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Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$

Statement

If $F$ is a finite field, then there is a unique prime $p$ and a unique positive integer $n$ such that

$∣ F ∣ = p^{n} .$

Here $p = char F$ and $n = [F : F_{p}]$ .

Facts & Assumptions

Given: A finite field $F$ .

[L1]

The order of a finite field is the cardinality of its underlying finite set (Finite fields and their order).

[L2]

The prime subfield in positive characteristic $p$ is isomorphic to $F_{p}$ (A field's prime subfield is isomorphic to $Q$ in characteristic zero and to $F_{p}$ in characteristic $p$ ).

[L3]

Extension degree is the size of a finite basis (The degree $[K : F] = \dim_{F} K$ of a finite field extension).

[L4]

Assuming the Axiom of Choice, if $S \subseteq V$ spans $V$ then there is a basis $B$ of $V$ with $B \subseteq S$ (Every spanning subset of a vector space contains a basis).

[L5]

Coordinates with respect to a finite ordered basis are unique (A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis).

[L6]

The set of functions from an $n$ -element set to a $p$ -element set has cardinality $p^{n}$ (The set $A^{B}$ of functions $B \to A$ between finite sets is finite, with $∣ A^{B} ∣ = ∣ A ∣^{∣ B ∣}$ , Exponentiation of natural numbers, $m^{n}$ , and its agreement with the integer power in $R$ ).

[L7]

If a prime $p$ divides a finite product of integers, it divides one of the factors (If a prime $p$ divides a finite product $\prod_{i < n} a_{i}$ of integers then $p ∣ a_{i}$ for some $i < n$ ; at $n = 0$ the product is $1$ and the hypothesis cannot hold).

Proof

technique · direct

1.1

The characteristic cannot be zero, because the distinct integer multiples of $1_{F}$ would give infinitely many elements. Hence it is a unique prime $p$ , and [L2] identifies the prime subfield with $F_{p}$ .

givenL2

1.2

The finite set $F$ spans itself over $F_{p}$ , so [L4] supplies a finite basis. Its size $n$ is positive because $F$ is not the zero vector space.

givenL3L4

2.1

By [L5], taking coordinates is a bijection from $F$ to the functions from an $n$ -element basis index set to $F_{p}$ . Thus [L6] gives $∣ F ∣ = p^{n}$ .

step 1.2L1L5L6

3.1

Steps 1.1 and 1.2 exhibit the pair $(p, n)$ with $p = char F$ and $n = [F : F_{p}]$ . For uniqueness, suppose also $∣ F ∣ = ℓ^{m}$ with $ℓ$ prime and $m \geq 1$ . Then $p$ divides $ℓ^{m}$ , so [L7] gives $p ∣ ℓ$ , and primality of $ℓ$ forces $p = ℓ$ . Now $p^{n} = p^{m}$ with $p \geq 2$ forces $n = m$ , since $n < m$ would give $1 = p^{m - n} \geq p \geq 2$ and symmetrically for $m < n$ .

step 1.1step 1.2L3L7algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 140 results over 24 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

K. Conrad, Finite Fields, Section 1 (standard reference, not scraped)

Every finite field has order pnpn for a unique prime characteristic pp and positive integer nn