4 YVES AUBRY AND FRANC ¸OIS RODIER then the polynomials Pf (x, y, z) and Pf (x, y, t) are homogeneous polynomials and we can consider the intersection X of the projective cones S1 and S2 of dimen- sion 2 defined respectively by Pf (x, y, z) = 0 and Pf (x, y, t) = 0 with projective coordinates (x : y : z : t) in the projective space P3(Fq). Even if X is now a projective algebraic subset of the projective space P3(Fq), Theorem 3 tells us also that: δ(f) 4 if and only if X(Fq) V, where V is the hypersurface of P3(Fq) defined by Equation (1). Indeed, the algebraic sets X and V in this section are closely related to but not equal to the sets X and V of the previous section. The set X of this section (resp. V ) is the set of lines through the origin of the set X (resp. V ) of the previous section which is invariant under homotheties with center the origin. For convenience, we keep the same notations. Lemma 4. The projective algebraic set X has dimension 1, i.e. it is a projective curve. Proof. We have to show that the projective surfaces S1 and S2 do not have common irreducible components. Since S1 and S2 are two cones, it is enough to prove that the vertex of one of the cones doesn’t lie in the other cone. The coordinates of the vertex of the cone S2 is (0 : 0 : 1 : 0). To show that it doesn’t lie in S1, we will prove that Pf (0 : 0 : 1 : 0) = 0. Indeed, S1 is defined by the polynomial Pf (x, y, z) = xd + yd + zd + (x + y + z)d (x + y)(x + z)(y + z) · Setting x + y = u, we obtain: Pf (x, y, z) = xd + (x + u)d + zd + (u + z)d u(x + z)(x + u + z) , which gives Pf (x, y, z) = xd−1 + zd−1 + uQ(x, z) (x + z)(x + u + z) , where Q is some polynomial in x and z. This expression takes the value 1 at the point (0 : 0 : 1 : 0). Now we know that X is a projective curve in P3(Fq), and in order to estimate its number of rational points over Fq, we must determine its irreducibility. We will prove that the curve C7, defined as the intersection of S2 with the projective plane H7 of equation x + y + z + t = 0, is an absolutely irreducible component of X, and hence that X is reducible. Proposition 5. The intersection of the curve X with the plane H7 with the equation x + y + z + t = 0 is equal to the curve C7 := S2 H7. Proof. Since X = S1 S2, it is enough to prove that C7 S1. Since t = x + y + z the points of intersection of the cone S2 with the plane x + y + z + t = 0 4
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