Intersection of irreducible curves and the Hermitian curve

Abstract

Let $H_q$ denote the Hermitian curve in $P^2$ over $F_{q^2}$ and $C_d$ be an irreducible plane projective curve in $P^2$ also defined over $F_{q^2}$ of degree d. Can $H_q$ and $C_d$ intersect in exactly $d(q+1)$ distinct $F_{q^2}$-rational points? Bézout's theorem immediately implies that $H_q$ and $C_d$ intersect in at most $d(q+1)$ points, but equality is not guaranteed over $F_{q^2}$. 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 $H_q$ defined over $F_{q^2}$ intersects $H_q$ in $q+1$ rational points. Moreover, all possible intersections of conics and $H_q$ were classified in [9] 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 $(q,d)\in \left\{\right(2,2),(3,2),(2,3\left)\right\}$, the answer is instead negative. We show that for $q\le d\le q^2-q+1$, $d=\lfloor (q+1)/2\rfloor$ and $d=3$, $q\ge 3$ the answer is again affirmative. Various partial results for the case d small compared to q are also provided.