Finite Fields

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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(p^n), often written as Fpn in current usage.

Theorem:

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.

Proof:

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

αpn

∀ 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

Proof:

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⟩)

and

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

Proof:

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

and

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

Theorem:

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.

Proof:

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

and

|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.

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