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Given a univariate polynomial of degree $n$ with coefficients from a field, the polynomial is uniquely defined by $n+1$ evaluation points. That is, given $n+1$ points $(i, f(i))$, one can uniquely determine $f$.

Is there a similar statement for multivariate polynomials? For example, given a polynomial $f(x,y)$, where the degree in each variable is at most $n$, how many evaluation points uniquely determine this polynomial?

My current intuition: If I fix the first variable to say 1, then I get that if polynomials $f$ and $g$ of degree $n$ agree on $(1 ,y_1), ~\dots~, (1 ,y_{n+1})$, then I have $f(1,x) = g(1,x)$ for all $x$ (since fixing the polynomials in one variable gives me univariate polynomials of degree $n$). Now I can define a polynomial $h(x,y) = f(x,y) - g(x,y)$, which is $0$ at any point of the form $(1,x)$.

Assume $f$ and $g$ agree on the points above, but $f \neq g$, then maybe I can say something about $h$ not being the constant $0$ polynomial and having too many roots? Is there a bound on how many roots a multivariate polynomial can have?

Furthermore, for this intuition I assumed that the polynomials agree on some specific points. Preferably, I'd like to have an argument that starts with given a set of arbitrary points.

Rodrigo de Azevedo
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1 Answers1

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These two articles would be helpful.
(1) A Simple Expression for Multivariate Lagrange Interpolation by Kamron Saniee
(2) On Multivariate Interpolation by Peter J. Olver

In short, consider $F[x_1, \dots, x_m]$ where $F$ is a field of characteristic zero. There are $\rho=\binom{m+n}{n}$ monomials of degree $\leq n$. In other words, there exists $\rho$ multi-indices $I_1, \dots, I_\rho$ with $|I_i| \leq n$.

Theorem. Let $x_1, \dots, x_\rho$ be a distinct points in $F^m$. Then the followings are equivalent.
(1) For given $y_1, \dots, y_\rho \in F$, there exists a unique polynomial $f$ in $F[x_1, \dots, x_m]$ of degree $\leq n$ such that $f(x_i) = y_i$ for all $ 1 \leq i \leq \rho$.
(2) The sample matrix $M=\left(x_i^{I_j}\right)_{1 \leq i, j \leq \rho}$ is invertible.

Note that $M$ is a Vandermonde matrix when $m=1$, so the univariate case follows.

In above theorem we need some conditions on our evaluation points in order to determine the polynomial. Unfortunately, this restriction is inevitable; we cannot obtain the uniqueness assertion by just considering much more arbitrary evaluation points. For example, let $m=n=2$ and consider $h_1, h_2 \in F[x, y]$ given by $h_1(x, y)=x^2-y^2$ and $h_2(x, y)=x^2-x-y^2+y$. Then $h_1$ and $h_2$ have the same value $0$ whenever $x=y$.

luxerhia
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