TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

For every prime $p$ and $n \geq 1$ , a field with $p^{n}$ elements exists

Statement

For every prime $p$ and every integer $n \geq 1$ , there exists a field with exactly $p^{n}$ elements.

Facts & Assumptions

Given: A prime $p$ , a positive integer $n$ , and $q = p^{n}$ .

[L1]

In characteristic $p$ , the roots of $t^{q} - t$ in a field form a subfield and are all simple (In characteristic $p$ , the roots of $x^{p^{n}} - x$ form a subfield and are all simple).

[L2]

Every nonzero polynomial over a field has a splitting field (Every nonzero polynomial over a field has a splitting field).

[L3]

The quotient $Z / p$ is a field (For every prime $p$ , the two operations on $Z / p$ make it a field).

[L4]

A splitting field is generated over the base by all roots of the polynomial (Polynomials that split and splitting fields of a polynomial or a family of polynomials).

[L5]

A field with finite underlying set is a finite field and its order is its cardinality (Finite fields and their order).

Proof

technique · constructive

1.1givenL2L3chooseconstruct

Over the field $F_{p}$ from [L3], use [L2] to choose a splitting field $E$ of $h (t) = t^{q} - t$ .

2.1step 1.1L1L4

Let $R$ be the root set of $h$ in $E$ . By [L1], $R$ is a subfield of $E$ and all roots are simple. By [L4], the roots generate $E$ , while the subfield $R$ already contains them and the base; hence $E = R$ .

3.1step 2.1L1algebra

The degree- $q$ polynomial $h$ splits in $E$ and has no repeated roots, so it has exactly $q$ distinct roots. Thus $∣ E ∣ = ∣ R ∣ = q = p^{n}$ .

4.1step 3.1L5discharge-construct∎

By [L5], $E$ is the required finite field.

Depends on

Used by

Every prime power order q admits q-1 mutually orthogonal Latin squares Corollary
For every finite field F_q and every n≥1, a monic irreducible polynomial of degree n exists Corollary
F₃(μ₅)capF₃(μ₇) is larger than F₃ although five and seven are coprime Counterexample
Reed-Solomon outer code and binary linear inner code Definition
Standard subgroups of finite general linear groups Definition
FALSE: degrees add in a tower of finite field extensions False statement
FALSE: F_pⁿ is the ring ℤ/pⁿℤ False statement
For every prime power q, the space PG(2,q) is a projective plane of order q Theorem
Over F_q, x^qⁿ-x is the product of all monic irreducibles whose degrees divide n Theorem

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Conrad, Finite Fields, Theorem 2.5 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Propositions 4.19-4.24 (standard reference, not scraped)

For every prime pp and n≥1n≥1, a field with pnpn elements exists