Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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n0Fp(t1/pn) is an infinite perfect field of characteristic p

Example

Inside an algebraic closure of Fp(t), let tn be the unique root of xpnt and put Fn=Fp(tn). Then

P:=n0Fn

is an infinite perfect field of characteristic p containing Fp(t).

Facts & Assumptions

Given: A prime p and an algebraic closure Ω of Fp(t).

[L1]

In characteristic p, 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).

[L2]

Frobenius is injective and respects field operations in characteristic p (Frobenius xxp is an injective endomorphism in characteristic p, and an automorphism for finite fields).

Verification

technique · direct
1.1

The given algebraic closure contains a root tn of every xpnt, and [L2] makes that root unique. Take t0=t. Uniqueness gives tn+1p=tn, so FnFn+1.

L2construct
2.1

A nested union of fields is a field, so P is a field of characteristic p containing F0=Fp(t). The distinct powers of the indeterminate t show that this fraction field, and hence P, is infinite.

step 1.1L3algebra
2.2

If z=r(tn)/s(tn)Fn, replace tn by tn+1 in the same rational expression. Since coefficients in Fp are fixed by Frobenius, [L2] shows that the resulting element of Fn+1 has pth power z. Thus Frobenius on P is surjective.

step 1.1L2algebra
3.1

By [L1], P is perfect.

step 2.2L1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources