5
$\begingroup$

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()
Improve this question
$\endgroup$
4
  • 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
    Commented May 26, 2019 at 12:36
  • $\begingroup$ @GerryMyerson Thanks. I edited with link and authors. $\endgroup$
    – joro
    Commented May 26, 2019 at 13:07
  • 6
    $\begingroup$ Perhaps the curve is not geometrically irreducible over the rationals. $\endgroup$
    – naf
    Commented May 26, 2019 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
    Commented May 27, 2019 at 12:30

1 Answer 1

Reset to default
14
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • 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
    Commented May 26, 2019 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
    Commented May 27, 2019 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
    Commented May 27, 2019 at 19:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .