**Problem:**



> 
> let $P\_1(x), P\_2(x), Q\_1(y), Q\_2(y)$ be some polynomials of degree $d$ in $\mathbb{F}\_p$. Let
> \begin{equation}
> A := \{ (x, y) \in \mathbb{F}\_p^2 : P\_1(x) = Q\_1(y) \},\\
> B := \{ (x, y) \in \mathbb{F}\_p^2 : P\_2(x) = Q\_2(y) \}.
> \end{equation}
> Let us also assume that equations $P\_1(x) = Q\_1(y)$ and $P\_2(x) = Q\_2(y)$ are independent in certain sense: for example, polynomials $P\_1(x) - Q\_1(y)$ and $P\_2(x) - Q\_2(y)$ do not have common divisors.
> Can one imply that
> $$|A \cap B| \ll \max{(|A|, |B|)} ?$$
> 
> 
> 


This is clear that each of the equations $P(x) = Q(y)$ may have at most $dp$ solutions, and this is nearly attained when, for example, $P(x) = x^d + 3, Q(y) = y^d + 3$, where $d | p - 1$.


There is the following variation of Bombieri's result by Chalk and Smith (see Theorem 2 in [here](http://matwbn.icm.edu.pl/ksiazki/aa/aa18/aa18121.pdf), page 202):


**Theorem.** Let $(b\_1, b\_2) \in \mathbb{F}\_p^2$ be nonzero and $f(x, y) ∈ \mathbb{F}\_p[x, y]$ be a polynomial of degree $d \geqslant 1$ with the following property: there is no $c \in \mathbb{F}\_p$ for which the polynomial $f(x, y)$ is divisible by $b\_1x + b\_2y + c$. Then
$$
\bigg|
\sum\_{(x, y):f(x,y)=0}
e^{2\pi i(b\_1x+b\_2y)/p}
\bigg| \leqslant 2d^2p^{1/2}.
$$


This is roughly saying, that as long as $P(x) - Q(y)$ does not have any linear divisors, then sets $A, B$ are equidistributed over $\mathbb{F\_p} \times \mathbb{F\_p}$.


This gives an intuition, that sets $A$ and $B$ should be somehow independent, and probably even an inequality $|A \cap B| \ll |A||B|/p^2 \ll d^2$ could have place, upon a certain condition of independence of polynomials $P\_1, P\_2, Q\_1, Q\_2$.


Are there any known results of this type?



Common roots to "independent" equations $P_1(x) = Q_1(y)$ and $P_2(x) = Q_2(y)$ in $\mathbb{F}_p \times \mathbb{F}_p$
The bound $d^2$ is, indeed, correct. It follows from Bézout's theorem, see [here](https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem#Plane_curves). In your case, affine plane curves are exactly $A$ and $B$, the condition on common divisor is the lack of common component. You can take $F=\mathbb F\_p$ and $E=\overline{\mathbb F\_p}$.


