# Bézout's theorem for arcs in the plane

Consider two polynomials $p,q \in {\mathbb R}[x,y]$, both of degree $d$. Let $\gamma_p$ and $\gamma_q$ be the two curves in ${\mathbb R}^2$ that are defined by these polynomials, and assume that these curves have no common components. As is well known, by Bézout's theorem the two curves intersect in at most $d^2$ points.

We cut $\gamma_p$ at each point where $\frac{\partial p}{\partial y} =0$, and similarly for $\gamma_q$. We obtain a set of arcs that behave like functions in the sense that each arc has at most one point for every $x$-coordinate.

Consider one such arc of $\gamma_p$ and one such arc of $\gamma_q$. My question is: Is the number of intersection points between two such arcs at most $d$? Or at least significantly smaller than $d^2$?

At first glance this may seem false. However, when $d=2$ it is clear that the two arcs have at most two intersection points. I played with the case of $d=3$ for a while and could not get more than three intersection points.

Let $f(x)$ be a cubic polymomial such that $df/dx$ has no real roots. Let $g(x)$ be a cubic polynomial with $3$ real roots on the half-line where $f(x)>0$.
For $\epsilon$ small enough, the derivative of $f(x)+\epsilon g(x)$ also has no real roots. Hence $x^2=f(y)$ and $x^2=f(y)+\epsilon g(y)$ have no critical points. But they intersect at six points, the three roots of $g(y)$ with two values of $x$ each.
• Very nice! Thank you. In general this symmetry trick gives $2d$ intersection points, which is still far from $d^2$. Do you think that it is possible to have more intersection points than that? Jul 23, 2015 at 23:45
• @AdamSheffer We can do $a(x) = b(y)$ where $b'(y)$ has no real and $a(x)=z$ has $d$ roots for $z$ in some interval. Then perturbing $b$ by some polynomial with $d$ roots in that interval gives $d^2$, for $d$ odd, or $d(d-1)$, for $d$ even. Jul 24, 2015 at 3:16