Corollary

— Let $K$ be a finite field and $\Omega$ an algebraically closed extension of $K$. The relative algebraic closure $\overline{K}$ of $K$ in $\Omega$ consists of () and roots of unity and is an algebraic closure of $K$.

We know $K$ to be an algebraic closure of $K$ (V, p. 22, Example 2) and it is clear that every root of unity in $\Omega$ belongs to $K$. Further, given $x \neq 0$ in $\overline{K}$, of degree $m$ over $K$, if $K$ has $q$ elements, then the field $K(x)$ has $q^m$, whence $x^{q^m - 1} = 1$, and so $x$ is a root of unity in $\Omega$.

Let $p$ be a prime number and let $\mathbf{F}_p = \mathbf{Z}/p\mathbf{Z}$ be the field of $p$ elements. We choose an algebraic closure $\Omega$ of $\mathbf{F}_p$, whose existence follows from Steinitz's theorem (V, p. 23, Th. 2). Let $f$ be a positive integer and $q = p^f$. By Prop. 3 there exists a unique subfield of $\Omega$ which is of degree $f$ over $\mathbf{F}_p$; we shall denote it by $\mathbf{F}_q(\Omega)$ or by abuse of notation $\mathbf{F}_q$. It is the unique subextension of $\Omega$ which has degree $f$ over $\mathbf{F}_p$. It is the unique subfield of $\Omega$ of cardinal $q$, and every field of cardinal $q$ is isomorphic (not canonically) to $\mathbf{F}_q$ (Cor. of Prop. 2). We note that $\mathbf{F}_q$ consists of those $x$ in $\Omega$ for which $x^q = x$ and that $F, \subset \mathbf{F}_{q'}$ if and only if $q'$ is a power of $q$.