Intersection of irreducible curves and the Hermitian curve

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 ≤ 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 ≥ 4 , as well. However, an exhaustive computer search quickly reveals that for ( q , d ) ∈ { ( 2 , 2 ) , ( 3 , 2 ) , ( 2 , 3 ) } , the answer is instead negative. We show that for q ≤ d ≤ q 2 − q + 1 , d = ⌊ ( q + 1 ) / 2 ⌋ and d = 3 , q ≥ 3 the answer is again affirmative. Various partial results for the case d small compared to q are also provided.