5
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Let $F(x,y,z)$ be the degree 12 homogeneous polynomial:

$$x^{12} - x^9 y^3 + x^6 y^6 - x^3 y^9 + y^{12} - 4 x^9 z^3 + 3 x^6 y^3 z^3 - 2 x^3 y^6 z^3 + y^9 z^3 + 6 x^6 z^6 - 3 x^3 y^3 z^6 + y^6 z^6 - 4 x^3 z^9 + y^3 z^9 + z^{12}$$

Over the rationals it is irreducible and $F=0$ is genus 1 curve.

Numerical evidence in Sagemath and Magma suggests that for infinitely many primes $p$, the curve $F=0$ is irreducible over $\mathbb{F}_p$ and $F=0$ has only one point over $\mathbb{F}_p$, the singular point $(1 : 0 : 1)$.

Q1 Is this true?

Set $p=50033$. Then we have only one point over the finite field and the curve is irreducible of genus 1. This appears to violate the bound on number of rational points over finite fields given in the paper "The number of points on an algebraic curve over a finite field", J.W.P. Hirschfeld, G. Korchmáros and F. Torres ,p. 23.

Q2 What hypothesis am I missing for this violation?

Sagemath code:

def tesgfppoints2():
   L1=5*10^4
   L2=2*L1
   for p in primes(L1,L2):
          K.<x,y,z>=GF(p)[]
          F=x^12 - x^9*y^3 + x^6*y^6 - x^3*y^9 + y^12 - 4*x^9*z^3 + 3*x^6*y^3*z^3 - 2*x^3*y^6*z^3 + y^9*z^3 + 6*x^6*z^6 - 3*x^3*y^3*z^6 + y^6*z^6 - 4*x^3*z^9 + y^3*z^9 + z^12
          C=Curve(F)
          ire=C.is_irreducible()
          if not ire:  continue
          rp=len(C.rational_points())

          print 'p=',p,';rp=',rp,'ir=',ire,'g=',C.genus()
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  • 1
    $\begingroup$ You have cited a paper by title and page number, but without author or journal, volume, year. Doesn't make it easy to find. $\endgroup$ – Gerry Myerson May 26 at 12:36
  • $\begingroup$ @GerryMyerson Thanks. I edited with link and authors. $\endgroup$ – joro May 26 at 13:07
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    $\begingroup$ Perhaps the curve is not geometrically irreducible over the rationals. $\endgroup$ – ulrich May 26 at 13:30
  • $\begingroup$ @ulrich If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right? $\endgroup$ – joro May 27 at 12:30
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The polynomial you wrote is the product of the four polynomials $x^3 - r y^3 - z^3$, where $r$ is a root of the polynomial $t^4 - t^3 + t^2 - t + 1$. I did not read your reference, but likely they assume that the curves are geometrically irreducible.

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  • 5
    $\begingroup$ Just to make it explicit that this answers Q1 as well: If none of the $r$ are elements of $\mathbb{F}_p$, then the only way $x^3-ry^3-z^3$ can be zero mod $p$ is if $y=0$, i.e., if you are at the singular point mentioned in the question. The roots of $t^4-t^3+t^2-t+1$ are the primitive $10$th roots of unity, so it will have no roots mod $p$ exactly when $p$ is not congruent to $1$ mod $10$. $\endgroup$ – dhy May 26 at 14:18
  • $\begingroup$ If we take G=F+p (x^12+y^12+z^12) then we need absolute reducibility over F_p, right? $\endgroup$ – joro May 27 at 12:29
  • $\begingroup$ Joro, if you know absolute irreducibility over F_p, then you can usually say something about number of points. Nevertheless, in your example, $G \equiv F \pmod{p}$, so this is still not absolutely irreducible modulo $p$. $\endgroup$ – dinamo May 27 at 19:17

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