LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The elements with a $p^{n}$ th power in the base form a perfect subfield carrying the one-step root condition

Statement

Let $L / F$ be an algebraic extension of characteristic $p > 0$ such that every nonconstant polynomial in $F [x]$ has a root in $L$ . Then

$F^{'} : = {a \in L : a^{p^{n}} \in F for some n \in N}$

is a perfect intermediate field, and every nonconstant polynomial in $F^{'} [x]$ has a root in $L$ .

Facts & Assumptions

Given: An algebraic root extension $L / F$ of characteristic $p > 0$ .

[L1]

Frobenius is injective and respects addition and multiplication in characteristic $p$ (Frobenius $x \mapsto x^{p}$ is an injective endomorphism in characteristic $p$ , and an automorphism for finite fields).

[L2]

In positive characteristic, a field is perfect exactly when Frobenius is surjective (A field is perfect exactly when it has characteristic zero or its Frobenius map is surjective).

[L3]

A subset containing $0, 1$ and closed under subtraction, multiplication, and nonzero inverses is a subfield (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

Proof

technique · direct

1.1L1L3

The set $F^{'}$ contains $F$ by taking $n = 0$ . For $a, b \in F^{'}$ , choose one exponent $N$ at least as large as exponents witnessing both memberships. Then [L1] gives $(a - b)^{p^{N}} = a^{p^{N}} - b^{p^{N}} \in F$ , $(a b)^{p^{N}} \in F$ , and, for $a \neq 0$ , $(a^{- 1})^{p^{N}} = (a^{p^{N}})^{- 1} \in F$ . Hence [L3] makes $F^{'}$ an intermediate field.

1.2L1L2choose

If $a \in F^{'}$ and $a^{p^{n}} = c \in F$ , the root hypothesis applied to $x^{p^{n + 1}} - c$ gives $b \in L$ with $b^{p^{n + 1}} = c$ . Then $(b^{p})^{p^{n}} = a^{p^{n}}$ , so injectivity in [L1] gives $b^{p} = a$ , and $b \in F^{'}$ by its displayed power. Thus Frobenius on $F^{'}$ is surjective and [L2] makes $F^{'}$ perfect.

1.3L1choose

Let $g (x) = \sum_{i = 0}^{d} a_{i} x^{i} \in F^{'} [x]$ be nonconstant. Choose one $n$ with every $a_{i}^{p^{n}} \in F$ . Then $g (x)^{p^{n}} = \sum_{i} a_{i}^{p^{n}} x^{i p^{n}}$ is a nonconstant polynomial over $F$ , so it has a root $u \in L$ .

2.1step 1.3L1∎

Since $g (u)^{p^{n}} = 0$ , injectivity of Frobenius gives $g (u) = 0$ . Thus every nonconstant polynomial over $F^{'}$ has a root in $L$ .

Depends on

Used by

An algebraic extension containing a root of every nonconstant base polynomial is algebraically closed Theorem

Dependency tree · two levels

20 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

P. L. Clark, Field Theory, Theorem 4.9 (standard reference, not scraped)
J. S. Milne, Fields and Galois Theory, Proposition 6.5 (standard reference, not scraped)

The elements with a pnpnth power in the base form a perfect subfield carrying the one-step root condition