Entry details for q = 31 = 3, g = 3
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Lower bound Nmin = 10

 Submitted by Gerrit Oomens Date 01/01/1900 Reference J-P. SerreNombre de points des courbes algébriques sur F_qSém. de Théorie des Nombres de Bordeaux, 1982/83, exp. no. 22. (= Oeuvres III, No. 129, p. 664-668). Comments Tags Explicit curves

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 Explicit example Everett Howe 05/20/2010 22:15 In his Harvard notes, Serre notes that the plane quartic y^3 - y = x^4 - x^2 has 10 points over F_3.
Upper bound Nmax = 10

 Submitted by Gerrit Oomens Date 01/01/1900 Reference Jean-Pierre SerreRational points on curves over finite fieldsNotes by Fernando Q. Gouvêa of lectures at Harvard University, 1985. Comments The Oesterlé bound. Tags None

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 A long but elementary argument Everett Howe 06/17/2010 18:13 Our reference is to the Oesterlé bound, but this result can also be obtained by more naïve methods. For example: We know a hyperelliptic curve over F_3 can have at most 8 points, so it will suffice to show that a nonsingular plane quartic can have at most 10 points. Suppose all 13 points of the projective plane over F_3 are solutions to a homogeneous quartic polynomial f. This gives 13 linear conditions on the 15 coefficients of f. Writing down these conditions, and solving the resulting linear algebra problem, we find that f must be a linear combination of the three polynomials x^3*z - x*z^3, x^3*y - x*y^3, and y^3*z - y*z^3. Thus, up to scalar multiples, there are (3^3 - 1)/(3 - 1) = 13 nonzero quartic polynomials f, corresponding to 13 quartic curves that pass through every point in the plane. For each point P of the projective plane, consider the singular plane quartic curve consisting of the four lines through P. This quartic clearly passes through every point of the plane. This construction gives us 13 singular quartics that pass through every point of the plane, so all of the quartics we found in the preceding paragraph must be singular. Likewise, if exactly 12 points of the projective plane are solutions to a quartic polynomial, we find that the polynomial must be a linear combination of the same three polynomials as before, and again every such combination gives a reducible curve. If a quartic passes through exactly 11 points of the plane, we can linearly transform the quartic so that the points it misses are our favorite two points P1 and P2. Writing down the linear system that expresses the condition that every point except these two are solutions to a quartic polynomial f, we find that f must be a linear combination of four specific polynomials (three of them being the ones from before). That means that up to scalars, there are (3^4 - 1)/(3 - 1) = 40 nonzero quartic polynomials to consider, corresponding to 40 quartic curves. 13 of these curves are the ones we considered before, leaving 27 that pass through every point of the plane except P1 and P2. Again, if we can construct 27 singular quartics that miss P1 and P2 but that pass through every other point, we will have shown that no nonsingular quartic has 11 points. Let Q be a point not on the line connecting P1 with P2; there are 9 choices for Q. Let L1 and L2 be the two lines through Q that do not pass through P1 or P2. Then there are 4 points of the plane that do not lie on L1 or L2 and that are not equal to P1 or P2. There is a 2-parameter family of quadratic polynomials that pass through these four points, giving 4 conics that pass through these four points. One of these is the degenerate conic consisting of the other two lines through Q. The other three conics are irreducible. If C is one of these irreducible conics, then the union of C with L1 and L2 is a quartic that does not pass through P1 or through P2 but that passes through every other point. This accounts for all 27 of the quartics we counted in the preceding paragraph.