## 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 the field in question. If has elements then which is commutative. Suppose now that any field with fewer elements than is commutative. The following lemma will be useful:

**Lemma** If is a finite field and is a subfield of which is commutative then there exists a positive integer such that . (where we denote by the number of elements of )

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

We use the usual notations , . We would like to prove that . Suppose that this is not the case. Denote . We can apply Lemma and find that there is such that . The class equation on the group gives us

where contains representants from the orbits (which of course are not in so .

Note that each is a subfield of with so by the hypothesys induction is commutative. Moreover is a subfield in every with . The lemma tells us that there exists such that . Since the ratio is an integer we have

fron where we can deduce that is a divisor of . If we denote the number of ‘s for which then we obtain the relation

We know the relation

where denotes the -ty cyclotomic polynomial. Since we have

and therefore for a divisor of we have

Since is an integer and the above polynomials are with integer coefficients, it follows that

for every with . Obviously so the relation given by the class equation shows us that

We know that , therefore if is a primitive root of unity of order then . This means that

which contradicts the previous division result.

Therefore the initial assumption, that 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.