Let $c,d\in\mathbb{N},\varepsilon>0$ and $p$ be a prime. Question: is it true that for all $\varepsilon>0$, if $p$ is sufficiently large depending on $c,d$ and $\varepsilon$, then for any varieties $X,Y\subseteq \mathbb{F}\_{p}^{d}$ of "complexity" at most $c$, either $\vert X\cap Y\vert< \varepsilon\vert X\vert$ or $X\subseteq Y$?


Here we say that a variety $X\subseteq \mathbb{F}\_{p}^{d}$ is of "complexity" at most $c$ if there exist $f\_{1},\dots,f\_{r}\in\mathbb{F}\_{p}[x\_{1},\dots,x\_{d}]$ of degree at most $c$ for some $r\leq c$ such that $X$ is the set of $x\in \mathbb{F}\_{p}^{d}$ with $f\_{1}(x)=\dots=f\_{r}(x)=0$.


For example, if $X=\mathbb{F}\_{p}^{d}$ and $Y$ is generated by a single polynomial of degree $c$, then I know the answer to the question is positive. I think this question is related to counting the number of points in a variety, but I am not aware of any related references, as I am not an expert in algebraic geometry.



Counting the number of points in a variety over a finite field
This is true if $X$ is geometrically irreducible by the [Lang-Weil bound](https://terrytao.wordpress.com/2012/08/31/the-lang-weil-bound/), which gives us that the size of $|X \cap Y|$ is $(c(X \cap Y) + O\_c(p^{-1/2})) p^{\dim (X \cap Y)|}$ where $c(X \cap Y)$ is the number of top-dimensional components of $X \cap Y$, which is bounded by some function of $c$. If we don't have $|X \cap Y| < \epsilon |X|$ then for sufficiently large $p$ it follows that $\dim (X \cap Y) = \dim X$, and then geometric irreducibility gives $X \cap Y = X$.


If $X$ is reducible then $X \cap Y$ can be a top-dimensional component of $X$ but not all of $X$. Between "geometrically irreducible" and "irreducible" it depends on what you mean by "$X \subseteq Y$."


