring has a basis (though its weakened edition "every module over a field has a basis" is much more commonly used). Therefore $\mathrm{Fix}(a)$ is a vector space over $Z$, and $D$ is a vector space over the division ring $C(a)$. The first assertion shows that $\mathrm{Fix}(a)$ has $q^r$ elements where $r=\dim_Z \mathrm{Fix}(a)$. Recall the transitivity of dimension: \begin{equation*} \dim_Z \mathrm{Fix}(a)\dim_{\mathrm{Fix}(a)}D=\dim_ZD\overset{\text{hypothesis}}{=}{r}. \end{equation*} Therefore ${\color{violet}{r \mid n}}$. Since the multiplicative group $(D-\{0\},\cdot)$ acts on $C_1$ transitively and $\mathrm{Fix}(a)-\{0\}$ is the stablizer of $a\in C_1$, we infer \begin{equation*} \text{card}(C_1)=\frac{\text{card(}D-{0})}{\text{card}(\mathrm{Fix}(a)-\{0\})}=\frac{q^n-1}{q^r-1}, \end{equation*} where we have proved ${\color{violet}{r \mid n}}$.

Now, the nth cyclotomic polynomial $\Phi_n$ is a factor of both the polynomial $X^n-1$ and $\frac{X^n-1}{X^{r_i}-1}$.

Since $\Phi_n$ is irreducible in $\mathbb{Z}[x]$, it is a prime element in the UFD $\mathbb{Z}[x]$. Therefore \begin{equation*} (\text{In }\mathbb{Z}[x])\ \Phi_n(x) \mid \frac{X^n-1}{X^{r_i}-1}\cdot {X^{r_i}-1},\ \Phi_n(x)\nmid {X^{r_i}-1} \implies \Phi_n(x) \mid \frac{X^n-1}{X^{r_i}-1}. \end{equation*}

I couldn't manage to understand $|a|=\prod |q-\xi|>q-1$. Why the product come in?

\begin{equation} {\color{violet}{q^n-1}}=q-1+{\color{violet}{\sum_{i=1}^k \frac{q^n-1}{q^{r_i}-1}}} \end{equation} All ${\color{violet}{\text{violet}}}$ terms are multiple of $\Phi_n(q)$ as we have proved. Then we should have $\Phi_n(q) \mid q-1$, which is absurd, since \begin{equation*} \Phi_n(x)=\prod_{\substack{1\le k\le n\\\gcd(k,n)=1}}(x-e^{\frac{2k\pi i}{n}}),\ \ (\text{a well-known equivalent definition of $\Phi_n$}) \end{equation*} and $|x-e^{\frac{2k\pi i}{n}}|>|x|-|e^{\frac{2k\pi i}{n}}|=x-1$.