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Let the base field be a finite field $\mathbb F_q$ and fix $d$ rational points that lie on a line in $\mathbb P^2$. Suppose $d$ is a large number (about the order of $q^{\alpha}$ for $\alpha$ some positive real number). Are there good estimates on the number of degree $d$ curves that pass through these fixed $d$ points?

I would be happy assuming the curves are irreducible if that makes the problem simpler (but I don't except this assumption to change the asymptotics anyway).

More generally, suppose we fix $r$ points but keep the same degree. Then how many curves of degree $d$ pass through $r$ points? I only mostly care about asymptotics.

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  • $\begingroup$ Are your points rational over $\mathbb{F}_q$? If not, then I guess you want to assume that your set of points is stable under Galois. $\endgroup$ Commented May 21, 2019 at 10:00
  • $\begingroup$ Yes, I am okay with assuming they are rational. $\endgroup$
    – Asvin
    Commented May 21, 2019 at 11:14
  • $\begingroup$ The space of plane curves of degree $d$ is a projective space, and passing through $r$ points gives you linear conditions on the coefficients of the equation, so what you get is a linear subspace, which has codimension $r$ if the points are in generic position. But maybe you also want to have control in every case? $\endgroup$ Commented May 21, 2019 at 17:47
  • $\begingroup$ Right, that would be ideal. Also, why does genericity imply the right codimension? $\endgroup$
    – Asvin
    Commented May 21, 2019 at 20:47
  • $\begingroup$ I forgot that your $r$ points are rational, so there are only finitely many possibilities for them and genericity doesn't make much sense. I was thinking geometrically (say over the algebraic closure). So the number of curves will depend on the position of these points; which number do you want to estimate? You have $r$ equations, so the codimension is at most $r$, but can be less e.g. if your points satisfy collinearity conditions. For conics ($5$-dim parameter space) passing through 5 points, "generic" means "no 4 points collinear", but in general I don't know. $\endgroup$ Commented May 21, 2019 at 20:55

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