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## Every finite field is commutative – Wedderbrun’s Theorem

Wedderbrun’s Theorem Every finite field is commutative.

Proof: The statement of the theorem is very simple to understand, yet it is hard (if you’re not an expert) to imagine why the axioms of a field and the finiteness assumption allow us to deduce that the field is commutative.

We will prove the theorem by induction over the number of elements of our field. Denote ${K}$ the field in question. If ${K}$ has ${2}$ elements then ${K=\{0,1\}}$ which is commutative. Suppose now that any field ${L}$ with fewer elements than ${K}$ is commutative. The following lemma will be useful:

Lemma If ${L}$ is a finite field and ${K\neq L}$ is a subfield of ${L}$ which is commutative then there exists a positive integer ${n \geq 2}$ such that ${|L|=|K|^n}$. (where we denote by ${|X|}$ the number of elements of ${X}$)

Proof of the lemma: Since ${K}$ is commutative we can regard ${L}$ as a vector space over ${K}$. This vector space has finite dimension ${n}$ (since ${K}$ and ${L}$ are finite), therefore ${L}$ is isomorphic to ${K^n}$ and the relation follows.

We use the usual notations ${Z(K)= \{a \in K : ax=xa,\ \forall x \in K\}}$, ${C(a)=\{x \in K : xa=ax\}}$. We would like to prove that ${Z(K)=K}$. Suppose that this is not the case. Denote ${q=|Z(K)|. We can apply Lemma and find that there is ${n \geq 2}$ such that ${|K|=q^n}$. The class equation on the group ${(K^*,\cdot)}$ gives us

$\displaystyle q^n-1 = |Z(K)\setminus\{0\}|+\sum_{x \in O} \frac{|K^*|}{|C(x)\setminus \{0\}|},$

where ${O}$ contains representants from the orbits ${C(x) \setminus\{0\}}$ (which of course are not in ${Z(K)}$ so ${C(x) \neq K}$.

Note that each ${C(x)}$ is a subfield of ${K}$ with ${|C(x)|<|K|}$ so by the hypothesys induction ${C(x)}$ is commutative. Moreover ${Z(G)}$ is a subfield in every ${C(x)}$ with ${|C(x)|>|Z(K)|}$. The lemma tells us that there exists ${d\geq 2}$ such that ${|C(x)|=q^d}$. Since the ratio ${|K^*|/|C(x)\setminus \{0\}|}$ is an integer we have

$\displaystyle q^d-1 |q^n-1$

fron where we can deduce that ${d}$ is a divisor of ${n}$. If we denote ${\lambda_d}$ the number of ${x}$‘s for which ${|C(x)\setminus \{0\}|=q^d}$ then we obtain the relation

$\displaystyle q^n - 1 = q-1 +\sum_{d | n,\ d

We know the relation

$\displaystyle X^n - 1 = \prod_{d|n} \phi_n(X),$

where ${\phi_n}$ denotes the ${n}$-ty cyclotomic polynomial. Since ${d we have

$\displaystyle X^n-1 = \prod_{l|d} \phi_l(X) \prod_{l|n, l>d}\phi_l(X)=(X^d-1)\phi_n(X)\prod_{d

and therefore for ${d a divisor of ${n}$ we have

$\displaystyle \phi_n(X) \text{ divides } \frac{X^n-1}{X^d-1} \text{ in }\Bbb{Z}[X].$

Since ${q}$ is an integer and the above polynomials are with integer coefficients, it follows that

$\displaystyle \phi_n(q) | (q^n-1)/(q^d-1)$

for every ${d with ${d|n}$. Obviously ${\phi_n(q)|q^n-1}$ so the relation given by the class equation shows us that

$\displaystyle \phi_n(q) |q-1.$

We know that ${q \geq 2}$, therefore if ${\zeta}$ is a primitive root of unity of order ${n}$ then ${|\zeta-q|> q-1\geq 1}$. This means that

$\displaystyle |\phi_n(q)|\geq |e^{\frac{2\pi i}{n}}-q|>q-1,$

which contradicts the previous division result.

Therefore the initial assumption, that ${K}$ is not commutative was false, and the inductive step is proved. The induction argument allows us to say that every finite field is indeed commutative.