Existence of Finite Fields - Imported Knowlpedia Content

A finite field is a finite field.

Theorem (prime power cardinality and existence).

If $K$ is a finite field, then its characteristic is a prime $p$ , and there exists an integer $n\ge 1$ such that $|K|=p^n.$ Concretely, $K$ contains a copy of the prime field $\mathbb{F}_p$ , and $K$ is a finite-dimensional vector space over $\mathbb{F}_p$ of dimension $n$ , so $|K|=p^n$ .

Conversely, for every prime $p$ and integer $n\ge 1$ there exists a field of order $p^n$ , usually denoted $\mathbb{F}_{p^n}$ . One construction is: choose an irreducible polynomial $f(x)\in \mathbb{F}_p[x]$ of degree $n$ and set $\mathbb{F}_{p^n}\ \cong\ \mathbb{F}_p[x]/(f).$ This gives a field extension of degree $n$ (see degree of an extension) generated by a root of $f$ , hence a simple extension.

Remarks

A deeper refinement is that $\mathbb{F}_{p^n}$ is unique up to isomorphism (see existence and uniqueness of finite fields).

Examples

The prime fields $\mathbb{F}_p$ . For any prime $p$ , the quotient ring $\mathbb{Z}/p\mathbb{Z}$ is a field, denoted $\mathbb{F}_p$ , and has $p$ elements.

A quadratic extension: $\mathbb{F}_4$ . Over $\mathbb{F}_2$ , the polynomial $x^2+x+1$ has no root (so it is irreducible). Thus $\mathbb{F}_4 \cong \mathbb{F}_2[x]/(x^2+x+1),$ whose elements can be written $\{0,1,\alpha,\alpha+1\}$ with $\alpha^2=\alpha+1$ .

Another quadratic extension: $\mathbb{F}_9$ . Over $\mathbb{F}_3$ , the polynomial $x^2+1$ is irreducible (since $-1\equiv 2$ is not a square in $\mathbb{F}_3$ ). Hence $\mathbb{F}_9 \cong \mathbb{F}_3[x]/(x^2+1),$ and every element has the form $a+b\beta$ with $a,b\in\mathbb{F}_3$ and $\beta^2=-1$ .

(As a similar cubic example, one may construct $\mathbb{F}_8$ as $\mathbb{F}_2[x]/(f)$ for any irreducible cubic $f$ over $\mathbb{F}_2$ .)