# Pointless, non-singular, absolutely irreducible affine plane curves over finite fields

We think the following is true:

For all sufficiently large primes $$p$$ and all natural $$g \ge 1$$, there exists affine plane curve $$f(x,y)=0$$ over $$\mathbb{F}_p$$ which is non-singular, absolutely irreducible, of genus $$g$$ and it doesn't have any rational points.

Is it true?

Is it known?

This doesn't violate Hasse-Weil bound, because the bound requires SMOOTH projective model and our examples have very few singular points on the projective model.

For $$p=13,g=1$$ check this question and comments

Added Example of pointless non-sigular affine curves for $$g=1$$ defined by two equations. Let $$f_0=x^3+y^2-1,f_1=z(x^p-x)-1$$ and the curve $$C : f_0=0,f_1=0$$. Then $$f_1$$ is linear in $$z$$ so $$C$$ is birationally equivalent to $$f_0=0$$.

The curve is pointless because $$x^p-x$$ is zero modulo $$p$$.

To get a single equation for the curve set $$f$$ the resultant of $$f_0$$ and $$f_1$$ wrt $$x$$, experimentally it is irreducible.

• Doesn't this contract the Hasse-Weil bound? You need small $p$ and large $g$ to get no points. – S. Carnahan May 29 '19 at 12:14
• A smooth geometrically connected genus $1$ curve over a finite field always has a rational point (by Hasse's estimate or the Lang-Steinberg theorem, say), so this can't be right. Maybe you meant a different set of quantifiers? – Gro-Tsen May 29 '19 at 12:23
• Aaaah, I'm thinking maybe you meant to ask about affine plane curves. If this is the case you should write it in big bold capital letters in the question, because everyone (and certainly the three people who reacted so far) will interpret “curve” as “projective curve”! – Gro-Tsen May 29 '19 at 12:29
• @S.Carnahan I edited. See example for $p=13,g=1$. I still believe this is true. – joro May 29 '19 at 12:33
• This is true: in fact, any plane curve over a finite field (possibly with the exception of $\mathbb{F}_2$) can be birationally transformed to another plane curve with all of its points contained in a single line. – dinamo May 29 '19 at 12:35

Yes, it is true.

If you have any smooth geometrically integral plane affine curve $$C/\mathbb{F}_q$$ of genus $$g$$ with $$|C(\mathbb{F}_q)|, then you can construct such a curve $$C'$$ birational to $$C$$ with $$|C'(\mathbb{F}_q)|=0$$:

Remove the rational points one by one by pushing them out to infinity: Pick $$P\in C(\mathbb{F}_q)$$ and $$Q\in\mathbb{A}^2(\mathbb{F}_q)\setminus C$$. By a linear transformation assume that $$Q=(0,0)$$ and $$P$$ lies on the line $$L=\{X=0\}$$. Now let $$C_0=C\setminus L$$ and observe that $$(x,y)\mapsto(x,x^{-1}y)$$ is an isomorphism of $$C_0$$ onto another plane affine curve. This curve is birational to the original curve and is still smooth but has at least one rational point less. Now repeat.

That a smooth geometrically integral plane affine curve $$C/\mathbb{F}_q$$ of genus $$g$$ with $$|C(\mathbb{F}_q)| (i.e. that is not space filling) exists is clear.

(Remark: I agree that the statement above sounds a bit odd, as we are pushing up to $${\rm min}(q^2-1,q+1+2g\sqrt{q})$$ many rational point to the line at infinity, which has only $$q+1$$ many rational points. But note that the points at infinity can be highly singular.)

Let $$f(x,y)=0$$ be smooth and absolutely reducible curve over $$\mathbb{F}_p$$.

To reduce the number of points while keeping the genus the same, for $$f_1, f_2 \in \mathbb{F}_p[x,y]$$ and a new variable $$z$$ set $$F(x,y,z)=f_1 z - f_2$$ and take the curve $$f=0,F=0$$. This is linear in $$z$$, so it preserves the genus of $$f=0$$.

If you want only one equation for the curve, take $$G$$ to be the resultant of $$f$$ and $$F$$ with respect to $$x$$.

We have the constraints $$f_1(x,y) = 0 \implies f_2(x,y)=0$$, which may fail by construction.

Extreme example $$p=13,f=x^3+y^3-2,f_1 = x^p-x, f_2 =1$$, the plane curve $$G=0$$ is pointless and smooth. since $$x^p-x=0$$ for all $$x$$. where

G=y^39*z^3 - 3*y^27*z^3 + 2*y^24*z^3 - 3*y^21*z^3 + y^18*z^3 + 2*y^12*z^3 - 5*y^9*z^3 - y^6*z^3 - 4*y^3*z^3 - 3*z^3 + 1

We think this $$G$$ is absolutely irreducible over $$\mathbb{F}_{13}$$.

If we want to exclude a set $$S$$ of $$x$$ coordinates, take $$f_2=1, f_1=z \prod_{a \in S}(x-a)$$.

Other $$f,f_1,f_2$$ may give non singular plane curves with few or no points, while keeping $$G$$ non-singular.