Finite Fields
Order p^n; F^× cyclic of order p^n - 1.
The idea
Theorem (Classification of finite fields).
If a field has finitely many elements, then it has $p^{n}$ of them for some prime $p$ and some integer $n \ge 1$. Conversely, for every prime power $p^{n}$ there is a field with $p^{n}$ elements, and any two finite fields of the same size are isomorphic.
We may therefore speak of the field of order $q = p^{n}$ and write it $\mathbb{F}_{q}$, or $\mathrm{GF}(q)$.
To see why only prime powers occur, add $1$ to itself repeatedly inside a finite field. The sums $1,\ 1 + 1,\ 1 + 1 + 1, \dots$ must eventually reach $0$, and the first one that does takes a prime number $p$ of steps: a composite count would factor into two smaller nonzero sums whose product is $0$, which a field forbids. The multiples of $1$ then form a copy of $\mathbb{Z}/p\mathbb{Z}$ inside the field, the field is a vector space over this copy, and a vector space of dimension $n$ over a field with $p$ elements has exactly $p^{n}$ elements.
One further theorem completes the picture.
Theorem (Multiplicative group of a finite field).
The nonzero elements of a finite field of order $q$ form a cyclic group of order $q - 1$ under multiplication: some single element has every nonzero element among its powers.
Ways to work on it
- Walkthrough. Prime-power order and the cyclic multiplicative group.
- Practice. Find the order of a finite field's multiplicative group.
- Hardest. Determine which subfields a finite field contains.
Not sure where to start? Take the ten-question placement test.