Intersection of irreducible curves and the Hermitian curve

Abstract

Let $\mathcal{H}_q$ denote the Hermitian curve in $\mathbb{P}^2$ over $\mathbb{F}_{q^2}$ and $\mathcal{C}_d$ be an irreducible plane projective curve in $\mathbb{P}^2$ also defined over $\mathbb{F}_{q^2}$ of degree $d$. Can $\mathcal{H}_q$ and $\mathcal{C}_d$ intersect in exactly $d(q+1)$ distinct $\mathbb{F}_{q^2}$-rational points? Bézout's theorem immediately implies that $\mathcal{H}_q$ and $\mathcal{C}_d$ intersect in at most $d(q+1)$ points, but equality is not guaranteed over $\mathbb{F}_{q^2}$. In this paper we prove that for many $d \le q^2-q+1$, the answer to this question is affirmative. The case $d=1$ is trivial: it is well known that any secant line of $\mathcal{H}_q$ defined over $\mathbb{F}_{q^2}$ intersects $\mathcal{H}_q$ in $q+1$ rational points. Moreover, all possible intersections of conics and $\mathcal{H}_q$ were classified by Donati et al. in 2009 and their results imply that the answer to the question above is affirmative for $d=2$ and $q \ge 4$, as well. However, an exhaustive computer search quickly reveals that for $(q,d) \in \{(2,2),(3,2),(2,3)\}$, the answer is instead negative. We show that for $q \le d \le q^2-q+1$, $d=\lfloor(q+1)/2\rfloor$ and $d=3$, $q \geq 3$ the answer is again affirmative. Various partial results for the case $d$ small compared to $q$ are also provided.