LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

How statement and proof provenance work

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.

In characteristic $p$ , the roots of $x^{p^{n}} - x$ form a subfield and are all simple

Statement

Let $E$ be a field of characteristic $p > 0$ , let $n \geq 1$ , and put $q = p^{n}$ . Then

$R_{q} : = {a \in E : a^{q} = a}$

is a subfield of $E$ . Every root of $t^{q} - t$ is simple.

Facts & Assumptions

Given: A field $E$ of characteristic $p$ , a positive integer $n$ , and $q = p^{n}$ .

[L1]

The $n$ -fold Frobenius iterate $a \mapsto a^{p^{n}}$ is an injective field endomorphism (Frobenius $x \mapsto x^{p}$ is an injective endomorphism in characteristic $p$ , and an automorphism for finite fields).

[L2]

A subset containing $1$ and closed under subtraction, multiplication, and inverses of nonzero elements is a subfield (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

[L3]

The formal derivative of $\sum a_{i} t^{i}$ is $\sum i a_{i} t^{i - 1}$ (The formal derivative of a polynomial).

[L4]

A root is repeated if and only if the derivative also vanishes there (A root is repeated exactly when it is also a root of the formal derivative).

Proof

technique · direct

1.1givenL1

The elements $0$ and $1$ lie in $R_{q}$ . Since the map $a \mapsto a^{q}$ is a field endomorphism by [L1], if $a, b \in R_{q}$ then $(a - b)^{q} = a^{q} - b^{q} = a - b$ and $(a b)^{q} = a^{q} b^{q} = a b$ .

1.2givenL3algebra

By [L3], the derivative of $t^{q} - t$ is $q t^{q - 1} - 1 = - 1$ , because $q = p^{n}$ is zero in characteristic $p$ . It vanishes nowhere.

2.1step 1.1L1L2

If $0 \neq a \in R_{q}$ , then $(a^{- 1})^{q} = (a^{q})^{- 1} = a^{- 1}$ , so $a^{- 1} \in R_{q}$ . Thus [L2] makes $R_{q}$ a subfield.

3.1step 1.2L4∎

Hence [L4] says every root of $t^{q} - t$ is simple.

Depends on

Used by

Dependency tree · two levels

21 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, Section 2 (standard reference, not scraped)

In characteristic pp, the roots of xpn−xxpn−x form a subfield and are all simple