Finite Fields

GF(p), GF(p^n) construction, and the Frobenius map

Finite Fields

A finite field (or Galois field) GF(q) exists if and only if q is a prime power pⁿ. For each prime power there is exactly one finite field (up to isomorphism). GF(p) = Z/pZ is the simplest case. For n > 1, GF(pⁿ) is constructed by quotienting Fₚ[x] by an irreducible polynomial of degree n.

In this lesson, you will explore GF(p) arithmetic, construct extension fields GF(pⁿ), and meet the Frobenius automorphism.

GF(p) Arithmetic

The field GF(p) is simply Z/pZ: integers modulo a prime. Its multiplicative group GF(p)* is cyclic of order p-1, meaning some element g generates all nonzero elements as powers g, g², ..., gᵖ⁻¹ = 1. This generator is called a primitive element.

p =

The primitive element g = 2 generates all 4 nonzero elements by repeated multiplication.

Key insight: The multiplicative group of any finite field is cyclic. This fundamental fact underlies discrete logarithm cryptography and the construction of extension fields.

Constructing GF(pⁿ)

To build GF(pⁿ), take an irreducible polynomial of degree n over GF(p) and form the quotient GF(p)[x]/(irreducible). The elements are polynomials of degree < n, and arithmetic is performed modulo the irreducible. Different irreducible polynomials give isomorphic fields.

Elements are polynomials of degree < 2 with coefficients in GF(2). Cell colors encode the product index. Every nonzero element has a multiplicative inverse.

Key insight: GF(4) is NOT Z/4Z. It is F₂[x]/(x²+x+1), a field with four elements where every nonzero element has an inverse. Z/4Z has zero divisors (2 × 2 = 0) and is not a field.

The Frobenius Automorphism

The map φ: x → xᵖ is an automorphism of GF(pⁿ) that fixes GF(p) pointwise. The Frobenius generates the Galois group Gal(GF(pⁿ)/GF(p)) ≅ Z/nZ. Its orbits reveal the subfield structure and connect finite field theory to Galois theory.

01αα+1

Frobenius: \u03B1 \u2192 \u03B1^2

0 (fixed)
1 (fixed)
α → α+1 → α

Fixed points: 2 = |GF(2)| (the prime subfield)

Frobenius order: 2

Gal(GF(22)/GF(2)) = \u27E8Frob\u27E9, order 2

Key insight: The Frobenius automorphism has order n, and its fixed field is exactly GF(p). The subfields of GF(pⁿ) correspond to divisors of n, matching the subgroups of the cyclic Galois group.

Key Takeaways

  • Finite fields exist for every prime power order and are unique up to isomorphism.
  • GF(p) = Z/pZ; GF(pⁿ) is constructed via irreducible polynomials over GF(p).
  • The multiplicative group of any finite field is cyclic.
  • The Frobenius map generates the Galois group and reveals the subfield lattice.