Erdős 699: effective nonvanishing from two coprime kernel polynomials

P699 research · Uniform jet nonvanishing

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The surjective-jet construction gives polynomials of degree at most $D+1$ which cannot all vanish when $X+Y$ is outside the grid rows. For effective finiteness, one can sometimes work in degree $D$ instead. The following criterion includes an explicit way to bound its exceptional zeros; it does not assume that a generic pair of polynomials is coprime.

Lemma. Let $A$ be an integer matrix expressing a finite collection of vanishing jets on polynomials of total degree at most $D$. Suppose that $A$ has full row rank modulo a prime $p$. Suppose its modular kernel contains polynomials $\overline F,\overline G$ which are coprime in $\mathbb F_p[X,Y]$, and that at least one has total degree exactly $D$. Then there are integer kernel polynomials $F,G$, of degree at most $D$, which are coprime over $\mathbb Q$. Their simultaneous integer zeros have effectively bounded $X+Y$.

Proof. Choose a maximal minor whose determinant is not divisible by $p$. Solving for its pivot variables gives a right inverse over $\mathbb Z_{(p)}$. Consequently every modular kernel vector lifts to a kernel vector over $\mathbb Z_{(p)}$: lift its free coordinates first, then solve for the pivot coordinates. Clearing denominators not divisible by $p$ produces integer polynomials reducing to nonzero scalar multiples of $\overline F,\overline G$.

Suppose the lifts had a nonconstant common rational factor. Gauss's lemma gives a common primitive integer factor $H$. Its reduction is nonzero. If its degree dropped modulo $p$, the degree of the reduction of the full-degree polynomial would also drop: degrees add in both polynomial rings, while reduction cannot increase the degree of either factor. This contradicts $\deg\overline F=D\ge\deg F$, after exchanging $F,G$ if needed. Hence $\overline H$ is nonconstant, contradicting modular coprimality. Thus $F,G$ are coprime over $\mathbb Q$.

Two coprime plane polynomials have only finitely many common affine zeros. Everything here is explicit integer algebra. For example, compute the elimination ideal of

$$F(X,Y),\quad G(X,Y),\quad Z-X-Y.$$

It contains a nonzero polynomial in $Z$ (or contains $1$ if there are no common zeros). Its roots have a computable Cauchy bound, bounding $X+Y$ at every simultaneous zero. $\square$

In a binomial application the fixed coordinate and row-line factors are nonzero on the relevant physical region. Above the finite bound in the lemma, at least one of $F(j,n-j),G(j,n-j)$ is nonzero. Thus the usual weighted divisibility argument may use residual degree $D$, saving one degree relative to the uniform nonvanishing family.

The required finite certificate consists of a full-row-rank check, two actual modular kernel vectors, their total degrees, and a polynomial gcd equal to a nonzero constant. Full row rank alone does not prove the coprimality condition: all degree-$D$ kernel polynomials can still share a curve. A numerical or genericity assertion about their gcd is not a substitute for this check.

There is also a direct numerical bound without computing an elimination polynomial. If $\|F\|_1,\|G\|_1\le C$, substitute $Y=N-X$ and take the resultant with respect to $X$, using the actual $X$-degrees. It is nonzero by coprimality, and its coefficient norm is at most

$$(2D)!\,(2^D C)^{2D}.$$

Every simultaneous zero consequently satisfies

$$\boxed{|X+Y|\le1+(2D)!\,(2^D C)^{2D}.}$$

If both transformed polynomials have $X$-degree zero, their coprimality already excludes simultaneous zeros. Otherwise the bound follows by expanding the Sylvester determinant and then applying the Cauchy root bound to its nonzero integer resultant. A drop in the $X$-degree does not invalidate the bound; one uses a smaller Sylvester matrix.

For completeness, if the jet matrix has $r$ rows and $s$ columns, its entries have absolute value at most $H\ge1$, and each basis polynomial has coefficient norm at most two, the lifting construction above permits

$$C=2s(s+1)p\,r!\,H^r.$$

Indeed choose free coordinates in $\{0,\ldots,p-1\}$, multiply them by the pivot determinant, and obtain the pivot coordinates with its adjugate. Expanding determinants gives this conservative bound. Thus explicit matrix data already suffice for a numerical common-zero bound. Evaluating it is separate from excluding any bad triples below it.