Home > Algebra, Group Theory, Higher Algebra, Linear Algebra > Every finite field is commutative – Wedderbrun’s Theorem

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)|<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<n } \lambda_d \frac{q^n-1}{q^d-1}.

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<n} 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<l<n}\phi_l(X),

and therefore for {d<n} 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<n} 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.

Advertisements
  1. No comments yet.
  1. No trackbacks yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: