Erdős 699: surjective interpolation gives nonvanishing polynomials

P699 research · Erdős problem 699

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This lemma supplies the nonvanishing step for the index-13 and subsequent interpolation certificates. It requires an exact rank calculation, but does not require a polynomial gcd or a bound for an exceptional algebraic zero set.

Lemma. Prescribe symmetric multiplicities $R_{xy}=R_{yx}\ge0$ at the integer points $x,y\ge0$, $x+y<i$. Suppose that the map taking symmetric polynomials of total degree at most $D$ to their prescribed Hasse jets is surjective over $\mathbb Q$. Then there is an effectively constructible finite family of integer polynomials of degree at most $D+1$, all having the prescribed multiplicities, that cannot vanish simultaneously at a point $(X,Y)$ with $X+Y>i-1$.

Proof. Let

$$I=\bigcap_{R_{xy}>0}(X-x,Y-y)^{R_{xy}},\qquad A=\mathbb Q[X,Y]/I,$$

and let $A^\sigma$ be the invariant subalgebra under $\sigma(X,Y)=(Y,X)$. The symmetric jets are exactly the coordinates of this finite-dimensional algebra. Let $\pi:A^\sigma\to\mathbb Q[X,Y]_{\le D}^\sigma$ be a rational linear right inverse to the jet map. Put $s=X+Y$.

For a rational basis $a_1,\ldots,a_r$ of $A^\sigma$, use the polynomials

$$G_0=1-\pi(1),\qquad G_j=s\pi(a_j)-\pi(sa_j)\quad(1\le j\le r). \tag{1}$$

They belong to $I$ and have degree at most $D+1$. Clear a common denominator to obtain integer polynomials with the same common zeros.

Suppose they all vanished at $P$. The functional $\varepsilon(a)=\pi(a)(P)$ would satisfy

$$\varepsilon(1)=1,\qquad \varepsilon(sa)=s(P)\varepsilon(a) \quad\text{for every }a\in A^\sigma.$$

Thus $s(P)$ would be an eigenvalue of multiplication by $s$ on $A^\sigma$. On the local algebra at $(x,y)$, this operator is $(x+y)$ times the identity plus a nilpotent operator. Its eigenvalues, including those on the invariant subalgebra, therefore belong to $\{0,1,\ldots,i-1\}$. This contradicts $s(P)>i-1$. $\square$

The right inverse is found by selecting a nonsingular maximal minor of the integer jet matrix. In particular, full row rank modulo any prime proves the required rational surjectivity. Adjugates of that minor give effective coefficient bounds for all polynomials (1), even if the rational polynomials are not explicitly computed.

For the usual symmetric Taylor-coordinate basis, multiplication by $s$ has integer entries. Its local constant term is $x+y$, and its remaining terms raise the jet order. A column has coefficient sum at most $x+y+3$: at a diagonal support point, for example, multiplying $U+V$ by $U+V$ gives invariant coordinates $1,2$ in $U^2+2UV+V^2$. These small coefficient bounds make the adjugate estimate quite direct.

If a known integer polynomial $H$ already supplies some of the desired Kummer multiplicities and is nonzero in the physical range, multiplying the family (1) by $H$ gives the complete certificate. The added degree needed to guarantee nonvanishing is only one.