Erdős 699: arithmetic nonvanishing for the two largest index-nine kernels

P699 research · The index-nine argument · The local arithmetic nonvanishing proof

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The common-zero thresholds for the polynomial certificates round3_000 and round3_001 can both be removed. Their degrees and prescribed multiplicities remain unchanged. The coefficient estimates, source thresholds, and other common-zero thresholds are separate parts of the index-nine argument.

The proof uses the elementary $251$-adic argument in the linked index-seven note: check all common residue points of two modular sections, use another section where necessary, and examine the tangent forms at each prescribed grid point. At an odd-multiplicity diagonal point symmetry forces the tangent $U+V=0$; an invertible matrix of the next linear terms excludes that direction as well.

Exact data

The retained certificate contains the kernel vectors and local matrices. The checker checks their jets, their independence, all residue points over $\mathbf F_{251}$, and the tangent calculations. It does not repeat either large jet rank computation or enumerate binomial rows.

Both prescribed supports contain 43 points. For round3_000, the first two sections have only two additional common residue points, $(29,112)$ and $(112,29)$. The newly retained third section has value 84 at both. For round3_001, the two existing sections already have no common residue point outside their support.

The tangent forms are coprime except at the following diagonal points. The matrices give the next linear terms in $t=U$, $w=(U+V)/U$ after dividing by $t^R$.

All determinants are nonzero. The constant terms of these strict transforms vanish exactly for symmetric lifts, since every displayed $R$ is odd. Therefore the same minimum-valuation argument as for index seven excludes every nonzero displacement in $251\mathbf Z_{251}^2$.

Consequence and lifting

The retained full row ranks provide maximal jet minors invertible over $\mathbf Z_{(251)}$. Every selected modular section consequently lifts to a symmetric rational section, with denominators prime to 251 and with the prescribed jets exactly. For either certificate, the common zeros of the selected lifts in $\mathbf Z_{251}^2$ are precisely the 43 support points. In particular, at least one section is nonzero at every physical integer point $(j,n-j)$ for index nine.

Integral monic Newton basis changes are unimodular and preserve this lifting argument, so the certificate is compatible with coefficient bounds in those bases. The assertion here concerns arithmetic zeros; it does not assert that the geometric base locus is empty over an algebraic closure.

The two common-zero contributions may thus be set to zero when combining the numerical thresholds. No additional polynomial degree is needed.