Existence and uniqueness of finite fields

A finite field is a field with finitely many elements.

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

1. (Existence) There exists a field $\mathbb{F}_q$ with exactly $q$ elements. It has characteristic $p$.
2. (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 $p^n$.

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

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

2. $q=4=2^2$. 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 $\mathbb{F}_4\cong \mathbb{F}_2[x]/(x^2+x+1)$.

3. $q=9=3^2$. 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 $\mathbb{F}_9\cong \mathbb{F}_3[x]/(x^2+1)$.