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Function-field vocabulary and Riemann's inequality: a formal genus bound for the Weil input #30

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@daira

The cited Weil input at the branch covers (#29) takes its per-cover constant from a genus fact whose statement needs vocabulary that Mathlib does not have. #28 deliberately scopes that vocabulary out; this issue tracks it, on the route that stops short of exact genus.

The route — all commutative algebra and valuation theory; no schemes, no differentials:

  • places of a function field F/K as discrete valuations trivial on K; place degrees; divisors;
  • Riemann spaces L(D): finiteness, and the one-place growth bound ℓ(D + P) ≤ ℓ(D) + deg P;
  • the pole divisor and the fundamental identity deg (x)_∞ = [F : K(x)] (Stichtenoth 1.4.11);
  • the genus, well-defined via Riemann's theorem (sup_D (deg D + 1 - ℓ(D)) is finite — the finiteness proof is essentially the same counting as the next item);
  • Riemann's inequality (Stichtenoth §3.11): for F = K(x, y), g ≤ ([F : K(x)] - 1)·([F : K(y)] - 1). At a branch cover F = F_q(u)(W) with W² = H(u) and deg H = 14, the 2r + 2 independent elements {uⁱ, uⁱ·W : i ≤ r} lie in L(r·(u)_∞ + (W)_∞), a divisor of degree 2r + 14; so g ≤ (2r + 14) + 1 - (2r + 2) = 13.

The payoff. The citation shrinks from "Weil/FFSTV Theorem 3 at covers of genus 6, computed on paper" to "Weil/FFSTV Theorem 3 at covers of formally bounded genus", leaving only the covering condition and Weil's theorem itself on paper. On constants: genus ≤ 13 gives the per-cover constant c = 2·13 - 2 = 24 (the parameter of weilBounded_zeroRepaired), so the recorded WeilBounded constant becomes C = c + 1/2 = 24.5, against the deployed C = 21/2 at c = 10. The chain is already parametric in c, so consumers change only at the final instantiation; the cost is (24.5/10.5)² ≈ 2^{2.44} in the regularity distance.

Out of scope here: exact genus 6 (the Riemann–Hurwitz tier), the no-unramified-subcover condition, and Weil's theorem itself (the Bombieri–Stepanov project).

🤖 Claude Fable 5

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