Odd-field coverings with multiplicity

20 September 2026 — First jets give the exact double-cover number over every finite field

Working proof, with exact finite controls; resolves the stated double-cover target internally. No novelty claim, independent review or formal verification. Produced by Codex (gpt-6-astra, medium reasoning).

Cite as: Kamil Braun, “First jets give the exact double-cover number over every finite field”, Odd-field coverings with multiplicity, Noemesis research notebooks, 20 September 2026, https://kbr.is-a.dev/math-research/branches/odd-field-double-covering/#entry-2026-09-20-double-cover-first-jets.

Assignment and stopping point. Pursue the odd-field double-covering question, stopping when solved or after five research cycles. The top obligation is multiplicity-sensitive-lower-bound: the inherited ternary gap is one plane. This cycle checks the source and first-jet constraints, stopping at a general implication or a falsification. Initial goal/status checking preceded instrumentation; the mathematical investigation and source retrieval are measured below.

Source comparison. The published Bishnoi–Boyadzhiyska–Das–Mészáros Question 4.2 predicts \(f_q(n,k,1)=(q-1)(n+1)+q(k-2)\) at sufficiently large dimension. Here only \(k=2\) is addressed. A bounded search found no later resolution of this exact case; that is not a novelty certificate. Ball–Serra (author manuscript dated 21 January 2009), Theorem 4.1 and its hyperplane application Theorem 5.3, already give the required lower bound when the origin is uncovered. Their nonvanishing premise must not be applied when the origin is covered once. The degree-controlled division underlying their Theorem 2.1 handles that case via first derivatives. The full elementary argument follows, independently of the heterogeneous frontier.

A first-jet degree bound, including the exceptional simple zero

Working lemma. Let \(q\) be any prime power, \(n\ge1\), and \(P\in\mathbb F_q[x_1,\ldots,x_n]\). Suppose \(\operatorname{mult}_a(P)\ge2\) for every nonzero \(a\in\mathbb F_q^n\), where multiplicity is Hasse multiplicity. Put \(D=n(q-1)\). If \(P(0)\ne0\), then \(\deg P\ge D+q-1\). If \(\operatorname{mult}_0(P)=1\), then \(\deg P\ge D+q\). In particular multiplicity at most one at zero implies degree at least \(D+q-1\). No restriction on dimension or on the characteristic relative to the degree is imposed.

Degree-controlled division and the one-point selector. Set \(F_i=x_i^q-x_i\). Repeated monic division in each variable writes every polynomial of total degree \(d\) as

The last bound follows directly during division: canceling a monomial divisible by \(x_i^q\) subtracts \(F_i\) times a monomial of degree at most \(d-q\), and the introduced term has lower degree. Iterating in all variables never increases total degree. If \(d<q\), the corresponding cofactors are zero. A polynomial of individual degrees below \(q\) is determined by its values on the full grid: induct on the number of variables, using the univariate root bound in the last variable. Thus the reduced representative of a function supported at zero is uniquely

The polynomial \(\Delta\) has degree \(D\) and a nonzero coefficient at \(\prod_jx_j^{q-1}\). Reduction never increases degree, so every polynomial nonzero at zero and zero elsewhere has degree at least \(D\). This proves the needed one-point bound directly rather than assuming it for a new multiplicity algebra.

First derivatives retain the missing information. Since \(q=0\) in \(\mathbb F_q\), we have \(\partial_j F_i=-\mathbf1_{i=j}\). At every grid point \(a\), the division identity therefore gives

Order at least two is equivalent to vanishing of the value and all first derivatives over any characteristic: these are exactly the constant and degree-one coefficients of \(P(a+z)\). No higher ordinary derivatives or factorial divisions are used. All first derivatives above are zero at nonzero grid points.

Case 1: origin value nonzero. Here \(c\ne0\), and the value bound first gives \(d\ge D\). Suppose for contradiction that \(d<D+q-1\). Then \(\deg H_i\le D-2\), while

is individually reduced and has degree exactly \(D-1\), with nonzero top coefficient. The polynomial \(H_i-c\partial_i\Delta\) vanishes at every nonzero grid point and has degree at most \(D-1\). By the one-point bound it must vanish at zero as well. Reduce it modulo the \(F_j\): its remainder must be zero by value uniqueness. But the remainder of \(H_i\) has degree at most \(D-2\) and cannot cancel the degree-\(D-1\) term of \(c\partial_i\Delta\). This is a contradiction. Negative degree bounds mean the zero polynomial, so the same reasoning includes \(q=2,n=1\).

Case 2: origin a simple zero. Now \(c=0\), so the value remainder \(R\) is zero. Some first derivative \(\partial_iP(0)\) is nonzero. The displayed derivative identity implies that \(H_i\) vanishes at every nonzero grid point and \(H_i(0)=-\partial_iP(0)\ne0\). Hence \(\deg H_i\ge D\) by the one-point bound. Since \(\deg H_i\le d-q\), we obtain \(d\ge D+q\). This closes the case not covered by a direct use of the nonvanishing-origin punctured bound.

Exact covering number

Working theorem. For every prime power \(q\ge2\) and every positive integer \(n\),

Lower bound. Let the cover contain \(m\) proper affine hyperplanes, with repetitions allowed, and choose a nonconstant affine defining form \(L_r\) for each occurrence. The formal product \(P=\prod_{r=1}^mL_r\) is nonzero and has total degree exactly \(m\): its top homogeneous part is a product of nonzero linear polynomials in an integral domain. At a point \(a\), each incident form has order one and every other form has order zero; products add orders because their lowest homogeneous parts have nonzero product. Consequently \(P\) has order at least two at every nonzero point and order zero or one at the origin. The lemma gives \(m\ge n(q-1)+q-1\); if the origin is covered once it gives the stronger \(m\ge n(q-1)+q\). Repeated planes are counted correctly by this argument.

Matching construction. Take every plane \(x_i=c\), for \(i\le n\) and \(c\in\mathbb F_q^\times\), and every plane \(\sum_i x_i=c\), for \(c\in\mathbb F_q^\times\). There are \((q-1)(n+1)\) occurrences and none covers zero. A vector with at least two nonzero coordinates is covered by at least two coordinate planes. A vector with exactly one nonzero coordinate is covered once by a coordinate plane and once by the sum plane with that same nonzero value. These exhaust nonzero vectors. When \(n=1\), the two families coincide but their repeated occurrences give the required double coverage. This is the source's construction, not a new upper bound. The stronger simple-origin bound is also attained by adding any plane through zero to this cover.

Scope. This proves the branch's odd-prime-power target for every positive dimension, stronger than the sufficiently-large-dimension quantifier. It does not settle Question 4.2 for \(k\ge3\). The argument does not use the frontier theorem or claim novelty for degree-controlled division, one-point interpolation or the previously known uncovered-origin bound.

Focused correctness review and finite controls

The review checked both origin cases separately, total-degree bounds during division, the nonzero top monomial of \(\partial_i\Delta\), first-order Hasse/ordinary-derivative equivalence, and additivity of orders for repeated affine factors. Full-field identities hold for prime powers, not just prime fields. The zero polynomial is excluded by the origin hypothesis. No reduction of the original covering product is mistaken for multiplicity preservation: values are reduced only to derive cofactor identities, and the original derivatives enforce multiplicity.

Reproducible exact rank controls tested \((q,n)=(2,2),(3,1),(3,2),(3,3),(5,2),(9,2)\). The extension field is represented as \(\mathbb F_3[\alpha]/(\alpha^2+1)\). At degree \(D+q-2\), adding all origin-jet constraints does not raise the rank of the nonzero-point constraints: all such polynomials already have a double zero at the origin. At degree \(D+q-1\), precisely the origin value becomes free, while the derivatives remain forced to zero; at degree \(D+q\), all first jets become available. The construction's coverage was checked on each grid. These finite controls attempt to falsify the theorem; the general proof is the argument above. Full ranks and matrix hashes are retained, with no floating-point rank or unbounded enumeration.

Assessment and process. The multiplicity-sensitive obligation and odd-field extension are discharged by the working proof. The chosen first-order test succeeded without building a thicker quotient framework or accumulating isolated cases. The short proof reuses classical tools more directly than the initial frontier proposal; no new framework mechanism was needed. Source novelty, independent review and any publication remain separate follow-up matters. The user-specified solved-or-five-cycles stopping condition is reached after this completed cycle, subject to checkpoint validation.

20 September 2026 — Higher jets refute the unrestricted polynomial route already at triple covering

Working general obstruction with finite exact checks; the affine hyperplane-covering conjecture remains open. Produced by Codex (gpt-6-astra, medium reasoning).

Cite as: Kamil Braun, “Higher jets refute the unrestricted polynomial route already at triple covering”, Odd-field coverings with multiplicity, Noemesis research notebooks, 20 September 2026, https://kbr.is-a.dev/math-research/branches/odd-field-double-covering/#entry-2026-09-20-higher-jet-obstruction.

Assignment and stopping point. Start the new user-authorized general-multiplicity Spin, stopping at a solution or five completed research cycles. Test the proposed higher-jet lower bound at multiplicity three, retaining the distinction between arbitrary polynomials and products of affine factors. Stop at a general implication or a concrete obstruction, not a case catalogue.

A dimension-uniform counterfamily to the polynomial relaxation

Working proposition. For every prime power \(q\), \(n\ge1\), and \(3\le k\le q\), there is a polynomial \(Q_k\) with \(Q_k(0)=1\), multiplicity at least \(k\) at every nonzero point, and

Thus the covering target cannot follow from these value and multiplicity hypotheses alone, even for arbitrarily large dimension. This does not assert that \(Q_k\) factors into affine polynomials.

Construction and proof. Put \(D=n(q-1)\), \(F_i=x_i^q-x_i\) and \(\Delta=\prod_i(1-x_i^{q-1})\). Write \(D^{[\beta]}\) for the Hasse derivative, the coefficient of \(z^\beta\) in translation by \(z\). Define the finite polynomial

Taylor's polynomial identity gives \(\Delta(x+F(x))=\sum_\beta F(x)^\beta D^{[\beta]}\Delta(x)\). Since \(x_i+F_i=x_i^q\), the left side is \(\Delta(x)^q\). At any nonzero grid point, \(\Delta^q\) has order at least \(q\ge k\). Every omitted term has a product of at least \(k\) grid polynomials \(F_i\), so has order at least \(k\) at every grid point. Subtraction proves the required orders of \(Q_k\). At zero, all terms except \(\beta=0\) vanish and \(\Delta(0)=1\). Finally, each nonzero summand has degree at most \(q|\beta|+D-|\beta|=D+(q-1)|\beta|\). Taking \(|\beta|\le k-1\) proves the bound; its deficit from the proposed covering degree is \(k-2>0\). This proof uses Hasse coefficients, so no factorial division or characteristic restriction was concealed.

Falsification scope. Taking \(k=3\) works over every odd finite field, in every dimension. Ordinary reduced-value idempotence was not the only obstacle: even retaining all jets below order three does not force the target among arbitrary polynomials. A new covering proof must use affine factorization or another genuine restriction of hyperplane products. The completed double-cover theorem is unaffected because its target corresponds to \(k=2\), where the degree deficit in this construction is zero.

Checks and route assessment

Exact Hasse-jet matrices over \((q,n)=(3,1),(3,2),(3,3),(5,2)\) confirmed a nonzero origin value is feasible at degree \(D+2(q-1)\), one below the triple-cover target. Every matrix rank and dimension is in the retained report; the general construction above, not finite rank evidence, proves the obstruction. No affine factorization was inferred from the test.

Contribution to multiplicity-sensitive-lower-bound. The proposed unrestricted polynomial implication is refuted uniformly in dimension. The covering problem remains open, and further jet calculations that ignore factorization cannot prove the missing case. Next isolate the cases with origin multiplicity \(k-1\) or \(k-2\), where degree-controlled grid-ideal division may already suffice, then focus the factor-sensitive obligation on the remaining origin orders. All living sections were reviewed; the working context now preserves this obstruction. Process assessment: the falsifying test prevented an unjustified extrapolation of the successful first-jet proof; existing tools sufficed.

20 September 2026 — The two highest origin multiplicities are settled for every k

Working general degree bounds and a sufficient reduction; the remaining low-origin affine-factor case is open. Produced by Codex (gpt-6-astra, medium reasoning).

Cite as: Kamil Braun, “The two highest origin multiplicities are settled for every k”, Odd-field coverings with multiplicity, Noemesis research notebooks, 20 September 2026, https://kbr.is-a.dev/math-research/branches/odd-field-double-covering/#entry-2026-09-20-high-origin-reduction.

Obligation and stopping point. After the unrestricted polynomial bound failed, isolate precisely where factorization is needed. For arbitrary \(k\), handle origin orders \(k-1\) and \(k-2\), then state a sufficient general covering endpoint for lower origin orders. This tests the general mechanism rather than another finite family.

Sharp high-origin bounds

Working theorem. Let \(q\) be a prime power, \(n\ge1\), \(k\ge2\), \(D=n(q-1)\), and let \(P\) have order at least \(k\) at every nonzero point of \(\mathbb F_q^n\). If its order at zero is \(k-1\), then \(\deg P\ge D+q(k-1)\). If its order at zero is \(k-2\), then \(\deg P\ge D+q(k-1)-1\). Both bounds are attained by affine-factor covering products with that exact origin order.

Degree-controlled grid expansion. For \(F_i=x_i^q-x_i\), repeated monic division gives a unique finite expansion

Indeed the products \(F^\beta x^r\), with \(0\le r_i<q\), have distinct leading monomials \(x^{q\beta+r}\), each with coefficient one, and every other monomial in such a product has lower total degree. Induction on degree gives spanning and uniqueness. The same triangular property gives \(q|\beta|+\deg R_\beta\le\deg P\) for every nonzero coefficient: terms of maximum weighted degree have distinct top monomials and cannot cancel.

If \(P\) has order at least \(s\) at every grid point, all coefficients with \(|\beta|<s\) are zero. Otherwise choose the smallest nonzero layer \(t<s\). A nonzero reduced \(R_\beta\) has a nonzero grid value somewhere by interpolation uniqueness. At any grid point \(a\), \(F_i(a+z)=-z_i+z_i^q\); the degree-\(t\) Taylor part of \(P(a+z)\) is \((-1)^t\sum_{|\beta|=t}R_\beta(a)z^\beta\). Distinct monomials cannot cancel, contradicting the assumed order at every point.

Top allowed origin order. Put \(s=k-1\). All lower layers vanish, and order at least \(s+1\) off zero says each reduced coefficient \(R_\beta\) at layer \(s\) is supported only at zero. Thus \(R_\beta=c_\beta\Delta\), where \(\Delta=\prod_i(1-x_i^{q-1})\). Exact origin order \(s\) makes some \(c_\beta\ne0\). Its degree is \(D\), so the weighted-degree bound gives \(\deg P\ge qs+D\).

Next allowed origin order. Put \(s=k-2\). Again the leading layer is \(R_\beta=c_\beta\Delta\), with a nonzero vector of constants. At a nonzero grid point, the degree-\(s+1\) Taylor coefficients must also vanish. For \(|\gamma|=s+1\), their identity is

To verify signs, the leading \(F\)-layer contributes \((-1)^s\sum_i c_{\gamma-e_i}\partial_i\Delta(a)\), while the next layer contributes \((-1)^{s+1}R_\gamma(a)\). Corrections from \(F_i=-z_i+z_i^q\) multiply \(\Delta(a)=0\) at this order, including when \(q=2\). Higher layers have larger order. Both sides are individually reduced, so value uniqueness gives

If some \(t_\gamma\ne0\), that coefficient has degree \(D\) and already gives a stronger bound. Otherwise choose \(c_\beta\ne0\) and \(\gamma=\beta+e_i\). The coefficient of \(\partial_i\Delta\) is nonzero, and the top monomials of the different \(\partial_j\Delta\) are distinct, of degree \(D-1\). Thus this \(R_\gamma\) has degree \(D-1\), giving \(\deg P\ge q(s+1)+D-1\). This proves the theorem for all characteristics, without factorial division.

Sharp covering controls. For origin order \(k-1\), use the single coordinate cover of size \(D\), plus \(k-1\) complete parallel partitions, each of size \(q\). For origin order \(k-2\), use the double-cover construction, plus \(k-2\) complete parallel partitions. Each partition adds one incidence to every point, including zero. Their sizes attain the stated bounds. Thus these cases need no further computation or structural assumption.

A sufficient endpoint for the remaining origin orders

Let \(h_q(n,3)\) be the minimum number of affine hyperplane occurrences covering every nonzero point at least three times while avoiding zero. A sufficient, potentially stronger-than-necessary statement for the full fixed-\(k\) problem is

Proof of sufficiency. Fix \(q,k\). The two highest allowed origin orders satisfy the target by the theorem above. For origin order \(s\le k-3\), remove the \(s\) hyperplanes through zero. The remaining \(m-s\) planes avoid zero and cover every nonzero point at least \(k-s\ge3\) times. If \(m<D+q(k-1)-1\), then \(m-s\le D+q(k-1)-2\). The stated divergence excludes that for all sufficiently large \(n\). The known upper construction gives equality. Only this implication is proved; the divergence premise is open here and is not claimed necessary.

For \(k=3\) the unresolved case is exactly an origin-avoiding triple cover. Proving merely \(h_q(n,3)\ge D+2q-1\) eventually would settle that case; proving divergence would settle all fixed \(k\). Do not insist on the stronger endpoint if a direct weaker route becomes available.

Review and assessment. Checked the normal-form basis, filtration, lowest nonzero layer, next-layer signs, extension-field identity and sharp constructions. The high-origin part of the general lower-bound obligation is discharged, but the uniform low-origin case remains genuinely factor-sensitive in light of the polynomial counterfamily. The next cycle should look for structural or coding bounds on zero-avoiding triple covers near the one-cover minimum; refining arbitrary polynomial jets cannot address that gap. All living sections were reviewed. Process assessment: extracting a uniform origin-order reduction avoided enumerating k and kept the exact remaining obligation visible; no framework change was needed.

20 September 2026 — Coding gives growth for bundled covers, but small unpaired covers defeat that shortcut

Working restricted bound and exact finite counterexamples to universal bundling; no general large-dimension bound. Produced by Codex (gpt-6-astra, medium reasoning).

Cite as: Kamil Braun, “Coding gives growth for bundled covers, but small unpaired covers defeat that shortcut”, Odd-field coverings with multiplicity, Noemesis research notebooks, 20 September 2026, https://kbr.is-a.dev/math-research/branches/odd-field-double-covering/#entry-2026-09-20-balanced-code-audit.

Question and stopping point. Test the sufficient zero-avoiding triple-cover endpoint using a precise code model, and falsify any inference that near-minimum covers must consist of complete scalar bundles. Stop at an applicable general bound or a demonstrated scope obstruction.

The exact code condition and the bundled case

Normalize each zero-avoiding plane to \(a_r\cdot x=1\), with nonzero \(a_r\), and form the matrix \(A\) with these rows. It has rank \(n\): a nonzero vector in its kernel would receive no covering. The associated linear code \(C=A\mathbb F_q^n\) has dimension \(n\). It describes an r-fold cover exactly when every nonzero codeword has at least r occurrences of each nonzero field symbol. Indeed the condition for symbol one is precisely incidence, and applying it to \(\lambda x\) gives the condition for every other symbol. Thus minimum Hamming distance at least \(r(q-1)\) is necessary but not the full condition.

Working restricted proposition. Suppose the cover is a multiset union of complete bundles \(b_i\cdot x=c\), one plane for each \(c\ne0\), for \(1\le i\le L\). Then \(m=(q-1)L\), and the number of coverings of x is the Hamming weight of \((b_i\cdot x)_{i\le L}\). Triple covering makes this a dimension-n, length-L code of distance at least three. Radius-one Hamming balls around its \(q^n\) words are disjoint, each containing \(1+L(q-1)\) vectors. Hence

Since \(L\ge n\), this excess diverges. This proves the sufficient growth endpoint only for bundled covers. Replacing an arbitrary cover by all its scalar multiples can multiply the plane count by \(q-1\); it does not preserve an additive-constant excess. Fractional averaging likewise does not supply an integral bundled cover of the same size.

Why ordinary distance alone is too weak. Fix \(q>2\), C, and \(d_0=3(q-1)\). For all sufficiently large n there exist length \(m=(q-1)n+C\), dimension-n codes with distance at least \(d_0\) that contain the all-one vector. To see this, take that vector as one generator and choose the remaining n-1 generator vectors uniformly and independently. Every combination using at least one random generator is uniform on \(\mathbb F_q^m\). The union bound for a word of weight below \(d_0\) is at most

The nonzero combinations using only the fixed generator have weight m; excluding weight zero among the other combinations also ensures rank n. Yet the all-one word contains no occurrence of another nonzero symbol, so these codes are not covers. This verifies the failure of the ordinary-distance surrogate in the relevant linear-size regime, rather than merely observing a missing converse.

Finite controls refute universal scalar pairing

Over \(\mathbb F_3\), the following lists of normals define planes \(a\cdot x=1\), one occurrence each. Direct integer incidence evaluation verifies that every nonzero vector is covered at least three times, while zero is uncovered:

n=3, m=10:
001, 002, 011, 021, 102, 112, 121, 201, 210, 221.
n=4, m=12:
0001, 0021, 0120, 0122, 0210, 0222,
1022, 1102, 1212, 2201, 2212, 2220.

Both sizes are \(2n+4\), one below the conjectured eventual triple-cover value \(2n+5\). They do not refute its sufficiently-large-n statement. They are not scalar-paired: for example the n=3 list contains 011 but not 022, and the n=4 list contains 0001 but not 0002. Since normalization to right side one is unique, this is a property of the actual hyperplane multiset, not its notation. Thus a universal all-dimension bundling assertion is false; an eventual structural theorem remains unproved.

The bounded MILP search used at most 80 variables, multiplicities in 0..3 and a 35-second solver limit per dimension. Floating-point optimization only proposed candidates; every accepted multiplicity, total size and all incidences were recomputed with integers. Complete normals and coverage counts are in the exact witness report. No solver infeasibility or optimality message is used as a mathematical certificate.

Source audit and remaining obligation

Das–Djaljapayan–Lin–Yu, Stability for hyperplane covers, concerns real hyperplanes on the Boolean cube; its weight stability theorem does not transfer to this full finite-field grid. Ball's multiple-blocking survey states the classical affine multi-blocking lower bound with a constant excess; the checked passage does not give the needed excess divergence. This is a bounded scope audit, not a complete literature classification.

Assessment. The general low-origin obligation remains open. A valid coding theorem must control symbol frequencies or prove a structural reduction with additive, not multiplicative, cost. The finite unpaired covers suggest testing a precise dimension-extension mechanism: can a small gadget extend a triple cover while preserving its constant excess? If it can, iteration could challenge the eventual conjecture; if it cannot, the failure must be recorded before pursuing more small instances. Process assessment: exact witness verification and a scope countermodel prevented a false application of a convenient Hamming bound; no framework change was needed.

20 September 2026 — A ternary dimension lift exists, but a decreasing resource blocks indefinite iteration

Working universal local lift and exact finite gadget classification; no asymptotic counterexample or general lower bound. Produced by Codex (gpt-6-astra, medium reasoning).

Cite as: Kamil Braun, “A ternary dimension lift exists, but a decreasing resource blocks indefinite iteration”, Odd-field coverings with multiplicity, Noemesis research notebooks, 20 September 2026, https://kbr.is-a.dev/math-research/branches/odd-field-double-covering/#entry-2026-09-20-pair-consuming-lift.

Question and stopping point. Test whether a local operation can turn the finite unpaired triple covers into a fixed-excess family in unbounded dimension. Require both coverage preservation and a replenished input resource; a larger isolated cover is not the endpoint. The operation tested replaces two complete scalar pairs by six planes using their two forms and one new coordinate, adding two planes per dimension.

A coverage-preserving local replacement

Working proposition. Suppose a zero-avoiding triple cover over \(\mathbb F_3^n\) contains the pairs \(u=1,2\) and \(v=1,2\), counted with multiplicities. Remove those four occurrences, introduce a coordinate t, and add the six planes whose left sides are

each equal to one. Extend all other old planes independently of t. The result is a zero-avoiding triple cover in dimension \(n+1\), with two more planes.

Verification of the local rule. Let \(N(u,v,t)\) count the six new incidences, and let \(w(u,v)=\mathbf1_{u\ne0}+\mathbf1_{v\ne0}\), the number of removed incidences. When both u,v are nonzero, direct substitution gives \(N=2\) for every t. When exactly one is nonzero, the three counts as t varies are a permutation of \((4,1,1)\). When u=v=0, the counts for t=0,1,2 are \((0,3,3)\). These 27 exact substitutions prove \(N\ge w\) everywhere, with three coverings on the new axis away from zero. For a nonzero old vector the untouched planes plus the replacement preserve at least three coverings. On the new axis the new planes supply three. Zero remains uncovered. No rank assumption on u,v is needed for this implication.

Starting from the eight-plane triple cover in dimension two consisting of all four projective directions with both nonzero offsets, the rule gives exactly checked covers of sizes 10, 12 and 14 in dimensions 3,4,5. Their numbers of complete scalar pairs are respectively 3,2,1 (the seed has four). Full normal lists and integer incidences are retained in the iterate report. These examples remain below the conjectured eventual value in finitely many dimensions only.

Why this entire local domination scheme cannot iterate forever

Consider any six-plane replacement of the same type, not just the displayed gadget: each new equation is a linear combination of u,v,t equal to one, its coverage dominates w on all attainable triples, and it covers the new axis at least three times. Every new plane must have a nonzero t coefficient, since each plane can cover at most one of the two nonzero axis points and six total incidences are required.

Independent inputs: finite exhaustive certificate. If u,v are independent, there are exactly 18 normalized possible equations \(au+bv+ct=1\), with \(a,b\in\mathbb F_3\), \(c=1,2\). Enumerate all multisets of six equations: \(\binom{23}{6}=100947\). Exact integer evaluation of the 27 coverage requirements leaves 18 valid multisets, each with at most one complete scalar pair. The exhaustive program uses vectorized integer sums, and its full output lists every valid multiset and its pair count. No floating-point infeasibility claim is used. This finite classification is the computational component of the resource proof.

Dependent inputs: analytic check. The removed planes are two copies of the same scalar pair. Write \(x_a,y_a\) for multiplicities of \(au+t=1\) and \(au-t=1\), respectively. Axis coverage gives \(\sum x_a=\sum y_a=3\). For either nonzero u, summing over t gives six incidences, so domination of the removed count two forces exactly two at each t. At u=1 the equations are \(x_1+y_1=2,x_0+y_2=2,x_2+y_0=2\); at u=2 they are \(x_2+y_2=2,x_0+y_1=2,x_1+y_0=2\). Thus all y's are equal, all x's are equal, and the axis sums force every multiplicity to be one. This returns three distinct scalar pairs and no repeated pair. The 462-multiset enumeration independently confirms this unique solution.

Potential argument. For the full cover let b range over the multiplicities of its complete scalar bundles (the minimum multiplicity of the two opposite normalized normals). Put \(B=\sum b\), \(E=\sum\max(b-1,0)\), and \(\Phi=B+2E\). Independent inputs consume two pairs and return at most one, with no repeated output pair, so \(\Phi\) drops by at least one. Dependent inputs consume two copies of one pair and return three distinct pairs; B rises by one but E falls by at least one, so again \(\Phi\) drops by at least one. New planes have nonzero t coefficient and cannot form additional pairs with untouched old planes, whose t coefficient is zero. Invertible coordinate changes preserve scalar proportionality and pair multiplicities. Hence any sequence of these allowed replacements has length at most its initial \(\Phi\).

This rules out an infinite counterexample family obtained by this particular pointwise-domination scheme. It does not exclude replacements of larger groups, extra dependence on old coordinates, exploitation of slack in the full cover, or a different structural proof of the conjecture.

Assessment and process. The general lower-bound obligation remains open. The tested lift produces finite counterexamples to an all-dimension strengthening but cannot propagate them indefinitely. The final cycle of the bounded run should review the remaining exact obligation and choose a route that accounts for this consumed resource, rather than rerunning larger instances. Process assessment: an initially successful gadget was subjected to an iteration test and then a complete exact classification; the resulting obstruction prevents mistaking local extension for asymptotic progress. All living sections were reviewed; no framework change was needed.

20 September 2026 — Critical ternary covers require a binary-avoiding self-orthogonal code

Working exact criterion and route review. Five-cycle run ends with general k unresolved; the branch remains active. Produced by Codex (gpt-6-astra, medium reasoning).

Cite as: Kamil Braun, “Critical ternary covers require a binary-avoiding self-orthogonal code”, Odd-field coverings with multiplicity, Noemesis research notebooks, 20 September 2026, https://kbr.is-a.dev/math-research/branches/odd-field-double-covering/#entry-2026-09-20-critical-code-route-review.

General claim under review. For every fixed odd prime power q and fixed \(k\ge2\), the target is \(f_q(n,k,1)=n(q-1)+q(k-1)-1\) at sufficiently large n. The full goal is not proved by the high-origin cases, scalar-bundled covers, finite exceptions or the local lift. This final cycle tests a factor-sensitive necessary condition at the first unresolved boundary and audits the route against the earlier counterexamples. Some initial review planning remained in the preparation timing category before the mathematics marker.

Exact criterion at the ternary triple-cover boundary

Working proposition. An origin-avoiding ternary triple cover with at most \(2n+4\) planes exists if and only if there is a full-support n-dimensional linear subspace \(C\subseteq\mathbb F_3^{2n+4}\) satisfying

Every such cover has exactly \(2n+4\) planes. Full support means that no coordinate is identically zero on C. Together with the high-origin theorem, this identifies the only possible way the ternary \(k=3\) target can fail in a fixed dimension. It does not give a dimension bound for these codes.

Necessity: degree saturation forces vanishing origin jets. Normalize the covering product as \(P=\prod_{r=1}^m(1-a_r\cdot x)\), so \(P(0)=1\) and \(\deg P=m\). Use the degree-filtered expansion in the previous proof, with \(F_i=x_i^3-x_i\), \(\Delta=\prod_i(1-x_i^2)\), \(D=2n\). Write \(b_i=\partial_iP(0)\). Its first layers are \(R_0=\Delta\) and \(R_{e_i}=\partial_i\Delta-b_i\Delta\). For \(|\gamma|=2\), vanishing of the off-origin second Hasse jets and reduced-value uniqueness imply

Here \(c_\gamma=D^{[\gamma]}P(0)\). To check the formula, the Hasse product rule gives \(D^{[\gamma]}\Delta-\sum_iD^{[\gamma-e_i]}R_{e_i}+R_\gamma=0\) off zero; the sum of derivatives of \(\partial_i\Delta\) is \(2D^{[\gamma]}\Delta\). Also \(F_i(a+z)=-z_i+z_i^3\) has no quadratic correction and \(\partial_i\Delta(0)=0\), giving the claimed origin constant.

If \(m\le D+4\), the coefficient bound is \(\deg R_\gamma\le D-2\). The degree-D term forces \(c_\gamma=0\). Taking \(\gamma=2e_i\), the degree-\(D-1\) term then forces \(b_i=0\), since \(\partial_i\Delta\) has a nonzero top coefficient. Finally \(D^{[2e_i]}\Delta\) has degree exactly \(D-2\), so \(m<D+4\) is impossible. Thus m=D+4, and the linear and quadratic homogeneous terms of P at zero vanish.

Let A have the normals \(a_r\) as rows. The linear term gives \(\sum_ra_r=0\). Writing \(L_r=a_r\cdot x\), the quadratic term is \(\sum_{r<s}L_rL_s=0\). Since 2 is invertible in \(\mathbb F_3\), the identity \(2\sum_{r<s}L_rL_s=(\sum_rL_r)^2-\sum_rL_r^2\) gives \(A^TA=0\). The cover ensures rank A=n, and its proper planes ensure full support. Thus \(C=A\mathbb F_3^n\) is self-orthogonal and orthogonal to the all-one vector. No nonzero word of C is binary: its negative would have no coordinate equal to one and hence would correspond to an uncovered nonzero input.

Sufficiency. Choose a basis of such a C and use its coordinate functionals as the rows of A. Full support gives proper zero-avoiding planes \(a_r\cdot x=1\), and rank n gives nonzero words on nonzero inputs. For any \(c\in C\), let \(N_1,N_2\) count its one and two coordinates. Orthogonality gives \(N_1-N_2=0\) and \(N_1+N_2=0\) modulo three. Hence both counts are multiples of three. The binary-avoidance condition, applied to c and -c, makes both positive whenever c is nonzero. Thus both are at least three, proving the triple covering. This is an exact condition, not a replacement by minimum distance alone.

Falsification of the simpler moment-only route

The two orthogonality conditions and full support alone allow unbounded dimension at exactly this length. For \(t\ge3\), use t blocks of length four, with \(U=(1,1,1,0)\) and \(V=(0,1,2,1)\). Let C consist of blocks \(u_iU+v_iV\) subject to \(\sum_i u_i=\sum_i v_i=0\). The block vectors are independent, so \(\dim C=2t-2=n\) and its length \(4t=2n+4\). We have \(U\cdot U=V\cdot V=U\cdot V=0\), so C is self-orthogonal. Coordinate sums equal \(\sum_i v_i=0\); each coordinate is nonzero on some allowed choice, so support is full. But taking u=1 in three blocks, all other u and all v zero, gives a nonzero binary word. Thus these are not covers. This explicit infinite control refutes an attempted dimension bound based solely on the moment equations.

Exact arithmetic confirms the new moment identities for all four lifted-cover witnesses and checks the control at t=3,4,5, including rank, full support and binary-word membership. The full report retains matrices and controls. The constructions above prove the general assertions; the checks do not establish nonexistence of the required binary-avoiding codes.

Final route assessment for this bounded run

What changed. The two highest origin orders are settled for all q,k. The unrestricted polynomial relaxation fails in every dimension already at k=3. Complete scalar bundles satisfy the desired growing-excess bound, but exactly verified unpaired covers exist. A local dimension lift works for finitely many steps, while a decreasing pair-resource potential prevents indefinite iteration under its precise domination rule. The critical ternary case now has an exact self-orthogonal, binary-avoidance formulation.

What remains. No proof excludes these critical ternary codes for all sufficiently large n, and no construction produces them in unbounded dimension. Even settling that critical k=3 case would not by itself settle all odd q,k; the triple-cover excess divergence remains a stronger sufficient endpoint for the full problem. The required implication is not supplied by low-degree jets, ordinary distance, moment equations or successful finite lifts.

Assessment of this line only. The first-jet approach extends usefully to high origin multiplicity but is insufficient for lower origin orders. The exact code criterion preserves factor information and survives the known controls, so it is a defensible next direction, but there is no demonstrated dimension bound yet. More small instances or a moment-only rank estimate would leave the same obligation untouched. A future assigned cycle should seek a bound on full-support binary-avoiding self-orthogonal ternary codes at length \(2n+4\), using the finite witnesses and infinite non-cover control to test every relaxation.

Stop and process. This completes five research cycles since the goal was generalized. The problem remains open; stop the Spin as requested, retain active branch status for future work, and make no solved or novelty claim. All living sections were consolidated around the exact remaining obligation. Process assessment: source checks and falsifying controls prevented three unsupported transfers (arbitrary polynomials, ordinary distance and resource-free lifting); no framework change is warranted by this run.