2019-03-13 19:42:35 8 Comments

Let $p$ be a prime. Let $f \in \mathbb{F}_p[x_1, \ldots, x_n]$ be a **homogenous** polynomial of degree $p$. Can $f$ have more than $(1-p^{-1}+p^{-2}) p^n$ zeroes in $\mathbb{F}_p^n$?

Basic observations: The lower bound (for $n \geq 2$) is achieved by $\prod_{c\in \mathbb{F}_p} (x_1 - c x_2)$. If nonhomogenous polynomials are allowed, we can get all $p^n$ points to be zeroes, using $\prod (x_1-c)$. It's easy to show a product of hyperplanes can't beat $(1-p^{-1}+p^{-2}) p^n$.

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## 1 comments

## @Sean Lawton 2019-03-13 23:43:27

As suggested by the OP, I am making my comment an answer, as it resolves his query in the affirmative.

Theorem 2.1 (Serre, 1991), cited in

Maximum Number of Common Zeros of Homogeneous Polynomials over Finite Fieldsby Peter Beelen, Mrinmoy Datta, Sudhir R. Ghorpadel, gives an explicit formula for the maximal number of solutions in projective $(n-1)$-space of a homogeneous polynomial of arbitrary degree $d$ over finite fields of order $q$.In particular, this maximum (which perhaps should be called the Tsfasman-Serre-Sørensen number) is: $$ dq^{n−2} + q^{n-3}+q^{n-4}+\cdots+q+1,$$ whenever $d \leq q.$

In the question, $d=p=q$ and so we have a maximum of $p^{n−1} + p^{n-3}+p^{n-4}+\cdots+p+1$ solutions in projective space. But the question was about solutions in affine space. So we scale by $\mathbb{F}_p^*$ then add 1 for the trivial solution to obtain:

$(p^{n−1} + p^{n-3}+p^{n-4}+\cdots+p+1)(p-1)+1=$ $$p^n-p^{n-1}+p^{n-2}=(1−p^{−1}+p^{−2})p^n,$$ as required.