A finite field is a field with a finite field order (i.e., number of elements), also called a Galois field. The order of a finite field is always a prime or a power of a prime. For each prime power, there exists exactly one (with the usual caveat that “exactly one” means “exactly one up to an isomorphism“) finite field GF(), often written as Fpn in current usage.
Let E be a finite field of characteristic p.
1. The cardinality of E is
|E| = pn, for some n ≥ 1. It is denoted E = Fpn
Furthermore, E is the splitting field for the separable polynomial f(X) = Xpn − X
over Fp, so that any finite field with pn elements is isomorphic to E. In fact, E coincides with the set of roots of f.
1. Let Fp be the finite field with p elements, given by the integers modulo p. Since E has characteristic p, it contains a copy of Fp. Thus E is a field extension of Fp, and we may see E as a vector space over Fp. If the dimension is n, then let α1,…,αn be a basis. Every x in E can be written as
x = x1α1 +···+ xnαn
and there are p choices for each xi, thus a total of pn different elements in E.
2. Let E× be the multiplicative group of non-zero elements of E. If α ∈ E×, then
αpn−1 = 1
by Lagrange’s Theorem, so that
∀ α in E (including α = 0). Thus each element of E is a root off, and f is separable.
Now f has at most pn distinct roots, and we have already identified the pn elements of E as roots of f.
Corollary: If E is a finite field of characteristic p, then E/Fp is a Galois extension, with cyclic Galois group, generated by the Frobenius automorphism
σ : x → σ(x) = xp, x ∈ E
By the above proposition, we know that E is a splitting field for a separable polynomial over Fp, thus E/Fp is Galois.
Since xp = x ∀ x in Fp, we have that
Fp ⊂ F(⟨σ⟩)
that is Fp is contained in the fixed field of the cyclic subgroup generated by the Frobenius automorphism σ. But conversely, each element fixed by σ is a root of Xp − X so F(⟨σ⟩) has at most p elements. Consequently
Fp = F(⟨σ⟩)
Gal(E/Fp) = ⟨σ⟩
This can be generalized when the base field is larger than Fp.
Corollary: Let E/F be a finite field extension with |E| = pn and |F| = pm. Then E/F is a Galois extension and m|n. Furthermore, the Galois group is cyclic, generated by the automorphism
τ : x → τ(x) = xpm, x ∈ E
If the degree [E : F] = d, then every x in E can be written as
x = x1α1 +···+ xdαd and there are pm choices for each xi, thus a total of
(pm)d = pn different elements in E, so that
d = m/n
The same proof as for the above corollary holds for the rest.
Thus a way to construct a finite field E is, given p and n, to construct E = Fpn as a splitting field for Xpn − X over Fp
If G is a finite subgroup of the multiplicative group of an arbitrary field, then G is cyclic. Thus in particular, the multiplicative group E× of a finite field E is cyclic.
The proof relies on the following fact: if G is a finite abelian group, it contains an element g whose order r is the exponent of G, that is, the least common multiple of the orders of all elements of G.
Assuming this fact, we proceed as follows: if x ∈ G, then its order divides r and thus
xr = 1
Therefore each element of G is a root of Xr − 1 and
|G| ≤ r
Conversely, |G| is a multiple of the order of every element, so |G| is at least as big as their least common multiple, that is
|G| ≥ r |G| = r
Since the order of |G| is r, and it coincides with the order of the element g whose order is the exponent, we have that G is generated by g, that is G = ⟨g⟩ is cyclic. Since E× is cyclic, it is generated by a single element, say α : E = Fp(α) and α is called a primitive element of E. The minimal polynomial of α is called a primitive polynomial.