Proposition 4

— *Let $\mathbf{K}$ be a finite field of $q$ elements and $\mathbf{K}_m$ an extension of finite degree $m$ of $K$.

a) The field $\mathbf{K}_m$ is a Galois extension of $K$ whose Galois group is the cyclic group of order $m$ generated by the automorphism $\sigma_q : x \mapsto x^q$.

b) For each $x \in \mathbf{K}_m$ the norm of $x$ with respect to $K$ is equal to $x^{(q^m - 1)/(q - 1)}$.

c) Every element of $K$ is the trace (resp. norm) of an element of $\mathbf{K}_m$.*

Let $\Gamma$ be the cyclic group of automorphisms of $\mathbf{K}_m$ generated by $\sigma_q$. The field of invariants of $\Gamma$ consists of the elements $x$ of $\mathbf{K}_m$ such that $x^q = x$, hence is equal to $K$. Therefore $\mathbf{K}_m$ is a Galois extension of $K$ with Galois group $\Gamma$, and this latter has order equal to $[K_m : K] = m$ (V, p. 66, Th. 3). Thus a) follows.

We have $\Gamma = \{1, \sigma_q, \sigma_q^2, \ldots, \sigma_q^{m-1}\}$; the norm of an element $x$ of $\mathbf{K}_m$ with respect to $K$ is thus $N(x) = \prod_{i=0}^{m-1} \sigma_q^i(x) = x^{1 + q + \cdots + q^{m-1}}$ and we have $1 + q + \cdots + q^{m-1} = \frac{q^m - 1}{q - 1}$. This proves b). Let $\xi$ be a generator of the cyclic group $\mathbf{K}_m^*$; the image of the norm $N : \mathbf{K}_m^* \to K^*$ is the cyclic subgroup of $K^*$ generated by the element $\xi = N(\xi) = \xi^{(q^m - 1)/(q - 1)}$; since $\xi$ is of order $q^m - 1$, $\xi$ is of order $q - 1$, and so generates $K^*$. This proves that every non-zero element of K is norm of a non-zero element of K ; moreover, we have $0 = N(0)$.

Finally since $K_m$ is a separable algebraic extension of K, the trace is a non-zero linear form on the vector space $K_m$ over K (V, p. 49, Cor.) ; so every element of K is the trace of an element of $K_m$.