Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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⋃n≥0Fp(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 xpn−t and put Fn=Fp(tn). Then

P:=⋃n≥0Fn

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 x↦xp is an injective endomorphism in characteristic p, and an automorphism for finite fields).

Verification

technique · direct
1.1L2construct

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

2.1step 1.1L3algebra

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.

2.2step 1.1L2algebra

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.

3.1step 2.2L1∎

By [L1], P is perfect.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources