2024/07/18 by Peter Beelen, Beelen, Peter, Mrinmoy Datta +5
Computer Science · Mathematics · #11G20 #14C17 #14G05 #14G15 #Algebraic Geometry (math.AG) #Algebraic Geometry and Number Theory #FOS: Mathematics #Polynomial and algebraic computation
paper · pdf · doi:10.48550/arxiv.2407.13521
openalex publication_date 2024/07/18 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/28
Let Hq denote the Hermitian curve in ℙ2 over \mathbbFq2 and Cd be an irreducible plane projective curve in ℙ2 also defined over \mathbbFq2 of degree d. Can Hq and Cd intersect in exactly d(q+1) distinct \mathbbFq2-rational points? Bézout's theorem immediately implies that Hq and Cd intersect in at most d(q+1) points, but equality is not guaranteed over \mathbbFq2. In this paper we prove that for many d ≤ q2-q+1, the answer to this question is affirmative. The case d=1 is trivial: it is well known that any secant line of Hq defined over \mathbbFq2 intersects Hq in q+1 rational points. Moreover, all possible intersections of conics and Hq 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 ≥ 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 ≤ q2-q+1, d=\lfloor(q+1)/2\rfloor and d=3, q ≥ 3 the answer is again affirmative. Various partial results for the case d small compared to q are also provided.