Entry details for q = 75 = 16807, g = 3
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Lower bound Nmin = 17582

Submitted by C. Ritzenthaler
Date 11-15-2009
Reference Not available
Comments
This number is reached bu the following curve

a1*t1^4+a2*t1^2*t2+2*t1*t3+t2^2=0;

where

t1:=X+Y+Z;t2:=X*Y+X*Z+Y*Z;t3:=X*Y*Z;
a1:=a^15527;a2:=a^577;

where a is a root of u^5+u-3=0 in F_{7^5}.

This curve has been found by J.-F. Mestre using his preprint

"courbes de genre 3 avec S_3 comme groupe d'automorphismes".
Tags Explicit curves

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Upper bound Nmax = 17582

Submitted by C. Ritzenthaler
Date 11-28-2009
Reference Kristin Lauter
Geometric methods for improving the upper bounds on the number of rational points on algebraic curves over finite fields
J. Algebraic Geom. 10 (2001) 19–36
Comments
Since m:=Floor(2*Sqrt(q)) is divisible by p=7 there does not exist an elliptic curve with trace -m over F_q. One can also check that (x^2+m*x+q)^3 cannot be the Weil polynomial of a simple abelian variety because this implies that q is a cube (see Prop. 2.5 in Maisner, Nart : Abelian surfaces over finite fi elds as Jacobians. With an appendix by Everett W. Howe, Exp. Math. 11 (2002), 321–337). So defect 0 is impossible.

Defect 1 is impossible as well due to the "resultant 1" method.

Now one can check that all the possibilities of Table 1 in the article for defect 2 are excluded (some because p divides m, some because of the "resultant 1" method and the last one because of the value of the fractional part of 2*Sqrt(q)).
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