Finite fields are super great. But for some reason it seems that many people regard them as completely mysterious except for \(\mathbb{Z}/ p\mathbb{Z}\).

Today. I partially fix that. A little bit.

Theorem. Let \(\mathbb{F}_q\) be a finite field with \(q\) elements. Then \(q\) must be a prime power.

Proof. Let \(p\) be the characterisitc of \(\mathbb{F}_q\), i.e. the smallest \(p\in \mathbb{N}\) such that \(p\cdot 1 = 1+ \cdots + 1\) \(p\) times is \(0\). Note that the characteristic must be prime (because inverses exist in a field so \(ab=0\) for \(a,b\in \mathbb{Z}\) would imply \(a=0\) or \(b=0\).)

Now, consider the action of \(\mathbb{F}_p\) on \(\mathbb{F}_q.\) We claim that \(\mathbb{F}_q\) is a \(\mathbb{F}_p\) vector space. This is clearly true, because check the axioms yourself.

But then the size of \(\mathbb{F}_q\) is \(p^{dim \mathbb{F}_q}\), as claimed.

Theorem. Let \(f\) be a degree \(k\) irreducible polynomial over \(\mathbb{F}_p\) for some prime \(p\). Then \(\mathbb{F}_p[x] / (f)\) is a finite field of size \(p^{k}.\)

Proof. The only suspect part is the existence of inverses really. Bezout’s theorem / euclidean algorithm says that for any polynomials \(f,g\) there exists \(a,b\) such that \(af + bg = gcd(f,g)\). But for irreducible polynomials this is saying there exists \(b\) such that \(bg \equiv 1 (f)\). As desired.

Remark. Apparently an equivalent characterization can be made by looking at the roots of \(x^{p^{n}}-x = 0\) with some appropriate definition of addition

todo later:

In fact, a much stronger result holds: the field we defined above for \(p^{n}\) is unique up to isomorphism.

Supposedly the following Lemma is helpful for establishing this:

Lemma. The multiplicative group of a field is cyclic.