Galois field

A Galois field is a field containing a finite number of elements. #m/def/ring The cardinality of a field is called its order, and finite fields only exist for orders of the form $𝑝 ℎ$ where $𝑝$ is prime. The Galois field of order $𝑝 ℎ$ , unique up to isomorphism, is denoted $𝔽 𝑝 ℎ$ or $G F (𝑝 ℎ)$ . #m/thm/ring Clearly every Galois field is in particular a Field of prime characteristic.

Construction and uniqueness

Let $𝑞 = 𝑝 ℎ$ be a where $ℎ, 𝑝 \in ℕ$ and $ℎ$ is prime. Then $𝑥 𝑞 - 𝑥 \in 𝔽 𝑝 [𝑥]$ is a separable polynomial. Moreover, a field $𝐹$ has precisely $𝑞$ elements iff it is the splitting field of $𝑥 𝑞 - 𝑥$ over $𝔽 𝑝$ , #m/thm/field whence follows uniqueness.¹

Proof

Let $𝐹$ be the splitting field of $𝑥 𝑞 - 𝑥$ over $𝕂 𝑝$ , and $𝐸 \subseteq 𝐹$ be the set of roots of $𝑥 𝑞 - 𝑥$ . Since the formal derivative $𝑓' (𝑥) = 𝑞 𝑥 𝑞 - 1 - 1 = - 1$ , we have $g c d {𝑓 (𝑥), 𝑓' (𝑥)} = 1$ , and thus $𝑓 (𝑥)$ is a separable polynomial of order $𝑞$ , so $| 𝐸 | = 𝑞$ . We show $𝐸$ is a field, whence $𝐸 = 𝐹$ , since by definition $𝐹$ is generated by $𝐸$ over $𝐾$ .

To this end, let $𝑎, 𝑏 \in 𝐸$ , whence $𝑎 𝑞 = 𝑎$ and $𝑏 𝑞 = 𝑏$ , so using the Freshman's dream

(𝑎 - 𝑏) 𝑝 ℎ = 𝑎 𝑞 + (- 1) 𝑞 𝑏 𝑞 = 𝑎 𝑞 - 𝑏 𝑞 = 𝑎 - 𝑏 .

If $𝑏 \neq 0$ ,

(𝑎 𝑏 - 1) 𝑞 = 𝑎 𝑞 (𝑏 𝑞) - 1 = 𝑎 𝑏 - 1

proving $𝐸$ is closed under subtraction and division, thus it indeed a subfield by the Tests for subfields.

For the converse, let $𝐹$ be a field such that $| 𝐹 | = 𝑞$ . Then $| 𝐹 \times | = 𝑞 - 1$ , so the multiplicative order of every element is a divisor of $𝑞 - 1$ . Therefore

𝑎 \neq 0 ⟹ 𝑎 𝑞 - 1 = 1 ⟹ 𝑎 𝑞 - 𝑎 = 0;

and we already have $0 𝑞 - 0 = 0$ . Thus, $𝑥 𝑞 - 𝑥$ has $𝑞$ roots in $𝐹$ , whence it is the splitting field, as stated.

Direct construction as quotient by a polynomial

A finite field of a given order can be constructed as a quotient of a polynomial ring. Given a polynomial ring $ℤ 𝑝 [𝑥]$ and an irreducible polynomial $𝑓$ of degree $ℎ$ , then $ℤ 𝑝 [𝑥] / ⟨ 𝑓 ⟩$ is a field of order $𝑝 ℎ$ . #m/thm/ring

Properties

Let $𝐾 = G F (𝑝 ℎ)$ . Then

$𝐾$ is a perfect field, and consequently, irreducible polynomials in $𝐾 [𝑥]$ are separable.

Proof of 1

Since the Frobenius endomorphism is injective (Field homomorphisms are injective), by the Pigeonhole principle it must also be surjective, proving ^P1.

Other results

Finite extension of a Galois field
By Wedderburn's little theorem these are the only finite division rings.

#state/tidy | #lang/en | #SemBr

2009. Algebra: Chapter 0, §VII.5.1, p. 441. ↩