1
$\begingroup$

Let $A_1$, $A_2$,...., $A_n$ be subsets of a field $F$ such that $|A_i| = N$ for all $i$. If $P$ is nonzero $n$-variable polynomial over $F$ of total degree $D$, show that the number of zeroes of $P$ in the box $A_1$ x $A_2$ x ...... x $A_n$ is bounded by $D{N^{n-1}}$.

My progress on the exercise:

Let $S$ be the set of zeroes of $P$ in $A_1$ x $A_2$ x ...... x $A_n$.

If $D \geq N$, then since clearly $|S| \leq N^n \leq D{N^{n-1}}$, it is done.

So, assume $D < N$. Let us write $P$ as a polynomial in $x_n$, i.e,

$P = \sum_{i = 0}^{t_n} P_i(x_1,x_2,.....,x_{n-1}){x_n}^i$

where each $P_i$ is a polynomial and $t_j$ is the maximum $x_j$ degree in $P$, for $1 \leq j \leq n$. For each fixed $n-1$ tuple, $(a_1,a_2,....,a_{n-1}) \in A_1$ x $A_2$ x ...... x $A_{n-1}$, the polynomial $P$ is a univariate polynomial in $x_n$ of degree atmost $D$. Therefore if it is nonzero then it can vanish on atmost $D$ many points of $A_n$. Since there are $N^{n-1}$ such $n-1$ tuples, we have $P$ can vanish on atmost $D{N^{n-1}}$ points, provided for each fixed ${n-1}$ tuple the univariate polynomial obtained is nonzero.

I don't know how to either show each univariate polyomial is nonzero or to find a bound on the number of $n-1$ tuples, such that the univariate polynomial is zero.

Moreover, this is my personal approach. I don't know whether in this direction this exercise can be solved or not.

One can read the paper Combinatorial Nullstellensatz by Noga Alon for this exercise. Thanks in advance!

  • 0
    It was a typo, I have corrected it.2017-01-31

1 Answers 1

1

Induction: Consider $P_i$ for $i=t_n$. It can be zero on at most $(D-t_n)N^{n-2}$ (n-1) tuples, which give at most $(D-t_n)N^{n-1}$ roots for P; if it is non-zero, the number of roots of $P$ is at most $t_nN^{n-1}$.