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The vanishing inequality behind (4.17) in the direct proof of Schneider–Lang holds for S₁ large, then E large

Proved
Transcendence.schneider_lang_vanishing_ineq

by carlok · Oct 2, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theorytranscendence

Let n≥1n \ge 1n≥1 be an integer and C≥1C \ge 1C≥1. There is an integer s0s_0s0​ such that for every integer S1≥s0S_1 \ge s_0S1​≥s0​ there is e0e_0e0​ with the following property. Let T≥1T \ge 1T≥1 and E≥e0E \ge e_0E≥e0​ be integers with log⁡T≤nlog⁡S1+nlog⁡E\log T \le n\log S_1 + n\log ElogT≤nlogS1​+nlogE, and put

U=S1ETlog⁡E2C⋅6n+1,N=U2C.U = \frac{S_1ET\log E}{2C\cdot6^{n+1}}, \qquad N = \frac{U}{2C}.U=2C⋅6n+1S1​ETlogE​,N=2CU​.

Then for every integer k<nETk < nETk<nET,

C(N+k(1+log⁡T)+T(S1+log⁡(1+k)))+log⁡k!<U.C\bigl(N + k(1 + \log T) + T(S_1 + \log(1 + k))\bigr) + \log k! < U.C(N+k(1+logT)+T(S1​+log(1+k)))+logk!<U.

It is condition 2 of Transcendence.exists_schneider_lang_parameters: set against (4.14) and (4.16), it makes every derivative of order less than nETnETnET of the auxiliary function vanish at the grid points.

It is one step of the proof of Baker's theorem formalised in this mission, which follows Chapter 4 of Waldschmidt's book: Baker's theorem from the criterion of Schneider–Lang for Cartesian products, as observed by Bertrand and Masser (1980).

Novelty. None: the inequality of step 4 of §4.6 of Waldschmidt's book (p. 139) that (4.17) is imposed to guarantee, for derivatives of total order k<nS0k < nS_0k<nS0​ with S0=ETS_0 = ETS0​=ET, and for the parameters of Transcendence.exists_schneider_lang_parameters. It is real arithmetic. The contribution of this node is the formal proof.

Preamble
import Mathlib
Formal statement
namespace Transcendence

/-- **The vanishing inequality (4.17) of the proof of Schneider–Lang** (Waldschmidt, *Diophantine Approximation
on Linear Algebraic Groups*, §4.6). Let `n ≥ 1` and `C ≥ 1`. For every large enough `S₁` there is `e₀` such that,
whenever `T ≥ 1`, `E ≥ e₀`, `log T ≤ n log S₁ + n log E`, `U = S₁ E T log E / (2C·6^{n+1})` and `N = U/(2C)`,
`C (N + k (1 + log T) + T (S₁ + log (1 + k))) + log k! < U` for every `k < n E T`. -/
theorem schneider_lang_vanishing_ineq (n : ℕ) (C : ℝ) (hn : 1 ≤ n) (hC : 1 ≤ C) :
    ∃ s₀ : ℕ, ∀ S₁ : ℕ, s₀ ≤ S₁ → ∃ e₀ : ℝ, ∀ (T E : ℕ) (U N : ℝ), 1 ≤ T → e₀ ≤ E →
      Real.log T ≤ n * Real.log S₁ + n * Real.log E →
      U = S₁ * (E * T) * Real.log E / (2 * C * 6 ^ (n + 1)) → N = U / (2 * C) →
      ∀ k : ℕ, k < n * (E * T) →
        C * (N + k * (1 + Real.log T) + T * (S₁ + Real.log (1 + k))) +
          Real.log (k.factorial : ℝ) < U := by
  sorry

end Transcendence
Source
M. Waldschmidt, Diophantine Approximation on Linear Algebraic Groups, Grundlehren 326, Springer, 2000, §4.6, step 4 and (4.17) (p. 139). Formal proof: Diaz modulus mission, 2 October 2026 (C. Perassi).

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