a) The characteristic of K is a prime number p, and there exists an integer f \geq 1 such that q = p^f.

b) The additive group of K is a direct sum of f cyclic groups of order p.

c) The multiplicative group of K is cyclic of order q - 1.

Since Z is infinite and K finite, the unique ring homomorphism \varphi : Z \to K is not injective and its kernel is a prime ideal of Z not equal to 0. Therefore the characteristic of K is a prime number and K is an algebra over the field \mathbf{F}_p of p elements (V, p. 2). Let f be the degree of K over \mathbf{F}_p. If f were infinite, K would contain, for each integer n \geq 0 a subspace of dimension n over \mathbf{F}_p, whence q \geq p^n, which is absurd. Therefore f is finite. The additive group of K is thus isomorphic to (\mathbf{F}_p)^f, whence the assertions a) and b).

Assertion c) follows by Lemma 1 (V, p. 78).

1 In agreement with the conventions of this chapter we shall only be concerned here with finite commutative fields. In fact, every finite field is commutative, as we shall see in Chapter VIII (cf. V, p. 170, Exercise 14).

For every $x \neq 0$ in $K$ we have $x^{q-1} = 1$ because $K^*$ is a finite group of order $q-1$ (I, p. 52). It follows that $x^q = x$ for each $x$ in $K$. The polynomial $X^q - X$ of $\mathbf{F}_p[X]$ is of degree $q$ and it has $q$ roots in $K$, whence

Proposition 2 follows at once from this.

Let $K'$ be a finite field of $q$ elements; its characteristic is a prime number $p'$ dividing $q = p^f$, whence $p' = p$. Therefore $K'$ is a splitting field of the polynomial $X^q - X$ in $\mathbf{F}_p[X]$ (Prop. 2), and so $K$ and $K'$ are isomorphic (V, p. 22, Cor.).

When $K = \mathbf{F}_p$, Formula (1) reduces to the relation

in the polynomial ring $\mathbf{Z}[X]$.

Formula (2) may also be written

In particular for $X = 0$ we obtain (« Wilson's formula »)

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.

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 $q$.

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 $K_m$.

Let $\Gamma$ be the subgroup of $\mathrm{Gal}(\bar{K}/K)$ generated by $\sigma_q$. For every integer $m > 0$ we have $\sigma_q^m(x) = x^{q^m}$ for all $x \in \bar{K}$, hence the set of fixed points of $\sigma_q^m$ is equal to $K$. Since $K \neq \bar{K}$, we have $\sigma_q^m \neq 1$. Hence there is an isomorphism $\pi_0$ of $\mathbf{Z}$ onto $\Gamma$ which maps 1 to $\sigma_q$.

The field of invariants of $\Gamma$ consists of the $x \in K$ such that $x^q = x$, hence is equal to $K$. Therefore (V, p. 67, Lemma 2) the group $\Gamma$ is dense in $\mathrm{Gal}(\bar{K}/K)$. Since every subextension of $\bar{K}$ of finite degree over $K$ is one of the fields $K_m$, the subgroups $\mathrm{Gal}(\bar{K}/K_m)$ form a fundamental system of neighbourhoods of 1 in $\mathrm{Gal}(\bar{K}/K)$. It is clear that $\Gamma \cap \mathrm{Gal}(\bar{K}/K_m)$ is the cyclic group generated by $\sigma_q^m$, hence equal to $\pi_0(m\mathbf{Z})$.

From the above remarks about the topology of $\mathbf{Z}$ the isomorphism $\pi_0 : \mathbf{Z} \to \Gamma$ extends in a unique fashion to an isomorphism of topological groups $\pi_K : \mathbf{Z} \to \mathrm{Gal}(\bar{K}/K)$.

Let $m \geq 1$ be an integer; it is clear that the Frobenius automorphism of $\bar{K}$ relative to $K$ is $\sigma_q^m$. Hence we obtain the relation

Therefore, taking into account Prop. 6 of V, p. 84 :