TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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Finite fields of the same order are isomorphic

Statement

Any two finite fields with the same order $q$ are isomorphic. More precisely, after identifying their prime subfields with $F_{p}$ , there is an isomorphism fixing $F_{p}$ pointwise. The isomorphism need not be unique.

Facts & Assumptions

Given: Finite fields $E$ and $E^{'}$ with $∣ E ∣ = ∣ E^{'} ∣ = q$ .

[L1]

Every finite field has prime-power order with base equal to its characteristic prime (Every finite field has order $p^{n}$ for a unique prime characteristic $p$ and positive integer $n$ ).

[L2]

A field of order $q$ is a splitting field of $t^{q} - t$ over its prime subfield (A field with $q$ elements is the splitting field of $x^{q} - x$ over its prime subfield).

[L3]

Two splitting fields of the same nonzero polynomial over a base field are isomorphic by an isomorphism fixing the base (Any two splitting fields of a polynomial are isomorphic over the base field).

[L4]

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

[L5]

Canonical prime factorisation makes the prime in a positive prime-power representation unique (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

Proof

technique · direct

1.1

By [L1], each field has prime-power order with base equal to its characteristic prime. Since the orders are the same, [L5] makes these primes equal, say to $p$ . Use [L4] to identify both prime subfields with one copy of $F_{p}$ .

givenL1L4L5

2.1

By [L2], $E$ and $E^{'}$ are splitting fields of the same polynomial $t^{q} - t$ over this base.

step 1.1L2

3.1

Apply [L3] to obtain a base-fixing field isomorphism $E \to E^{'}$ . Splitting-field uniqueness asserts existence, not uniqueness, so no stronger claim follows.

step 2.1L3∎

Depends on

Used by

For every finite field F_q and every n≥1, a monic irreducible polynomial of degree n exists Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 141 results over 26 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, Theorem 2.5 (standard reference, not scraped)