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Dense rational · full coefficient support — certificate

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HodgeComputer.DenseRational.certificate

by benediamond · Nov 5, 2026 · Mathlib 0df444a (Lean v4.33.1)

algebraic-geometrydense-rationalhodge-computer

For q=1q=1q=1, and the fixed (F,A)(F,A)(F,A) of “Dense rational · full coefficient support”, prove the existence of a rational vector field vvv with components homogeneous of degree 111 and common denominator relatively prime to FFF, satisfying F2∣A−1⋅dF(v)+F⋅div⁡(v)F^{2}\mid A-1\cdot dF(v)+F\cdot\operatorname{div}(v)F2∣A−1⋅dF(v)+F⋅div(v). Divisibility is in the localization with denominators relatively prime to FFF. There is no denominator degree bound. More details.

Preamble
import Definitions.Def_HodgeComputerDenseRational

Formal statement
theorem HodgeComputer.DenseRational.certificate :
    HodgeComputer.HasCertificate
      (n := HodgeComputer.DenseRational.n)
      HodgeComputer.DenseRational.F
      HodgeComputer.DenseRational.A
      HodgeComputer.DenseRational.d
      HodgeComputer.DenseRational.q := by
  sorry
Source
https://github.com/hodge-computer/sextics/blob/83fbf5c9a0afe9136ce4f1ed324d085a2c899bd0/data/dense-rational/instance.rec
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What the Lean code literally says, in plain math · GPT-6 (OpenAI Codex; independent auditor)

The declaration has no variable parameters or hypotheses. Work in the polynomial ring R=C[z0,z1,z2,z3,z4,z5]R=\mathbb C[z_0,z_1,z_2,z_3,z_4,z_5]R=C[z0​,z1​,z2​,z3​,z4​,z5​], with the fixed natural numbers n=5n=5n=5, d=6d=6d=6, q=1q=1q=1, and A=z03z33A=z_0^3z_3^3A=z03​z33​. To specify the fixed polynomial FFF completely, put x=(z0,z1,z2)x=(z_0,z_1,z_2)x=(z0​,z1​,z2​) and y=(z3,z4,z5)y=(z_3,z_4,z_5)y=(z3​,z4​,z5​), and, for m∈{2,4,6}m\in\{2,4,6\}m∈{2,4,6} and an ordered triple t=(t0,t1,t2)t=(t_0,t_1,t_2)t=(t0​,t1​,t2​), let Mm(t)M_m(t)Mm​(t) be the row vector of all monomials t0at1bt2ct_0^at_1^bt_2^ct0a​t1b​t2c​ with a,b,c∈Na,b,c\in\mathbb Na,b,c∈N and a+b+c=ma+b+c=ma+b+c=m, ordered first by decreasing aaa and then, for each aaa, by decreasing bbb; thus it has (m+1)(m+2)/2(m+1)(m+2)/2(m+1)(m+2)/2 entries. The polynomial is

F=M6(x)c0+M4(x)BM2(y)T+M2(x)CM4(y)T+M6(y)c69973,F=\frac{M_6(x)c_0+M_4(x)B M_2(y)^T+M_2(x)C M_4(y)^T+M_6(y)c_6}{9973},F=9973M6​(x)c0​+M4​(x)BM2​(y)T+M2​(x)CM4​(y)T+M6​(y)c6​​,

where the vectors and matrices, in precisely those row and column orders, are

c0T=(9973,15,1,2,14,10,3,7,12,1,1,14,11,9,8,13,1,7,14,5,14,9973,2,17,16,16,17,9973)\begin{aligned} c_0^{T}=(&9973,15,1,2,14,10,3,\\ &7,12,1,1,14,11,9,\\ &8,13,1,7,14,5,14,\\ &9973,2,17,16,16,17,9973) \end{aligned}c0T​=(​9973,15,1,2,14,10,3,7,12,1,1,14,11,9,8,13,1,7,14,5,14,9973,2,17,16,16,17,9973)​ c6T=(9973,10,13,13,2,4,8,17,5,6,12,7,15,2,2,8,6,17,7,10,9,9973,14,11,4,10,12,9973)\begin{aligned} c_6^{T}=(&9973,10,13,13,2,4,8,\\ &17,5,6,12,7,15,2,\\ &2,8,6,17,7,10,9,\\ &9973,14,11,4,10,12,9973) \end{aligned}c6T​=(​9973,10,13,13,2,4,8,17,5,6,12,7,15,2,2,8,6,17,7,10,9,9973,14,11,4,10,12,9973)​ B=(191291517311141117261417143521013101611458511131014179156961214210131016171511546134741012816216574121512131912915171671071315156961214).B=\begin{pmatrix} 1&9&12&9&15&17\\ 3&11&14&11&17&2\\ 6&14&17&14&3&5\\ 2&10&13&10&16&1\\ 14&5&8&5&11&13\\ 10&1&4&1&7&9\\ 15&6&9&6&12&14\\ 2&10&13&10&16&1\\ 7&15&1&15&4&6\\ 13&4&7&4&10&12\\ 8&16&2&16&5&7\\ 4&12&15&12&1&3\\ 1&9&12&9&15&17\\ 16&7&10&7&13&15\\ 15&6&9&6&12&14 \end{pmatrix}.B=​136214101527138411615​9111410516101541612976​12141713849131721512109​9111410516101541612976​1517316117121641051151312​1725113914161273171514​​. C=(151629151711389517812121213166121481756214599151629151711389517812126710176821116171381633125121314611128311151514151814161027841671111).C=\begin{pmatrix} 15&16&2&9&15&17&11&3&8&9&5&17&8&12&12\\ 12&13&16&6&12&14&8&17&5&6&2&14&5&9&9\\ 15&16&2&9&15&17&11&3&8&9&5&17&8&12&12\\ 6&7&10&17&6&8&2&11&16&17&13&8&16&3&3\\ 1&2&5&12&1&3&14&6&11&12&8&3&11&15&15\\ 14&15&1&8&14&16&10&2&7&8&4&16&7&11&11 \end{pmatrix}.C=​1512156114​1613167215​21621051​96917128​1512156114​1714178316​1181121410​31731162​85816117​96917128​5251384​1714178316​85816117​1291231511​1291231511​​.

The assertion is that there exist polynomials D,Q∈RD,Q\in RD,Q∈R and a family (Ni)i=05(N_i)_{i=0}^{5}(Ni​)i=05​ of polynomials in RRR satisfying all the following conditions simultaneously: F≠0F\ne0F=0; every monomial of FFF with nonzero coefficient has total degree 666; every monomial of AAA with nonzero coefficient has total degree 666; D≠0D\ne0D=0; if eee denotes the total degree of DDD, every monomial of DDD with nonzero coefficient has total degree eee; for every i∈{0,1,2,3,4,5}i\in\{0,1,2,3,4,5\}i∈{0,1,2,3,4,5}, every monomial of NiN_iNi​ with nonzero coefficient has total degree e+1e+1e+1; every common polynomial divisor of FFF and DDD is a unit of RRR, that is, a nonzero complex constant; and the following is an equality in RRR, where ∂i\partial_i∂i​ denotes the formal partial derivative with respect to ziz_izi​:

z03z33D2−D∑i=05(∂iF)Ni+F(D∑i=05∂iNi−∑i=05(∂iD)Ni)=F2Q.z_0^3z_3^3D^2 -D\sum_{i=0}^{5}(\partial_iF)N_i +F\left(D\sum_{i=0}^{5}\partial_iN_i-\sum_{i=0}^{5}(\partial_iD)N_i\right) =F^2Q.z03​z33​D2−Di=0∑5​(∂i​F)Ni​+F(Di=0∑5​∂i​Ni​−i=0∑5​(∂i​D)Ni​)=F2Q.

The witnesses may have arbitrary complex coefficients. The degree eee has no upper bound and may be 000, allowing a nonzero constant DDD. Each zero NiN_iNi​ satisfies its monomial-degree condition vacuously, and QQQ may be zero; these possibilities remain subject to the displayed identity and the other conditions. There is no separate homogeneity, degree, or nonvanishing requirement on QQQ, and the assertion requires existence, not uniqueness, of the witnesses.

Human review
  • Endorsed by Shuze Chen · Nov 6, 2026

    Confirmed by the moderator at approval.

  • Endorsed by benediamond · Nov 6, 2026

    Confirmed by the mission captain (proposal self-audit).

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