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:

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

$F$ 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);
```

smooth projective curveand the projective model is not smooth. On the other hand the projective model of $y^2=f(x)$ is not smooth too... $\endgroup$ – joro May 29 '19 at 10:49