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Determinantal descent in dimension two: the pencil determinant is a multiple of x1x2−ρx02x_1x_2-\rho x_0^2x1​x2​−ρx02​

Proved
Diaz.det_pencil_eq_conic

by carlok · Sep 8, 2026 · Mathlib 0df444a (Lean v4.33.1)

diaz-modulus-leannumber-theory

Determinantal descent in dimension two: a singular pencil over the candidate's hull descends to the Diaz conic.

Let K⊆CK \subseteq \mathbb{C}K⊆C be a subfield, u≠0u \neq 0u=0 transcendental over KKK with ρ=uuˉ∈K\rho = u\bar u \in Kρ=uuˉ∈K, and let A,B,CA, B, CA,B,C be 2×22\times22×2 matrices with entries in KKK. If

det⁡(A+uB+uˉ C)=0,\det\bigl(A + uB + \bar u\,C\bigr) = 0,det(A+uB+uˉC)=0,

then there is c∈Kc \in Kc∈K such that for all x,y,z∈Cx, y, z \in \mathbb{C}x,y,z∈C

det⁡(xA+yB+zC)=c (yz−ρ x2).\det\bigl(xA + yB + zC\bigr) = c\,\bigl(yz - \rho\,x^{2}\bigr).det(xA+yB+zC)=c(yz−ρx2).

Why. Polarizing the 2×22\times22×2 determinant writes det⁡(xA+yB+zC)\det(xA+yB+zC)det(xA+yB+zC) as a ternary quadratic form with six coefficients in KKK. Substituting uˉ=ρ/u\bar u = \rho/uuˉ=ρ/u into the hypothesis and clearing u2u^2u2 turns it into the vanishing of a polynomial of degree four in uuu with coefficients in KKK; transcendence of uuu kills all five of its coefficients. What survives is det⁡B=0\det B = 0detB=0, det⁡C=0\det C = 0detC=0, the vanishing of the xyxyxy- and xzxzxz-coefficients, and det⁡A+ρ β=0\det A + \rho\,\beta = 0detA+ρβ=0 where β\betaβ is the yzyzyz-coefficient. The displayed identity is then exactly what is left, with c=βc = \betac=β.

Role. This is the 2×22\times22×2 case of the manuscript's Determinantal descent of the Diaz conic, combined with the coefficient-matching half of Uniqueness of the 2×22\times22×2 obstruction. It says that a singular matrix over the candidate's hull Wu=Q‾⊕Q‾u⊕Q‾uˉW_u = \overline{\mathbb{Q}} \oplus \overline{\mathbb{Q}}u \oplus \overline{\mathbb{Q}}\bar uWu​=Q​⊕Q​u⊕Q​uˉ cannot be singular by accident: its determinant form, as a form in the three hull coordinates, is forced to be a scalar multiple of Qρ=x1x2−ρx02\mathcal{Q}_\rho = x_1x_2 - \rho x_0^2Qρ​=x1​x2​−ρx02​ — the conic on which the candidate lives. Two special cases are worth reading off: taking x=z=0x = z = 0x=z=0 gives det⁡B=0\det B = 0detB=0, taking x=y=0x = y = 0x=y=0 gives det⁡C=0\det C = 0detC=0; and c=0c = 0c=0 is exactly the degenerate branch in which the whole pencil is singular, which by the Dasgupta--Kakde alternative is the branch with a vanishing algebraic matrix coefficient.

Note that the statement is unconditional given the transcendence hypothesis — no unproved transcendence conjecture enters — because uˉ=ρ/u\bar u = \rho/uuˉ=ρ/u makes the candidate's hull a rational function field in one transcendental.

Source: Carlo Perassi's private manuscript on Diaz's modulus conjecture, section The precise open boundary, theorem Determinantal descent of the Diaz conic (case m=n=t=2m = n = t = 2m=n=t=2) together with the coefficient comparison in the theorem Uniqueness of the 2×22\times22×2 obstruction. No novelty is claimed.

Preamble
import Mathlib

open ComplexConjugate
Formal statement
theorem Diaz.det_pencil_eq_conic {K : Subfield ℂ} {u : ℂ} (hT : Transcendental K u) (hu0 : u ≠ 0)
    (hρ : u * conj u ∈ K) (A B C : Matrix (Fin 2) (Fin 2) ℂ)
    (hA : ∀ i j, A i j ∈ K) (hB : ∀ i j, B i j ∈ K) (hC : ∀ i j, C i j ∈ K)
    (hdet : (A + u • B + (conj u) • C).det = 0) :
    ∃ c : ℂ, c ∈ K ∧ ∀ x y z : ℂ,
      (x • A + y • B + z • C).det = c * (y * z - (u * conj u) * x ^ 2) := by sorry

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