# Smooth absolutely irreducible (?) genus 1 plane pointless curve over $\mathbb{F}_{13}$

We got a family of genus 1 plane curves that may violate a bound in a paper.

Explicitly: Let $$F(x,y)$$ be the degree 39 polynomial with integer coefficients:

x^3*y^36 - 6*x^3*y^33 - 3*x^3*y^30 + 5*x^3*y^27 + 6*x^3*y^24 - 3*x^2*y^25 - 5*x^2*y^24 - 3*x^3*y^21 - x^2*y^22 - 6*x^2*y^21 + 4*x^3*y^18 + 5*x^2*y^19 + 4*x^2*y^18 + x^3*y^15 - 5*x^2*y^16 - 4*x^2*y^15 + x^3*y^12 + x^2*y^13 + 3*x*y^14 + 6*x^2*y^12 - 3*x*y^13 + 4*x*y^12 + 4*x^3*y^9 + 3*x^2*y^10 - 6*x*y^11 + 5*x^2*y^9 + 6*x*y^10 + 5*x*y^9 - 5*x^3*y^6 + 2*x^2*y^7 - 2*x*y^8 - x^2*y^6 + 2*x*y^7 + 6*x*y^6 - 5*x^3*y^3 - 2*x^2*y^4 + 5*x*y^5 + x^2*y^3 - 5*x*y^4 - 2*x*y^3 + 5*x^3 + 3*x^2*y - 2*x*y^2 + 5*x^2 + 2*x*y - 5*y^2 + 6*x - 4*y - 2


Let $$p=13$$. Over $$\mathbb{F}_{13}$$, $$F=0$$ has no rational points and hence no singular points. The genus is $$1$$.

We believe that it is also absolutely irreducible over the finite field for the following reasons:

1. Magma's commands k:=FieldOfGeometricIrreducibility(C);IrreducibleComponents(BaseChange(C,k)); suggests it is absolutely irreducible.

2. $$F$$ is irreducible over $$\mathbb{F}_{13^k}$$ for $$1 \le k \le 39$$.

The projective closure has only two singular points.

Q1 Is $$F$$ absolutely irreducible over $$\mathbb{F}_{13}$$?

In case of positive answer 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. 6 Corollary 3.6

Magma online code:

p:=13;
Kp:=GF(p);
K<x,y>:=AffineSpace(Kp,2);
f:= x^3*y^36 - 6*x^3*y^33 - 3*x^3*y^30 + 5*x^3*y^27 + 6*x^3*y^24 - 3*x^2*y^25 - 5*x^2*y^24 - 3*x^3*y^21 - x^2*y^22 - 6*x^2*y^21 + 4*x^3*y^18 + 5*x^2*y^19 + 4*x^2*y^18 + x^3*y^15 - 5*x^2*y^16 - 4*x^2*y^15 + x^3*y^12 + x^2*y^13 + 3*x*y^14 + 6*x^2*y^12 - 3*x*y^13 + 4*x*y^12 + 4*x^3*y^9 + 3*x^2*y^10 - 6*x*y^11 + 5*x^2*y^9 + 6*x*y^10 + 5*x*y^9 - 5*x^3*y^6 + 2*x^2*y^7 - 2*x*y^8 - x^2*y^6 + 2*x*y^7 + 6*x*y^6 - 5*x^3*y^3 - 2*x^2*y^4 + 5*x*y^5 + x^2*y^3 - 5*x*y^4 - 2*x*y^3 + 5*x^3 + 3*x^2*y - 2*x*y^2 + 5*x^2 + 2*x*y - 5*y^2 + 6*x - 4*y - 2;
C:=Curve(K,f);
Genus(C);
k:=FieldOfGeometricIrreducibility(C);
//IsIrreducible(C);
IrreducibleComponents(BaseChange(C,k));
pc:=ProjectiveClosure(C);
Points(C);

• Dear joro, the curve is absolutely irreducible. Nevertheless, I interpret the last paragraphs of the introduction of the paper that you cite to mean that for smooth curves, the bounds are correct, otherwise, there might be some small discrepancy. Your example shows that there can indeed be discrepancies. – dinamo May 29 '19 at 10:40
• IMHO, it would be a counterexample to Hasse-Weil theorem... – Dima Pasechnik May 29 '19 at 10:45
• @DimaPasechnik No, it won't be counterexample since Hasse-Weil wants smooth projective curve and the projective model is not smooth. On the other hand the projective model of $y^2=f(x)$ is not smooth too... – joro May 29 '19 at 10:49
• well, the 1st comment correctly, I think, points out that Cor 3.6 you cite is only applicable to smooth curves, and your (projective) curve is not. – Dima Pasechnik May 29 '19 at 10:59
• The discrepancy involves understanding the contribution of the singular points. An upper bound can be constructed using the arithmetic genus of the curve, instead of the geometric genus. For a plane curve of degree $d$, the arithmetic genus is $\binom{d-1}{2}$. – dinamo May 29 '19 at 11:36