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
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:
F/Kas discrete valuations trivial onK; place degrees; divisors;L(D): finiteness, and the one-place growth boundℓ(D + P) ≤ ℓ(D) + deg P;deg (x)_∞ = [F : K(x)](Stichtenoth 1.4.11);sup_D (deg D + 1 - ℓ(D))is finite — the finiteness proof is essentially the same counting as the next item);F = K(x, y),g ≤ ([F : K(x)] - 1)·([F : K(y)] - 1). At a branch coverF = F_q(u)(W)withW² = H(u)anddeg H = 14, the2r + 2independent elements{uⁱ, uⁱ·W : i ≤ r}lie inL(r·(u)_∞ + (W)_∞), a divisor of degree2r + 14; sog ≤ (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 ofweilBounded_zeroRepaired), so the recordedWeilBoundedconstant becomesC = c + 1/2 = 24.5, against the deployedC = 21/2atc = 10. The chain is already parametric inc, 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