On the mutual position of two irreducible conics in PG(2, q), q odd

Abstract

Given an irreducible conic C in PG(2, q) with q odd, how many points of another irreducible conic D in PG(2, q) are external to C? It is not hard to find D such that either all its points off C are external to C, or none of its points are. Apart from these cases, we prove that for q large enough, the number we seek differs from 1 2 (q − 1) by at most √ q + 3.

Infinite family of large complete arcs in PG(2, q n ), with q odd and n > 1 odd

“…For n > 1 odd, there exists an arc K of size 1 2 (q n − q) + |K | sharing exactly 1 2 (q n − q) points with the irreducible conic C containing C. We point out that if k ≥ 5 and K is not contained in a proper subplane of PG(2, q) then = G and = H are the only possibilities in the above construction. This is actually a consequence of the Dickson's classification of subgroups of PGL(2, q), see [15] and [11,Theorem A.8].…”

“…The genus g of F is equal to 1 2 (deg F −2) = 2n −1. Let N q (F) be the number of G F(q)-rational points of F .…”

“…(1) The bound is achieved by a conic for q odd and by a conic plus the meet of its tangents for q even. Conversely, Segre showed that for q odd every (q + 1)-arc is a conic.…”

“…We do not discuss here other good choices of K i . However, we observe that some more examples can be constructed bearing in mind that there are irreducible conics containing 1 2 N external (or 1 2 N internal) points to a given irreducible conic where N is the number of points of a non-singular plane cubic curve in PG(2, q), see [1].…”

Simultaneous diagonalization of conics in $$PG(2,q)$$ P G ( 2 , q )

“…We start with two disjoint conics such that their pencil is of (q − 1)-form, which means that exactly two degenerate conics occur, namely one point P and one line g. We can now apply a collineation to the whole pencil, such that, without loss of generality, the point P is given by [1,0,0] T and the line g is incident with [0, 1,0] T and [0, 0, 1] T . So we can consider the pencil given by the two elements g = V(x 2 ) and P = V(y 2 + cz 2 + f yz), where c and f are chosen such that y 2 + cz 2 + f yz is irreducible.…”