Proposition 3

— *Let $K$ be a finite field of $q$ elements, $\Omega$ an algebraically closed extension of $K$ and $m$ an integer $\geqslant 1$. a) There exists a unique subextension $K_m$ of $\Omega$ which is of degree $m$ over $K$. b) The field $K_m$ has $q^m$ elements and is the set of fixed points of the automorphism $x \mapsto x^{q^m}$ of $\Omega$. c) We have $K_m = K(\zeta)$ for every generator $\zeta$ of the cyclic group $K_m^*$. Let $p$ be the characteristic of $K$ and $f$ the degree of $K$ over $\mathbf{F}_p$. We have $q^m = p^{fm}$ and the mapping $x \mapsto x^{q^m}$ is thus an automorphism of the perfect field $\Omega$ (V, p. 7, Prop. 4). Therefore the set $K_m$ of roots of the polynomial $X^{q^m} - X$ of $K[X]$ is a subfield of $\Omega$. Since the derivative of $X^{q^m} - X$ is equal to $-1$, all the roots of this polynomial are simple (IV, p. 17, Prop. 7) and so $K_m$ has $q^m$ elements. It follows that $[K_m : K] = m$.

Now let L be a subextension of $\Omega$, of degree $m$ over $K$. As vector space over $K$, $L$ is isomorphic to $K^m$ and so has $q^m$ elements. We thus have $x^{q^m} = x$ for all $x \in L$ (Prop. 2), whence $L \subset K_m$. Since $[L : K] = [K_m : K] = m$, we finally have $L = K_m$.

We have thus proved Assertions a) and b), and c) is trivial.