Let $X \subseteq\mathbb F_3^n$. Let $M(n,d)$ denote the set of polynomials of degree at most $d$, where we are requiring that degree of each variable is at most $2$. Let $V$ denote the vector space of polynomials of degree $\le d$ that vanish on $X$. I assume, but he does not say, that the degree of each variable in a polynomial in $V$ is required to be at most $2$ (i.e. there is no $x_2^3$).

Here, Zeilberger claims that $\dim V \ge \dim M(n,d) - \vert X \vert$. I don't see why this should be the case. Clearly $\dim V \ge \dim M(n, d - \vert X \vert)$, as we can multiply together linear polynomials that vanish on each point in $X$ and multiply by a polynomial of degree at most $d - \vert X \vert$, but this isn't good enough.

EDIT: I guess you don't even get this bound, as you might get terms with degree more than $2$ by doing this.

  • 1
    $\begingroup$ What is $F_3^n$? The n-fold Cartesian product $\mathbb{Z}/3\mathbb{Z} \times \dots \times \mathbb{Z}/3\mathbb{Z}$? $\endgroup$ – Chill2Macht Aug 15 '16 at 0:57
  • $\begingroup$ @William You are correct. $\endgroup$ – vukov Aug 15 '16 at 1:04

The idea, I believe, is that the cardinality of $X$ gives an upper bound on the lost degrees of freedom resulting from the requirement that the polynomials are to vanish on $X$.

If the polynomials vanish on the finite set $X$, then they are constrained to equal $0$ everywhere on $X$.

In other words, $V$ is the subset of $M(n,d)$ such that for all $x \in X$, we have that $$x_1^{\alpha_1} \cdots x_n^{\alpha_n} = 0, \quad 0 \le \alpha_i \le 2, \quad \alpha_1 + \dots + \alpha_n =d $$

The claimed inequality is equivalent to (as the paper notes I just realized) $$ |X| \ge \dim M(n,d) - \dim V$$ This is true in the trivial case that $X=( 0, \dots, 0)$, since then $M(n,d)=V$ (since the paper is assuming that the polynomials have no constant terms, so they all vanish at the origin).

We can do all of our dimension computations on the finite set of basis polynomials which have the form (as mentioned in the paper) $$\{x_1^{\alpha_1} \cdots x_n^{\alpha_n},0 \le \alpha_i \le 2,\alpha_1 + \dots + \alpha_n =d \}$$ Combining this with the depiction of $V$ I mentioned above, we get that the dimension of $V$ is equal to $\dim M(n,d)$ minus the number of basis polynomials which vanish on $X$, which I will denote $\beta$.

Since $\beta$, which is number of basis polynomials which vanish on $X$, is bounded above by $|X|$ (see below) we get the desired result: $$\beta \le |X| \implies -\beta \ge -|X| \implies \dim M(n,d) - \beta \ge \dim M(n,d) - |X| \\ \dim V = \dim M(n,d) - \beta \implies \dim V \ge \dim M(n,d) - |X| $$

In other words, the crux of what I am saying is the claim that the number of basis polynomials which vanish on $X$ is bounded above by the cardinality of $X$, i.e. the claim that $\beta \le |X|$.

This claim (that $\beta \le |X|$) should follow from the Schwartz-Zippel Lemma, see Theorem 1 of this page (and multiply by $|S|$ - then my suspicion is that $|S|\mathbb{P}(P(r_1, \dots, r_n))=\beta$ in this case). It should also follow from the Lagrange interpolation theorem.

The essential idea is that the condition defining $V$ can be written as a set of $|X|$ equations (let $X=\{\chi_1, \dots, \chi_m\}$): $$P(\chi_1) = 0 \\ \vdots \\ P(\chi_m)=0 $$ Since there are $|X|$ equations, there should be at most $|X|$ linearly independent solutions to them, their span will equal $V$, which implies immediately that $\beta \le |X|$.

Related: Identically zero multivariate polynomial function

Counting roots of a multivariate polynomial over a finite field



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