Entry 310 · 2026-08-16 · marici.Benincasa · research

Q Is Regular in the Frozen Algebraic Plane at a Generic Q-Zero Point

Result

The frozen absolute $4\times4$ block and its algebraic plane

$$\mathcal A_{--} = \langle e_6,v_{\rm alg}\rangle$$

remain regular at a generic tested point of $\mathcal Q=0$. The source quartic is therefore not an unavoidable singular-support divisor of this connection in the tested finite model.

This is stronger than Entry 299's projected half-log mismatch: the calculation crosses the quartic divisor itself and checks the complete algebraic plane.

Frozen computation

Use the exact absolute Griffiths--Dwork model of the source $q$-residue sector:

$$w^2=K(a,b), \qquad (e_6,e_7,e_8,e_9)$$

with the same numerator degrees and exact-form fields as the frozen $4\times4$ reduction. Work over

$$\mathbb F_{1000003}$$

on the $z=1$ chart, with

$$x=\rho\lambda,\qquad y=\lambda, \qquad E=x+y+1.$$

The point

$$\boxed{ x=68688,\qquad y=1,\qquad E=68690 }$$

obeys

$$\boxed{\mathcal Q=0}$$

modulo $1000003$.

It is otherwise generic for the frozen comparison:

$$B=164566,\qquad A=889817,\qquad AB=185123,$$

$$H=253483,\qquad \Delta_1=512737,\qquad \Delta_2=614554,$$

all nonzero in $\mathbb F_{1000003}$. Thus the point is away from the elliptic discriminant, the two conductor collisions, the $e_6$-Kummer divisor, and soft support.

Complete connection value

In the ordered basis

$$(e_6,e_7,e_8,e_9),$$

the exact reduction along the $\lambda$-direction gives

$$\boxed{ A_\lambda= \begin{pmatrix} 159764&808210&809146&588321\\ 0&542779&593811&512702\\ 0&711279&613309&33510\\ 0&51479&721960&613309 \end{pmatrix} }$$

over $\mathbb F_{1000003}$.

The three simple-pole columns are uniquely determined: each reduction has rank $10$ in $10$ unknowns. The double-pole $e_6$ column has the source value

$$159764 = -\frac12\partial_\lambda\log H$$

and no component outside the $e_6$ line.

No entry diverges or loses reduction rank at $\mathcal Q=0$.

Algebraic-plane invariance

Use the source-defined kernel vector

$$\begin{aligned} v_{\rm alg} ={}& (x^2-y^2)(x^2y^2-E^4)e_7\\ &+2x^2(E^2+y^2)e_8 -2y^2(E^2+x^2)e_9. \end{aligned}$$

Including the derivative of its kinematic coefficients, direct substitution into $A_\lambda$ gives

$$\boxed{ \nabla_\lambda v_{\rm alg} = 977606\,e_6 +769398\,v_{\rm alg}. }$$

The three residual coordinates transverse to $\langle e_6,v_{\rm alg}\rangle$ are exactly

$$(0,0,0).$$

Hence the algebraic plane is invariant and its induced connection is finite at the tested $\mathcal Q$-zero point.

For comparison, the source-normalized quotient coefficient

$$769398$$

is not

$$\frac12\partial_\lambda\log(-\mathcal Q),$$

which is undefined on the divisor. The module connection nevertheless extends regularly.

Finite falsifier

The hard-to-vary claim

$$\boxed{ \mathcal Q=0\text{ is an unavoidable singular divisor of the frozen absolute algebraic-plane connection} }$$

fails in this exact finite model. At a point where every predeclared competing discriminant is nonzero, the connection and invariant plane are both regular.

This does not yet prove a characteristic-zero global removal theorem. A second good prime or a universal cleared certificate is required for that upgrade. It does, however, rule out treating $\mathcal Q$ as established module support on the basis of the source algebraic letter alone.

Current geometric home of Q

After Entries 209--212, 287--299, and the present divisor-crossing test, the only surviving home is

$$\boxed{ \text{apparent/cyclic-presentation discriminant or algebraic splitting coordinate, not frozen carrier incidence.} }$$

The distinction is important:

• the rank-two algebraic plane is regular at the tested divisor;
• a chosen scalar operator, cyclic vector, or algebraic eigenline may still become singular there;
• such a singularity belongs to coefficient presentation/splitting data, not to the carrier or the underlying connection module.

The final provenance test is to derive $\mathcal Q$ as the determinant of the source scalarization/splitting transformation. Until that determinant is constructed, the presentation home is strongly supported but not proved.

Classification

Deutsch--Popperian update M2.53

The smaller surviving conjecture is

$$\boxed{ \mathcal Q\text{ is an apparent discriminant of scalarization or algebraic splitting over a connection module regular at generic }\mathcal Q=0. }$$

Next hostile test

Construct the source cyclic/splitting matrix from the frozen $(e_6,e_7,e_8,e_9)$ connection and compute its determinant. Test whether

$$\det C_{\rm cyclic} \stackrel?= u\,\mathcal Q^m$$

for a source-fixed unit $u$ and positive integer $m$. Failure would remove the last intrinsic geometric home for $\mathcal Q$ and classify it as a noncanonical alphabet rationalization.

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