Existence and uniqueness of finite fields - Imported Knowlpedia Content

A finite field is a field with finitely many elements.

Theorem (Existence and uniqueness). Let $q=p^n$ where $p$ is prime and $n\ge1$ .

(Existence) There exists a field $\mathbb{F}_q$ with exactly $q$ elements. It has characteristic $p$ .
(Uniqueness) Any two fields with $q$ elements are isomorphic (so $\mathbb{F}_q$ is well-defined up to unique isomorphism).

A concrete construction is:

Start with $\mathbb{F}_p$ .
Choose an irreducible polynomial $f(x)\in \mathbb{F}_p[x]$ of degree $n$ .
Form the quotient $\mathbb{F}_p[x]/(f)$ , which is a field of size $p^n$ .

Remarks

Uniqueness can be seen by noting that every field with $q$ elements is the splitting field of $x^q-x$ over $\mathbb{F}_p$ , and splitting fields are unique up to $\mathbb{F}_p$ -isomorphism.

Examples

$q=p$ . Then $\mathbb{F}_p\cong \mathbb{Z}/p\mathbb{Z}$ is the unique field of order $p$ .

$q=4=2^2$ . Take $f(x)=x^2+x+1\in\mathbb{F}_2[x]$ , which has no root in $\mathbb{F}_2$ and hence is irreducible. Then $\mathbb{F}_4\cong \mathbb{F}_2[x]/(x^2+x+1)$ .

$q=9=3^2$ . The polynomial $f(x)=x^2+1\in\mathbb{F}_3[x]$ has no root in $\mathbb{F}_3$ (since $0^2+1=1$ , $1^2+1=2$ , $2^2+1=2$ ), so it is irreducible. Then $\mathbb{F}_9\cong \mathbb{F}_3[x]/(x^2+1)$ .