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Exercise 1871Algebra › Rings and fieldsM1Added 3 Nov 2026

Finite fields

#finite field#order#prime power

Statement

Show that a finite field has pnp^n elements for some prime pp and integer n≥1n\ge1.

Hint

The characteristic is a prime pp, so Fp⊆F\mathbb{F}_p\subseteq F. Then FF is a finite-dimensional vector space over Fp\mathbb{F}_p.

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Solution

Characteristic. The characteristic of a finite field FF is pp (prime): otherwise, Z/nZ⊆F\mathbb{Z}/n\mathbb{Z}\subseteq F for composite nn, which has zero divisors, contradicting that FF is a field.

Vector space. FF contains Fp=Z/pZ\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}, and is a finite-dimensional Fp\mathbb{F}_p-vector space of dimension n=[F:Fp]n=[F:\mathbb{F}_p].

Order. ∣F∣=∣Fp∣n=pn|F|=|\mathbb{F}_p|^n=p^n.

Conclusion. Every finite field has order pnp^n. Conversely, for every prime power pnp^n, there exists a finite field of order pnp^n (unique up to isomorphism, denoted Fpn\mathbb{F}_{p^n}).