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Finite fields of the same order are isomorphic
Statement
Any two finite fields with the same order are isomorphic. More precisely, after identifying their prime subfields with , there is an isomorphism fixing pointwise. The isomorphism need not be unique.
Facts & Assumptions
Given: Finite fields and with .
Every finite field has prime-power order with base equal to its characteristic prime (Every finite field has order for a unique prime characteristic and positive integer ).
A field of order is a splitting field of over its prime subfield (A field with elements is the splitting field of over its prime subfield).
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).
The prime subfield of a characteristic- field is isomorphic to (A field's prime subfield is isomorphic to in characteristic zero and to in characteristic ).
Canonical prime factorisation makes the prime in a positive prime-power representation unique (For and any injective list of primes containing every prime divisor of , one has ; the exponents are determined by , and for every prime outside the list).
Proof
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 . Use [L4] to identify both prime subfields with one copy of .
By [L2], and are splitting fields of the same polynomial over this base.
Apply [L3] to obtain a base-fixing field isomorphism . Splitting-field uniqueness asserts existence, not uniqueness, so no stronger claim follows.
Depends on
- Every finite field has order $p^n$ for a unique prime characteristic $p$ and positive integer $n$
- A field with $q$ elements is the splitting field of $x^q-x$ over its prime subfield
- Any two splitting fields of a polynomial are isomorphic over the base field
- A field's prime subfield is isomorphic to $\mathbb Q$ in characteristic zero and to $\mathbb F_p$ in characteristic $p$
- For $n \ge 1$ and any injective list $p : r \to \mathbb{Z}$ 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
Used by
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Sources
- K. Conrad, Finite Fields, Theorem 2.5 (standard reference, not scraped)