# Does not Hasse-Weil theorem hold in curves with no rational points?

According to Hasse-Weil theorem (see, for example, Silverman, Tate, Rational Points on Elliptic Curves, Theorem 4.1) we have:

If $$C$$ is a non-singular irreducible curve of genus $$g$$ defined over a finite field $$F_p$$, then the number of points on $$C$$ with coordinates in $$F_p$$ is equal to $$p + 1 - \epsilon$$, where the "error term" $$\epsilon$$ satisfies $$|\epsilon| \leq 2g\sqrt{p}$$.

For my surprise, I found very easily curves satisfying (in principle) the conditions of the theorem but have no rational points. For example, I have used MAPLE to check that in $$F_7$$, the curve $$C := {x_{1}}^{3} - 2\,{x_{1}}^{2}\,{x_{3}} + {x_{1}}\,{ x_{2}}^{2} + 3\,{x_{1}}\,{x_{2}}\,{x_{3}} + {x_{1}}\,{x_{3}}^{2} - {x_{2}}^{3} - {x_{2}}\,{x_{3}}^{2} + {x_{3}}^{3}$$ (with $$x_1$$, $$x_2$$, $$x_3$$ the projective coordinates), has no rational points. However $$C$$ is irreducible, non-singular, and of genus $$g=1$$, and accordingly it seems to meet the conditions in the thorem above, so it should have at least two points, since $$8 - 2\,\sqrt{7} > 2$$. It has none, though.

What am I missing?

• According to my recollection and Wikipedia the formal definition of an elliptic curve is a non-singular projective algebraic curve over K with genus 1 and endowed with a distinguished point defined over K. In other words, for the curve to be called elliptic, it needs to have at least one $K$-rational point. In Weierstrass form the point at infinity is often that distinguished point. I'm afraid I don't remember the proper terminology for curves without $K$-rational points. Jan 15, 2020 at 22:23
• @JyrkiLahtonen very late on this one, but in case you were still interested, the terminology for a genus $1$ curve $C$ without a point is a "torsor under $E$" where $E$ is the jacobian of $C$ (equiv. a "principal homogeneous space" for $E$). If you know the order $n$ of $[C] \in WC(E/K)$ then you could call it an $n$-covering. Mar 26, 2021 at 22:03

According to Magma, and assuming I typed it correctly, your curve is singular, hence the theorem doesn't apply.

> P<x,y,z>:=ProjectiveSpace(GF(7), 2);
> f:=x^3-2*x^2*z+x*y^2+3*x*y*z+x*z^2-y^3-y*z^2+z^3;
> Curve(P, f);
Curve over GF(7) defined by
x^3 + x*y^2 + 6*y^3 + 5*x^2*z + 3*x*y*z + x*z^2 + 6*y*z^2 + z^3
> RationalPoints(Curve(P, f));
{@ @}
> IsSingular(Curve(P,f));
true
> Genus(Curve(P,f));
0