Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Ramified frobenius has no canonical lift

Statement refuted

The residue Frobenius congruence does not determine a unique element of D at a ramified prime. At 2 in Q(i), with P=(1+i), one has D=I=C2 and κ(P)=F2. Identity and complex conjugation are distinct lifts of the same arithmetic Frobenius coset in D/I.

Facts & Assumptions

Given: The data and hypotheses of the statement.

[F1]

Gaussian and eisenstein frobenius: In Q(i), an odd prime p splits if p≡1(mod4) and is inert if p≡3(mod4); arithmetic Frobenius sends i↦ip. The prime 2 ramifies. In Q(ζ3), a prime p≠3 splits if p≡1(mod3) and is inert if p≡2(mod3); arithmetic Frobenius sends ζ3↦ζ3p. The prime 3 ramifies.

[F2]

Arithmetic frobenius coset: For finite Galois L/K and nonzero P∣p, the arithmetic Frobenius coset is the unique element of D(P/p)/I(P/p) corresponding under the residue isomorphism to x↦xNp on κ(P), where Np=∣κ(p)∣. It is defined also when P is ramified. Its inverse is called geometric Frobenius. The quotient element is distinguished; a representative in D need not be unique.

Counterexample

1.1F1

The Gaussian calculation gives (2)=P2 and residue field F2. There is only one prime above 2, so both automorphisms stabilize it. Modulo P, i=-1=1, and conjugation also sends i to -i=1. As every integral element is a+bi, both automorphisms act identically on every residue.

2.1F2step 1.1∎

Thus D=I=C2 and D/I is trivial. The arithmetic map on F2 is x squared, which is identity on its two elements 0 and 1. Both identity and conjugation satisfy its congruence, but they differ on i in the number field. This refutes uniqueness from residue data; it does not preclude an additional external convention from selecting a representative.

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