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A polynomial $p$ can be specified by its coefficient function, a finitely supported function $c:mathbb N^d_0tomathbb R.$ Here $mathbb N_0={0,1,2,dots}$ and $dinmathbb N_0.$ The value of $p$ at a point $xinmathbb R^d$ is $p(x)=sum_{alphainmathbb N^d_0}c(alpha)x_1^{alpha_1}dots x_d^{alpha_d}$ (the sum makes sense by the assumption that $c$ has finite support). We call $p$ non-zero if $c(alpha)neq 0$ for some $alpha.$

For all non-zero $p$ does there exist $alpha$ such that $c(alpha)neq 0$ and $p(alpha)neq 0$?

Equivalently: for all finite $Asubsetmathbb N_0^d,$ is the $|A|times |A|$ matrix defined by $M_{alpha,beta}=(alpha_1^{beta_1}dotsalpha_d^{beta_d})$ non-singular? (In one direction. take $A$ to be the support of a counterexample $c,$ which is then in the kernel of $M.$ In the other direction, take $c$ to be a vector in the kernel of a counterexample $M.$) Call $A$ "good" if this holds. I have checked some randomly generated sets $A$ are good. Also:

• If $Asubset mathbb N_0^d$ and $Bsubset mathbb N_0^e$ are both non-empty and good then the Cartesian product $Atimes Bsubsetmathbb N_0^{d+e}$ is good. In terms of matrices this is because the Kronecker product of two positive-dimensional square matrices is non-singular iff the two matrices are non-singular.




• Let $Asubset mathbb N_0^{d+1}.$ If the sets defined by $A_0={alphain Amid alpha_{d+1}=0}$ and $A_+={alphain Amid alpha_{d+1}>0}$ are good then $A$ is good. Proof: assume $A_0$ and $A_+$ are good and consider a polynomial $p$ with coefficients $c$ zero outside $A.$ If $p(alpha)=0$ for $alphain A_0$ then $c(alpha)=0$ for all $alphain A_0,$ because $A_0$ is good and the $A_+$ coefficients don't contribute to $p(x)$ when $x_{d+1}=0.$ So $c$ is zero outside $A_+,$ and hence $p$ must also be zero on $A_+$ because $A_+$ is good.




• If $Asubset mathbb N_0^1$ then $A$ is good. Proof: by the last point we can assume $0notin A.$ By Descartes' rule of signs a univariate polynomial with at most $|A|$ non-zero coefficients has at most $|A|-1$ positive zeroes.




• $Asubset mathbb N_0^d$ is good if it is downwards-closed, i.e. for all $betain A$ and all $alpha$ such that $alpha_ileqbeta_i$ for all $1leq ileq d$ we have $alphain A.$ Proof: apply the forward difference operator $(Delta p)(x)=p(x_1,dots,x_{d-1},x_d+1)-p(x).$ By induction $A'={alpha in Amid (alpha_1,dots,alpha_d+1)in A}$ is good. If $p$ had zero coefficients outside $A$ and also vanished on $A,$ then $Delta p$ would have zero coefficients outside $A'$ and vanish on $A',$ which forces $Delta p$ to be the zero polynomial. This means $p$ has zero coefficients outside $A_0$ (as defined in the last point) and we can apply induction on dimension.




• A small variation: if $Asubset mathbb N_0^d$ is downwards closed and $alphainmathbb N_0^d$ then the shifted set $alpha+A={alpha+betamid betain A}$ is good. This follows from the same argument but using the modified forwards difference operator defined by $Delta' p=x^alpha Delta x^{-alpha} p.$

R. Zippel's "Interpolating polynomials from their values" calls similar questions "zero avoidance problems", but I couldn't find anything answering this question.